ﺣﺪﻭﺩ ﺍﻟﻔﻴﺰﻳﺎء ﺍﻟﻜﻼﺳﻴﻜﻴﺔ THE LIMITS OF CLASSICAL PHYSICS ﻤﻴﻼﺩ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻜﻤﻴﺔ The Dawn of the Quantum Theory
ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻘﺭﻥ ﺍﻟﺘﺎﺴﻊ ﻋﺸﺭ ،ﺍﻋﺘﻘﺩ ﻜﺜﻴﺭ ﻤﻥ ﺍﻟﻌﻠﻤﺎﺀ ﺃﻥ ﻜل ﺍﻻﻜﺘـﺸﺎﻓﺎﺕ
ﺍﻟﻌﻠﻤﻴﺔ ﻗﺩ ﺘﻡ ﺇﻨﺠﺎﺯﻫﺎ ﻭﻓﻬﻤﻬﺎ ﻭﺃﻨﻪ ﻟﻡ ﻴﺒﻘﻰ ﺇﻻ ﺒﻌﺽ ﺍﻟﻤـﺴﺎﺌل ﺍﻟﺒـﺴﻴﻁﺔ ﺍﻟﺘـﻲ
ﺘﺤﺘﺎﺝ ﻟﻤﺯﻴﺩ ﻤﻥ ﺍﻹﻴﻀﺎﺡ .ﺇﻥ ﻫﺫﻩ ﺍﻟﻘﻨﺎﻋﺔ ﻭﻜﺎﻨﺕ ﻨﺎﺸﺌﺔ ﻤﻥ ﺍﻟﺘﻘﺩﻡ ﺍﻟﻌﻠﻤﻲ ﻓـﻲ
ﻤﺠﺎﻻﺕ ﺸﺘﻰ ﻭﺍﻟﺫﻱ ﺘﻤﺜل – ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل – ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﻨﻴـﻭﺘﻥ Newton
ﻭﺍﻟﺘﻲ ﻁﹸﻭﺭﺕ ﺒﻭﺍﺴﻁﺔ ﺍﻟﻌﺎﻟﻤﺎﻥ ﻻﺠـﺭﺍﻨﺞ ﻭﻫـﺎﻤﻠﺘﻭﻥ J. LaGrange and W.
.Hamiltonﺤﻴﺙ ﺘﻡ ﺍﺴﺘﺨﺩﺍﻡ ﻫﺫﻩ ﺍﻟﻨﻅﺭﻴﺔ ﻟﻭﺼﻑ ﺤﺭﻜﺔ ﺍﻟﻜﻭﺍﻜﺏ ﻭﻜﺫﻟﻙ ﻓﻬـﻡ
ﻜﺜﻴﺭ ﻤﻥ ﺍﻟﻅﻭﺍﻫﺭ ﺍﻟﻤﻌﻘﺩﺓ ﻤﺜل ﻨﻅﺭﻴﺔ ﺍﻟﻤﺭﻭﻨـﺔ elasticity theoryﻭﺩﻴﻨﺎﻤﻴﻜـﺎ ﺍﻟﻤﻭﺍﺌﻊ hydrodynamicﺇﻨﺠﺎﺯﺍﺕ ﺍﻟﻌﺎﻟﻡ ﺠﻭل ﻭﺒﻴﺎﻥ ﺘﻜﺎﻓﺅ ﺍﻟـﺸﻐل ﻭﺍﻟﺤـﺭﺍﺭﺓ،
ﺃﺒﺤﺎﺙ ﻜﺎﺭﻨﻭﺕ Carnotﻭﺍﻟﺘﻲ ﺃﺩﺕ ﻟﻔﻬﻡ ﺍﻹﻨﺘﺭﻭﺒﻲ ﻭﺍﻟﻘﺎﻨﻭﻥ ﺍﻟﺜـﺎﻨﻲ ﻟﻠـﺩﻴﻨﺎﻤﻴﻜﺎ ﺍﻟﺤﺭﺍﺭﻴﺔ ،ﻭﻤﺎ ﻴﺘﺒﻊ ﻫﺫﻩ ﺍﻷﺒﺤﺎﺙ ﻤﻥ ﺘﻁﻭﻴﺭ ﻋﻠﻰ ﻴﺩ ﺍﻟﻌـﺎﻟﻡ ﺠـﺒﺱ J. Gibbs
ﻹﺭﺴﺎﺀ ﺃﺴﺱ ﻋﻠﻡ ﺍﻟﺩﻴﻨﺎﻤﻴﻜﺎ ﺍﻟﺤﺭﺍﺭﻴﺔ.
ﻭﺸﻬﺩﺕ ﻤﺠـﺎﻻﺕ ﺃﺨـﺭﻯ ﻤـﻥ ﺍﻟﻔﻴﺯﻴـﺎﺀ )ﻤﺜـل ﺍﻟـﻀﻭﺀ ﻭﺍﻟﻨﻅﺭﻴـﺔ
ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻴﺔ (optics and electromagnetic theoryﺇﻨﺠﺎﺯﺍﺕ ﻤﻠﺤﻭﻅـﺔ. ﻓﻤﺜﻼﹰ ،ﺍﻻﺴﺘﻨﺘﺎﺠﺎﺕ ﺍﻟﻬﺎﻤﺔ ﺍﻟﺘﻲ ﺘﻭﺼل ﺇﻟﻴﻬﺎ ﺍﻟﻌﺎﻟﻡ ﻤﺎﻜﺴﻭﻴل J. Maxwellﻤﺘﻤﺜﻠﺔ
ﺒﻤﻌﺎﺩﻻﺘﻪ ﺍﻟﺸﻬﻴﺭﺓ "ﻭﺍﻟﺒﺴﻴﻁﺔ" ﻭﺍﻟﺘـﻲ ﻭﺤـﺩﺕ ﻤﺠـﺎﻻﺕ ﺍﻟـﻀﻭﺀ ﻭﺍﻟﻜﻬﺭﺒﻴـﺔ
ﻭﺍﻟﻤﻐﻨﺎﻁﻴﺴﻴﺔ ﻭﻤﺎ ﻴﺘﺒﻊ ﻫﺫﻩ ﺍﻷﺒﺤﺎﺙ ﻤﻥ ﺍﻟﺘﺠﺎﺭﺏ ﺍﻟﻤﻌﻤﻠﻴﺔ ﺒﻭﺍﺴﻁﺔ ﺍﻟﻌﺎﻟﻡ ﻫﻴﺭﺘﺯ H. Hertzﻓﻲ ﻋﺎﻡ 1887ﻭﺍﻟﺘﻲ ﺃﺩﺕ ﺇﻟﻰ ﺇﺜﺒﺎﺕ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﻭﺠﺒﺔ ﻟﻠﻀﻭﺀ.
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ﻜل ﻫﺫﻩ ﺍﻹﻨﺠﺎﺯﺍﺕ ﻓﻲ ﺍﻟﻤﺠﺎﻻﺕ ﺍﻟﻤﺨﺘﻠﻔﺔ ﻟﻠﻔﻴﺯﻴﺎﺀ ﻜﻭﻨﺕ ﻤﺎ ﻴﻌـﺭﻑ ﺍﻵﻥ ﺒﺎﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ Classical physicsﻭﻤﻊ ﺒﺩﺍﻴﺔ ﺍﻟﻘﺭﻥ ﺍﻟﻌﺸﺭﻴﻥ ،ﻭﺠﺩﺕ ﺒﻌﺽ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﺍﻟﺠﺩﻴﺩﺓ ﻭﺍﻟﺘﻲ ﺍﺴﺘﻠﺯﻡ ﺘﻔﺴﻴﺭﻫﺎ ﻤﻔﺎﻫﻴﻡ ﻓﻴﺯﻴﺎﺌﻴﺔ ﺠﺩﻴﺩﺓ ﺘﺘﻨـﺎﻗﺽ
ﻤﻊ ﻤﺒﺎﺩﺉ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺠﺩﻴﺩﺓ ﻭﻟﺩ ﻤﺎ ﻴﺴﻤﻰ ﺍﻵﻥ ﺒﺎﻟﻨﻅﺭﻴﺔ ﺍﻟﻜﻤﻴﺔ quantum theory
ﻭﺴﻨﺤﺎﻭل ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺃﻥ ﻨﺼﻑ ﺨﻠﻔﻴﺔ ﻫﺫﻩ ﺍﻷﺯﻤﺎﺕ ﻟﻨـﺼل ﻤـﻥ ﺨﻼﻟﻬـﺎ ﻟﻤﻌﺭﻓﺔ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻜﻤﻴﺔ.
ﻭﻴﻤﻜﻨﻨﺎ ﺘﻠﺨﻴﺹ ﺍﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﺠﺩﻴـﺩﺓ ﻓـﻲ :ﺍﻟﺨـﻭﺍﺹ ﺍﻟﺠـﺴﻴﻤﻴﺔ ﻟﻺﺸﻌﺎﻉ ،the particle properties of radiationﺍﻟﺨﻭﺍﺹ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﻤﺎﺩﺓ the
،wave properties of matterﻭﺘﻜﻤﻴﻡ ﺍﻟﻜﻤﻴﺎﺕ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ the quantization of
.physical quantitiesﻭﺴﻨﻘﻭﻡ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺒﻤﻨﺎﻗﺸﺔ ﻫﺫﻩ ﺍﻟﻤﻔﺎﻫﻴﻡ. .1-1ﺇﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ "ﺍﻷﺴﻭﺩ"
1-1 Blackbody Radiation
ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻟﻡ ﺘﺘﻤﻜﻥ ﻤﻥ ﺸﺭﺡ ﺇﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ- Blackbody Radiation could not be explained by classical physics
ﻤﻥ ﺃﻫﻡ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﺍﻟﺘﻲ ﺃﺤﺩﺜﺕ ﺜﻭﺭﺓ ﻓﻲ ﺍﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺘﻠﻙ ﺍﻟﻤﺘﻌﻠﻘﺔ ﺒﺎﻹﺸﻌﺎﻉ ﺍﻟﺼﺎﺩﺭ ﻤﻥ ﺍﻷﺠﺴﺎﻡ ﻋﻨﺩ ﺘﺴﺨﻴﻨﻬﺎ .ﻓﻤـﻥ ﺍﻟﻤﻌﻠـﻭﻡ ﻋﻨـﺩ
ﺘﺴﺨﻴﻥ ﺠﺴﻡ ﻤﺎ ،ﻨﺠﺩ ﺃﻥ ﻟﻭﻨﻪ ﻴﺘﻐﻴﺭ ﻤﻊ ﺯﻴﺎﺩﺓ ﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ ﺤﻴﺙ ﻴﺒﺩﺃ ﺒﺎﻷﺤﻤﺭ
ﺜﻡ ﺍﻷﺒﻴﺽ ﺜﻡ ﺍﻷﺯﺭﻕ .ﻭﺒﺩﻻﻟﺔ ﺍﻟﺘﺭﺩﺩ ،ﻨﻘﻭل ﺃﻥ ﺍﻹﺸﻌﺎﻉ ﺍﻟﻤﻨﺒﻌﺙ ﻤﻥ ﻫﺫﺍ ﺍﻟﺠﺴﻡ
ﻴﺒﺩﺃ ﺒﺘﺭﺩﺩﺍﺕ ﻤﻨﺨﻔﻀﺔ ،ﻭﻋﻨﺩ ﺍﺭﺘﻔﺎﻉ ﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ ،ﺘﺯﺩﺍﺩ ﺍﻟﺘﺭﺩﺩﺍﺕ ،ﺤﻴﺙ ﺃﻥ ﺍﻟﻠﻭﻥ ﺍﻷﺤﻤﺭ ﺫﻭ ﺘﺭﺩﺩ ﻗﻠﻴل ﻓﻲ ﻤﻨﻁﻘﺔ ﻁﻴﻑ ﺍﻹﺸﻌﺎﻉ ﻭﺫﻟـﻙ ﻤﻘﺎﺭﻨـﺔ ﺒـﺎﻟﻠﻭﻥ
ﺍﻷﺯﺭﻕ .ﺇﻥ ﻁﻴﻑ ﺍﻟﺘﺭﺩﺩ ﻟﻺﺸﻌﺎﻉ ﺍﻟﻤﻨﺒﻌﺙ ﻤﻥ ﺠﺴﻡ ﻤﺎ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﻁﺒﻴﻌﺔ ﺍﻟﺠﺴﻡ ﻨﻔﺴﻪ ،ﻭﻟﻜﻥ ﺍﻟﺠﺴﻡ ﺍﻟﻤﺜﺎﻟﻲ ،deal bodyﻭﺍﻟﺫﻱ ﻴﻤﺘﺹ ﺃﻭ ﻴﺒﻌﺙ ﻜل ﺍﻟﺘـﺭﺩﺩﺍﺕ
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ﻴﺴﻤﻰ ﺒﺎﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ ﻭﻴﻌﺘﺒﺭ ﺤﺎﻟﺔ ﻤﺜﺎﻟﻴﺔ ﻷﻱ ﻤﺎﺩﺓ ﺘﹸـﺼﺩﺭ ﺇﺸـﻌﺎﻉ .ﺍﻹﺸـﻌﺎﻉ ﺍﻟﻤﻨﺒﻌﺙ ﻤﻥ "ﺠﺴﻡ ﺃﺴﻭﺩ" ﻴﺴﻤﻰ ﺇﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ. an ideal body, which absorbs and emits all frequencies, is called a blackbody and serves as an idealization for any radiating material, the radiation emitted by a blackbody is called blackbody radiation.
ﺸﻜل 1-1ﻴﻭﻀﺢ ﺘﻐﻴﺭ ﺸﺩﺓ ﺍﻹﺸﻌﺎﻉ ﺍﻟﺼﺎﺩﺭ ﻤﻥ ﺠﺴﻡ ﺃﺴﻭﺩ ﻤﻊ ﺍﻟﺘـﺭﺩﺩ
ﻭﺫﻟﻙ ﻋﻨﺩ ﺩﺭﺠﺎﺕ ﺤﺭﺍﺭﺓ ﻤﺨﺘﻠﻔﺔ .ﻭﻗﺩ ﺤﺎﻭل ﺍﻟﻌﺩﻴﺩ ﻤﻥ ﺍﻟﻔﻴﺯﻴـﺎﺌﻴﻴﻥ ﺍﺴـﺘﻨﺘﺎﺝ ﻤﻌﺎﺩﻟﺔ ﺭﻴﺎﻀﻴﺔ ﺘﺸﺭﺡ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ )ﻓﻲ ﺸﻜل (1-1ﻭﻟﻜﻥ ﺒﺩﻭﻥ ﺘﻭﺍﻓﻕ ﻜﺎﻤل
ﻭﺃﻭﻟﻰ ﺍﻟﻤﺤﺎﻭﻻﺕ ﻟﻭﺼﻑ ﻫﺫﻩ ﺍﻟﻨﺘﺎﺌﺞ ﻗﺎﻡ ﺒﻬﺎ ﻜل ﻤﻥ ﺍﻟﻌﺎﻟﻤﻴﻥ ﺭﺍﻟـﻲ ﻭﺠﻴﻨـﺯ Rayleigh and Jeansﻭﺍﻟﺘﻲ ﺍﺸﺘﻘﺕ ﺒﻨﺎﺀﺍﹰ ﻋﻠﻰ ﻗﻭﺍﻨﻴﻥ ﺍﻟﻘﺭﻥ ﺍﻟﺘﺎﺴﻊ ﻋﺸﺭ ﻭﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﺒﺎﻟﺼﻴﻐﺔ: )(1-1
ν2
ﺤﻴﺙ
)u (ν , T
8D K B T 3
C
= )u (ννT
ﻜﺜﺎﻓﺔ ﺍﻟﻁﺎﻗﺔ energy densityﻭﻭﺤﺩﺘﻬﺎ ﺠـﻭل ﻟﻜـل ﻤﺘـﺭ
ﻤﻜﻌﺏ ) .(J/m3ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ )T ،(1-1ﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ ﺒﺎﻟﻜﻴﻠـﭭﻥ c ،ﺴﺭﻋﺔ ﺍﻟﻀﻭﺀ،
KBﺜﺎﺒﺕ ﺒﻭﻟﺘﺯﻤﺎﻥ .ﺍﻟﺨﻁ ﺍﻟﻤﺘﻘﻁﻊ ﻓﻲ ﺸﻜل 1-1ﻴﺒﻴﻥ ﺍﻟﻌﻼﻗﺔ ﺤـﺴﺏ ﻤﻌﺎﺩﻟـﺔ
ﺭﺍﻟﻲ -ﺠﻴﻨﺯ .ﻻﺤﻅ ﺍﻟﺘﻭﺍﻓﻕ ﺒﻴﻥ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﻭﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﻋﻨﺩ ﺍﻟﺘـﺭﺩﺩﺍﺕ ﺍﻟﻤﻨﺨﻔﻀﺔ .ﻋﻨﺩ ﺍﻟﺘﺭﺩﺩﺍﺕ ﺍﻟﻌﺎﻟﻴﺔ ،ﻭﺤﺴﺏ ﻤﻌﺎﺩﻟﺔ ﺭﺍﻟـﻲ-ﺠﻴﻨـﺯ ،ﻓـﺈﻥ ﻜﺜﺎﻓـﺔ ﺍﻹﺸﻌﺎﻉ ﺘﺯﺩﺍﺩ ﻁﺒﻘﺎﹰ ﻟـ
ν2
ﻭﺘﺼل ﺇﻟﻰ ﻤﺎﻻ ﻨﻬﺎﻴﺔ ﻭﺫﻟﻙ ﻋﻨﺩﻤﺎ ﺘﺼل ﺍﻟﺘـﺭﺩﺩﺍﺕ
ﺇﻟﻰ ﻤﺎﻻ ﻨﻬﺎﻴﺔ ﻭﻫﺫﺍ ﻴﺤﺩﺙ ﻓﻲ ﻤﻨﻁﻘﺔ ﺍﻷﺸﻌﺔ ﻓﻭﻕ ﺍﻟﺒﻨﻔﺴﺠﻴﺔ ultravioletﻭﻫـﺫﺍ
ﻤﺎ ﻴﻌﺭﻑ ﺒﺎﻻﻨﻬﻴﺎﺭ ﻓﻭﻕ ﺒﻨﻔﺴﺠﻲ .ultraviolet catastropheﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ )(1-1
ﻭﻟﻜل ﺍﻟﺘﺭﺩﺩﺍﺕ ﻤﻥ ﺼﻔﺭ ﺇﻟﻰ ﻤﺎﻻ ﻨﻬﺎﻴﺔ.
8 Dk B T 2 ∞ → ν dν C3
∞
∞
°
°
∫ = ∫ u (ν1T) dν
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ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺘﺘﻨﺎﻗﺽ ﻤﻊ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﺤﻴﺙ ﺃﻥ ﺸﺩﺓ ﺍﻹﺸﻌﺎﻉ ﺘﺯﺩﺍﺩ ﻤﻊ ﺜﻡ ﺘﻘل ﺇﻟﻰ
ν max
ﺯﻴﺎﺩﺓ ﺍﻟﺘﺭﺩﺩ ﻟﺘﺼل ﺇﻟﻰ ﺃﻗﺼﻰ ﻗﻴﻤﺔ ﻋﻨﺩ ﺘﺭﺩﺩ ﻤﻌﻴﻥ ﻨﺭﻤﺯ ﻟﻪ ﺒـ
ﻭﻤﻥ ﺍﻟﺠـﺩﻴﺭ ﺒﺎﻟـﺫﻜﺭ.ﺍﻟﺼﻔﺭﺓ ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﻗﻴﻤﺔ ﺍﻟﺘﻜﺎﻤل ﻻ ﺘﺴﺎﻭﻱ ﻤﺎﻻ ﻨﻬﺎﻴﺔ .ﺘﺘﻐﻴﺭ ﻤﻊ ﺘﻐﻴﺭ ﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﺒﺎﻟﺸﻜل
ν max
ﻜﺫﻟﻙ ﻤﻼﺤﻅﺔ ﺃﻥ
FIGURE 1.1 Spectral distribution of the intensity of blackbody radiation as a function of frequency for several temperatures. The intensity is given in arbitrary units. The dashed line is the prediction of classical physics. As the temperature increases. The maximum shifts to higher frequencies and the total radiated energy (the area under each curve) increases significantly. Note that the horizontal axis is labeled by 14 -1 ν /10 s . This notation means that the dimensionless numbers on that axis re frequencies divided by 1014 s-1. We shall use this notation to label columns in tables and axes in figures because of its unambiguous nature and algebraic convenience.
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ﻜﻴﻑ ﺘﻡ ﺤل ﻫﺫﺍ ﺍﻹﺸﻜﺎل ﺒﻴﻥ ﺍﻟﻨﻅﺭﻴﺔ ﻭﺍﻟﺘﺠﺎﺭﺏ ﺍﻟﻌﻤﻠﻴﺔ؟ 2-1ﺘﻭﺯﻴﻊ ﺒﻼﻨﻙ ﻭﺘﻜﻤﻴﻡ ﺍﻟﻁﺎﻗﺔ 1-2 The Planck Distribution and the Quantum of Energy
ﺇﻥ ﺃﻭل ﻤﻥ ﻗﺩﻡ ﺘﻔﺴﻴﺭ ﺼﺤﻴﺢ ﻹﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ ﻫﻭ ﺍﻟﻌﺎﻟﻡ ﺍﻷﻟﻤـﺎﻨﻲ
ﻤﺎﻜﺱ ﺒﻼﻨﻙ Max Planckﻓﻲ ﻋﺎﻡ .1900ﻭﻓﻲ ﻨﻅﺭﻴﺘﻪ ،ﺍﻓﺘـﺭﺽ ﺒﻼﻨـﻙ ﺃﻥ ﺍﻹﺸﻌﺎﻉ ﺍﻟﻤﻨﺒﻌﺙ ﻤﻥ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ ﻤﻥ ﺍﻫﺘﺯﺍﺯ ﺍﻹﻟﻜﺘﺭﻭﻨـﺎﺕ ﺍﻟﻤﻜﻭﻨـﺔ ﻟﻤـﺎﺩﺓ ﺍﻟﺠﺴﻡ .ﻭﻟﻜﻭﻥ ﻫﺫﻩ ﺍﻻﻫﺘﺯﺍﺯﺍﺕ ﺫﺍﺕ ﺘﺭﺩﺩﺍﺕ ﻋﺎﻟﻴﺔ ﻓﺈﻨﻨﺎ ﻨﺠﺩ ﻓﻲ ﻁﻴﻑ ﺍﻹﺸﻌﺎﻉ
ﺍﻟﻤﻨﺒﻌﺙ ﺘﺭﺩﺩﺍﺕ ﻓﻲ ﻤﻨﻁﻘﺔ ﺍﻟﻀﻭﺀ ﺍﻟﻤﺭﺌﻲ ﻭﺍﻷﺸﻌﺔ ﺘﺤـﺕ ﺍﻟﺤﻤـﺭﺍﺀ ﻭﻓـﻭﻕ
ﺍﻟﺒﻨﻔﺴﺠﻴﺔ ﺒﻴﻨﻤﺎ ﻻ ﻨﺠﺩ ﺃﻱ ﻤﻥ ﺘﺭﺩﺩﺍﺕ ﺍﻟﺭﺍﺩﻴﻭ ﻓﻲ ﻫﺫﺍ ﺍﻟﻁﻴﻑ .ﻁﺒﻘـﺎﹰ ﻟﻨﻅﺭﻴـﺔ
ﺭﺍﻟﻲ-ﺠﻴﻨﺯ ،ﻓﺈﻨﻪ ﻤﻔﻬﻭﻡ ﻅﻤﻨﺎﹰ ﺃﻥ ﻁﺎﻗﺔ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻬﺘﺯﺓ ،ﻭﺍﻟﺘﻲ ﻫﻲ ﺴـﺒﺏ ﺍﻨﺒﻌﺎﺙ ﺍﻹﺸﻌﺎﻉ ﻤﻥ ﺍﻟﻤﺎﺩﺓ -ﻤﺴﻤﻭﺡ ﻟﻬﺎ ﺃﻥ ﺘﺄﺨﺫ ﺃﻱ ﻗﻴﻤﺔ ﻤﻥ ﺍﻟﻁﺎﻗـﺔ .ﻭﻫـﺫﻩ
ﺍﻟﻔﺭﻀﻴﺔ ﻫﻲ ﺇﺤﺩﻯ ﺍﻷﺴﺎﺴﻴﺎﺕ ﺍﻟﻔﺭﻀﻴﺔ ﻓﻲ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴـﺔ .ﻓـﻲ ﺍﻟﻔﻴﺯﻴـﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ،ﺍﻟﻜﻤﻴﺎﺕ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﻤﺘﻐﻴﺭﺓ ﻭﺍﻟﺘﻲ ﺘﹸﻤﺜـل ﻤـﺸﺎﻫﺩﺍﺕ )ﻤﺜـل ﺍﻟﻤﻭﻗـﻊ
،Positionﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ Momentumﻭﺍﻟﻁﺎﻗﺔ (energyﻴﻤﻜﻥ ﺘﻤﻠﻙ ﻗﻴﻡ ﻤﺘﺼﻠﺔ. In classical physics, the variables that represent observables (such as position, momentum and energy) can take on a continuum of values.
ﻭﻟﻘﺩ ﺃﺩﺭﻙ ﺍﻟﻌﺎﻟﻡ ﺒﻼﻨﻙ – ﺒﻌﻤﻕ ﺘﻔﻜﻴﺭﻩ – ﺒﻀﺭﻭﺭﺓ ﺇﺤﺩﺍﺙ ﺘﻐﻴﻴﺭ ﺠﺫﺭﻱ ﻭﺠﻭﻫﺭﻱ ﻓﻲ ﻫﺫﺍ ﺍﻟﻤﻔﻬﻭﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻲ ﻓﻜﺎﻨﺕ ﻓﺭﻀﻴﺘﻪ ﺍﻻﻨﻘﻼﺒﻴـﺔ ﻓـﻲ ﺍﻟﻔﻴﺯﻴـﺎﺀ
ﺍﻟﺤﺩﻴﺜﺔ :ﻁﺎﻗﺔ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻬﺘﺯﺓ ﻤﻜﻤﻤﺔ ﻭﻗﻴﻤﻬﺎ ﻏﻴﺭ ﻤﺘﺼﻠﺔ ﻭﺘﺘﻨﺎﺴﺏ ﺒـﺭﻗﻡ
ﻜﻤﻲ ﺼﺤﻴﺢ ﻤﻊ ﺍﻟﺘﺭﺩﺩ ﻭﺫﻟﻙ ﻤﻥ ﺨﻼل ﺍﻟﻤﻌﺎﺩﻟﺔ
E = nhν
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Energies of the oscillators were discrete and had to be proportional to an integral multiple of the frequently in an equation of the form E = nhv
ﺤﻴﺙ Eﻫﻲ ﻁﺎﻗﺔ ﺍﻟﻤﺘﺫﺒﺫﺏ n ،ﻫﻭ ﺭﻗﻡ ﺼﺤﻴﺢ h ،ﺜﺎﺒﺕ ﺍﻟﺘﻨﺎﺴﺏ ﻭﻴﻌﺭﻑ ﺒﺜﺎﺒﺕ ﺒﻼﻨﻙ ،ﻭ vﻫﻭ ﺘﺭﺩﺩ ﺍﻟﻤﺘﺫﺒﺫﺏ. ﻭﺒﻨﺎﺀﺍﹰ ﻋﻠﻰ ﻤﺒﺩﺃ ﺘﻜﻤﻴﻡ ﺍﻟﻁﺎﻗﺔ ﻭﻤﻔﺎﻫﻴﻡ ﺩﻴﻨﺎﻤﻴﻜﺎ ﺤﺭﺍﺭﻴﺔ ﺇﺤﺼﺎﺌﻴﺔ ،ﺘﻤﻜـﻥ
ﺒﻼﻨﻙ ﻤﻥ ﺍﺴﺘﻨﺘﺎﺝ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ:
8 Dh ν3 u (ν1 T) = 3 hv/k T c e B −1
)(1-2
ﻭﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﺘﺘﻔﻕ ﺘﻤﺎﻤﺎﹰ ﻤﻊ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﻋﻨﺩ ﻜل ﺍﻟﺘﺭﺩﺩﺍﺕ ﻭﺩﺭﺠﺎﺕ ﺍﻟﺤﺭﺍﺭﺓ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1-2ﺘﹸﻌﺭﻑ ﺒﺘﻭﺯﻴﻊ ﺒﻼﻨﻙ ﻹﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ Planck distribution lance for black body radiation.
) (iﻋﻨﺩ ﺍﻟﺘﺭﺩﺩﺍﺕ ﺍﻟﻤﻨﺨﻔﻀﺔ ،ﺘﺅﻭل ﻤﻌﺎﺩﻟﺔ ﺒﻼﻨﻙ ﺇﻟﻰ ﻤﻌﺎﺩﻟﺔ ﺭﺍﻟﻲ-ﺠﻴﻨﺯ ﺒﺎﻟﻨﻅﺭ ﺇﻟﻰ ﺍﻟﻨﺴﺒﺔ
hv k BT
ﻓﻌﻨﺩﻤﺎ ﺘﻜﻭﻥ vﺼﻐﻴﺭﺓ ﻓﺈﻥ
ﻤﻥ ﻤﻜﻨﻭﻥ ﺘﺎﻴﻠﻭﺭ ﻟﻠﺩﺍﻟﺔ ﺍﻷﺴﻴﺔ
v2 + ... 2i
hv 〈〈 1 k BT e v =1+ v +
ﺇﺫﺍ ﻜﺎﻨﺕ vﺼﻐﻴﺭﺓ ،ﻴﻤﻜﻨﻨﺎ ﺇﻫﻤﺎل ﺍﻟﺤﺩﻭﺩ ﺫﺍﺕ ﺍﻷﺴـﺱ ﺍﻟﻌﻠﻴـﺎ ﻭﻋﻠﻴـﻪ e v ≈1+ v
ﻭﺫﻟﻙ ﺇﺫﺍ ﻜﺎﻨﺕ
v 〈〈 1
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺔ ﻤﻌﺎﺩﻟﺔ ﺒﻼﻨﻙ
8Dh v3 8Dh v 3 u (v1 T) = 3 = c 1 + hν − 1 c 3 hv vT k BT 8D K T v3 c3 B
)(1-3
= )u (v1 T
ﻭﻫﺫﻩ ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺭﺍﻟﻲ -ﺠﻴﻨﺯ. ) (iiﺍﻟﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ
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∞
∞
v3 8Dh u (t) = ∫ u (v1T) dv = 3 ∫ dv hv/k T c ° e B −1 °
ﻭﻹﺠﺭﺍﺀ ﻫﺫﺍ ﺍﻟﺘﻜﺎﻤل ﻨﻌﻭﺽ ﻋـﻥ ﻭﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺔ dvﺒﺩﻻﻟﺔ dvﻭ ﻋﻥ vﺒـ
v
k BT h
dv
k BT h
hν k BT
= ، vﻫـﺫﺍ ﻴﻌﻨـﻲ ﺃﻥ
h dv k BT
= dv
= . dvﺤﺩﻭﺩ ﺍﻟﺘﻜﺎﻤل ﻟﻡ ﺘﺘﻐﻴﺭ ،ﺒـﺎﻟﺘﻌﻭﻴﺽ
ﻭﻜﺫﻟﻙ dvﺒﺩﻻﻟﺔ dvﻨﺘﺤﺼل ﻋﻠﻰ ∞
8Dh kT v 3 kT u (v, T) = 3 ∫ ( ) 3 v ( ) dv c ° h e −1 h ∞ ∞ 8Dh kT 4 v 3 8Dh k B T 4 v 3 ( ) dv = ( ) ∫° e v −1 dv D c 3 ∫° h e v − 1 c3
ﺍﻟﺘﻜﺎﻤل ∫ v dvﻗﻴﻤﺘﻪ ﺘﺴﺎﻭﻱ e −1 ∞
3
v
°
D4 15
8Dh k B T 4 D 4 8D 5 k B 4 4 ( ) T ( ) = h 15 15 c 3 h c3 )(1 − 4
ﺤﻴﺙ:
=
= )∴u (T
= aT 4 ............. 8D 5 k B 4 ) ( 15c 3 h
=a
ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1-4ﺘﻤﺜل ﻤﺎ ﻴﻌـﺭﻑ ﺒﻘـﺎﻨﻭﻥ ﺴـﺘﻴﻔﺎﻥ -ﺒﻭﻟﺘﺯﻤـﺎﻥ Stefan-
Boltzmanﻟﻠﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ ﻟﻜل ﻭﺤﺩﺓ ﺍﻟﺤﺠﻭﻡ.
) (iiiﻤﻌﺎﺩﻟﺔ ﺒﻼﻨﻙ ﺘﻤﻜﻨﻨﺎ ﻤﻥ ﺇ – ﺝ ﻗﺎﻨﻭﻥ ﻭﺍﻴﻥ Wienﻭﺍﻟﺫﻱ ﻴﻨﺹ ﻋﻠﻰ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ
λ max
ﻀﺭﺏ )..(1-5
ﺘﻤﺜل ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﺍﻟﻤﻘﺎﺒل ﻷﻗﺼﻰ ﻗﻴﻤﺔ ﻟﻜﺜﺎﻓﺔ ﺍﻟﻁﺎﻗﺔ uﻓﺈﻥ ﺤﺎﺼل
λ max
ﻓﻲ ﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ Tﻴﻌﻁﻴﻨﺎ ﻗﻴﻤﺔ ﺜﺎﺒﺘﺔ ،ﺃﻱ: λ max T = 2.90 × 10 −3 m.K
ﻭﻟﻠﺤﺼﻭل ﻋﻠﻰ ﻤﻌﺎﺩﻟﺔ ) (1-5ﻤﻥ ﻤﻌﺎﺩﻟﺔ ﺒﻼﻨﻙ. ﻟﻜﻲ ﻨﻭﺠﺩ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻘﺼﻭﻯ ﻟـ vﻭﺘﺴﺎﻭﻱ ﻫﺫﺍ ﺍﻟﺘﻔﺎﻀل ﺒﺎﻟﺼﻔﺭ ،ﺃﻱ
) u (ν
ﻴﻠﺯﻡ ﺃﻥ ﺘﻔﺎﻀل ﺍﻟﺩﺍﻟﺔ
) u (ν
ﺒﺎﻟﻨﺴﺒﺔ ﻟـ
) dv (ν =0 dv
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or d 8Dh v3 =0 3 hv/k BT dv c e − 1
ﻭﻤﺭﺓ ﺃﺨﺭﻯ ﻨﻌﻭﺽ ﻋﻥ
hv k BT
= dv , v
k BT h
vν
ﻭﺒﻘﺴﻤﺔ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻋﻠﻰ ﻜل ﺍﻟﺜﻭﺍﺒﺕ ﻨﺘﺤﺼل ﻋﻠﻰ. 3 (e v − 1) − ve v = o 3 e v − 3 = ve v
ﺒﺎﻟﻘﺴﻤﺔ ﻋﻠﻰ ev 3 − 3 e −v = v or 3 - v = 3e - v
ﻭﻴﻤﻜﻥ ﺤل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺨﻼل ﻨﻘﻁﺔ ﺍﻟﺘﻘﺎﻁﻊ ﺒﻴﻥ ﺍﻟﺩﺍﻟﺘﻴﻥ ) (3-vﻭ ) (3e-vﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﺒﺎﻟﺭﺴﻡ ﻓﻲ ﺸﻜل )(١-٤
ﺸﻜل :1-2ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ 3 – x = 3 e-xﺒﺎﻟﺭﺴﻡ
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ﻤﻥ ﺍﻟﺸﻜل ،ﻨﺠﺩ ﺃﻥ v = 2.82 or hv mar = 2.82 k BT or k BT
= v max
)(2.82 h h T c ∴ )= b (2.82 λ max h hc )(6.63 × 10 −34 J.s) (3.0 × 10 8 m/s = = 2.9 × 10 −3 m.K − 23 −1 2.82K B ) (2.82) (1.38× 10 J.v
= or λ max T
ﻭﻤﻥ ﺍﻟﺠﺩﻴﺭ ﺒﺎﻟﺫﻜﺭ ،ﺃﻥ ﻨﻅﺭﻴﺔ ﺇﺸﻌﺎﻉ ﺍﻟﺠﺴﻡ ﺍﻷﺴﻭﺩ ﺘﺴﺘﺨﺩﻡ ﻓﻲ ﻋﻠﻡ ﺍﻟﻔﻠﻙ ﻭﺫﻟﻙ ﻟﺘﺤﺩﻴﺩ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺍﻟﻨﺠﻭﻡ .ﺒﺎﻟﻨﻅﺭ ﻟﺸﻜل ﻭﺍﻟﺫﻱ ﻴﻭﻀﺢ ﻁﻴﻑ ﺇﺸـﻌﺎﻉ ﺍﻟﺸﻤﺱ ﻤﻘﺎﺱ ﻤﻥ ﻁﺒﻘﺎﺕ ﺍﻟﺠﻭ ﺍﻟﻌﻠﻴﺎ ﻟﻸﺭﺽ ،ﻨﺠﺩ ﺃﻥ ﻗﻴﻤﺔ
λ max = 500nm
،(1nm = 10-9 mﻭﺒﺎﺴﺘﺨﺩﺍﻡ ﻗﺎﻨﻭﻥ ﻭﺍﻴﻥ ﻴﺘﺒﻴﻥ ﺃﻥ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺍﻟﺸﻤﺱ. 2.9 ×10 −3 m.k 2.9 × 10 −3 m.k = = 5800k λ max 500 × 10 −3 m
)ﺤﻴﺙ
=T
ﺍﻟﻁﻴﻑ ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻲ ﻷﺸﻌﺔ ﺍﻟﺸﻤﺱ.
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FIGURE 1.3 The electromagnetic spectrum of the sun as measured in the upper atmosphere of the earth. A comparison of this figure with Figure 1.2 shows that the sun's surface radiates as a blackbody at a temperature of about 6000 k.
ﻭﺒﻌﺩﻤﺎ ﻋﺭﻓﻨﺎ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﻜﻤﻴﺔ ﻟﻠﻁﺎﻗﺔ )ﻤﻤﺜﻠﺔ (quantaﻭﻫﻲ nvﻓﺈﻥ ﻫـﺫﺍ
ﻴﻌﻨﻲ ﺃﻥ ﺸﻌﺎﻉ ﻁﻭﻟﻪ ﺍﻟﻤﻭﺠﻲ
°
λ = 6000 A
)ﺤﻴﺙ
m
−10
(1 A =10ﻁﺎﻗﺘﻪ ﺘﺴﺎﻭﻱ. °
)hc (6.63 × 10 34 J.s) (3 × 10 8 m/s = = E = hv λ (6000 ×10 −10 m = 3.3 × 10 −19 J = 2.06 ev
ﻓﺈﺫﺍ ﻜﺎﻥ ﻫﺫﺍ ﺍﻟﺸﻌﺎﻉ ﺼﺎﺩﺭ ﻤﻥ ﻤﺼﺩﺭ ﻗﺩﺭﺘـﻪ 100ﻭﺍﻁ )) (Wﻭﺤـﺩﺓ
ﺍﻟﻭﺍﻁ Wattﻫﻲ ﺠﻭل ﻟﻜل ﺜﺎﻨﻴﺔ ،(W: Joul/secﻫﺫﺍ ﻤﻌﻨﺎﻩ ﺃﻥ ﺇﺫﺍ ﻜﺎﻨﺕ Nﺘﺭﻤﺯ
ﻟﻌﺩﺩ ﺍﻟﻜﻤﺎﺕ ﻓﻲ ﺍﻟﺜﺎﻨﻴﺔ ﺍﻟﻭﺍﺤﺩﺓ ،ﻓﺈﺫﺍ ﻜﺎﻨﺕ ﻜل ﻜﻡ ﻤﻥ ﺍﻟﻁﺎﻗﺔ ﺘـﺴﺎﻭﻱ hvﻓـﺈﻥ ﺍﻟﻘﺩﺭﺓ ).Power = N (hv ∴ P = 100 = W = N hv P 100W = =∴N = 3×10 20 guanta/sec −19 E 3.3× 10 J
ﻭﻫﺫﺍ ﺒﻼ ﺸﻙ ﻋﺩﺩ ﻜﺒﻴﺭ ﺠﺩﺍﹰ ﻤﻥ ﺍﻟﻜﻤﺎﺕ ﻤﻤﺎ ﻴﺅﺩﻱ ﺇﻟﻰ ﻋـﺩﻡ ﺇﺤـﺴﺎﺴﻨﺎ ﺒﺎﻟﻁﺒﻴﻌﺔ ﺍﻟﺠﺴﻴﻤﻴﺔ ﻟﻠﻀﻭﺀ )ﻋﻠﻰ ﺍﻓﺘﺭﺍﺽ ﺃﻥ ﻜل ﻜﻡ ﻤﻥ ﺍﻟﻁﺎﻗﺔ ﻫﻭ ﺠﺴﻴﻡ(. 3-1ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﻜﻬﺭﻭﻀﻭﺌﻲ
1-3 The Photoelectric Effect
ﻓﻲ ﻋﺎﻡ 1887, 1886ﻭﺒﻴﻨﻤﺎ ﻜﺎﻥ ﻴﺠﺭﻱ ﺘﺠﺎﺭﺒﻪ ﺍﻟﺘـﻲ ﺃﻜـﺩﺕ ﻨﻅﺭﻴـﺔ
ﻤﺎﻜﺴﻭﻴل ﺍﻟﺨﺎﺼﺔ ﺒﺎﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﻀﻭﺀ ،ﺍﻜﺘﺸﻑ ﺍﻟﻔﻴﺯﻴﺎﺌﻲ ﺍﻷﻟﻤﺎﻨﻲ ﻫﻴﻨـﺭﻙ
ﻫﻴﺭﺘﺯ Heinrich Hertzﺃﻥ ﺍﻷﺸﻌﺔ ﺍﻟﻀﻭﺌﻴﺔ ﺍﻟﻔﻭﻕ ﺒﻨﻔـﺴﺠﻴﺔ ultraviolet light
ﺘﺴﺒﺏ ﺍﻨﻁﻼﻕ )ﺍﻨﺒﻌﺎﺙ( emissionﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻤﻥ ﺴﻁﺢ ﻤﻌﺩﻥ ﺒﺎﻹﺸﻌﺎﻉ ﻴﺴﻤﻰ
ﺒﺎﻟﺘﺄﺜﻴﺭ ﺍﻟﻜﻬﺭﻭﻀﻭﺌﻲ.
ﻭﻁﺒﻘﺎﹰ ﻟﻘﻭﺍﻨﻴﻥ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ،ﻓﺈﻥ ﺍﻹﺸﻌﺎﻉ ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻲ ﻋﺒﺎﺭﺓ ﻋﻥ ﻤﺠﺎل ﻜﻬﺭﺒﻲ ﻴﺘﺫﺒﺫﺏ ﻋﻤﻭﺩﻴﺎﹰ ﻋﻠﻰ ﺍﺘﺠﺎﻩ ﺍﻨﺒﻌﺎﺙ ﺍﻹﺸﻌﺎﻉ )ﻫﻨﺎ ﺃﻫﻤﻠﻨـﺎ ﺍﻟﻤﺠـﺎل - ١١PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﻟﻤﻐﻨﺎﻁﻴﺴﻲ( .ﻭﺍﻟﺫﻱ ﻨﺭﻴﺩ ﺃﻥ ﻨﺭﻜﺯ ﻋﻠﻴﻪ ﻫﻨﺎ ﺃﻥ ﺸﺩﺓ ﺍﻷﺸﻌﺔ intensity of the
radiationﺘﺘﻨﺎﺴﺏ ﻤﻊ ﻤﺭﺒﻊ ﺴﻌﺔ ﺍﻟﻤﺠﺎل ﺍﻟﻜﻬﺭﺒﻲ. The intensity of the radiation is proportioned to the square of the amplitude of the .oscillating electric field.
ﻭﻴﻤﻜﻥ ﻟﻼﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﺘﻲ ﻋﻠﻰ ﺴﻁﺢ ﺍﻟﻤﻌﺩﻥ ﺃﻥ ﺘﺘﺫﺒـﺫﺏ ﻤـﻊ ﺍﻟﻤﺠـﺎل ﺍﻟﻜﻬﺭﺒﻲ ﺍﻟﺴﺎﻗﻁ ﻋﻠﻴﻬﺎ ،ﻭﻜﻠﻤﺎ ﺍﺯﺩﺍﺩﺕ ﺸـﺩﺘﻪ )ﺴـﻌﺘﻪ( ﺘـﺯﺩﺍﺩ ﺴـﻌﺔ ﺘﺫﺒـﺫﺏ
ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻜﺜﻴﺭﺍﹰ ﻤﻤﺎ ﻴﺅﺩﻱ ﻓﻲ ﺍﻟﻨﻬﺎﻴﺔ ﺇﻟﻰ ﻜﺴﺭ ﺍﺭﺘﺒﺎﻁﻬﺎ ﺒﺎﻟﺴﻁﺢ ﻭﺍﻨﻁﻼﻗﻬـﺎ
ﺒﻁﺎﻗﺔ ﺤﺭﻜﻴﺔ kinetic energyﻭﺍﻟﺘﻲ ﺴﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺴﻌﺔ )ﺸﺩﺓ( ﺍﻟﻤﺠﺎل ﺍﻟﻜﻬﺭﺒﻲ ﻟﻺﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ .ﺇﻥ ﻫﺫﺍ ﺍﻟﺘﻔﺴﻴﺭ ﺍﻟﻔﻴﺯﻴﺎﺌﻲ )ﺍﻟﺘﻘﻠﻴـﺩﻱ( ﻴﺘﻌـﺎﺭﺽ ﺘﻤﺎﻤـﺎﹰ ﻤـﻊ ﺍﻟﻤﺸﺎﻫﺩﺍﺕ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﻟﻬﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﻭﺍﻟﺘﻲ ﺘﻤﺜﻠﺕ ﻓﻲ:
-1ﻭﺠﺩ ﺃﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻨﺒﻌﺜﺔ ﻤﻥ ﺍﻟﺴﻁﺢ ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺸﺩﺓ ﺍﻻﻨﺒﻌﺎﺙ .independent of the intensity of incidental -2ﻭﺠﺩ ﺘﺠﺭﻴﺒﻴﺎﹰ ﺃﻥ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻻ ﺘﻨﺒﻌﺙ ﻤﻥ ﺍﻟﺴﻁﺢ ﺇﻻ ﺇﺫﺍ ﻜﺎﻥ ﺘﺭﺩﺩ ﺍﻹﺸﻌﺎﻉ
ﺍﻟﺴﺎﻗﻁ ﺃﻜﺒﺭ ﻤﻥ ﺘﺭﺩﺩ ﻤﻌﻴﻥ ﺍﻟﺘﺭﺩﺩ
ν°
ν°
ﺒﻐﺽ ﺍﻟﻨﻅﺭ ﻋﻥ ﺸﺩﺓ ﺍﻹﺸـﻌﺎﻉ ﺴﻨـﺴﻤﻲ ﻫـﺫﺍ
ﺒﻌﺘﺒﺔ ﺍﻟﺘﺭﺩﺩ threshold frequencyﻭﻗﻴﻤﺔ
ν°
ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﻨﻭﻉ ﺍﻟﻤﻌﺩﻥ.
ﻭﻜﺫﻟﻙ ﻭﺠﺩ ﺃﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻟﻼﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻨﺒﻌﺜﺔ ﺘﺘﻨﺎﺴﺏ ﺨﻁﻴﺎﹰ ﻤﻊ ﺍﻟﺘﺭﺩﺩ ﻭﺫﻟﻙ ﺇﺫﺍ ﻜﺎﻨﺕ
ν
ﺃﻜﺒﺭ
ﻤﻥ ν °
ν
ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﻓﻲ ﺸﻜل .1-4
ﺸﻜل :1-4ﺘﻐﻴﺭ ﻗﻴﻡ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ﻤﻊ ﺍﻟﺘﺭﺩﺩ. - ١٢PDF created with pdfFactory Pro trial version www.pdffactory.com
FIGURE 1.4 The kinetic energy of electrons ejected from the surface of
sodium metal versus the frequency of the incident ultraviolet radiation. The )threshold frequency here is 4.40 × 1014 Hz (1 Hz = 1 s-1
ﻭﻟﺘﻔﺴﻴﺭ ﻫﺫﻩ ﺍﻟﻨﺘﺎﺌﺞ ،ﻁﻭﺭ ﺍﻟﻌﺎﻟﻡ ﺍﻴﻨﺸﺘﺎﻴﻥ ﻓﺭﻀﻴﺔ ﺒﻼﻨﻙ ﻤﻥ ﺘﻜﻤﻴﻡ ﺍﻟﻁﺎﻗﺔ.
ﻭﻫﻨﺎ ﻨﺫﻜﺭ ﺃﻥ ﺒﻼﻨﻙ ﻁﺒﻕ ﻤﻔﻬﻭﻡ ﺘﻜﻤﻴﻡ ﺍﻟﻁﺎﻗﺔ ﻓﻘﻁ ﻋﻠﻰ ﺍﻟﻤﺘﺫﺒﺫﺏ ﺍﻟﺫﻱ ﻴﻤـﺘﺹ ﺃﻭ ﻴﺒﻌﺙ ﺍﻟﻁﺎﻗﺔ )ﻓﺈﺫﺍ ﻤﺼﺩﺭ ﺘﺭﺩﺩﻩ
ν
ﻓﺈﻨﻪ ﻴﻤﻠﻙ ﻁﺎﻗﺔ
E = nhν
( ﺃﻤﺎ ﺇﺫﺍ ﻤﺎ ﺍﻨﺒﻌﺙ
ﺍﻹﺸﻌﺎﻉ ﻤﻥ ﻫﺫﺍ ﺍﻟﻤﺘﺫﺒﺫﺏ ﻓﺈﻨﻪ ﻴﺴﻠﻙ )ﻜﻤﺎ ﺍﻗﺘﺭﺡ ﺒﻼﻨﻙ( ﻜﻤﻭﺠﺔ ﻋﺎﺩﻴﺔ ﺘﺨـﻀﻊ
ﻟﻠﻤﻔﺎﻫﻴﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ .ﺃﻤﺎ ﺍﻟﻌﺎﻟﻡ ﺍﻴﻨﺸﺘﺎﻴﻥ ﻓﻘﺩ ﺫﻫﺏ ﺇﻟﻰ ﻤﺎ ﻫﻭ ﺃﺒﻌﺩ ﻤﻥ ﻫﺫﺍ،
ﻓﻘﺎل ﺇﻥ ﺍﻟﺘﻜﻤﻴﻡ ﻻ ﻴﻨﻁﺒﻕ ﻓﻘﻁ ﻋﻠﻰ ﺍﻟﻤﺘﺫﺒﺫﺏ ﺒل ﺇﻥ ﺍﻹﺸﻌﺎﻉ ﺍﻟﻤﻨﺒﻌﺙ ﻤﻨﻪ ﻴﻨﺒﻌﺙ
ﻜﻤﺎﺕ ﻤﻥ ﺍﻟﻁﺎﻗﺔ ) ﺃﻭ ﺤﺯﻡ ﻤﻥ ﺍﻟﻁﺎﻗﺔ( ﻤﻨﻔﺼﻠﺔ ﻋﻥ ﺒﻌﻀﻬﺎ ﻭﻁﺎﻗـﺔ ﻜـل ﻜـﻡ E = hν
ﻭﺃﻁﻠﻕ ﺍﺴﻡ ﺍﻟﻔﻭﺘﻭﻥ photonﻋﻠﻴﻪ.
Einstein proposed that the radiation itself existed as smell packers of energy, E = hν , now known as photons.
ﻭﺒﻨﺎﺀﺍﹰ ﻋﻠﻰ ﻤﺒﺩﺃ ﺤﻔﻅ ﺍﻟﻁﺎﻗﺔ ،Conservation of energyﻭﻀﺢ ﺍﻴﻨﺸﺘﺎﻴﻥ ﺃﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ
1 mv 2 2
= K. E
)ﺍﻟﺘﻲ ﻴﻨﺒﻌـﺙ ﺒﻬـﺎ ﺍﻹﻟﻜﺘـﺭﻭﻥ ﺫﻭ ﺍﻟﻜﺘﻠـﺔ m
ﻭﺍﻟﺴﺭﻋﺔ (vﻤﻥ ﺴﻁﺢ ﺍﻟﻤﻌﺩﻥ ﻋﺒﺎﺭﺓ ﻋﻥ ﺍﻟﻔﺭﻕ ﺒﻴﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟـﺴﺎﻗﻁﺔ ﻟﻠﻔﻭﺘـﻭﻥ hν
ﻭﺃﻗل ﻁﺎﻗﺔ ﻻﺯﻤﺔ ﻟﻨﺯﻉ ﺍﻹﻟﻜﺘﺭﻭﻥ )ﻭﺘﺤﺭﻴﺭﻩ( ﻤﻥ ﻁﺎﻗﺔ ﺭﺒﻁﺔ ﺒﺎﻟﻤﻌـﺩﻥ W
)ﻭﻴﺭﻤﺯ ﻟﻬﺎ ﺃﺤﻴﺎﻨﺎﹰ ﺒـ ( ο/ﻭﺍﻟﺘﻲ ﺘﺴﻤﻰ ﺩﺍﻟﺔ ﺍﻟﺸﻐل .Work function ﻭﻴﻤﻜﻨﻨﺎ ﺼﻴﺎﻏﺔ ﻤﺎ ﺴﺒﻕ ﺫﻜﺭﻩ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ: 1 mv 2 = hv − φ 2
)(1-5
ﻻﺤﻅ ﺃﻥ ﻗﻴﻤﺔ
1 2 mc 2
= K. E
ﻤﻭﺠﺒﺔ ﻭﻋﻠﻴﻪ ﻓـﺎﻟﻔﺭﻕ
ﺴﺎﻟﺏ .ﺃﻱ ﺃﻨﻪ . hν ≥ φﺇﻥ ﺍﻗل ﻗﻴﻤﺔ ﻟـ
hν
hν − φ
ﻫـﻲ ﺍﻟﻁﺎﻗـﺔ ﺍﻟﻼﺯﻤـﺔ ﻟﺘﺤﺭﻴـﺭ
ﺍﻹﻟﻜﺘﺭﻭﻥ ﻤﻥ ﺭﺒﻁ ﺍﻟﻨﻭﺍﺓ ﻭﻫﺫﺍ ﻴﺤﺩﺙ ﻋﻥ ﻋﺘﺒﺔ ﺍﻟﺘﺭﺩﺩ )(1-6
ﻻ ﻴﻤﻜـﻥ ﺃﻥ ﻴﻜـﻭﻥ
ν°
ﺃﻱ ﺃﻥ:
hν ° = φ
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ﻭﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺔ ﻤﻌﺎﺩﻟﺔ ) (1-5ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ. 1 mc 2 = hv − hv ° 2
)(1-7
ﻭﻫﺫﻩ ﻤﻌﺎﺩﻟﺔ ﺨﻁ ﻤﺴﺘﻘﻴﻡ )ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ (y = mx-cﻭﻫﻭ ﺘﻤﺎﻤﺎﹰ ﻤﺎ ﻴﺸﺎﻫﺩ
ﻤﻥ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﺍﻟﻤﻭﻀﺤﺔ ﻓﻲ ﺸﻜل .1-5
Figure 1-5. Photoelectric effect data showing a plot of retarding potential necessary to stop electron flow from a metal (lithium), or equivalently, electron kinetic energy, as a function of frequency of the incident light, the slope of the line is h/e.
ﻭﻗﺒل ﺃﻥ ﻨﻨﺘﻘل ﻟﻤﻭﻀﻭﻉ ﺁﺨﺭ ،ﻤﻥ ﺍﻟﻤﻬﻡ ﺃﻥ ﻨﻭﻀﺢ ﺍﻟﻭﺤﺩﺍﺕ ﺍﻟﻤـﺴﺘﺨﺩﻤﺔ ﻓـﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ .ﺒﺎﻟﻨﺴﺒﺔ ﻟـ
φ
ﻋﺎﺩﺓ ﺘﻌﻁﻲ ﺒﻭﺤﺩﺓ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻓﻭﻟﺕ ،eVﻭﻫـﺫﻩ ﻭﺤـﺩﺓ
ﻁﺎﻗﺔ ،ﻭﻟﺤﺴﺎﺏ )ﻤﺜﻼﹰ( ﺴﺭﻋﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺍﻟﻤﻨﺒﻌﺙ ﺒﻭﺤﺩﺓ m/sﻓﻤـﻥ ﺍﻟﻤﻬـﻡ ﺃﻥ ﻴﻌﺭﻑ ﺍﻟﻁﺎﻟﺏ ﻜﻴﻑ ﻴﻘﻭﻡ ﺒﻌﻤﻠﻴﺔ ﺍﻟﺘﺤﻭﻴل ﺒﻁﺭﻴﻘﺔ ﺼﺤﻴﺤﺔ. - ١٤ -
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ﺃﻭﻻﹰ :ﻤﻥ ﺘﻌﺭﻴﻑ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻓﻭﻟﺕ (1 coulomb) (1 volt) = 1 Joule
ﻭﺍﻟﺸﺤﻨﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻠﺒﺭﻭﺘﻭﻥ ﺘﺴﺎﻭﻱ ﻭﺍﻟﺸﺤﻨﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻠﺒﺭﻭﺘﻭﻥ ﺘﺴﺎﻭﻱ
1.6 ×10 −19 C
ﻋﻨﺩﺌﺫ
) Jﺘﺭﻤﺯ ﻟﻠﺠﻭل (
)1 eV = (1.6 × 10 −19 C) (1 V = 1.6 ×10 −19 J
ﻤﺜﺎل :2ﺩﺍﻟﺔ ﺍﻟﺸﻐل ﻟﻠﺼﻭﺩﻴﻭﻡ ﺘﺴﺎﻭﻱ ،1.82 enﺍﺤﺴﺏ ﻋﺘﺒﺔ ﺍﻟﺘـﺭﺩﺩ
ν°
ﻟﻠﺼﻭﺩﻴﻭﻡ؟ ﺍﻟﺤل :ﺃﻭﻻﹰ :ﻴﻠﺯﻡ ﺘﺤﻭﻴل
φ
ﻤﻥ eVﺇﻟﻰ ﺠﻭل: φ = 1.82 eV
J ) eV
= (1.82 eV) (1.6 × 10 −19 = 2.92 ×10 −19 J
ﻭﻤﻥ ﻤﻌﺎﺩﻟﺔ )(1-6
]= φ
°
[hνﻴﻤﻜﻨﻨﺎ
ﺤﺴﺎﺏ ν °
)φ (2.92 ×10 −19 J = )h (6.63×10 −3 J.S 1 = 4.40 ×1014 H 2 5
ﺤﻴﺙ Hzﺘﺭﻤﺯ ﻟﻠﻬﺭﺘﺯ ﻭﻫﻭ
= ν°
= 4.40 ×1014
1 ec 5
ﻤﺜﺎل:3 ﺃﺸﻌﺔ ﻓﻭﻕ ﺒﻨﻔﺴﺠﻴﺔ ﻁﻭﻟﻬﺎ ﺍﻟﻤﻭﺠﻲ 3500 Åﺘﺴﻘﻁ ﻋﻠﻰ ﺴﻁﺢ ﺒﻭﺘﺎﺴﻴﻭﻡ. ﺃﻗﺼﻰ ﻁﺎﻗﺔ ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻀﻭﺌﻴﺔ ﺘﺴﺎﻭﻱ .1.6 eV
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ﺍﺤﺴﺏ ﺩﺍﻟﺔ ﺍﻟﺸﻐل ﻟﻠﺒﻭﺘﺎﺴﻴﻭﻡ؟ ﺍﻟﺤل: ﻤﻥ ﻤﻌﺎﺩﻟﺔ )(1-5 K.E = hν − ο/ ∴φ = hν − k.E
ﻨﺤﺴﺏ ﺃﻭﻻﹰ ﻗﻴﻤﺔ
hν )hc (6.63×10 −34 J.s) (3 × 10 8 m/s = = hν λ )(3500 × 10 −10 m
ﻭﻟﺘﺤﻭﻴل ﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ ﻤﻥ ﺍﻟﺠﻭل ﻟﻺﻟﻜﺘﺭﻭﻥ ﻓﻭﻟﺕ ∴φ = 3. eV − 1.6 eV = 1.95 eV
ﻤﺜﺎل :ﻋﻨﺩ ﺘﻌﺭﺽ ﺴﻁﺢ ﻤﺎﺩﺓ ﺍﻟﻠﻴﺜﻴﻭﻡ ﻟﻺﺸﻌﺎﻉ ،ﻓﺈﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻨﺒﻌﺜﺔ
2.935 ×10 −19 J
ﻭﺫﻟﻙ ﺇﺫﺍ ﻜﺎﻨﺕ . λ = 300.0nm =
ﺇﻤﺎ ﺇﺫﺍ ﻜﺎﻨﺕ
λ
ﻓﺈﻥ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﺘﺴﺎﻭﻱ
1.28 × 10 −19 J
ﺍﺤﺴﺏ ) (aﺜﺎﺒﺕ ﺒﻼﻨﻙ؟ ) (bﻋﺘﺒﺔ ﺍﻟﺘﺭﺩﺩ؟ ) (cﺍﻟﺸﻐل؟ ﺍﻟﺤل :ﻤﻥ ﻤﻌﺎﺩﻟﺔ ) (1-5ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﺍﻷﻭل ﻭﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ
ﺍﻟﺜﺎﻨﻲ ﻨﺘﺤﺼل ﻋﻠﻰ ﺒﺎﻟﺘﻌﻭﻴﺽ:
1 1 ) − λ1 λ 2
( (K.E)1 − (K.E) 2 = h (ν1 − ν 2 ) = hc
1 1 2.935×10 −19 J − 1.280 × 10 −19 J = h (3×10 8 m/s) − −9 −9 3 ×10 m 400 ×10 m
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1.655 ×1019 J ∴h = 6.625 ×10 −34 J.S 14 −1 2.498 ×10 s
) (bﻟﺤﺴﺎﺏ ﻋﺘﺒﺔ ﺍﻟﺘﺭﺩﺩ ،ﻨﺄﺨﺫ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ )ﻤﺜﻼﹰ(
λ = 300 nm
∴ = hv − hv °
hc − hv ° 300 ×10 −9 m
= ∴ 2.935 ×10 −19 J
ﻭﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ ﻗﻴﻡ c,hﻨﺠﺩ ﺃﻥ ﻗﻴﻤﺔ
v°
v ° = 5.564 × 1014 Hz
) (٢ﺩﺍﻟﺔ ﺍﻟﺸﻐل ﺘﺤﺴﺏ ﻤﺒﺎﺸﺭﺓ ﻤﻥ ﺍﻟﻌﻼﻗﺔ
φ = hv°
= 3.687 ×10 −19 J 3.687 ×1019 J/ = = 2.30/eV J/ 1.6 ± ×10 −19 eV
4-1ﺘﺄﺜﻴﺭ ﻜﻭﻤﺒﺘﻭﻥ: 1-4 The Compton Effect ﺍﻟﺘﺠﺭﺒﺔ ﺍﻟﺘﻲ ﺘﺅﻜﺩ ﺒﻭﻀﻭﺡ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﺠـﺴﻴﻤﻴﺔ ﻟﻺﺸـﻌﺎﻉ the particle
nature of radiationﺘﺴﻤﻰ ﺒﺘﺄﺜﻴﺭ )ﻅﺎﻫﺭﺓ( ﻜﻭﻤﺒﺘﻭﻥ ﻨﺴﺒﺔ ﻟﻠﻌـﺎﻟﻡ ﻜﻭﻤﺒﺘـﻭﻥ
Arthur H. Comptonﻟﻘﺩ ﺍﻜﺘﺸﻑ ﻜﻭﻤﺒﺘﻭﻥ ﺃﻨﻪ ﺇﺫﺍ ﺍﺨﺘﺭﻕ ﺇﺸـﻌﺎﻉ ﺫﻭ ﻁـﻭل
ﻤﻭﺠﻲ )ﻓﻲ ﻤﻨﻁﻘﺔ ﺍﻷﺸﻌﺔ ﺍﻟﺴﻴﻨﻴﺔ (X-rayﺸﺭﻴﺤﺔ ﻤﻌﺩﻨﻴﺔ ﻓـﺴﻴﺘﺒﻌﺜﺭ scattered
ﺒﻁﺭﻴﻘﺔ ﻻ ﻴﻤﻜﻥ ﺘﻔﺴﻴﺭﻫﺎ ﺤﺴﺏ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻟﻺﺸﻌﺎﻉ. ﺍﻟﺫﻱ ﺘﺨﺒﺭﻨﺎ ﺒﻪ ﻗﻭﺍﻨﻴﻥ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺃﻥ ﺸﺩﺓ ﺍﻹﺸﻌﺎﻉ Iﺍﻟﻤﻨﺒﻌﺙ ﻤـﻥ
ﻤﺎﺩﺓ ﻨﺘﻴﺠﺔ ﺘﺄﺜﺭﻫﺎ ﺒﺈﺸﻌﺎﻉ ﺴﻘﻁ ﻋﻠﻴﻬﺎ )ﻤﻤﺎ ﻴﺅﺩﻱ ﺇﻟﻰ ﺍﻫﺘﺯﺍﺯ ﺇﻟﻜﺘﺭﻭﻨﺎﺘﻬﺎ ﻭﺍﻟﺘـﻲ ﺒﺩﻭﺭﻫﺎ ﺴﺘﺒﻌﺙ ﺇﺸﻌﺎﻉ( ﻋﻨﺩﻤﺎ ﺘﻘﺎﺱ ﻋﻨﺩ ﺯﺍﻭﻴﺔ ﺍﻟﺴﺎﻗﻁﺔ( ﻓﺈﻥ Iﺘﺘﻐﻴﺭ ﻤﻊ
θ
ﺤﺴﺏ ﺍﻟﻌﻼﻗﺔ
θ
)ﺒﺎﻟﻨـﺴﺒﺔ ﻻﺘﺠـﺎﻩ ﺍﻷﺸـﻌﺔ
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)Ι ≈ (1 + cos 2 θ
)(1-8
ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ Iﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻟﻸﺸﻌﺔ ﺍﻟـﺴﺎﻗﻁﺔ ،ﻭﻫـﺫﺍ ﻴﺘﻌﺎﺭﺽ ﺒﻭﻀﻭﺡ ﻤﻊ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ )ﺸﻜل (1-6ﻭﺍﻟﺘﻲ ﻴﺘﺒﻴﻥ ﺒﻭﻀﻭﺡ ﺘﻐﻴﺭ I
ﺒﺘﻐﻴﺭ . λ
Figure 1-6. The spectrum of radiation scattered by carbon, showing the unmodified line at 0.7078 Å on the left and the shifted line at 0.7314 Å on the right. The former is the wave-length of the primary radiation.
ﻨﺘﺎﺌﺞ ﺘﺠﺭﺒﺔ ﻜﻭﻤﺒﺘﻭﻥ: ﻟﻘﺩ ﻭﺠﺩ ﻜﻭﻤﺒﺘﻭﻥ ﺃﻥ ﺍﻹﺸﻌﺎﻉ ﺍﻟﻤﺘﺒﻌﺜﺭ ﻟﻪ ﻤﺭﻜﺒﺘﻴﻥ؛ ﻤﺭﻜﺒﺔ ﻁﻭﻟﻬﺎ ﺍﻟﻤﻭﺠﻲ
ﻤﺴﺎﻭٍ ﻟﻁﻭل ﻤﻭﺠﺔ ﺍﻹﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ ﻭﻤﺭﻜﺒﺔ ﺃﺨﺭﻯ ﺘﺨﺘﻠﻑ ﻓﻲ ﻁﻭﻟﻬﺎ ﺍﻟﻤـﻭﺠﻲ ﻋﻥ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻟﻺﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ ﻭﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺯﺍﻭﻴﺔ ﺍﻟﺒﻌﺜـﺭﺓ ﻭﻗـﺩ ﺘﻤﻜـﻥ ﻜﻭﻤﺒﺘﻭﻥ ﻤﻥ ﺸﺭﺡ ﻭﺠﻭﺩ ﻤﺭﻜﺒﺔ ﺍﻹﺸـﻌﺎﻉ ﺍﻟﻤﺘﺒﻌﺜـﺭﺓ ﺫﺍﺕ ﺍﻟﻁـﻭل ﺍﻟﻤـﻭﺠﻲ
ﺍﻟﻤﺨﺘﻠﻑ ﻋﻥ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﺍﻟﺴﺎﻗﻁ ﻭﺫﻟﻙ ﺒﺎﻋﺘﺒﺎﺭ ﺍﻟﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ ﻋﺒﺎﺭﺓ ﻋـﻥ
ﺸﻌﺎﻉ ﻤﻥ ﺍﻟﻔﻭﺘﻭﻨﺎﺕ ﺒﻁﺎﻗﺔ
hv
ﺤﻴﺙ ﻴﻌﺎﻨﻲ ﻜل ﻓﻭﺘﻭﻥ ﻤﻥ ﺘﺒﻌﺜﺭ )ﺘﺸﺘﺕ( ﻤـﺭﻥ
elastic scatteringﻤﻊ ﻜل ﺇﻟﻜﺘﺭﻭﻥ.
ﻭﻜﻤﺎ ﻫﻭ ﻤﻌﻠﻭﻡ ،ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﺘﺸﺘﺕ ﺍﻟﻤﺭﻥ ﻓﺈﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜـﺔ momentum ﻭﺍﻟﻁﺎﻗﺔ energyﻜﻤﻴﺎﺕ ﺘﺨﻀﻊ ﻟﻘﺎﻨﻭﻥ ﺍﻟﺤﻔﻅ )ﺍﻟﺒﻘﺎﺀ( momentum and energy
.must be conserved
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ﻭﻟﺘﻔﺴﻴﺭ ﻫﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﺭﻴﺎﻀﻴﺎﹰ ،ﺍﻓﺘﺭﺽ ﻜﻭﻤﺒﺘﻭﻥ ﺃﻥ ﺍﻟﻔﻭﺘﻭﻥ ﻟـﻪ ﻜﻤﻴـﺔ ﺤﺭﻜﺔ pﺘﻌﻁﻲ ﺒﺎﻟﻌﻼﻗﺔ hv c
)(1-9
=p
ﺤﻴﺙ ﺘﻡ ﺍﻋﺘﺒﺎﺭ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺩﻴﻨﺎﻤﻴﻜﻴﺔ ﻟﻠﻔﻭﺘﻭﻥ ﻜﺠﺴﻴﻡ ﻴﺨﻀﻊ ﻟﻘﻭﺍﻨﻴﻥ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻨﺴﺒﻴﺔ ﻭﺍﻟﺘﻲ ﺘﻭﻀﺢ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻟﻁﺎﻗﺔ ﻭﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ.
)] (1-10 2
1
[
E = (m o c 2 ) 2 + (pc) 2
ﺤﻴﺙ moﻫﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﺴﻜﻭﻨﻴﺔ rest massﻟﻠﺠﺴﻴﻡ ،ﻭﺴﺭﻋﺔ ﺍﻟﺠﺴﻴﻡ vﻋﻨﺩ
ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ pﺘﻌﻁﻰ ﺒﺎﻟﻌﻼﻗﺔ 2
]
1
[
aE d = (m ο c 2 ) 2 + (pc) 2 ap dp
d )(pc ap
1 2
−
(pc)c 2
)(1-11
]
[
1 = = (m ο c 2 ) 2 + (pc) 2 2
]
1
=v
1 1 2 (m c 2 ) 2 + (pc) 2 °
[
pc 2 pc 2 = E (m 2 c 4 + p 2 C 2 ) 12 °
ﻭ
ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻔﻭﺘﻭﻥ ،ﻨﻌﻭﺽ ﻋﻥ
m° = o
)(1-12
pc 2 ⇒E=pc E
v= c
=
= ∴v
ﻓﻲ ﻤﻌﺎﺩﻟﺔ )(1-11
= ∴C/
ﻤﻥ ﻤﻌﺎﺩﻟﺔ ) (1-12ﻨﺘﺤﺼل ﻋﻠﻰ ﻤﻌﺎﺩﻟﺔ ) (1-9ﺤﻴﺙ E hv = C C
E = hv
=p
ﺩﻋﻨﺎ ﻨﻔﺘﺭﺽ ﺍﻵﻥ ﻭﺠﻭﺩ ﻓﻭﺘﻭﻥ ﺒﻜﻤﻴﺔ ﺤﺭﻜﺔ ﺍﺒﺘﺩﺍﺌﻴـﺔ
p°
ﺴـﺎﻗﻁ ﻋﻠـﻰ
ﺇﻟﻜﺘﺭﻭﻥ ﺴﺎﻜﻥ .ﻭﺒﻌﺩ ﺍﻟﺘﺼﺎﺩﻡ ،ﻨﻔﺘﺭﺽ ﺃﻥ ﻜﻤﻴـﺔ ﺍﻟﺤﺭﻜـﺔ ﻟﻠﻔﻭﺘـﻭﻥ
p
ﺃﻤـﺎ
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ﺍﻹﻟﻜﺘﺭﻭﻥ ﻓﻴﺤﺩﺙ ﻟﻪ ﺍﺭﺘﺩﺍﺩ recoilﺒﻜﻤﻴﺔ ﺤﺭﻜﺔ . pﻭﺒﺘﻁﺒﻴﻕ ﻗﺎﻨﻭﻥ ﺒﻘﺎﺀ ﻜﻤﻴـﺔ e
ﺍﻟﺤﺭﻜﺔ )ﺍﻨﻅﺭ ﺸﻜل .(1-7 )(1-13
pe = p ° + p
ﺒﺘﺭﺒﻴﻊ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ.
2 2 r r 2 p e = p ° − p − 2p ° p
)(1-14
ﻨﻁﺒﻕ ﺍﻵﻥ ﻗﺎﻨﻭﻥ ﺒﻘﺎﺀ ﺍﻟﻁﺎﻗﺔ
ﺃﻭﻻﹰ ،ﻁﺎﻗﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ Eﻗﺒل ﺍﻟﺘﺼﺎﺩﻡ ] 2
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ
m° = 0
ﺘﺴﺎﻭﻱ ﺼﻔﺭ(.
1
[
E = (m ° c 2 ) 2 + (pc) 2
ﻷﻥ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺴـﺎﻜﻥ ﻗﺒـل ﺍﻟﺘـﺼﺎﺩﻡ )ﻷﻥ ﺃﻥ P ∴E = m ° c = mc 2
)(1-15
ﻗﺎﻨﻭﻥ ﺒﻘﺎﺀ ﺍﻟﻁﺎﻗﺔ :ﻁﺎﻗﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻭﺍﻟﻔﻭﺘﻭﻥ ﺍﻟﺴﺎﻗﻁ ﻗﺒل ﺍﻟﺘﺼﺎﺩﻡ = ﻁﺎﻗﺔ
ﺍﻹﻟﻜﺘﺭﻭﻥ ﻭﺍﻟﻔﻭﺘﻭﻥ ﺒﻌﺩ ﺍﻟﺘﺼﺎﺩﻡ 2
1
) hv ° + mc 2 = hv + (m 2 c 4 + p e c 2 2
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ﺒﻨﻘل hvﻟﻠﻁﺭﻑ ﺍﻷﻴﺴﺭ ﻭﺘﺭﺒﻴﻊ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (hv ° + mc 2 − hv) 2 = (m 2c 4 + pe c 2 2
or m 2c 4 + pe c 2 = (hv ° − hv + mc 2 )2 2
= (hν ° − hv) 2 + 2m c 2 (hv ° − hv) + m 2c 4
)(1-16
ﻤﻥ ﻤﻌﺎﺩﻟﺔ ) ،(1-14ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ
hv hv , p° = ° c c
hv ° 2 hv 2 hv hv ) + ( ) − 2 ( ° ) ( ) cosθ c c c c
ﺤﻴﺙ
θ
=p
( = ∴p e 2
ﻫﻲ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﺤﺼﻭﺭﺓ ﺒﻴﻥ ﺍﺘﺠﺎﻩ ﺍﻟـﺸﻌﺎﻉ ﺍﻟﻤـﺸﺘﺕ ﻭﺍﻟـﺸﻌﺎﻉ
ﺍﻟﺴﺎﻗﻁ .ﺒﻀﺭﺏ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻓﻲ C2
pe c 2 = (hv° )2 + (hv)2 − 2 (hv ° ) (hv) cosθ 2
)(1-17
ﻭﻟﻠﺤﺼﻭل ﻋﻠﻰ ﻤﺭﺒﻊ ﻜﺎﻤل ،ﻨﻀﻴﻑ ﻭﻨﻁﺭﺡ ] [2hv° hvﻟﻤﻌﺎﺩﻟﺔ )(1-17 pe c 2 (hv ° ) 2 + (hv) 2 − 2hv °hv + 2(hv ° ) (hv) − 2(hv° ) (hv) cosθ 1444 424444 3 14444 4244444 3 2
)2hv ° hv (1− cos θ
)∴pe c 2 = (hv° − hv) 2 + 2hv °hv (1 − cos θ 2
)(1-18
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ ﻗﻴﻤﺔ
ﻋﻠﻰ
(hv ° − hv) 2
(hv ° − hv) 2
ﻤﻥ ﻤﻌﺎﺩﻟﺔ ) (1-18ﻓﻲ ﻤﻌﺎﺩﻟﺔ ) (1-18ﻨﺘﺤـﺼل
m 2 c 4 + pe c 2 = p e c 2 − 2hv °hv (1 − cos θ) + 2mc 2 (hv ° − hv) + m 2c 4 2
2
) ..(1-19ﺒﺤﺫﻑ ﺍﻟﺤﺩﻭﺩ ﺍﻟﻤﺘﺸﺎﺒﻬﺔ ﻤﻥ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ:
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2hv °hv (1 − cos θ) = 2mc 2 (hv ° − hv) − )= 2mc 2 h (v° − v 1 1 c c ) − ) 2mc 2 hc ( − λ° λ λ° λ
( = 2mc 2 h
λ − λ° c c h (1 − cos θ) = 2mc 2 h λ°λ λ° λ
2h or
λ − λ° c/ 2 (1 − cos θ) 2/ mc 3/ h/ 2/ h λ/ °λ/ λ °λ 2/
) h (1 − cos θ) = mc (λ − λ °
)...(1-20 ﻻﺤﻅ ﺃﻥ
h )(1 − cos θ cm h mc
= or λ − λ °
ﻓﻲ ﻤﻌﺎﺩﻟﺔ ) (1-20ﻟﻪ ﺒﻌﺩ ﺍﻟﻁﻭل ﻭﻫﺫﺍ ﺍﻟﺤﺩ ﻴـﺴﻤﻰ ﻁـﻭل
ﻤﻭﺠﺔ ﻜﻭﻤﺒﺘﻭﻥ Compton wavelengthﻟﻼﻟﻜﺘﺭﻭﻥ ﻭﻤﻘﺩﺍﺭﻩ h ≅ 2.4 × 10 −10 cm mc
)(1-21
ﻭﻗﺩ ﺘﺒﻴﻥ ﻤﻥ ﺍﻟﻘﻴﺎﺴﺎﺕ ﺍﻟﻤﻌﻤﻠﻴﺔ ﺃﻥ ﻁﻭل ﻤﻭﺠﺔ ﺍﻟﻔﻭﺘـﻭﻥ ﻭﺍﻟﻤﺘـﺸﺘﺕ ﺘﺘﻁﺎﺒﻕ ﻤﻊ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻨﻅﺭﻴﺔ .ﺃﻤﺎ ﺍﻟﻤﺭﻜﺒﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻟـ
ﻴﺒﻴﻥ ﻤﺭﻜﺒﺘﻴﻥ ﻟـ
λ
ﺇﺤﺩﺍﻫﻤﺎ ﺘﺨﺘﻠﻑ ﻋﻥ
λ°
λ
)ﺍﻨﻅﺭ ﺸﻜل
λ
ﻭﺍﻟـﺫﻱ
ﻭﺍﻷﺨﺭﻯ ﻤﺴﺎﻭﻴﺔ ﻟـ ( λﻭﺍﻟﺘﻲ ﻻ °
ﺘﺘﻐﻴﺭ ﺒﺎﻟﻨﺴﺒﺔ ﻓﺈﻥ ﻤﻨﺸﺄﻫﺎ ﻫﻭ ﺘﺼﺎﺩﻡ ﺍﻟﻔﻭﺘﻭﻥ ﺍﻟﺴﺎﻗﻁ ﻤﻊ ﺍﻟﺫﺭﺓ ﻜﻜل ،ﻓﻠﻭ ﻋﻭﻀﻨﺎ
ﻋﻥ mﺒﻜﺘﻠﺔ ﺍﻟﺫﺭﺓ )ﺒﺩﻻﹰ ﻤﻥ ﻜﺘﻠﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ( ﻭﺤﻴﺙ ﺃﻥ ﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ ﻓـﻲ ﺍﻟﻤﻘـﺎﻡ )ﻭﻫﻲ ﻜﺒﻴﺭﺓ ﺠﺩﺍﹰ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻜﺘﻠﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ( ﻓﺈﻥ ﺍﻟﺤﺩ ﺠﺩﺍﹰ ﻗﺭﻴﺒﺔ ﻤﻥ ﺍﻟﺼﻔﺭ ،ﺃﻱ ﺃﻥ
λ − λ° ≅ ο
ﻭﻫﺫﺍ ﻴﻌﻨﻲ
h mc
ﺴﺘﻜﻭﻥ ﻗﻴﻤﺘﻪ ﺼـﻐﻴﺭﺓ
λ ≈ λ°
ﻭﺃﺨﻴـﺭﺍﹰ ﻤـﺎ ﺍﻟـﺫﻱ
ﻨﺴﺘﻨﺘﺠﻪ ﻤﻥ ﺘﺄﺜﻴﺭ ﻜﻭﻤﺒﺘﻭﻥ؟ ﺇﻥ ﺍﻟﻘﻴﺎﺴﺎﺕ ﺍﻟﺘﻲ ﺃﺠﺭﻴﺕ ﻋﻠﻰ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺍﻟﻤﺭﺘﺩ ﻭﺍﻟﻔﻭﺘﻭﻥ ﺍﻟﻤﺘﺒﻌﺜﺭ ﻤﻨـﻪ
ﺘﺅﻜﺩ -ﺒﻤﺎ ﻻ ﻴﺩﻉ ﻤﺠﺎﻻﹰ ﻟﻠﺸﻙ – ﺒﺄﻥ ﻫﺫﺍ ﺍﻟﺘﺼﺎﺩﻡ ﻤﻤﺎﺜل ﻟﻠﺘﺼﺎﺩﻡ ﺍﻟﺫﻱ ﻴﺤـﺩﺙ
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ﺒﻴﻥ ﻜﹸﺭﺘﻲ ﺒﻠﻴﺎﺭﺩﻭ ،ﺃﻱ ﺃﻥ ﺍﻟﻔﻭﺘﻭﻥ )ﺃﻭ ﺍﻟﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ( ﻴﺠﺏ ﺃﻥ ﺘﺘﻌﺎﻤـل ﻤﻌـﻪ ﻋﻠﻰ ﺃﺴﺎﺱ ﺃﻨﻪ ﺠﺴﻴﻡ ،ﻭﻫﺫﺍ ﻴﺅﻜﺩ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﺠﺴﻴﻤﻴﺔ ﻟﻺﺸﻌﺎﻉ. 1-5ﺍﻟﺨﺼﺎﺌﺹ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﻤﺎﺩﺓ ﻭﺤﻴﻭﺩ ﺍﻹﻟﻜﺘﺭﻭﻥ 1-5 Wave Prosperities and Electron Diffraction
ﻭﺍﺠﻪ ﺍﻟﻌﻠﻤﺎﺀ ﻜﺜﻴﺭﺍﹰ ﻤﻥ ﺍﻟﺼﻌﻭﺒﺎﺕ ﻓﻲ ﻭﺼﻑ ﻁﺒﻴﻌـﺔ ﺍﻟـﻀﻭﺀ .ﺒﻌـﺽ ﺍﻟﺘﺠﺎﺭﺏ ﺘﺒﻴﻥ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﻀﻭﺀ )ﻤﺜـل؟ dispersionﺍﻟـﻀﻭﺀ ﺍﻷﺒـﻴﺽ
ﻟﻤﺭﻜﺒﺎﺕ ﻁﻴﻔﻪ ﻋﻨﺩ ﻤﺭﻭﺭﻩ ﺩﺍﺨل ﻤﻨﺸﻭﺭ (prismﻭﺍﻟﺒﻌﺽ ﺍﻵﺨﺭ ﻤﻥ ﺍﻟﺘﺠـﺎﺭﺏ
ﻴﺜﺒﺕ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﺠﺴﻴﻤﻴﺔ ﻟﻠﻀﻭﺀ )ﻤﺜل ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﻜﻬﺭﻭﻀﻭﺌﻲ(..
ﻭﻫﺫﺍ ﻤﺎ ﻴﻌﺭﻑ ﺒﺎﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﺯﺩﻭﺠﺔ ﻟﻠـﻀﻭﺀ ware- particle duality of
.lights
ﻓﻲ ﻋﺎﻡ ،1924ﻗﺩﻡ ﻋﺎﻟﻡ ﻓﺭﻨﺴﻲ ﻴﺩﻋﻰ ﺩﻴﺒﺭﻭﻟـﻲ – Louis de Broglie ﻭﺒﻨﺎﺀﺍﹰ ﻋﻠﻰ ﻤﺎ ﺴﺒﻕ ﺫﻜﺭﻩ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻀﻭﺀ -ﻨﻤﻭﺫﺠﺎﹰ ﻴﺒﻴﻥ ﻓﻴﻪ ﺍﻟﻁﺒﻴﻌـﺔ ﺍﻟﻤﻭﺠﻴـﺔ ﻟﻠﻤﺎﺩﺓ؛ ﻓﺈﺫﺍ ﻜﺎﻥ ﺍﻟﻀﻭﺀ -ﺫﻭ ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﻭﺠﻴﺔ – ﻴﺴﻠﻙ ﺃﺤﻴﺎﻨﺎﹰ ﻜﻤـﺎ ﻟـﻭ ﻜـﺎﻥ
ﺠﺴﻴﻤﺎﺕ ،ﻓﻠﻤﺎﺫﺍ ﻻ ﻴﻜﻭﻥ ﻟﻠﻤﺎﺩﺓ ﻁﺒﻴﻌﺔ ﻤﻭﺠﻴﺔ؟!! ﻭﻗﺩ ﺼﺎﻍ ﺩﻴﺒﺭﻭﻟﻲ ﻓﻜﺭﺘﻪ ﻫﺫﻩ ﺒﺼﻴﻐﺔ ﺭﻴﺎﻀﻴﺔ -ﺍﻗﺘﺒﺴﻬﺎ ﻤﻥ ﻋﻼﻗﺔ ﺍﻴﻨﺸﺘﺎﻴﻥ ﺍﻟﺘﻲ ﺘﺭﺘﺒﻁ ﺒـﻴﻥ ﻁـﻭل ﻤﻭﺠـﺔ ﺍﻟﻔﻭﺘﻭﻥ
ﺤﺭﻜﺘﻪ ﻤﻭﺠﻲ
λ
ﻭﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ P
p = mv
λ
)...(1-22
)(١
،ﻓﺈﺫﺍ ﻜﺎﻥ ﺠﺴﻴﻡ ﻜﺘﻠﺘﻪ mﻭﺴﺭﻋﺘﻪ vﻓﺈﻥ ﻜﻤﻴﺔ
ﻋﻨﺩﺌﺫ ﻓﺈﻥ ﺍﻟﺠﺴﻴﻡ – ﻭﺤﺴﺏ ﻨﻤﻭﺫﺝ ﺩﻴﺒﺭﻭﻟﻲ – ﺴﻴﻜﻭﻥ ﻟﻪ ﻁـﻭل
ﻴﻌﻁﻲ ﺒﺎﻟﻌﻼﻗﺔ. h p
λ
) (١ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﻲ ﻗﺩﻤﻬﺎ ﺍﻴﻨﺸﺘﺎﻴﻥ ﻟﻠﻔﻭﺘﻭﻥ ﺍﻟﺫﻱ ﺘﺭﺩﺩﻩ vﻭﻁﻭل ﻤﻭﺠﺘﻪ λﻭﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ p c hc hc h h = = = = v hv E E p ) ( c
=λ
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ﺤﻴﺙ hﻫﻭ ﺜﺎﺒﺕ ﺒﻼﻨﻙ. ﻤﺜﺎل :5 ﺍﺤﺴﺏ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻟﻜﺭﺓ ﻜﺘﻠﺘﻬﺎ 0.14 kgﻭﺴﺭﻋﺘﻬﺎ 40 m/sﻭﻗﺎﺭﻥ ﻫﺫﺍ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻤﻊ ﻁﻭل ﻤﻭﺠﺔ ﺇﻟﻜﺘﺭﻭﻥ ﺴﺭﻋﺘﻪ 1.00%ﻤﻥ ﺴﺭﻋﺔ ﺍﻟﻀﻭﺀ؟ ﺍﻟﺤل: ﻨﻭﺠﺩ ﺃﻭﻻﹰ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻟﻠﻜﺭﺓ )p = mco = (0.14kg) (40m/s = 5.6 kg.m.s -1
ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ
λ
ﺤﺴﺏ ﻤﻌﺎﺩﻟﺔ ﺩﻴﺒﺭﻭﻟﻲ h 6.63×10 −34 J.s = =λ = 1.2 × 10 −34 m −1 p 5.6 kg ms
!!!
ﻻﺤﻅ ﺃﻥ ﻫﺫﺍ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻤﺘﻨﺎﻫﻲ ﻓﻲ ﺍﻟﺼﻐﺭ ﻨﻭﺠﺩ ﺍﻵﻥ ﻜﻤﻴﺔ ﺤﺭﻜﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ) p = m e v = (9.1× 10 −31 kg) (2.998× 10 6 ms −1
)ﻻﺤﻅ ﺃﻥ ﺍﻟﺴﺭﻋﺔ vﻟﻺﻟﻜﺘﺭﻭﻥ ﻫﻲ 1%ﻤﻥ ﺴﺭﻋﺔ ﺍﻟـﻀﻭﺀ ﻜﻤـﺎ ﻫـﻭ ﻤﻌﻁﻰ ﻓﻲ ﺍﻟﺴﺅﺍل(. ∴ p = 2.73× 10 −24 kg.m.s −1
ﻁﻭل ﻤﻭﺠﺔ ﺩﻴﺒﺭﻭﻟﻲ ﻟﻺﻟﻜﺘﺭﻭﻥ. h 6.63 ×10 −34 J.s = =λ = 2.43× 10 −10 m p 2.73× 10 − 24 kg.m.s −1 = 243 pm
)ﺤﻴﺙ pmﻴﺴﺎﻭﻱ
m
−12
..(10ﻭﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ 243ﻤﻘﺎﺭﻨﺔ ﻟﻸﺒﻌﺎﺩ ﺍﻟﺫﺭﻴﺔ..
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ﻤﻥ ﺨﻼل ﻫﺫﻩ ﺍﻟﻘﻴﻡ ،ﻴﺘﻀﺢ ﻟﻨﺎ ﺃﻥ ﻁﻭل ﻤﻭﺠﺔ ﺩﻴﺒﺭﻭﻟﻲ ﻟﻺﻟﻜﺘﺭﻭﻥ ﻤﻘﺎﺭﺒﺔ ﻟﻁﻭل ﻤﻭﺠﺔ ﺍﻷﺸﻌﺔ ﺍﻟﺴﻴﻨﻴﺔ .ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺴﻴﺴﻠﻙ ﻜﻤﺎ ﻟﻭ ﻜﺎﻥ ﺃﺸﻌﺔ
ﺴﻴﻨﻴﺔ!!.
ﺃﻤﺎ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻜﺭﺓ ﻓﺈﻥ ﻁﻭل ﻤﻭﺠﺔ ﺩﻴﺒﺭﻭﻟﻲ ﻟﻬﺎ ﻗﺼﻴﺭ ﺠﺩﺍﹰ ﻤﻘﺎﺭﻨﺔ ﺒﺎﻷﺒﻌﺎﺩ ﺍﻟﺫﺭﻴﺔ... ﻤﻭﺠﺎﺕ ﺩﻴﺒﺭﻭﻟﻲ ﻴﻤﻜﻥ ﻤﺸﺎﻫﺩﺘﻬﺎ ﺘﺠﺭﻴﺒﻴﺎﹰ de Broglie Waves are observed Experimentally
ﻋﻨﺩ ﺍﺼﻁﺩﺍﻡ ﺃﺸﻌﺔ Xﺒﻤﺎﺩﺓ ﺒﻠﻭﺭﻴﺔ ﻓﺈﻨﻪ ﻴﺘﺸﺘﺕ ﺒﻁﺭﻴﻘـﺔ ﻤﻌﻴﻨـﺔ ﺘﻌﻜـﺱ ﻁﺒﻴﻌﺔ ﺍﻟﺘﺭﺘﻴﺏ ﺍﻟﺩﻭﺭﻱ ﺍﻟﺒﻠﻭﺭﻱ ﻟﻠﻤﺎﺩﺓ ﻫﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﺘﹲﻌﺭﻑ ﺒﺤﻴـﻭﺩ ﺃﺸـﻌﺔ .X
X-ray diffractionﻭﺴﺒﺏ ﺤﺩﻭﺙ ﻫﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﻫﻭ ﺃﻥ ﺍﻷﺒﻌﺎﺩ ﺍﻟﺒﻠﻭﺭﻴﺔ )ﺍﻟﻤﺴﺎﻓﺔ
ﺒﻴﻥ ﺍﻟﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﺒﻠﻭﺭﻴﺔ( ﻤﺘﻘﺎﺭﺒﺔ ﻤﻊ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻷﺸﻌﺔ .X
ﻭﻤﻥ ﺍﻟﻨﺎﺤﻴﺔ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ،ﻓﻘﺩ ﻭﺠﺩ ﺃﻥ ﺘﺒﻌﺜﺭ ﺸﻌﺎﻉ ﻤﻥ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺒﻤـﺎﺩﺓ
ﺒﻠﻭﺭﻴﺔ ﻴﻨﺸﺄ ﻋﻨﻪ ﺤﻴﻭﺩ ﻤﺸﺎﺒﻪ ﻟﺤﻴﻭﺩ ﺃﺸﻌﺔ Xﻤﻤﺎ ﺒﻴﻥ -ﺒﻤﺎ ﻻ ﻴﺩﻉ ﻤﺠﺎﻻﹰ ﻟﻠﺸﻙ- ﺍﻟﻁﺒﻴﻌﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﺠﺴﻴﻤﺎﺕ.
ﺸﻜل 1-8ﻴﻭﻀﺢ ﺘﺸﺘﺕ ﻤﻭﺠـﺎﺕ ﻨﺘﻴﺠـﺔ ﺍﺼـﻁﺩﺍﻤﻬﺎ ﺒﺘﺭﻜﻴـﺏ ﺩﻭﺭﻱ
) periodic Structureﻤﺜﻼﹰ ﺸﻜل ﺒﻠﻭﺭﻱ( .ﻭﻤﻥ ﺍﻟﺸﻜل ﻴﺘﻀﺢ ﺃﻨﻪ ﺴﻴﻜﻭﻥ ﻓـﺭﻕ ﻓﻲ ﺍﻟﻁﻭﺭ phase differenceﺒﻴﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﻤﻨﻌﻜﺴﺔ ﻤﻥ ﻤـﺴﺘﻭﻴﺎﺕ ﺒﻠﻭﺭﻴـﺔ ﻤﺘﺠﺎﻭﺭﺓ ﻭﻤﻘﺩﺍﺭ ﻓﺭﻕ ﺍﻟﻁﻭﺭ . ( 2D ) 2a sin θ λ
ﻭﻋﻨﺩﻤﺎ ﻴﺴﺎﻭﻱ ﻓﺭﻕ ﺍﻟﻁﻭﺭ
2 Dn
)ﺤﻴﺙ nﻫﻭ ﺭﻗﻡ ﺼـﺤﻴﺢ( ﻓـﺴﻴﺘﺤﻘﻕ
ﺸﺭﻁ ﺍﻟﺘﺩﺍﺨل ﺍﻟﺒﻨﱠﺎﺀ ،ﺃﻱ ﺃﻥ. 2D 2a sinθ = 2D n λ
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Figure 1-8 the scattering of X-rays by crystal planes when the angle 0 between the scattered X-rays and the crystal plane equals the angle between the incident X-rays and the plane. λ=
2a sin θ n
... (1-23)
ﺇﻥ ﺸﻜل ﺍﻟﺘﺩﺍﺨل ﺍﻟﻤﺸﺎﻫﺩ ﺘﺠﺭﻴﺒﻴﺎﹰ ﻤﻥ ﺘﺸﺘﺕ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻨﺘﻴﺠﺔ ﺍﺴـﺘﺩﺍﻤﻬﺎ (Davisson and ﺒﺒﻠﻭﺭﺓ )ﻜﻤﺎ ﺘﺤﻘﻕ ﻋﻠﻰ ﻴﺩﻱ ﺍﻟﻌـﺎﻟﻤﻴﻥ ﺩﺍﻓﻴـﺴﻭﻥ ﻭ ﺠﻴﺭﻤـﺭ (1- ﺘﻌﻁﻲ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ
λ
( ﻭﺫﻟﻙ ﺒﺸﺭﻁ ﺃﻥ1-23) ﻴﻤﻜﻥ ﺘﻔﺴﻴﺭﻩ ﺒﺎﻟﻤﻌﺎﺩﻟﺔGermer
ﺇﻥ ﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﺴﺎﻫﻤﺕ ﺒﺨﻁـﻭﺓ.1-9 ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﻓﻲ ﺸﻜل22) .Wave mechanics ﻜﺒﻴﺭﺓ ﻓﻲ ﺘﻁﻭﻴﺭ ﻋﻠﻡ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻤﻭﺠﺎﺕ
FIGURE 1.9 (a) the X-ray diffraction pattern of aluminum foil. (b) the electron diffraction pattern of aluminum foil. The similarity of these two patterns shows that electrons can behave like X-rays and display wavelike properties.
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1-6ﺫﺭﺓ ﺒﻭﺭ 1-6 The Bohr Atom "The Bohr Theory of the Hydrogen Atom Can be Used "Drive the Rydberg For mole
ﻓﻲ ﻋﺎﻡ ،1911ﻗﺩﻡ ﺍﻟﻌﺎﻟﻡ ﺍﻟﺩﻨﻤﺎﺭﻜﻲ ﻨﻴﻠﺱ ﺒـﻭﺭ Niels Bohrﻨﻅﺭﻴﺘـﻪ
ﺍﻟﺸﻬﻴﺭﺓ ﻟﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﻭﺍﻟﺘﻲ ﺘﺸﺭﺡ ﻭﺘﻔﺴﺭ ﺒﺒﺴﺎﻁﺔ ﺍﻟﻁﻴـﻑ ﺍﻟﻤﻨﺒﻌـﺙ ﻤـﻥ ﺍﻟﺫﺭﺓ. ﻁﺒﻘﺎﹰ ﻟﻠﻨﻤﻭﺫﺝ ﺍﻟﻨﻭﻭﻱ ﻟﻠﺫﺭﺓ ،ﻭﺍﻟﺫﻱ ﻴﻘﻭﻡ ﻋﻠﻰ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﻟﺘﻁـﺎﻴﺭ
ﺠﺴﻴﻤﺎﺕ ، αﻴﻤﻜﻥ ﺍﻋﺘﺒﺎﺭ ﺃﻥ ﻜﺘﻠﺔ ﺍﻟﺫﺭﺓ ﻤﺘﺭﻜﺯﺓ ﻓﻲ ﺍﻟﻨﻭﺍﺓ ﻭﺍﻟﺘﻲ ﺘﻌﺘﺒـﺭ ﺜﺎﺒﺘـﺔ ﻭﻴﺩﻭﺭ ﺤﻭﻟﻬﺎ ﺇﻟﻜﺘﺭﻭﻥ .ﺍﻟﻘﻭﺓ Fﺍﻟﺘﻲ ﻴﺭﺘﺒﻁ ﺒﻬﺎ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻓﻲ ﻤﺩﺍﺭ ﺩﺍﺌﺭﻱ ﻫـﻲ ﻗﻭﺓ ﻜﻭﻟﻭﻡ ﺤﺴﺏ ﻗﺎﻨﻭﻥ ﻜﻭﻟﻭﻡ.
)... (1-24
)1 (Ze) (e 4Dε ° r 2
=F
ﺤﻴﺙ ) (Zeﻫﻲ ﺸﺤﻨﺔ ﺍﻟﻨﻭﺍﺓ )ﻟﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ( Z =1ﻭ ) (eﻫﻲ ﺸﺤﻨﺔ
ﺍﻹﻟﻜﺘﺭﻭﻥ.
ε ° = 8.85× 10 −12 c 2 N −1 m −2
,
r
ﻨﺼﻑ ﻗﻁﺭ ﺍﻟﺫﺭﺓ .ﺘﺘﻭﺍﺯﻥ ﻗﻭﺓ ﻜﻭﻟﻭﻡ ﻤﻊ ﺍﻟﻘﻭﺓ
ﺍﻟﻁﺎﺭﻗﺔ ﺍﻟﻤﺭﻜﺯﻴﺔ. )...(1-25
mv 2 r
=F
ﺤﻴﺙ vﻫﻲ ﺍﻟﺴﺭﻋﺔ ﺍﻟﺨﻁﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻥ. ﺒﻤﺴﺎﻭﺍﺓ ﺍﻟﻘﻭﺘﺎﻥ ﻨﺠﺩ: )...(1-26
1 e 2 mv 2 = 4Dε ° r 2 r
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ﻁﺒﻘﺎﹰ ﻟﻘﻭﺍﻨﻴﻥ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﻜﻼﺴﻴﻜﻴﺔ ،ﻓﺈﻥ ﺍﻟﺠﺴﻡ ﺍﻟﻤﺸﺤﻭﻥ ﺍﻟﻤﺘﺴﺎﺭﻉ .ﻴﺼﺩﺭ ﺇﺸﻌﺎﻉ ﻤﻤﺎ ﻴﺅﺩﻱ ﺇﻟﻰ ﻓﻘﺩﺍﻥ ﻟﻁﺎﻗﺘﻪ ،ﻭﻟﻬﺫﺍ ﺍﻟﺴﺒﺏ ﻓﺈﻥ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺴـﻴﻔﻘﺩ ﻁﺎﻗﺘـﻪ
ﺨﻼل ﺩﻭﺭﺍﻨﻪ ﺤﻭل ﺍﻟﻨﻭﺍﺓ ﻭﺴﻴﺩﻭﺭ ﻓﻲ ﺸﻜل ﺤﻠﺯﻭﻨﻲ ﻭﻴﺘﻼﺸﻰ ﺩﺍﺨـل ﺍﻟﻨـﻭﺍﺓ،
ﻭﻋﻠﻴﻪ ﻓﻼ ﻴﻤﻜﻥ ﺃﻥ ﻴﻭﺠﺩ ﻤﺩﺍﺭ ﻤﺴﺘﻘﺭ stable orbitﻟﻼﻟﻜﺘﺭﻭﻥ .ﻭﻟﻠﺨﺭﻭﺝ ﻤـﻥ ﻫﺫﺍ ﺍﻹﺸﻜﺎل ﺍﻗﺘﺭﺡ ﺒﻭﺭ ﻓﺭﻀﻴﺎﺘﻪ ﺍﻟﺘﻲ ﺘﺨﺎﻟﻑ ﻗﻭﺍﻨﻴﻥ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ. ﺍﻟﻔﺭﻀﻴﺔ ﺍﻷﻭﻟﻰ: ﺍﻟﻤﺩﺍﺭﺍﺕ ﺍﻹﻟﻜﺘﺭﻭﻨﻴﺔ ﺍﻟﻤﺴﺘﻘﺭﺓ )ﺍﻟﺜﺎﺒﺘﺔ( Stationary electron orbits
ﻭﻫﺫﻩ ﺍﻟﻔﺭﻀﻴﺔ ﻫﻲ ﺘﹶﺤﺩٍ ﺒﺎﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻟﻠﻔﻴﺯﻴﺎﺀ .ﻭﻟﻘﺩ ﺤﺩﺩ ﺒـﻭﺭ ﻫـﺫﻩ
ﺍﻟﻤﺩﺍﺭﺍﺕ ﺒﺎﺴﺘﺤﺩﺍﺙ ﺸﺭﻁ ﺍﻟﺘﻜﻤﻴﻡ quantization conditionﻭﺍﻓﺘﺭﺽ ﺃﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺯﺍﻭﻴﺔ angular momentum Lﻤﻜﻤﻠﺔ ﺤﺴﺏ ﺍﻟﻌﻼﻗﺔ. L = mvr = nh , n = 1, 2, ........
)...(1-27 ﻤﻥ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﻨﺠﺩ ﺃﻥ
hh mr
=v
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )(1-26 1 e 2 m nh 2 m n 2 h 2 ) ( = 4Dε ° r 2 r mr m2 r3
ﻭﻓﻴﻬﺎ ﻨﺠﺩ ﺃﻥ. )...(1-28
4D ε ° h 2 2 n2 h2 1 2 .n = e =⇒ r 4Dε ° mr me 2
ﻭﻤﻥ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﻨﺠﺩ ﺃﻥ ﺃﻨﺼﺎﻑ ﺃﻗﻁﺎﺭ ﺍﻟﻤﺩﺍﺭﺍﺕ ﻟﻬﺎ ﻤﻘﺎﺩﻴﺭ ﻤﺤـﺩﺩﺓ ﺃﻭ ﻤﻜﻤﻤﺔ .ﺇﻥ ﺃﻗل ﻨﺼﻑ ﻗﻁﺭ ﻟﻺﻟﻜﺘﺭﻭﻥ ﻴﻭﺠﺩ ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ .n = 1 - ٢٨PDF created with pdfFactory Pro trial version www.pdffactory.com
( u D ) ( 8.85×10−12 c 2 N −1 m −2 ) (1.055× 10−34 J.s) 2 =r (9.11× 10 − 31 kg) (1.602 × 10 −19 c) 2 °
= 5.29 ×10 −11 m = 52.9 pm ≈ 0.53 A
ﻭﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ ﻴﺭﻤﺯ ﻟﻬﺎ ﻋﺎﺩﺓ ﺒـ
a°
ﺍﻟﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻥ Eﻫﻲ ﻋﺒﺎﺭﺓ ﻋﻥ ﻤﺠﻤﻭﻉ ﻁـﺎﻗﺘﻲ ﺍﻟﺤﺭﻜـﺔ k.E
ﻭﺍﻟﺘﻲ ﺘﺴﺎﻭﻱ )...(1-29
1 mv 2 2
e2 1 4D ε ° r
ﻭﻁﺎﻗﺔ ﺍﻟﻭﻀﻊ ) V(rﻭﺼﻴﻐﺘﻬﺎ.
V (r) = −
ﺍﻹﺸﺎﺭﺓ ﺍﻟﺴﺎﻟﺒﺔ ﻓﻲ ) (1-29ﺘﻌﻨﻲ ﺃﻥ ﺍﻟﺒﺭﻭﺘﻭﻥ ﻭﺍﻹﻟﻜﺘﺭﻭﻥ ﻴﺠـﺫﺏ ﻜـل
ﻤﻨﻬﻤﺎ ﺍﻵﺨﺭ.
ﻻﺤﻅ ﺃﻥ ﻁﺎﻗﺔ ﺍﻟﺘﺠﺎﺫﺏ ﺒﻴﻥ ﺍﻟﺒﺭﻭﺘﻭﻥ ﻭﺍﻹﻟﻜﺘﺭﻭﻥ ﺘﻘل ﻜﻠﻤﺎ ﺯﺍﺩﺕ ﺍﻟﻤﺴﺎﻓﺔ
ﺒﻴﻨﻬﻤﺎ ﻭﻋﻨﺩ )...(1-30
∞= r
ﻓﺈﻥ
V (∞ ) = 0
1 e2 1 mv 2 + (− ) 2 4Dε ° r
ﻤﻥ ﺍﻟﻌﻼﻗﺔ )(1-26
= )E = KE + V (r
1 e 2 mv 2 = 2 r 4Dε ° r
ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ mv2ﺒﺎﻟﺼﻴﻐﺔ
1 e2 4Dε ° r
= mv 2
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )(1-30 1 1 e2 1 e2 ( = ∴E )− 2 4D ε ° r 4Dε ° r )......(1 − 3
1 1 e2 1 e2 =− 2 4Dε ° r 8Dε ° r
=−
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ rﻤﻥ ) (1-28ﻓﻲ ) (1-31ﻨﺠﺩ: h 2D me 4 1 2 2 2 2 32 D ε ° h n
=− n2
1 e2 8Dε ° 4Dε ° h 2 2
=h
E=−
me
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ﺍﻻﺨﺘﺼﺎﺭ
h 2h
=h 1 n2 n =1,2,.............
me 4
E=−
2
h 4D 2 me 4 1 ∴E n = − 2 2 2 8ε ° h n 2
(32)D 2 ε °
)...(1-32 ﺍﻹﺸﺎﺭﺓ ﺍﻟﺴﺎﻟﺒﺔ ﻓﻲ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﺩل ﻋﻠﻰ ﺃﻥ ﺤﺎﻻﺕ ﺍﻟﻁﺎﻗﺔ ﺤﺎﻻﺕ ﻤﻘﻴﺩﺓ
.bound states
ﻻﺤﻅ ﺃﻥ ﻓﻲ ﺤﺎﻟﺔ n = 1ﻓﺈﻥ ﻤﻌﺎﺩﻟﺔ ) (1- 32ﺘﻌﻁﻴﻨﺎ ﺍﻟﺤﺎﻟﺔ ﺫﺍﺕ ﺃﻗل ﻁﺎﻗﺔ state of lowest energyﻭﺘﺴﻤﻰ ﻫﺫﻩ ﺍﻟﻁﺎﻗﺔ ﺒﺎﻟﻁﺎﻗﺔ ﺍﻷﺭﻀﻴﺔ ﺃﻭ ﻁﺎﻗﺔ ﺍﻟﺤﺎﻟـﺔ
ﺍﻷﺭﻀﻴﺔ .ground-state energyﺃﻤﺎ ﺤﺎﻻﺕ ﺍﻟﻁﺎﻗﺔ ﺍﻷﻋﻠﻰ ﺘـﺴﻤﻰ ﺍﻟﺤـﺎﻻﺕ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ excited statesﻭﻋﺎﺩﺓ ﻤﺎ ﺘﻜﻭﻥ ﻏﻴﺭ ﻤﺴﺘﻘﺭﺓ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺤﺎﻟﺔ ﺍﻷﺭﻀﻴﺔ،
ﻭﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ ﺍﻟﺫﺭﺓ )ﺃﻭ ﺍﻟﺠﺯﻱﺀ( ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ ﻓﺈﻨﻬﺎ ﺘـﺴﺘﺭﺨﻲ ﻭﺘﺭﺠـﻊ
ﻟﻠﺤﺎﻟﺔ ﺍﻷﺭﻀﻴﺔ ﻭﺘﻌﻁـﻲ ﺃﻭ ﺘـﺘﺨﻠﺹ ﻤـﻥ ﻁﺎﻗﺘﻬـﺎ ﻓـﻲ ﺼـﻭﺭﺓ ﻤﻭﺠـﺎﺕ
ﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻴﺔ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﻓﻲ ﺍﻟﺸﻜل ).(1-10
ﺍﻟﻔﺭﻀﻴﺔ ﺍﻟﺜﺎﻨﻴﺔ :ﻴﻤﻜﻥ ﻟﻼﻟﻜﺘﺭﻭﻨﺎﺕ ﺃﻥ ﺘﻨﺘﻘل )ﺃﻭ ﺘﻘﻔـﺯ( ﻤـﻥ ﻤـﺩﺍﺭﺍﺘﻬﺎ
ﺒﻁﺭﻴﻘﺔ ﻏﻴﺭ ﻤﺘﺼﻠﺔ discontinuous transitionsﻭﺃﻥ ﺍﻟﺘﻐﻴﺭ ﻓـﻲ ﺍﻟﻁﺎﻗـﺔ ΔE
ﻴﺅﺩﻱ ﻻﻨﺒﻌﺎﺙ ﺇﺸﻌﺎﻉ ﻟﻪ ﺘﺭﺩﺩ . hvﻭﻟﻬﺫﺍ ﺍﻟﺴﺒﺏ ﻟﻭ ﺍﻨﺘﻘل ﺇﻟﻜﺘﺭﻭﻥ ﻤـﻥ ﺍﻟﻤـﺩﺍﺭ ﺍﻟﺫﻱ ﻟﻪ n2 =1ﺇﻟﻰ ﺍﻟﻤﺩﺍﺭ ﺫﻭ n1 = 2ﻓﺈﻥ ﺍﻟﻔﺭﻕ ﻓﻲ ﺍﻟﻁﺎﻗﺔ. me 4 1 me 4 1 − − ( ) 2 2 2 2 8ε ° h 2 n 2 8 ε ° h 2 n1 )...............(1 − 33
ΔE = E 2 − E 1 = −
me 4 1 1 ΔE = 2 2 ( 2 − 2 ) = hv 8ε ° h n 1 n 2
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FIGURE 1.10 The energy level diagram for the hydrogen atom, showing how transitions from higher states into some particular state lead to the observed spectral series for hydrogen.
( ﻴﻤﻜﻨﻨﺎ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﻟﻠﻌﻼﻗﺔ ﺍﻟﻌﺩﺩﻴﺔ1-33) ﻤﻥ ﻤﻌﺎﺩﻟﺔ
.ﺍﻟﻤﻌﺭﻭﻓﺔ ﺒﻤﻌﺎﺩﻟﺔ ﺭﻴﺩﺒﺭﺝ ﻭﺍﻟﺘﻲ ﺘﺼﻑ ﺠﻤﻴﻊ ﺨﻁﻭﻁ ﺍﻟﻁﻴﻑ ﻟﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ .( ﻨﺠﺩ1-33) ﻓﻲ
hc λ
ﺒـhv ﺒﺎﻟﺘﻌﻭﻴﺽ ﻋﻥ
hc me 4 1 1 = 2 2 ( 2 − 2) λ 8ε ° h n 1 n 2 1 λ 1 me 4 1 1 = 2 3 ( 2 − 2) λ 8ε ° h c n 1 n 2 1 1 1 = R H ( 2 − 2 ) ..............(1 − 34) λ n1 n2
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ﻭﻓﻴﻬﺎ ﻨﺠﺩ
ﺤﻴﺙ RHﻫﻭ ﺜﺎﺒﺕ ﺭﻴﺩﺒﺭﺝ Rydberg constant me 4 (9.11× 10 −31 kg) (1.60× 10−19 c)4 = 2 )8ε ° h 3c (8) (8.85× 10 −12 c 2 N −1m − 2 ) (6.63 × 10 − 34 J.s)3 (2.0 × 108 m/s
= RH
R H =1.089 × 107 m −1 )(1
ﺒﺎﻟﻨﻅﺭ ﻟﺨﻁﻭﻁ ﺍﻟﻁﻴﻑ ﻓﻲ ﺸﻜل 1-10ﻴﻤﻜﻥ ﺃﻥ ﻨﻼﺤﻅ ﺍﻟﺘﻭﺍﻓﻕ ﺒﻴﻥ ﻨﻤﻭﺫﺝ ﺒﻭﺭ ﻭﻫﺫﻩ ﺍﻟﺨﻁﻭﻁ ﺍﻟﻤﺘﻔﻘﺔ ﻤﻊ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ .ﻓﻤﺜﻼﹰ ﺨﻁﻭﻁ ﺴﻠـﺴﻠﺔ ﻟﻴﻤـﺎﻥ Lymanﺘﻨﺸﺄ ﻤﻥ ﺭﺠﻭﻉ )ﺍﺴﺘﺭﺨﺎﺀ( ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﻌﻠﻴﺎ
ﻟﻠﻤﺩﺍﺭ ﺍﻷﻭل ) ،(n = 1ﻭﻜﺫﻟﻙ ﺨﻁﻭﻁ ﺴﻠﺴﻠﺔ ﺒﺎﻟﻤﺭ Balmer seriesﺘﺤﺩﺙ ﻤـﻥ ﺍﺴﺘﺭﺨﺎﺀ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ ﻤﻥ ﻜل ﺍﻟﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﻌﻠﻴـﺎ ﻟﻠﻤـﺴﺘﻭﻯ ﺍﻟﺜـﺎﻨﻲ
).(n=2
٢
ﻤﺜﺎل 6
ﺍﺤﺴﺏ ﻁﺎﻗﺔ ﺍﻟﺘﺄﻴﻥ ﻟﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ؟ ?Calculate the ionization energy of the hydrogen atom
ﺍﻟﺤل: ﻁﺎﻗﺔ ﺍﻟﺘﺄﻴﻥ ﻫﻲ ﺍﻟﻁﺎﻗﺔ ﺍﻟﻼﺯﻤﺔ ﻹﺯﺍﻟﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻤـﻥ ﻤـﺴﺘﻭﻯ ﺍﻟﺤﺎﻟـﺔ
ﺍﻷﺭﻀﻴﺔ n1 = 1ﺇﻟﻰ ﺍﻟﺤﺎﻟﺔ ﻏﻴﺭ ﺍﻟﻤﻘﻴﺩﺓ ﺃﻱ
∞= n2
me 4 1 1 )( 2 − 2 2 2 8ε ° h n 1 n 2
= ΔE
(9.1 × 10 - 34 kg) (1.9 × 10 −19 c) 4 1 1 = )( 2 − 2 − 12 2 −1 −2 − 34 2 ∞ (8) (8.85 × 10 c N m ) (6.63 × 10 J.s) 1 = 2.18 × 10 -18 J = 13.6 eV
ﻓﻲ ﻋﺎﻡ ،1925ﻅﻬﺭﺕ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﺤﺩﻴﺜﺔ ﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻭﺫﻟﻙ ﺒﻨﺎﺀﺍﹰ ﻋﻠـﻰ ﺍﻷﺒﺤﺎﺙ ﺍﻟﺘﻲ ﺭﺴﺦ ﺩﻋﺎﺌﻤﻬﺎ ﻜل ﻤﻥ ﺍﻟﻌﺎﻟﻡ ﻫﻴﺯﻨﺒﺭﺝ W. Heisenbergﻭﺍﻟﻌـﺎﻟﻡ (2 ) The exact value of RH = 1.09736 × 107 m-1 = 109.736 cm-1
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ﻤﺎﻜﺱ ﺒﻭﺭﻥ ،M. Bornﻭﺍﻟﻌﺎﻟﻡ ﺸـﺭﻭﺩﻨﺠﺭ E. Shrödingerﻭﺍﻟﻌـﺎﻟﻡ ﺩﻴـﺭﺍﻙ .P. Dirac ﻭﺘﺸﺭﺡ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻜﻤﻴﺔ ﺍﻟﺤﺩﻴﺜﺔ ﺍﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﻤﺤﻴﺭﺓ ﺒﺸﺭﻁ ﺃﻥ ﻨﺘﺨﻠـﻰ ﻋـﻥ
ﺒﻌﺽ ﺍﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﺘﻲ ﺘﺭﺴﺨﺕ ﻓﻲ ﺃﺫﻫﺎﻨﻨﺎ ﻤﻥ ﻋﻠﻡ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ!. (Q1ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ﻜﺜﺎﻓﺔ ﺍﻹﺸﻌﺎﻉ ﺘﻌﻁﻰ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ
8Dh v3 c 3 e hv/KT − 1
) (aﺍﺤﺴﺏ ﻜﺜﺎﻓﺔ ﺍﻟﻁﺎﻗﺔ ﻓﻲ ﻤﺩﻯ ﻁﻭل ﻤﻭﺠﻲ ، Δλﺃﻱ ) (bﺍﺴﺘﺨﺩﻡ ﺍﻟﻨﺘﻴﺠﺔ ﻓﻲ ﺠﺯﺀ ) (aﻹﻴﺠﺎﺩ ﻗﻴﻤﺔ
max
λ =λ
ﺍﻹﺸﻌﺎﻉ ﺃﻗﺼﻰ ﻤﺎ ﻴﻤﻜﻥ؟
) (cﻭﻀﺢ ﺃﻥ
λ max
ﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺘﻬﺎ ﻋﻠﻰ ﺍﻟﺼﻴﻐﺔ
b T
max
, T)dλ
= )u (v, T
u (λ؟
ﻭﺍﻟﺘﻲ ﻋﻨﺩﻫﺎ ﺘﻜﻭﻥ ﻜﺜﺎﻓﺔ
، λﻭﺍﺤﺴﺏ ﻗﻴﻤـﺔ bﻓـﻲ
ﺤﺎﻟﺔ ﺴﻁﺢ ﺍﻟﺸﻤﺱ ﻋﻠﻤﺎﹰ ﺒﺄﻥ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺴﻁﺢ ﺍﻟﺸﻤﺱ ﺘﺴﺎﻭﻱ 5620k؟ )ﺘﻨﺒﻴﻪ :ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ s-x = se-xﺒﺎﻟﺭﺴﻡ ﺍﻟﺒﻴﺎﻨﻲ(. ) (dﻴﻨﺒﻌﺙ ﻤﻥ ﺃﺸﺩ ﺍﻟﻨﺠﻭﻡ ﺤﺭﺍﺭﺓ )ﻨﺠﻡ ﺍﻟﺸﹼﻌﺭﻱ (Siriusﻁﻴﻑ ﺇﺸﻌﺎﻉ ﺍﻟﺠـﺴﻡ ﺍﻷﺴﻭﺩ ﻭﺍﻟﺫﻱ ﻟﻪ
= 260nm
max
. λﺍﺤﺴﺏ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺴﻁﺢ ﻫﺫﺍ ﺍﻟﻨﺠﻡ؟
) (eﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ﻤﺘﻭﺴﻁ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺴﻁﺢ ﺍﻷﺭﺽ .288 kﺍﺴﺤﺏ ﺍﻟﻁـﻭل ﺍﻟﻤﻭﺠﻲ ﻷﻗﺼﻰ ﻜﺜﺎﻓﺔ ﺇﺸﻌﺎﻉ ﻟﻠﺠﺴﻡ ﺍﻷﺴﻭﺩ ﻟﻸﺭﺽ .ﺤﺩﺩ ﻷﻱ ﺠﺯﺉ ﻤـﻥ
ﺍﻟﻁﻴﻑ ﻴﻘﺎﺒل ﻫﺫﺍ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ؟ (Q2ﺃﻗﺼﻰ ﻁﺎﻗﺔ ﺤﺭﻜﻴﺔ ﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻤﻨﺒﻌﺜﺔ ﻤﻥ ﺴﻁﺢ ﺃﻟﻭﻤﻨﻴﻭﻡ ﺘﺴﺎﻭﻱ 2.3 eV
ﻭﺫﻟﻙ ﻋﻨﺩ ﺘﻌﺭﺽ ﻫﺫﺍ ﺍﻟﺴﻁﺢ ﻷﺸﻌﺔ ﺫﺍﺕ ﻁﻭل ﻤﻭﺠﻲ .2000Åﺃﻤﺎ ﻋﻨـﺩ ﺘﻌﺭﺽ ﺍﻟﺴﻁﺢ ﻷﺸﻌﺔ ﻁﻭﻟﻬﺎ ﺍﻟﻤﻭﺠﻲ 2580 Åﻓـﺈﻥ ﺍﻟﻁﺎﻗـﺔ ﺍﻟﺤﺭﻜﻴـﺔ
ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﻨﺒﻌﺜﺔ ﺘﺴﺎﻭﻱ .0.9 eVﺍﺤﺴﺏ ﻗﻴﻤﺔ ﺜﺎﺒـﺕ ﺒﻼﻨـﻙ ﻭﺩﺍﻟـﺔ ﺍﻟﺸﻐل ﻟﻸﻟﻤﻭﻨﻴﻭﻡ؟
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(Q3ﺍﺤﺴﺏ ) (aﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻭﺍﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻹﻟﻜﺘﺭﻭﻥ ﻤﻌﺠل ﺘﺤﺕ ﺘـﺄﺜﻴﺭ ﻓﺭﻕ ﺠﻬﺩ 100 Vﻭ ) (bﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻟﻪ ﻁﻭل ﻤﻭﺠﻲ ﺩﻴﺒﺭﻭﻟﻲ
) 200 pmﺤﻴﺙ (1 pm = 10-12 m؟
(Q4ﺃﺸﻌﺔ Xﺍﺴﺘﻁﺎﺭﺕ ﺒﺈﻟﻜﺘﺭﻭﻥ ﺴﺎﻜﻥ .ﺍﺤﺴﺏ ﻁﺎﻗﺔ ﺃﺸـﻌﺔ Xﺍﻟـﺴﺎﻗﻁﺔ ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ﻁﻭل ﻤﻭﺠﺔ ﺍﻷﺸﻌﺔ ﺍﻟﻤﺴﺘﻁﺎﺭﺓ ﻋﻨﺩ ﺯﺍﻭﻴﺔ
60°
ﺘﺴﺎﻭﻱ 0.035 A؟ °
(Q5ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﻤﺴﺘﻭﻴﻴﻥ ﻤﺘﺠﺎﻭﺭﻴﻥ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﺒﻠﻭﺭﻴـﺔ ﻴـﺭﺍﺩ ﻗﻴﺎﺴـﻬﺎ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺃﺸﻌﺔ Xﻁﻭﻟﻬﺎ ﺍﻟﻤﻭﺠﻲ 0.5 Åﻭﺍﻟﺘﻲ ﺘﻡ ﻗﻴﺎﺴﻬﺎ ﻋﻨﺩ ﺯﺍﻭﻴـﺔ . 5 °
ﺍﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﻫﺫﻴﻥ ﺍﻟﻤﺴﺘﻭﻴﻴﻥ؟ ﻋﻨﺩ ﺃﻱ ﺯﺍﻭﻴﺔ ﻴﻤﻜﻨﻨـﺎ ﻗﻴـﺎﺱ
ﺍﻟﻘﻴﻤﺔ ﺍﻟﺜﺎﻨﻴﺔ؟
ﺇﺠﺎﺒﺎﺕ ﺍﻷﺴﺌﻠﺔ :ﺍﻹﺠﺎﺒﺔ ﺍﻷﺨﻴﺭﺓ ﻤﻥ ﻜل ﺴﺅﺍل: 8Dhc hc/λc/ (e − T) −1 λs
= )u (λ , T
)Q1) (a
ﻨﻅﺭﻱ λ max = b / Tﺍﻟﻤﻁﻠﻭﺏ ﺇﺜﺒﺎﺕ ﺍﻟﻘﺎﻨﻭﻥ )(b °
λ mex = 5160 A
)(c
T = 1.12 ×10 4 k 1
)(d
λ = 1.01× 10 −5 m
)(e
W = 3.92 eV
)(Q2
K.E = 6.02 × 10-18 J
)(Q3
E = 5.4× 105 eV
)(Q4
θ =10 °
)(Q5
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ﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ ﻭﻋﻼﻗﺎﺕ ﺍﻟﻼﺘﺤﺩﻴﺩ WAVE PACKETS AND THE UNCERTAINTY RELATIONS
ﺇﻥ ﻋﻠﻡ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻴﺯﻭﺩﻨﺎ ﺒﻔﻬﻡ ﺼﺤﻴﺢ ﻟﻜل ﺍﻟﻅﻭﺍﻫﺭ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﺘﻲ ﺘـﻡ
ﻤﻨﺎﻗﺸﺘﻬﺎ ﻓﻲ ﺍﻟﻔﺼل ﺍﻷﻭل .ﻭﻫﺫﺍ ﺍﻟﻌﻠﻡ ﻀﺭﻭﺭﻱ ﻟﻔﻬﻡ ﺴﻠﻭﻙ ﺍﻟﺫﺭﺍﺕ ،ﺍﻟﺠﺯﻴﺌﺎﺕ، ﺃﻨﻭﻴﺔ ﺍﻟﺫﺭﺍﺕ ﻭﺘﺠﻤﻌﺎﺕ ﻤﻥ ﻫﺫﻩ ﻭﺘﻠﻙ .ﻭﻋﺎﺩﺓ ﻤﺎ ﺘﺘﻡ ﺍﻟﺩﺭﺍﺴﺔ ﺒﺎﺴﺘﺨﺩﺍﻡ ﻤﻌﺎﺩﻟـﺔ
ﺸﺭﻭﺩﻨﺠﺭ Shrodinger equationﻭﺍﻟﺘﺄﻭﻴﻼﺕ ﺍﻟﺼﺤﻴﺤﺔ ﻟﺤﻠﻭﻟﻬﺎ .ﻭﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻻ ﻴﻭﺠﺩ ﻟﻬﺎ ﺇﺜﺒﺎﺕ! ﻭﻟﻜﻥ ﺍﺴﺘﻁﺎﻉ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﺘﻭﺼل ﺇﻟﻴﻬﺎ ﻋﻥ ﻁﺭﻴـﻕ ﺇﺘﺒـﺎﻉ
ﺘﻭﺠﻴﻬﺎﺕ )ﺃﻭ ﻓﺭﺍﺴﺎﺕ (insightﺍﻟﻌﺎﻟﻡ ﺍﻟﻔﺭﻨﺴﻲ ﺩﻴﺒﺭﻭﻟﻲ .ﻭﻤﻥ ﺍﻟﺠﺩﻴﺭ ﺒﺎﻟﺫﻜﺭ ﺃﻥ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻭﺠﺩ ﺨﺎﺭﺝ ﻨﻁﺎﻕ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ،ﺍﻷﻤﺭ ﺍﻟﺫﻱ ﻴﺠﻌل ﺍﺴـﺘﻨﺒﺎﻁﻬﺎ
ﺼﻌﺏ .ﻭﺨﻼل ﺍﻟﻤﻨﺎﻗﺸﺔ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺴﻨﺤﺎﻭل ﺍﻟﺘﻭﻓﻴﻕ ﺒﻴﻥ ﺍﻟﺨﻭﺍﺹ ﺍﻟﺠﺴﻴﻤﻴﺔ ﻭﺍﻟﻤﻭﺠﻴﺔ ﻟﻼﻟﻜﺘﺭﻭﻨﺎﺕ.
ﺇﻨﻪ ﻤﻥ ﺍﻟﺼﻌﺏ ﺘﺨﻴل ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻋﻠﻰ ﺃﺴﺎﺱ ﺃﻥ ﻟﻬﺎ ﺴـﻠﻭﻙ ﺍﻟﻤﻭﺠـﺎﺕ
ﻭﻟﻜﻥ ﺘﺠﺎﺭﺏ ﺍﻟﺤﻴﻭﺩ ﺍﻟﺘﻲ ﺃﺠﺭﺍﻫﺎ ﻜـل ﻤـﻥ Freshet and Youngﺃﺩﺕ ﺇﻟـﻰ ﺍﻹﺠﻤﺎﻉ )ﺃﻭ ﺍﺘﺤﺎﺩ ﺍﻵﺭﺍﺀ( ﺒﻘﺒﻭل ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﻀﻭﺀ .ﻭﻤﻥ ﺠﺎﻨـﺏ ﺁﺨـﺭ
ﺒﺎﻹﻤﻜﺎﻥ ﺃﻥ ﻨﺘﺼﻭﺭ ﺘﺠﻤﻌﺎﺕ )ﺃﻭ ﻫﻴﺌﺎﺕ( ﻟﻠﻤﻭﺠﺎﺕ ﺍﻟﻤﺤﺼﻭﺭﺓ ﺃﻭ ﺍﻟﻤﺘﺤﻴﺯﺓ )ﺃﻭ
ﺍﻟﻤﺘﻤﺭﻜﺯﺓ ﺠﺩﺍﹰ( Very localizedﻓﻌﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل ﺘﻌﺘﺒﺭ ﻗﺭﻗﻌﺔ ﺍﻟﺭﻋﺩ a clap
of thunderﻤﺜﺎل ﻟﺘﺩﺍﺨل ﺍﻟﻤﻭﺠﺎﺕ ﻭﺘﺭﺍﻜﺒﻬﺎ ﻤﻤﺎ ﻴﺅﺩﻱ ﺇﻟﻰ ﺘﺄﺜﻴﺭ ﻤﺘﻤﺭﻜﺯ ﻤـﻊ
ﺍﻟﺯﻤﻥ .localized in timeﻤﺜل ﻫﺫﻩ ﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴـﺔ ﺍﻟﻤﺘﻤﺭﻜـﺯﺓ localized
" "wane packetsﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻴﻬـﺎ ﻋـﻥ ﻁﺭﻴـﻕ ﺘﺭﺍﻜـﺏ ﺍﻟﻤﻭﺠـﺎﺕ superposing wavesﺒﺘﺭﺩﺩﺍﺕ ﻤﺨﺘﻠﻔﺔ ﺒﺤﻴﺙ ﻴﺘﻡ ﺘﺩﺍﺨل ﻜل ﻤﻭﺠﺔ ﻤﻊ ﺍﻷﺨـﺭﻯ ﺨﺎﺭﺝ ﻤﻨﻁﻘﺔ ﻤﻜﺎﻨﻴﺔ. They interfere with each other almost completely outside of a given spatial region.
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ﻭﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﺇﺘﺒﺎﻋﻬﺎ )ﺃﻭ ﺍﻷﺩﻭﺍﺕ ﺍﻟﺭﻴﺎﻀﻴﺔ ﺍﻟﻤﺴﺘﺨﺩﻤﺔ( ﻫﻲ ﻋﻤل ﺘﻜﺎﻤﻼﺕ ﻓﻭﺭﻴﺭ .Fourier integrals ﻭﻗﺒل ﺍﻟﺒﺩﺀ ﻓﻲ ﺍﻟﻤﻌﺎﻟﺠﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﻟﻠﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ ،ﻟﻌﻠﻪ ﻤﻥ ﺍﻟﻤﻔﻴﺩ ﻤﺭﺍﺠﻌﺔ
ﺒﻌﺽ ﺃﺴﺎﺴﻴﺎﺕ ﺍﻟﺩﻭﺍل ﺍﻟﻤﺭﻜﺒﺔ ﻭﺘﻜﺎﻤﻼﺕ ﻓﻭﺭﻴﺭ ﻭﻜﻴﻔﻴﺔ ﺇﻴﺠﺎﺩ ﻋﺭﺽ ﺍﻟـﺩﻭﺍل ﻭﻜﺫﻟﻙ ﻤﺭﺍﺠﻌﺔ ﺒﻌﺽ ﺍﻟﺘﻜﺎﻤﻼﺕ. ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ﺃﻱ ﺭﻗﻡ ﻤﺭﻜﺏ zﺒﺎﻟﺼﻴﻐﺔ. z= x+i y
)......(A : 2 − 1
ﺤﻴﺙ xﺘﻤﺜل ﺍﻟﺠﺯﺀ ﺍﻟﺤﻘﻴﻘﻲ ﻟـ zﺃﻱ
z
) Re ( z
ﺍﻟﻭﺤﺩﺓ ﺍﻟﺘﺨﻴﻠﻴﺔ Iﻭﻫﻲ:
)(A : 2 − 2
ﻭ yﺘﻤﺜل ﺍﻟﺠﺯﺀ ﺍﻟﺘﺨﻴﻠﻲ Im
i = −1 ⇒ i2 = −1
ﻟﻘﺴﻤﺔ ﺩﻭﺍل ﺘﺨﻴﻠﻴﺔ )ﺃﻭ ﺃﺭﻗﺎﻡ ﺘﺨﻴﻠﻴﺔ( ﻴﺤﺴﻥ ﺒﻨﺎ ﺍﺴﺘﺤﺩﺍﺙ ﺍﻟﻤﺭﺍﻓﻕ ﺍﻟﻤﺭﻜﺏ
complex conjugateﻭﻴﺭﻤﺯ ﻟﻪ * zﺤﻴﺙ:
z * = v − iy
)(A : 2 − 3
ﻻﺤﻅ ﺃﻨﻨﺎ ﺍﺴﺘﺒﺩﻟﻨﺎ iﺏ –iﺍﻟﺭﻗﻡ ﺍﻟﺘﺨﻴﻠﻲ ﻤﻀﺭﻭﺏ ﻓﻲ ﻤﺭﺍﻓﻘﻪ ﺃﻱ * zzﻴﻌﻁﻴﻨﺎ )........... (A : 2 − 4
ﻻﺤﻅ ﺃﻥ
*z z
zz * = (x + iy) (x − iy) = x 2 − i 2 y 2 = x 2 + y 2
ﺘﺴﺎﻭﻱ ﻤﺭﺒﻊ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ
2
z
ﻭﺍﻵﻥ ﻹﻴﺠﺎﺩ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﻘﻴﻘﻴﺔ ﻭﺍﻟﺘﺨﻴﻠﻴﺔ ﻟـ
1 1 = z x + iy
1 1 x − iy x − iy x y = = 2 2 = 2 2 −i 2 2 z x + iy x − iy x + y x + y x +y 1 x 1 y ⇒ Re ( ) = 2 2 and In ( ) = 2 2 z x +y z x +y
ﻤﻥ ﺍﻟﻤﻔﻴﺩ ﻜﺘﺎﺒﺔ
z
ﺒﺩﻻﻟﺔ
θ, r
ﺒﺎﺴﺘﺨﺩﺍﻡ ﻤﻌﺎﺩﻟﺔ Euler
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eiθ = cos θ + i sin θ
(A : 1 − 5)
x = r cos θ and y = r sin θ ⇒ z = x + iy = r cos θ + i r sin θ = r (cos θ + i sin θ ) = reiθ
ﻫﻭ
e iθ . e − iθ = e ° = 1⇐ e − iθ
e iθ
ﻻﺤﻅ ﺃﻥ ﺍﻟﻤﺭﺍﻓﻕ ﻟـ ﻭﻜﺫﻟﻙ
z* = r e − iθ
ﻭﻤﻨﻬﺎ ﻨﺠﺩ
zz * = (re−iθ ) (re−iθ ) = r 2
ﻭﻤﻥ ﺍﻟﻤﻔﻴﺩ ﺘﺫﻜﺭ ﺃﻥ e iθ + e − iθ 2
(A : 2 − 6)
eiθ − e − iθ sinθ = 2i
............ (A : 2 − 7)
cos θ = and
ϕ (θ ) = 0 ≤ θ ≤ 2D
1 imθ e 2D
ﻫﻲ
ϕ ( θ) θ
ﻭ ﺤﺩﻭﺩ
ﺇﺫﺍ ﻜﺎﻨﺕ ﻟﺩﻴﻨﺎ ﺩﺍﻟﺔ
m = 0 ± 1 ± 2 .......
:ﻓﺈﻥ 1 imθ 1 eimθ dθ (θ ) dθ e ϕ = = ∫° ∫° 2D 2D im ° let m ≠ o 1 1 i2 Dm ° = e −e 2D im 2D
2D
[
2D
]
1 1 cos 2Dm + i sin 2Dm − 1 424 3 1 424 3 2D im 1 1 0 1 1 = [1 − 1]= 0 2D im =
let m = 0 2D
2D
°
°
2D
2D 1 i(0) θ 1 1 e = dθ [ θ ] = ° 2D 2D ∫° 2D 1 2D . (2D − 0) = = 2D 2D 2D
∫ dθ ϕ (θ ) = ∫ dθ =
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ﺤﻴﺙ
ﻴﻤﻜﻥ ﺇﺠﺭﺍﺀ ﺍﻟﺘﻜﺎﻤل ﺍﻷﻭل ﺒﺼﻭﺭﺓ ﺃﺴﻬل ﻜﺎﻟﺘﺎﻟﻲ: 2D
2D
2D
i 1 1 imθ sin mθ dθ = 0 ∫ cos m θ dθ + ∫ = e dθ 2D 2D 2D ° °
∫ °
ﻜل ﺤﺩ ﻤﻥ ﺍﻟﺘﻜﺎﻤل ﻴﺴﺎﻭﻱ ﺼﻔﺭ ﻷﻨﻨﺎ ﻨﺠﺭﻱ ﺤﺩﻭﺩ ﺍﻟﺘﻜﺎﻤـل ﻋﺒـﺭ ﺩﻭﺭﺓ
ﻜﺎﻤﻠﺔ ﻭﻨﺭﺠﻊ ﻟﻨﻔﺱ ﺍﻟﺒﺩﺍﻴﺔ ﻭﻟﻬﺫﺍ ﻨﺴﺘﺨﻠﺹ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ m = 0ﻓـﺈﻥ ﺍﻟﺘﻜﺎﻤـل
ﻴﺴﺎﻭﻱ
2D
ﺃﻤﺎ ﻟﻠﻘﻴﻡ ﺍﻷﺨﺭﻯ ﻟـ mﻓﺈﻥ ﺍﻟﺘﻜﺎﻤل ﻴﺴﺎﻭﻱ ﺼﻔﺭ ﻭﻴﻤﻜﻥ ﺼـﻴﺎﻏﺔ
ﺍﻟﻨﺘﻴﺠﺔ ﺒﻬﺫﻩ ﺍﻟﺼﻭﺭﺓ. 2D
∫ dθϕ (θ ) = 0
)for all values of m ≠ 0 ..................(A : 2 - 8
°
for m = 0
)(A : 2 - 9
= 2D
ﻭﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ ﻴﻤﻜﻨﻨﺎ ﺍﺴﺘﻨﺘﺎﺝ ﺍﻟﻌﻼﻗﺔ 2D
)..................(A : 2 - 10
m≠n
(θ )ϕ n (θ ) = 0
)(A : 2 -11
m= n
=1
* m
∫ dθϕ °
1 − imθ 1 inθ e . e 2D 2D 2D
∫ dθ °
2D
2D 1 1 1 i (n − m)θ ° [ ] ⇒m=n dθ e = dθ e θ° = 2D ∫° 2D ∫° 2D
1 (2D − 0) = 1 2D
2D
Now, Let
= 2D
1 dθ ei (n − m) θ 2D ∫°
⇒ Let m ≠ n
2D
1 ] dθ [cos ( n − m ) θ + i sin ( n − m ) θ = 2D ∫°
ﻗﻴﻤﺔ ﻫﺫﺍ ﺍﻟﺘﻜﺎﻤل ﺘﺴﺎﻭﻱ ﺼﻔﺭ ﻭﺫﻟﻙ ﻷﻥ ﺍﻟﺘﻜﺎﻤل ﻴﺠﺭﻱ ﻋﻠﻰ ﺩﻭﺭﺓ ﻜﺎﻤﻠﺔ
ﻟﻠﺩﺍﻟﺔ )ﺃﻱ ﺃﻨﻪ ﻴﺒﺩﺃ ﺜﻡ ﻴﻨﺘﻬﻲ ﻋﻨﺩ ﻨﻔﺱ ﺍﻟﻘﻴﻤﺔ(.
ﻭﻤﻥ ﺍﻟﺩﻭﺍل ﺍﻟﻤﺭﻜﺒﺔ ﺍﻟﻤﻬﻤﺔ sin h x :ﻭ cos h xﻭﺘﻌﺭﻑ ﺒـ )..................(A : 2 - 12 )(A : 2 - 13
ex − e− x ⇒ sin h i x = i sin x 2 ex + e− x cos h x ⇒ cos h i x = cos x 2
= sin h x
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ﻨﺒﺫﺓ ﻤﺨﺘﺼﺭﺓ ﻋﻥ ﺘﺤﻠﻴﻼﺕ ﻓﻭﺭﻴﺭ ﻭﺘﻜﺎﻤﻼﺘﻪ Short Introduction about Fourier Series and its integral.
ﺇﻥ ﺃﻱ ﺩﺍﻟﺔ ﻟﻬﺎ ﺼﻭﺭﺓ ﻤﻭﺠﻴﺔ ﺃﻭ ﺘﻜﺭﺭ ﻨﻔﺴﻬﺎ ﺒﺩﻭﺭﺓ ﻤﻌﻴﻨﺔ ،ﻤﺜل ﺍﻟـﺩﻭﺍل
ﺍﻟﻤﻭﺠﻴﺔ ﺍﻟﻤﻭﻀﺤﺔ ﻓﻲ ﺸﻜل ،A:2.1ﻴﻤﻜﻥ ﻭﺼﻔﻬﺎ ﺭﻴﺎﻀـﻴﺎﹰ ﺒﺎﺴـﺘﺨﺩﺍﻡ ﻤﺒـﺩﺃ ﺍﻟﺘﺭﺍﻜﺏ ،Principle of super positionﺃﻱ ﻴﻤﻜﻥ ﺍﻟﺘﻌﺒﻴﺭ ﻋﻨﻬﺎ ﻋﻥ ﻁﺭﻴﻕ ﺠﻤﻊ
)ﺃﻭ ﺘﺭﺍﻜﺏ( ﻤﻭﺠﺎﺕ ﺘﻜﺘﺏ ﺭﻴﺎﻀﻴﺎﹰ ﺒﺩﻻﻟﺔ ﺍﻟـ :ﺠﺎ sinﻭﺍﻟـ ﺠﺘـﺎ .cosﻫـﺫﻩ
ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﺘﹸﻌﺭﻑ ﺒﺘﺤﻠﻴل ﻓﻭﺭﻴﺭ Fourier analysisﻨﺴﺒﺔ ﻟﻠﻌﺎﻟﻡ ﻓـﻭﺭﻴﺭ.
ﻭﺘﻌﺘﻤﺩ ﻫﺫﻩ ﺍﻟﻁﺭﻴﻘﺔ ﻋﻠﻰ ﺃﺴﺎﺱ ﺃﻨﻬﺎ ﻴﻤﻜﻥ ﺼﻴﺎﻏﺔ ﺃﻱ ﺩﺍﻟﺔ )) f (xﺤﻴﺙ xﻴﻤﻜﻥ ﺃﻥ ﺘﻤﺜل ﺍﻟﻤﻭﻗﻊ ﺃﻭ ﺍﻟﺯﻤﻥ( ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﺘﺏ ﻓﻲ ﺼﻭﺭﺓ ﻤﺘﺴﻠﺴﻠﺔ ﻤﻥ ﺩﻭﺍل ﺩﻭﺭﻴﺔ
)ﺘﻌﺭﻑ ﺒﺴﻠﺴﻠﺔ ﻓﻭﺭﻴﺭ( .Fourier series
1 f (x) = a ° + a 1 cos x + a 2 cos 2x + .....+ a n cos n x + b1 sin x + b 2 sin 2x + .....+ b n sin n x + ........ 2 ∞ ∞ 1 )= a ° + ∑ a u cos n x + ∑ b n sin v x ..............(A : 2 − 14 2 n =1 n =1
ﺘﺘﻜﻭﻥ ﻫﺫﻩ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﻤﻥ ﺤﺩ ﺜﺎﺒﺕ ﻭﺠﻴﻭﺏ ﺍﻟﺘﻤﺎﻡ
1 a 2 °
ﺒﺎﻹﻀﺎﻓﺔ ﺇﻟﻰ ﺤﺩﻭﺩ ﻤﻥ ﺍﻟﺠﻴـﺏ
(sine and cosine termsﻟﻬـﺎ ﺴـﻌﺎﺕ ﻤﺨﺘﻠﻔـﺔ a1, a2, ….
.b1, b2, ……… Different amplitudesﺘﺭﺩﺩﺍﺕ ﺍﻟﺠﺎ ﻭﺍﻟﺠﺘﺎ ﻓﻲ ﻜل ﺤﺩ ﻤـﻥ ﺍﻟﺤﺩﻭﺩ ﻋﺒﺎﺭﺓ ﻋﻥ harmonicsﻭ ﻤﻀﺭﻭﺒﺎﺕ multiplesﺍﻟﺘﺭﺩﺩ ﺍﻷﺴﺎﺴﻲ .x
.Fundamental frequency xﻟﻠﺘﻌﺒﻴﺭ ﻋﻥ ﻤﻌﻴﻨﺔ ﻤﻥ ﺨﻼل ﺘﺤﻠﻴل ﻓﻭﺭﻴﺭ ﻓـﺈﻥ
ﺍﻟﻤﻌﺎﻤﻼﺕ ……… a1, a2,ﻭ b1, b2, ……..ﻴﻠﺯﻡ ﺘﺤﺩﻴﺩﻫﺎ ﻭﺫﻟﻙ ﺒﺘﻜﺎﻤل ﺍﻟﺩﺍﻟﺔ
) f(xﺨﻼل ﺩﻭﺭﺓ ﻜﺎﻤﻠﺔ ﺃﻱ ﻤﻥ ﺼﻔﺭ ﺇﻟﻰ
2D
ﻜﻤﺎ ﻴﻠﻲ:
2D
1 )b n = ∫ f (x) sin n x dx .....................(A : 2 − 15 D °
2D
,
1 a n = ∫ f (x) cos n x dx D °
ﺍﻟﺸﻜل ﻴﻭﻀﺢ ﺃﻥ ﺘﻭﻟﻴﺩ ﺩﺍﻟﺔ ﻤﻭﺠﻴﺔ ﻤﺭﺒﻌﺔ: - ٣٩PDF created with pdfFactory Pro trial version www.pdffactory.com
ﻴﺘﻡ ﻋﻥ ﻁﺭﻴﻕ ﺘﺩﺍﺨل )ﺃﻭ ﺘﺭﺍﻜﺏ( ﺜﻼﺙ ﻤﻭﺠﺎﺕ ﺠﻴﺒﻴﺔ ﻫﻲ ﺃﻭل ﺍﻟﺤﺩﻭﺩ ﻓﻲ ﻤﺘﺴﻠﺴﻠﺔ ﻓﻭﺭﻴﺭ ﺒﺎﻟﺼﻭﺭﺓ.
1 a 2 °
1 1 sin 2x , sin 3x , sinx 2 3
ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻪ )ﻤﻥ ﻤﻌﺎﺩﻟـﺔ (A: 2-15
2D
1 f (x) dx 2D ∫°
ﻭﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻤﺜل ﺒﺒﺴﺎﻁﺔ ﻤﺘﻭﺴﻁ ﺍﻟﺩﺍﻟﺔ ) f(xﻋﺒﺭ ﺍﻟﻔﺘﺭﺓ . 2D
ﺘﺤﻠﻴل ﻓﻭﺭﻴﺭ ﻭﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ: ﻴﻤﻜﻥ ﺍﻻﺴﺘﻔﺎﺩﺓ ﻤﻥ ﺘﺤﻠﻴل ﻓﻭﺭﻴﺭ ﻟﺘﻤﺜﻴل ﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴـﺔ Wave packets
ﻭﻫﻲ ﻋﺒﺎﺭﺓ ﻋﻥ ﻨﺒﻀﺎﺕ ﻤﻭﺠﻴﺔ Wave pulsesﻋﻨﺩ ﺩﺭﺍﺴﺔ ﺍﻟﺤﺭﻜـﺔ ﺍﻟﻤﻭﺠﻴـﺔ ﻟﻨﺘﺤﺼل ﻋﻠﻰ ﻨﺒﻀﺔ ﻤﻭﺠﻴﺔ ﻴﻠﺯﻡ ﺘﺭﺍﻜﺏ ﻤﻭﺠﺎﺕ ﺠﻴﺒﻴﺔ ﻋﻠﻰ ﻤﺩﻯ ﻤﺘﺼل ﻤـﻥ
ﺍﻷﺭﻗﺎﻡ ﺍﻟﻤﻭﺠﻴﺔ kﻓﻲ ﻤﺩﻯ ﻗﺩﺭﺘﻪ . Δkﻴﻤﻜﻨﻨﺎ ﻓﻬﻡ ﻜﻴﻔﻴﺔ ﺘﻜﻭﻥ ﺤﺯﻤﺔ ﻤﻭﺠﻴـﺔ،
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ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل ،ﺨﺫ ﺘﺭﺍﻜﺏ ﻋﺩﺩ ﺴﺒﻌﺔ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﺠﻴﺒﻴﺔ ﺍﻟﺘﻲ ﻟﻜـل ﻤﻨﻬـﺎ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ. ).......... .......... .(A : 2 − 16
)y = A k cos ( k x − ω t
ﻭﻋﻨــﺩ ،t = 0ﺍﻷﺭﻗــﺎﻡ ﺍﻟﻤﻭﺠﻴــﺔ ) k (Wave umbersﺘــﺴﺎﻭﻱ 45, 42, 39, 36, 33, 30 , 27ﻭﻜل ﻤﻥ ﻫﺫﻩ ﺍﻷﺭﻗﺎﻡ ﻴﻘﺎﺒﻠﻬﺎ ﺴـﻌﺔ Akﺘـﺴﺎﻭﻱ
0.25 , 0.33 , 0.5 , 1.0, 0.5 , 0.33 , 0. 25ﺸﻜل A: 2-3ﻴﻭﻀﺢ ﺘﻐﻴﺭ ﺍﻟﺴﻌﺔ
Akﻤﻊ kﻭﻫﻲ ﻤﺎ ﻴﺴﻤﻰ ﺒﻁﻴﻑ ﺍﻟﺭﻗﻡ ﺍﻟﻤﻭﺠﻲ .Wave umber spectrum
Figure A:2-3 the frequency spectrum used to generate the we packet shown in Figure 12.84.
ﺍﻟﺸﻜل ﺍﻟﻤﻭﺠﻲ ﻟﻜل ﻤﻭﺠﺔ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﺴﺒﻌﺔ ﻭﺤﺎﺼـل ﺘﺭﺍﻜﺒﻬـﺎ )ﺃﻭ
ﻤﺤﺼﻠﺔ ﺘﺭﺍﻜﺒﻬﺎ( ﻤﻭﻀﺤﺔ ﻓﻲ ﺸﻜل ).(A: 2-4
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ﺸﻜل ) (A: 2-4ﻜﻴﻔﻴﺔ ﺘﺅﻜﺩ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ )ﺃﺴﻔل ﺍﻟﺸﻜل( ﻤﻥ ﺘﺩﺍﺨل ﺴـﺒﻌﺔ
ﻤﻭﺠﺎﺕ ﻜﺎﻟﻤﻭﻀﺤﺔ ﻓﻲ ﺃﻋﻠﻰ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ.
ﺃﻤﺎ ﻓﻲ ﺤﺎﻟﺔ ﺘﺭﺍﻜﺏ ﻋﺩﺩ ﻜﺒﻴﺭ ﺠﺩﺍﹰ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺒﺤﻴﺙ ﻴﻤﺘﺩ ﺘﻐﻴﺭ ﺍﻷﺭﻗﺎﻡ ﺍﻟﻤﻭﺠﻴﺔ ﻋﺒﺭ ﻋﺭﺽ ﻗﺩﺭﻩ
ﻤﻭﻀﺢ ﻓﻲ ﺍﻟﺸﻜل .A: 2-5
ΔK
ﺴﻨﺘﺤﺼل ﻋﻠﻰ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﻭﺍﺤﺩﺓ ﻜﻤـﺎ ﻫـﻭ
ﻻﺤﻅ ﺃﻥ ﻋﺭﺽ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ...... ΔK
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ﻭﻴﻤﻜﻥ ﺃﻥ ﻨﻼﺤﻅ ﻨﻘﻁﺔ ﻫﺎﻤﺔ ﺠﺩﺍﹰ ﻭﻫﻲ ﺃﻨﻪ ﻜﻠﻤﺎ ﺍﺯﺩﺍﺩﺕ ﺍﻟﺤﺯﻤﺔ
ΔK
ΔK
ﻓﺈﻥ ﻋﺭﺽ
ﻴﻘل .ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﻫﺎﻤﺔ ﻷﻨﻬﺎ ﺍﻷﺴﺎﺱ ﺍﻟﺫﻱ ﻴﺒﻨﻰ ﻋﻠﻴﻪ ﻤﺒﺩﺃ ﺍﻟﻼﺘﺤﺩﻴﺩ
ﻓﻲ ﻋﻠﻡ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ.
Figure A: 2-5 (a) the wave number spectrum and (b) the wave packet. Produced when ∆k the wave number spread shown in Figure A:2-4 becomes broader.
ﺇﻥ ﺘﺭﺍﻜﺏ ﻋﺩﺩ ﻻﻨﻬﺎﺌﻲ infinite numberﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﻴﻤﻜﻥ ﺼـﻴﺎﻏﺘﻪ
ﺭﻴﺎﻀﻴﺎﹰ ﺒﺘﻜﺎﻤل ﻓﻭﺭﻴﺭ Fourier integralﺒﺩﻻﹰ ﻤﻥ ﻤﺘﺴﻠﺴﻠﺔ ﻓﻭﺭﻴﺭ .ﻭﻫﻜﺫﺍ ،ﻓﻌﻨﺩ ﺯﻤﻥ ﺜﺎﺒﺕ ،ﻓﺈﻥ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺘﻌﻁﻲ ﺒـ: )..........(A : 2 − 17
∞
y (x) ∫ A (k) cos k x dk °
ﺤﻴﺙ ) A(kﻫﻲ ﺍﻟﺴﻌﺔ ﻭﺍﻟﺘﻲ ﺘﺘﻐﻴﺭ ﻤﻊ .kﻭﺴﻨﻨﺎﻗﺵ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺒﻌـﺩ
ﻫﺫﻩ ﺍﻟﻤﻘﺩﻤﺔ ،ﻜﻴﻑ ﻴﻤﻜﻨﻨﺎ ،ﺒﺎﺴﺘﺨﺩﺍﻡ ﺘﻜﺎﻤﻼﺕ ﻓـﻭﺭﻴﺭ ،ﺃﻥ ﻨﻭﻀـﺢ ﺃﻥ
Δx , Δk
ﻤﺭﺘﺒﻁﺘﺎﻥ ﺒﺎﻟﺩﺍﻟﺔ.
)....................(A : 2 − 18
(Δ x) (Δ k) = constant
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ﻭﻟﻌﻠﻪ ﻤﻥ ﺍﻟﻤﻔﻴﺩ ﺼﻴﺎﻏﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (A: 2-17ﻟﻠﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺒﺩﻻﻟﺔ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺭﻜﺒﺔ ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ ﺍﻟﻌﺎﻤﺔ. ∞
).............(A : 2 − 19
ikx
∫ dk A (k) e
∞−
1 2D
=)f (x
ﻭﺇﺫﺍ ﺃﺩﺨﻠﻨﺎ ﺍﻟﺯﻤﻥ ﻜﻤﺘﻐﻴﺭ ﻟﺩﺭﺍﺴﺔ ﺘﺄﺜﻴﺭ ﺍﻟﺯﻤﻥ ﻋﻠـﻰ ﺍﻨﺘـﺸﺎﺭ ﺍﻟﺤﺯﻤـﺔ
ﺍﻟﻤﻭﺠﻴﺔ ﻴﻤﻜﻥ ﺼﻴﺎﻏﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (A:2-19ﺒﺎﻟﺼﻭﺭﺓ.
∞
).............(A : 2 − 20
ikx - ω (k) t
∫ dk A (k) e
= ) f ( x, t
∞−
ﻋﻨﺩ ﺘﺩﺍﺨل ﻤﻭﺠﺎﺕ ﺴﺭﻋﺔ ﻜل ﻤﻨﻬﺎ vﻓﺈﻥ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺍﻟﻤﺘﻜﻭﻨﺔ ﺘﻜﻭﻥ
ﺴﺭﻋﺘﻬﺎ ﺘﻤﺜل ﺴﺭﻋﺔ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﻤﻜﻭﻨﺔ ﻟﻠﺤﺯﻤﺔ ﻭﺘـﺴﻤﻰ ﺒـﺴﺭﻋﺔ ﺍﻟﻤﺠﻤﻭﻋﺔ group velocityﻭﻴﺭﻤﺯ ﻟﻬﺎ ﺒـ vgﻭﺘﻌﻁﻰ ﺒﺎﻟﻌﻼﻗﺔ. )...................... (A : 2 − 21
dω dk
= vg
ﻜﻴﻔﻴﺔ ﺇﻴﺠﺎﺩ ﻋﺭﺽ ﺩﺍﻟﺔ: ﻤﻥ ﺍﻟﻨﻘﺎﻁ ﺍﻟﺘﻲ ﺘﻬﻤﻨﺎ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺇﻴﺠﺎﺩ ﻋﺭﺽ ﺩﺍﻟـﺔ ﻭﺫﻟـﻙ ﻷﻫﻤﻴـﺔ
ﺍﻟﻤﻭﻀﻭﻉ ﺒﻤﺒﺩﺃ ﺍﻟﻼﺘﺤﺩﻴﺩ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ.
ﺨﺫ ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﻀﺤﺔ ﻓﻲ ﺍﻟﺸﻜل A: 2-6
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ﻜﻠﺘﺎ ﺍﻟﺩﺍﻟﺘﺎﻥ ﻟﻬﻤﺎ ﻨﻔﺱ ﺍﻟﻌﺭﺽ ،ﻭﻟﻜﻥ ﺇﺤﺩﺍﻫﻤﺎ ﻤﺘﻤﺎﺜﻠﺔ ﺤﻭل x = oﻭﺍﻷﺨـﺭﻯ ﻤﺘﻤﺎﺜﻠﺔ ﺤﻭل .xoﺍﻷﻭل ﻓﻲ ﺍﻟﺸﻜل ) (aﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻬﺎ ﺭﻴﺎﻀﻴﺎﹰ ﺒﺎﻟﺼﻴﻐﺔ
2
y = A e− α x
ﺤﻴﺙ αﺜﺎﺒﺕ ﻭﻤﺭﺘﺒﻁ ﺒﻌﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻭ Aﻴﻤﺜل ﺴﻌﺔ ﺍﻟﺩﺍﻟـﺔ ﺃﻱ ﻗﻴﻤـﺔ y
ﻋﻨﺩﻤﺎ .x = o
ﺇﺤﺩﻯ ﺍﻟﻁﺭﻕ ﻹﻴﺠﺎﺩ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ﺃﻥ ﻨﺄﺨﺫ ﺍﻟﻌﺭﺽ ﻋﻨﺩﻤﺎ ﺘﻘل ﺍﻟﺩﺍﻟﺔ )ﺃﻱ
ﻋﻨﺩﻤﺎ ﺘﻘل ﻗﻴﻤﺔ .yﻤﻥ ﺃﻗﺼﻰ ﻗﻴﻤﺔ )ﻋﻨﺩﻤﺎ (y = Aﺇﻟﻰ ﺃﻥ ﺘـﺼل ﺇﻟـﻰ ، y = A 2
ﻋﻨﺩﻫﺎ ﻨﺄﺨﺫ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ . ∆x ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﺒﺎﻟﺸﻜل. ln y = ln A − α x 2 ⇒ α x 2 = ln A − ln y ﻧﻌﻮض ﻋﻦ
1 A 2
=y
A = ln 2 = 0.693 A/2 0.693 0.693 = ⇒ Δx ° ⇒ Δx = 2 α α
1 α x 2 = ln A − ln A 2 = ln 1 ⇒ x 2 = 0.693 α
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ﻴﻤﻜﻥ ﺇﻴﺠﺎﺩ ﺘﻌﺭﻴﻑ ﺁﺨﺭ ﻟﻌﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻭﻫﻭ ﺃﺴﻬل ﻤﻥ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟـﺴﺎﺒﻘﺔ. ﻨﺄﺨﺫ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻋﻨﺩﻤﺎ ﺘﻘل ﻗﻴﻤﺔ yﻤﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻘﺼﻭﻯ ﺇﻟﻰ ﺃﻥ ﺘﺼل ﺇﻟـﻰ ﺤﻴﺙ e = 2.718ﻭﻋﻠﻴﻪ
1 = 0.368 e
1 e
ﺃﻱ ﺒﺩل ﺃﻥ ﻨﺄﺨﺫ ﺍﻟﻌﺭﺽ ﻋﻨﺩﻤﺎ ﺘﻘـل yﺇﻟـﻰ
ﺍﻟﻨﺼﻑ )ﻜﻤﺎ ﻓﻲ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺴﺎﺒﻘﺔ( ﻨﺄﺨﺫ ﺍﻟﻌﺭﺽ ﻋﻨﺩﻤﺎ ﺘﻘل yﺇﻟﻰ 0.368ﻤـﻥ
ﻗﻴﻤﺘﻬﺎ ﺍﻟﻘﺼﻭﻯ ..ﻭﺴﻬﻭﻟﺔ ﻫﺫﻩ ﺍﻟﻁﺭﻴﻘﺔ ﻤﻥ ﺍﻟﻨﺎﺤﻴﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ.
2
y = A e− α x
ﺨﺫ ﻤﺜﻼﹰ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺴﺎﺒﻘﺔ
ﻗﻴﻤﺔ yﺍﻟﻘﺼﻭﻯ ﻫﻲ y = Aﻟﻜﻲ ﺘﻘل yﺇﻟﻰ ﻫﻲ ﻗﻴﻤﺔ x2ﺍﻟﺘﻲ ﺘﺠﻌل ﺍﻷﺱ
ﻭﺍﺤﺩ ﻓﺈﻥ
) (α x 2
1 e
ﻤﻥ ﻗﻴﻤﺘﻬﺎ ﺍﻟﻘﺼﻭﻯ ،ﺃﻱ ﻤﺎ
ﻴﺴﺎﻭﻱ ﺍﻟﻭﺍﺤﺩ .ﻋﻨﺩﻤﺎ ﻴﺴﺎﻭﻱ ﺍﻷﺱ ﻗﻴﻤـﺔ
1 y = ( ) A ⇐ y = A e −1 e 1 2 = ⇒ Δx = 2Δ x ° α α
= ⇒ α x 2 =1 ⇒ Δ x °
ﻭﻫﺫﻩ ﻫﻲ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺘﻲ ﺴﻨﺴﺘﺨﺩﻤﻬﺎ ﻹﻴﺠﺎﺩ ﻋﺭﺽ ﺍﻟﺩﻭﺍل ﺍﻟﺘﻲ ﻟﻬﺎ ﺍﻟﺼﻴﻐﺔ
ﺍﻷﺴﻴﺔ .ﻭﻫﻨﺎﻙ ﻁﺭﻕ ﺃﺨﺭﻯ ﻹﻴﺠﺎﺩ ﻋﺭﺽ ﺍﻟﺩﻭﺍل ﺍﻟﺠﻴﺒﻴﺔ.
ﻭﺒﻌﺩ ﻫﺫﺍ ﺍﻻﺴﺘﻁﺭﺍﺩ ﻨﻌﻭﺩ ﻟﻤﻨﺎﻗﺸﺔ ﻤﻭﻀﻭﻉ ﻫﺫﺍ ﺍﻟﻔـﺼل ﻭﻫـﻭ ﺍﻟﺤـﺯﻡ
ﺍﻟﻤﻭﺠﻴﺔ ﻭﻋﻼﻗﺎﺕ ﺍﻟﻼﺘﺤﺩﻴﺩ..
ﺨﺫ ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل ﺍﻟﺩﺍﻟﺔ ) f (xﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﻼﻗﺔ. - ٤٦PDF created with pdfFactory Pro trial version www.pdffactory.com
∞
f (x) = ∫ dk g (k) e ikv
)....(2 − 1
∞−
ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﺘﹸﻌﺒﺭ ﻋﻥ ﺘﺭﺍﻜﺏ ﺨﻁﻲ ﻤﻥ ﻤﻭﺠﺎﺕ ﺫﺍﺕ ﻁﻭل ﻤﻭﺠﻲ ، λ = 2D k
ﻭﻟﻜل ﻗﻴﻤﺔ ﻟـ kﻓﻜل ﻤﻭﺠﺔ ﺘﻜﺭﺭ ﻨﻔﺴﻬﺎ ﻋﻨﺩﻤﺎ ﺘﺘﻐﻴﺭ xﺇﻟﻰ . v + 2D k
ﻭﻟﺘﻭﻀﻴﺢ ﻁﺒﻴﻌﺔ ﻫﺫﻩ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ،ﺩﻋﻨﺎ ﻨﺨﺘﺎﺭ ﺼـﻴﻐﺔ ﻟﻠﺩﺍﻟـﺔ )g (k
ﺤﻴﺙ.
2
) g (k) = e −α (k − k °
)...(2-2 Let k 1 = k − k ° ⇒ k = k1 + k ° ⇒ dk = dk 1 i (u1 + k ° )v
12
⇒ g (k) → g (k 1 ) = e −αk and e ikv = e
ﺒﻌﺩ ﻫﺫﻩ ﺍﻟﺘﻌﻭﻴﻀﺎﺕ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ) (2-1ﺒﺎﻟﺼﻴﻐﺔ i (k1 + k ° )v
∞
⇒ f (x) = ∫ dk 1 e −αk . e 12
∞−
1
∞
° {= ∫ dk e − αk e ik v . e ik v
1
12
ﺛﺎﺑﺖ
1
∞−
12
e − αk e ik v
∞
1
∫ dk
= e ikov
∞−
ﻟﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﺘﻜﺎﻤل ﻴﺠﺏ ﺘﺤﻭﻴﺭ ﺼﻴﻐﺘﻪ ﺍﻟﺭﻴﺎﻀﻴﺔ ﻟﻠـﺼﻭﺭﺓ ﻭﺫﻟﻙ ﻷﻥ ﻗﻴﻤﺔ ﺍﻟﺘﻜﺎﻤل
D α
).................(2 − 3
1
dk 1
−αk − 2
∞
∫e
∞−
∞
− αk 1 ∫ e dk = 2 12
°
ﻭﻟﻠﻭﺼﻭل ﻟﻬﺫﻩ ﺍﻟﺼﻭﺭﺓ ﻴﻠﺯﻡ ﺃﻥ ﻨﺠﻌل ﺍﻷﺱ ﻤﺭﺒﻊ ﻜﺎﻤل ﺤﺴﺏ ﺍﻟﺨﻁﻭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ. 12 + ik iv
∞
1 − αk ik v 1 − αk ∫ dk e . e = ∫ dk e 1
∞−
ﺨﺫ
) ( −α
∞
12
∞−
ﻋﺎﻤل ﻤﺸﺘﺭﻙ ﻓﻲ ﺍﻷﺱ. - ٤٧PDF created with pdfFactory Pro trial version www.pdffactory.com
i )− α(k12 − k1v α
∞
= ∫ dk e
− αk12 + ik1v
1
∞
1
e
∫ dk
∞−
∞−
ﻟﻜﻲ ﻨﺘﺤﺼل ﻋﻠﻰ ﻤﺭﺒﻊ ﻜﺎﻤل ﻟﻸﺱ
ﻨﻁﺭﺡ ﻭﻨـﻀﻴﻑ ﺍﻟﺤـﺩ
i )(k 12 − u 1 v α
v2 4α 2 v2 v2 i i u 12 − k 1 v = k 12 − 2 − k 1 v + 2 α 4α 4α { 1 23 α v2 v2 i 1 k )v − + 4α 2 4α 2 α
= (k 12 −
2
v2 iv = k 1 − ( ) + 2 2α 4α
ﻓﻴﺼﺒﺢ ﺍﻟﺘﻜﺎﻤل ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ 2 v 2 1 iv −α k − ( ) + 4α 2 1422α 4 3 q
∞
1
e
∫ dk
∞−
∞ v2 4α
1 − αq ∫−∞dk e 14243 2
−
=e
v2 4α 2
−α
− αq 2
.e
= ∫ dk e 1
∞−
ﺍﻟﺩﺍﻟﺔ ﺯﻭﺠﻴﺔ ﻭﻋﻠﻴﻪ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ﺍﻟﺘﻜﺎﻤل ﻭﻗﻴﻤﺔ ﺍﻟﺘﻜﺎﻤل ﻴﺴﺎﻭﻱ
∞
2
∞
2∫ du 1 e −αq °
∫ 1 2 α
)..............(2 − 4
v2 4α
−
.e
D ∴f (x) = e{ . α phase factor ikov
ﻻﺤﻅ ﺃﻥ ﺍﻟﺩﺍﻟﺔ ) f (xﻓﻲ ) (2-4ﺩﺍﻟﺔ ﺘﺨﻴﻠﻴﺔ .ﻭﻟﻜﻲ ﺘﺒﻌﺩ ﻫﺫﻩ ﺍﻟﺩﺍﻟﺔ ﻋﻠـﻰ ﺠﺴﻡ ﻴﻠﺯﻡ ﺃﻥ ﺘﻜﻭﻥ ﺤﻘﻴﻘﻴﺔ .ﻭﻟﻬﺫﺍ ﺍﻟﺴﺒﺏ ﻓﺎﻟﺫﻱ ﻴﻬﻤﻨﺎ ﻟﻴﺱ ) f (xﺒل ﻤﺭﺒﻊ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻁﻠﻘﺔ ﻷﻨﻬﺎ ﺤﻘﻴﻘﻴﺔ.
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)f (x) = f * (x) f (x 2
x2
x2
D − 4α D − 4α e ) (e +ikox ) e α α
− kox
= (e
v2
D − f ( x ) = e 2α α
)........... (2 − 5
2
ﻫﺫﻩ ﺍﻟﺩﺍﻟﺔ ﺘﻤﺜل ﺠﺴﻴﻡ. ﺍﻟﺩﺍﻟﺔ ) (2-5ﻟﻬﺎ ﻗﻴﻤﺔ ﻗﺼﻭﻯ )ﺃﻭ ﻗﻴﻤﺔ( ﻋﻨﺩ v = 0ﻭﺘﻌﺘﻤﺩ ﻋﻠـﻰ ﻗﻴﻤـﺔ ﺍﻟﺜﺎﺒﺕ ، αﺇﺫﺍ ﻜﺎﻨﺕ
α
ﻜﺒﻴﺭﺓ ﺘﻜﻭﻥ ﺍﻟﺩﺍﻟﺔ ﻋﺭﻴﻀﺔ ﻭﺍﻟﻌﻜﺱ ﺼﺤﻴﺢ ﺇﺫﺍ ﻜﺎﻨﺕ
α
ﺼﻐﻴﺭﺓ ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ ﺘﻌﺒﺭ ﻋﻥ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﻀـﻴﻘﺔ harrowﻭﻜﻤـﺎ ﻗﻠﻨـﺎ ﻓـﺈﻥ 2
)f (x
ﺘﻌﺒﺭ ﻋﻥ ﺠﺴﻴﻡ.
ﺒﺈﻤﻜﺎﻨﻨﺎ ﺃﻥ ﻨﻭﺠﺩ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ
)ﻤﻠﺤﻕ (Aﻨﺄﺨﺫ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ
∆v
ﻟﻜﻲ ﺘﻨﺫل ﺍﻟﺩﺍﻟﺔ ﻤﻥ ﺍﻟﻘﻴﻤﺔ ﺇﻟﻰ
1
e
2
)f (x
،ﻜﻤﺎ ﺸﺭﺤﻨﺎ ﻓﻲ ﺍﻟﺠـﺯﺀ ﺍﻟـﺴﺎﺒﻕ
ﻋﻨﺩﻤﺎ ﺘﻘل ﻗﻴﻤﺘﻬﺎ ﺇﻟﻰ
e
1
ﻋﻥ ﺃﻗﺼﻰ ﻗﻴﻤﺔ ﻟﻬﺎ
ﻫﺫﺍ ﻴﺤﺩﺙ ﻋﻨـﺩﻤﺎ ﻴـﺴﺎﻭﻱ ﺍﻷﺱ ﻭﺍﺤـﺩ ﺃﻱ
v2 ﻋﻨﺩﻤﺎ . =1 2α
⇒ v 2 = 2α ⇒ Δv ° = ± 2α )................(2 − 6
⇒ Δv = 2 2α
ﻨﻭﺠﺩ ﺍﻵﻥ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ) g (kﻭﺍﻟﺘﻲ ﺘﻌﻁﻲ ﺒــ
ﻴﻬﻤﻨﺎ ﻫﻭ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ) g2 (kﻭﻟـﻴﺱ ﺍﻟﺩﺍﻟـﺔ
)2 g (k
2
) g (k) = e − α (k − k °
) g 2 (k) = e − 2 α (k − k °
ﻭﺍﻟـﺫﻱ
ﻫـﺫﻩ ﺍﻟﺩﺍﻟـﺔ
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ﻤﺘﻤﺭﻜﺯﺓ ﺤﻭل
k°
ﻭﻟﻴﺱ ﺤﻭل k = 0ﺒل ﺤﻭل
ﺍﻟﻁﺭﻴﻘﺔ ،ﻨﻭﺠﺩ ﺍﻟﻌﺭﺽ ﻋﻨﺩﻤﺎ ﺘﻘل ﺍﻟﺩﺍﻟﺔ ﺇﻟﻰ )...................(2 − 7
)
1 2α
e
1
k =k°
ﻤﻥ ﻗﺴﻤﺘﻬﺎ ﺍﻟﻘﺼﻭﻯ.
( )Δk = 2ΔΔ° = (2
ﻻﺤﻅ ﺍﻟﺘﻨﺎﺴﺏ ﺍﻟﻌﻜﺴﻲ ﻓﻲ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ
ﻓﻲ ) (2-6ﻭﺍﻟﻌﺭﺽ ﻓﻲ
∆v
ﻓﻲ ﺍﻟﻌﻼﻗﺔ ) (2-7ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻨﻪ ﻜﻠﻤﺎ ﺘﻤﻴﺯﺕ ﺍﻟﺩﺍﻟﺔ
)f (x
ﺍﻨﺒﺴﻁﺕ ﻭﺍﺴﺘﻌﺭﻀﺕ ﻓﻲ ﻓﻀﺎﺀ kﻭﺍﻟﻌﻜﺱ ﺼﺤﻴﺢ. ﻭﻤﻥ ﺍﻟﻤﻬﻡ ﺃﻥ ﻨﻨﻅﺭ ﺇﻟﻰ ﺤﺎﺼل ﻀﺭﺏ ﺍﻟﻌﺭﻀﻴﻥ )...............(2 − 8
2 2α = 4
ﻋﺭﻀﻬﺎ ﻴﻭﺠـﺩ ﺒـﻨﻔﺱ
.
1 2α
∆k
ﻓـﻲ ﻓـﻀﺎﺀ xﻜﻠﻤـﺎ ∆ u , ∆v
Δ u . Δv = 2
ﺍﻟﺫﻱ ﻴﻬﻤﻨﺎ ﻓﻲ ﺍﻟﻌﻼﻗﺔ ) (2-8ﻟﻴﺱ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺩﺩﻴﺔ ،ﺍﻟﺫﻱ ﻴﻬﻡ ﻫﻭ ﺃﻥ ﺤﺎﺼل
ﺍﻟﻀﺭﺏ
∆ u , ∆v
ﻻ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺒﺕ .αﻫﺫﻩ ﺨﺎﺼﻴﺔ ﻋﺎﻤﺔ ﻟﻜـل ﺍﻟـﺩﻭﺍل
ﺘﺭﺘﺒﻁ ﺒﺒﻌﻀﻬﺎ ﺒﺘﺤﻭﻴﻼﺕ ﻓﻭﺭﻴﺭ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﻓﻲ ﺸﻜل .2-1 This is a general property of functions that are Fourier trans forms of each other.
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ﻭﻨﺨﻠﺹ ﻤﻥ ﻫﺫﺍ ﺍﻟﺠﺯﺀ ﺒﺎﻻﺴﺘﻨﺘﺎﺝ ﺃﻨﻪ ﻤﻥ ﺍﻟﻤﺴﺘﺤﻴل ﺃﻥ ﻨﺠﻌل ﻜـﻼﹰ ﻤـﻥ ﺍﻷﻤﺭ ﺍﻟـﺫﻱ، ﻭﻫﺫﻩ ﺨﺎﺼﻴﺔ ﻋﺎﻤﺔ ﻟﻠﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ.ﺫﺍﺕ ﻗﻴﻡ ﺼﻐﻴﺭﺓ
∆k
ﻭ
∆v
.ﻴﺘﺭﺘﺏ ﻋﻠﻴﻪ ﻨﺘﺎﺌﺞ ﻫﺎﻤﺔ ﺠﺩﺍﹰ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ
It is impossible to make both Δ v and Δk small. This is a general feature of wouae packets, but we shall soon see rat it has some very deep implications for quantum mechanics.
Figure 2-4 Relation between wave packet and its Fourier transform for a square-shaped wave packet.
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ﺍﻨﺘﺸﺎﺭ ﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ THE PROPAGATION OF WAVE PACKETS
ﺒﻌﺩﻤﺎ ﺩﺭﺴﻨﺎ ﺘﺭﺍﻜﺏ ﻤﻭﺠﺎﺕ ﻤﻥ ﻨﻭﻉ
e ikx
ﻟﻨﺘﺤﺼل ﻋﻠﻰ ﺩﺍﻟﺔ ) ،f (xﻨـﻭﺩ
ﺍﻵﻥ ﻤﻨﺎﻗﺸﺔ ﻜﻴﻔﻴﺔ ﺇﻨﺘﺸﺎﺭ ﻫﺫﻩ ﺍﻟﻤﻭﺠﺎﺕ؟ ﺒﻼ ﺸﻙ ﺇﻥ ﺍﻨﺘﺸﺎﺭ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﻴﻌﺘﻤﺩ
ﻋﻠﻰ ﺇﻨﺘﺸﺎﺭ ﻜل ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﻤﻜﻭﻨﺔ ﻟﻬﺎ .ﻓﻠﻭ ﺃﺨﺫﻨﺎ ﻤﻭﺠﺔ ﺘﺘﻐﻴﺭ ﻤﻜﺎﻨﻴﺎﹰ ﻋﻠـﻰ
ﻤﺤﻭﺭ ) Xﺒﻴﻨﻤﺎ ﻻ ﺘﻌﺘﻤﺩ ﺍﻟﻤﻭﺠﺔ ﻤﻥ ﺤﻴﺙ ﺍﻹﺤﺩﺍﺜﻴﺎﺕ t , yﻋﻠﻰ ﺍﻟﺯﻤﻥ( .ﻫـﺫﻩ ﺍﻟﻤﻭﺠﺎﺕ ﺘﺴﻤﻰ ﻤﻭﺠﺎﺕ ﻤﺴﺘﻭﻴﺔ ،plane wanesﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ ﻫﺫﻩ ﺍﻟﻤﻭﺠﺎﺕ ﻜﺩﺍﻟﺔ
ﻓﻲ ﺍﻟﺯﻤﻥ tﻭﺍﻟﻤﻭﺠﺔ Xﺒﺎﻟﺼﻴﻐﺔ. )...... (2-9 ﺤﻴﺙ
e iuv − iwt
w = 2Dv 2D =k λ
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )(2-9 )................(2 − 10
v ) 2 Di ( − vt D
=e
2D v − i 2 Dvt λ
ei
ﻟﻭ ﺃﺨﺫﻨﺎ ﻜﺤﺎﻟﺔ ﺨﺎﺼﺔ؛ ﺇﻨﺘﺸﺎﺭ ﻤﻭﺠﺎﺕ ﺍﻟﻀﻭﺀ ﻓﻲ ﺍﻟﻔـﻀﺎﺀ ﺤﻴـﺙ ﻟﻠﻀﻭﺀ ،ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ).(2-10 ).................(2 − 12
) = e ik ( v − ct
2D ) i ( v − ct λ
=e
v c )− t λ λ
( 2 Di
c λ
=v
e
ﺨﺫ ﺍﻵﻥ ﺘﺭﺍﻜﺏ ﺍﻟﻤﻭﺠﺎﺕ ﺴﻌﺘﻬﺎ ) g (kﻤﻥ ﻫﺫﺍ ﺍﻟﻨﻭﻉ ﻤﻥ ﺍﻟﻤﻭﺠﺎﺕ ،ﻓﺒﻌﺩ
ﻤﺭﻭﺭ ﺯﻤﻥ ﻗﺩﺭﻩ t
∞
)f (x 1 t) = ∫ du g (u) e iu (v −ct) = eik (v − ct ∞−
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ﺇﻥ ﻫﺫﻩ ﺍﻟﺩﺍﻟﺔ ) f (x – ctﻟﻬﺎ ﻨﻔﺱ ﺍﻟﺸﻜل ﺍﻟﺫﻱ ﺒﺩﺃﺕ ﺒﻪ ﻗﺒل ﺍﻻﻨﺘﺸﺎﺭ ﻤـﺎ ﻋﺩﺍ ﺃﻥ ﻤﻭﻗﻌﻬﺎ ﺒﺩﻻﹰ ﻤﻥ ﺃﻥ ﻴﻜﻭﻥ Xﺃﺼـﺒﺢ ﺍﻵﻥ ﻤﺘﺭﻜـﺯ )ﺃﻭ ﻤﺘﺤﻴـﺯ( ﻋﻨـﺩ v = ct ⇐ v − ct = o
ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺤﺯﻡ ﺍﻟﻀﻭﺀ ﺍﻟﻤﻭﺠﻴﺔ ﺘﻨﺘﺸﺭ ﻓﻲ ﺍﻟﻔﻀﺎﺀ ﺒﻼ ﺘﺸﻭﻴﻪ. Thus the wave packet go light waves propagates, without any distortion, with velocity, the velsal'y of light.
ﻭﻟﻜﻥ ﺍﻟﺫﻱ ﻴﻌﻨﻴﻨﺎ ﻫﻨﺎ ﺍﻨﺘﺸﺎﺭ ﻤﻭﺠﺎﺕ ،ﻴﻔﺘﺭﺽ ﺃﻨﻬﺎ ﺘﹸﻌﺒﺭ ﻋﻥ ﺠﺴﻴﻤﺎﺕ ،ﻓﻲ
ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻟﻥ ﻨﺴﺘﻁﻴﻊ ﺍﺴﺘﺨﺩﺍﻡ ﺍﻟﻌﻼﻗﺔ. )... (2-13
w = kc W (k) = kc
ﻻﺤﻅ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﺼﺤﻴﺤﺔ ﻓﻘﻁ ﻓﻲ ﺤﺎﻟﺔ ﺍﻨﺘﺸﺎﺭ ﺤﺯﻤﺔ ﻀﻭﺀ ﻤﻭﺠﻴﺔ ﻓـﻲ
ﺍﻟﻔﻀﺎﺀ .ﻭﻟﻜﻲ ﺘﻨﻁﺒﻕ ﺩﺭﺍﺴﺘﻨﺎ ﻋﻠﻰ ﺍﻨﺘﺸﺎﺭ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ )ﺠﺴﻴﻡ( ﻴﻠﺯﻡ ﺍﺴـﺘﺨﺩﺍﻡ
ﺍﻟﻌﻼﻗﺔ ﺍﻟﻌﻼﻤﺔ ﻟـ ) w (kﺤﻴﺙ
w = (k) ≠ kc or kv
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻨﻌﻴﺩ ﻜﺘﺎﺒﺔ ﻤﻌﺎﺩﻟﺔ ) (2-11ﺒﺎﻟﺼﻴﻐﺔ. )ik (v − ct )f (x 1 t) = ∫ dk g (k) e ik (v − ct
ﻟﺘﺼﺒﺢ ﻓﻲ ﺤﺎﻟﺔ ﺠﺴﻡ ﺤﺭ ﺍﻟﺤﺭﻜﺔ = dk g(k) e ikv − iw (k)t
)..........(2 − 14
ﻻﺤﻅ ﺃﻨﻨﺎ ﻟﻡ ﻨﻌﺭﻑ ﺒﻌﺩ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺤﻘﻴﻘﻴﺔ ﻟﻁﺒﻴﻌﺔ ﺍﻟﺩﺍﻟﺔ ) ،w (kﻭﻟﻜﻥ ﺨـﺫ ﺍﻵﻥ ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﻤﺘﻤﺭﻜﺯﺓ ﻭﻤﺘﺤﺒﺯﺓ ﺠﺩﺍﹰ ﺤﻭل ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ αﻓﻲ ﺍﻟﻌﻼﻗﺔ
2
) g (k) = e −α(k − k °
k°
ﻓﻲ ﻓﻀﺎﺀ .u
ﺴﺘﻜﻭﻥ ﻜﺒﻴﺭﺓ .ﻫﺫﺍ ﻗﺩ ﻻ ﻴﻌﻨﻲ ﺃﻥ
) f (xﺴﺘﻜﻭﻥ ﻤﺘﺤﺒﺯﺓ ﻓﻲ ﻓﻀﺎﺀ vﻭﻟﻜﻥ ﻫﺫﺍ ﺍﻟﺘﻘﺭﻴﺏ ﺴﻴﺴﻬل ﺤﺴﺎﺒﺎﺘﻨﺎ. - ٥٣ -
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ﺒﻤﺎ ﺃﻥ ﺍﻟﺘﻜﺎﻤل ﻓﻲ ) (2- 14ﻤﺘﻤﺭﻜﺯ ﺤﻭل
ﻤﻊ ﺍﻓﺘـﺭﺍﺽ ﺃﻥ
k°
)w (k
ﻻ
ﺘﺘﻐﻴﺭﻩ ﺒﺸﺩﺓ ﻤﻊ ﺘﻐﻴﺭ .k )............(2 − 15
ﺍﻟﺤﺩ ﺍﻷﻭل
) w (k °
)dw(k 1 )d 2 w (k ( ) k + (k − k ° ) 2 ) ° dk 2 dk 2 k°
( ) w (k) ≅ w (k ° ) + (k − k °
ﺜﺎﺒﺕ ﻷﻨﻪ ﻻ ﻴﻌﺘﻤﺩ ﻋﻠﻰ .kﺍﻟﺤﺩ ﺍﻟﺜﺎﻨﻲ
dw ) dk k = k °
(
ﻴﻤﺜل ﺴﺭﻋﺔ
ﺍﻟﻤﺠﻤﻭﻋﺔ -group velocityﻜﻤﺎ ﺸﺭﺤﻨﺎ ﻓﻲ ﺍﻟﺘﻤﻬﻴﺩ -ﻫﻲ ﺍﻟﺴﺭﻋﺔ ﺍﻟﺘﻲ ﺘـﺴﻴﺭ
ﺒﻬﺎ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ.
dw ) = vg dk k = k°
⇒
(
)(2-16
and 1 d2w ) =β ( 2 dk 2 k = k °
ﺒﺎﻓﺘﺭﺍﺽ ﺃﻥ: k1 = k − k ° ⇒ dk 1 = dk
ﻓﺈﻥ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﺒﺎﻟﺼﻴﻐﺔ. i (k1 + k ° ) v − i w (k ° ) + k1vg + k12β t
∞
f (x 1 t) = ∫ dk 1 e −αk e 12
∞−
ﺨﺫ ﺍﻟﺤﺩﻭﺩ ﺍﻟﺜﺎﺒﺘﺔ ﺨﺎﺭﺝ ﺍﻟﺘﻜﺎﻤل. 1 1
12βt
12
1
e −αk . e ik v. e −ik vgt. e −ik 12βt
e −ik
)1 (v − vgt
e ik
∞
∫ dk
∞−
ik ° v − i w(k ° )t
∞
∫ dk e
1 − αk12
=e
ikov − iw(k ° )t
=e
∞−
)........(2-17 ﻭﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺘﻲ ﺃﺠﺭﻴﻨﺎﻫﺎ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ ،ﻴﻤﻜﻥ ﻋﻤل ﺍﻟﺘﻜﺎﻤل ﻋﻥ
ﻁﺭﻴﻕ ﺇﻜﻤﺎل ﺍﻟﻤﺭﺒﻊ ﻓﻲ ﺍﻷﺱ ،ﺨﺫ ﺍﻷﺱ ﻓﻘﻁ. )− (α + iβ t ) k12 + ik1 (v − μ g t
=e
12 + ik1 (v − μgt) − ik12 βt
e − αk
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.ﻋﺎﻤل ﻤﺸﺘﺭﻙ =e
− (α + iβ t)
ﺨﺫ
(v − ν g t) − (α + iβ t) k12 − ik1 (α = iβ t )
.ﻟﻠﺤﺼﻭل ﻋﻠﻰ ﻤﺭﺒﻊ ﻜﺎﻤل ﻨﻁﺭﺡ ﻭﻨﻀﻴﻑ ﺍﻟﺤﺩ (v − ν g t)2 4(α + iβ t)
e
iu1 (u − v g t) (x − v g t) 2 ( x − v g t) 2 + − (α + iβ t) u12 − − α + iββ 4 (α + iββ) 4 (α + iβ t)
ﺨﺫ ﻫﺫﻩ ﺍﻹﺸﺎﺭﺓ ﻤﺸﺘﺭﻙ ﻤﻊ ﺍﻟﺤﺩ ﺍﻷﻭل ik 1 (u − v g t) (u − v g t) 2 − (α + iβ t) k12 − − α + iβ t 4 (α + iββ)
e ⇒ e
i (u − v g t) − (α + iβ t) k1 − 2 ( α + iβ t
2 +
( x − v t) 2 g − 4 (α + iβ t)
2 1 ( x − v g t) 4 (α + iβ t)
............(2 − 18)
:( ﻨﺘﺤﺼل ﻋﻠﻰ2-17) ( ﻓﻲ2-18) ﺒﺘﻌﻭﻴﺽ f ( x1t ) = e
∞
∫ dk
i ( k° x − w ( k° ) t )
1
e
(v − µ g t ) − (α + i β t ) k1 − 2 (α = i β t )
−∞
= e i ( k° x − w ( k° ) t e
−
( x− y g t ) 2 4 (α + iβ t )
2 −
ﺛﺎﺑﺖ
( x − vg t ) 2
e 4 ( α + i βt ) 1424 3 اﻟﺤﺪ
ھﺬا
اﺑﺖ
i ( x −u gt) 2 − ( α + i β t ) k1 − 2 (α + iβ t ) 1
∞
∫ dk e
−∞
1444424444 3
− (x − u g t) 2
f (x 1 t) = e
i (u ° x − w (k ° ) t
e
D (α + iβ t)
4 (α + iβ t)
2
f (x 1 t) = f (n 1 t) f * (x 1 t) = e ° . e
−
(v − u g t) 2 4 (α + iβ t)
−
( v − u g t) 2 4 (α − iβ t)
D D (α + i β t ) ( α − i β t)
− (v − u g t) 2 ( 2α ) D 2 α 2 + β 2t 2 = 1. e (α 2 + β 2 t 2 ) =e
−
(v − u g t) 2
D
2 (α + β 2 t 2 /αα (α 2 + β 2 t 2 )
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ﺒﻘﺴﻤﺔ ﺍﻷﺱ ﻓﻲ ﺍﻟﺒﺴﻁ ﻭﺍﻟﻤﻘﺎﻡ ﻋﻠﻰ . α ﺒﻤﻘﺎﺭﻨﺔ ﺍﻟﺼﻴﻐﺔ ﻓﻲ ) (2-19ﺒـ )(2-5
v2 D − 2α 2 f (x) = e α
ﻨﺠﺩ ﺃﻥ ﻫﻨﺎﻙ ﺘﻤﺎﺜل ﻤﻊ ﻤﻼﺤﻅﺔ ﺃﻥ vﻓﻲ ) (2-5ﺘﻐﻴﺭﺕ ﻟـ ﺘﺤﻭﻟﺕ ﻟـ
β2 t 2 α
ﻭﻟﻬﺫﺍ ﻓﺈﻥ
)(v − u g t
ﻭα
α+ 2
) f (x 1 t
ﺘﻤﺜل ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﺘﺴﻴﺭ ﺒﺴﺭﻋﺔ
ug
ﻭﻟﻜـﻥ ﻋـﺭﺽ
ﻫﺫﻩ ﺍﻟﺤﺯﻤﺔ ﻏﻴﺭ ﻤﺤﺩﺩ ﻭﻟﻜﻥ ﻴﺘﻐﻴﺭ ﻤﻊ ﺍﻟﺯﻤﻥ ،ﻻﺤﻅ ﺃﻥ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ).(2-19 Δv = 2. Δ v ° = 2. 2 (α + β 2 t 2 /αα or )...............(2 − 20
β 2 t 2 12 ) α2
Δv = 8α (1 +
ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (2-20ﻴﺘﻀﺢ ﺃﻥ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻴﺯﺩﺍﺩ ﻤﻊ ﺍﻟﺯﻤﻥ ﺃﻱ ﺃﻥ ﺍﻟﺩﺍﻟﺔ ﺘﻨﺒﺴﻁ ﻤﻊ ﻤﺭﻭﺭ ﺍﻟﺯﻤﻥ ﻭﺫﻟﻙ ﻤﻘﺎﺭﻨﺔ ﺒﻌﺭﻀﻬﺎ ﻋﻨﺩ t = 0 Δv = ( t = 0 ) = 8α
ﺃﻤﺎ ﻋﻨﺩ ﺃﻱ ﺯﻤﻥ ﺁﺨﺭ ﻓﺈﻥ ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻴﻌﻁﻲ ﺒﺎﻟﺼﻴﻐﺔ ) (2- 20ﺇﻥ ﻤﻌﺩل
ﺍﻻﻨﺒﺴﺎﻁ ﻓﻲ ﺍﻟﺩﺍﻟﺔ ﺴﻴﻜﻭﻥ ﺼﻐﻴﺭﺍﹰ ﺇﺫﺍ ﻜﺎﻨﺕ αﻜﺒﻴﺭﺓ ﺃﻱ ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ ﻤﻨﺒﺴﻁﺔ ﻜﺜﻴﺭﺍﹰ ﻋﻨﺩ ﺍﻟﺒﺩﺍﻴﺔ.
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ﻤﻥ ﺍﻟﺤﺯﻡ ﺍﻟﻤﻭﺠﻴﺔ ﺇﻟﻰ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ: FROM WAVE PACKETS TO THE SCHRöDINGER EQUATION
ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ }
.........
iku −iw ( k ) t
{f ( x t ) = ∫ dk g (u) e 1
ﺘﻤﺜل ﺠﺴﻴﻡ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻟﻪ
ﺘﺴﺎﻭﻱ pﻭﻁﺎﻗﺘﻪ ﺍﻟﺤﺭﻜﻴﺔ ) ،(p2 /2mﻓﺈﻥ ﻫﺫﺍ ﻴﺴﺘﻠﺯﻡ ﺃﻥ ﺴﺭﻋﺔ ﺍﻟﻤﺠﻤﻭﻋـﺔ ug
ﺘﻌﻁﻰ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ. dw p )= particle velocity ⇒ = .................(2 − 21 dk m
= ug
ﻭﻤﻥ ﺍﻓﺘﺭﺍﻀﺎﺕ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ،ﻓﺈﻥ ﺃﻱ ﺇﺸﻌﺎﻉ ﻟﻪ ﺘﺭﺩﺩ wﻓﺈﻨﻪ ﻴﻘﺎﺒﻠﻪ ﻁﺎﻗﺔ
Eﺘﻌﻁﻲ ﺒﺎﻟﻌﻼﻗﺔ.
E=hw or E h
=w
ﺇﺫﺍ ﻜﺎﻨﺕ Eﻫﻲ ﻁﺎﻗﺔ ﺠﺴﻴﻡ ﺤﺭ،ﻓﺈﻨﻬﺎ ﺘﻤﺜل ﻁﺎﻗﺘﻪ ﺍﻟﺤﺭﻜﻴﺔ p2/2mﺃﻱ ﺃﻥ. ( p 2 / 2m) p 2 =w = 2mh h
)(2 − 23
ﻭﻋﻠﻴﻪ ﻓﻤﻥ ﺍﻟﻤﻤﻜﻥ ﺃﻥ ﻨﻜﺘﺏ ﻤﺘﺠﻪ ﺍﻟﻤﻭﺠﺔ ) kﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ( ﻭﻤﻥ ﻤﻌﺎﺩﻟﺔ
ﺩﻴﺒﺭﻭﻟﻲ . λ = n p
2D 2 D p p = = = λ (h ) ( h ) h p 2D
).................(2 − 24
=k
ﻭﻟﻬﺫﺍ ﺍﻟﺴﺒﺏ ﻴﻤﻜﻨﻨﺎ ﺇﻋﺎﺩﺓ ﻜﺘﺎﺒﺔ ﺍﻟﺼﻴﻐﺔ ) (2-14ﺒﺩﻻﻟﺔ ) pﺒﺩﻻﹰ ﻤﻥ (k ψ (p ) e i (px −Et)/h
)............(2 − 25
ﺤﻴﺙ ﺃﻥ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ
)ψ (x 1 t
ﺍﻟﺠﺯﺌﻴﺔ .partial differentiable.
1
∫ dp 2Dh
= )ψ (x 1 t
ﺘﻤﺜل ﺍﻟﺤل ﺍﻟﻌـﺎﻡ ﻟﻠﻤﻌﺎﺩﻟـﺔ ﺍﻟﺘﻔﺎﻀـﻠﻴﺔ
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iD
∂ ψ (x 1 t) ∂t
=
1
∫ dp φ (p) E e 2Dh
i (px − Et)/t
p 2 i( px − Et) /D = ∫ dp φ (p) 2m e 2Dh 1
.x ﻤﺭﺘﻴﻥ ﺒﺎﻟﻨﺴﺒﺔ ﻟـ
ψ (x 1 t)
............(2 − 26)
( ﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻴﻬﺎ ﺒﺘﻔﺎﻀل2-26) ﺤﻴﺙ ﺃﻥ
∂ ∂ψ ∂ 1 ∂ ( )= dp φ (p) e i (px − Et)/D ∫ ∂x ∂x ∂x 2Dh ∂x ∂ 1 ip = dp φ (p) . e i (px − Et)/h ∫ h ∂x 2Dh = =
1 2Dh 1 2Dh
ip ip i (px − Et)/t e h
∫ dp φ (p) h . ∫ dp φ ( p ) ( −
p 2 i (px − Et) /t )e ]................(2 − 27) D2
ﻴﻠﺯﻡ ﻀﺭﺏ،( ﻤﻥ ﺍﻟﻁﺭﻑ ﺍﻷﻴﻤﻥ2-26) ( ﺒـ2-27) ﻭﻟﻜﻲ ﺘﺘﺴﺎﻭﻯ ﻤﻌﺎﺩﻟﺔ
ﺃﻱ، − D ( ﻓﻲ2-27) ﻁﺭﻓﻲ 2
2m
−
h2 ∂2ψ 1 D2 p 2 i (px − Et)/D = dp φ (p) ( − ) ( − )e ∫ 2m ∂x 2 2m D2 2Dh 1 p 2 i (px − Et)/D ] = dp φ (p) ( )e ∫ 2m 2Dh
]
:ﻭﻋﻠﻴﻪ ﻓﺈﻥ ih
∂ψ (x 1 t) ∂t
=−
2 h 2 ∂ ψ (v1 t) 2m ∂v 2
................(2 − 28)
.ﻭﻫﺫﻩ ﺼﺤﻴﺤﺔ ﻟﺠﺴﻡ ﺤﺭ (ﻋﻼﻗﺎﺕ ﺍﻟﻼﺘﺤﺩﻴﺩ )ﺍﻟﻼﺘﻌﻴﻴﻥ
The uncertainty Relations
ﺃﺤﺩ ﺍﻟﻨﺘﺎﺌﺞ ﺍﻟﻤﻬﻤﺔ ﺍﻟﺘﻲ ﺘﻭﺼﻠﻨﺎ ﻟﻬﺎ ﻤﻥ ﻤﻨﺎﻗﺸﺘﻨﺎ ﻟﻠﺤﺯﻡ )ﺍﻟﺭﺯﻡ( ﺍﻟﻤﻭﺠﻴـﺔ .
k
ﻭ ﻓﺭﺍﻍ
v
ﻫﻲ ﺍﻟﻤﻌﻜﻭﺴﺔ ﻟﻠﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻷﻋﺭﺍﺽ ﻓﻲ ﻓﺭﺍﻍ Δ k Δ v ≥1
p=h k
ﻭﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﻌﻼﻗﺔ
h
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ﺒﻀﺭﺏ ﺍﻟﻁﺭﻓﻴﻥ ﺒـ
ﺴﻨﺘﺤﺼل ﻋﻠﻰ ﻋﻼﻗﺔ ﻫﺎﻴﺯﻨﺒﺭﺝ Heisenberg ancernrainrg clarion Δ pΔ v≥h
)........(1
ﻭﺒﻤﺎ ﺃﻥ ﺍﻟﻌﺭﺽ ﻴﻤﺜل ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﺘﻲ ﻤﻥ ﺍﻟﻤﺤﺘﻤل ﺃﻥ ﻴﻜﻭﻥ ﻓـﻲ ﺍﻟﺠـﺴﻡ
)( p- , v - Iﻓﻀﺎﺀ ﻓﺈﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﺘﺒﻴﻥ:
ﻟﻭ ﺃﺭﺩﻨﺎ ﺃﻥ ﻨﻌﻤل ﺤﺯﻤﺔ )ﺭﺯﻤﺔ( ﻤﺘﻤﺭﻜﺯﺓ )ﻤﺘﻤﻴﺯﺓ ﺠﺩﺍﹰ( ﻭﻤﻭﺠﻴﺔ ﻓـﻲ x
ﻓﻀﺎﺀ ،ﻓﺈﻨﻪ ﻤﻥ ﺍﻟﻤﺴﺘﺤﻴل ﻋﻠﻴﻨﺎ ﺃﻥ ﻨﺸﺎﺭﻙ ﺒﻬﺎ ﻜﻤﻴﺔ ﺤﺭﻜﺔ ﻤﻌﺭﻓﺔ ﺠﻴﺩﺍﹰ) .ﺍﻷﻤﺭ
ﺍﻟﺫﻱ ﻴﻌﺘﺒﺭ ﺒﺩﻴﻬﻲ ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﺭﺯﻤﺔ ﻤﻭﺠﻴﺔ ﺫﺍﺕ ﻜﻤﻴﺔ ﻤﺤﺩﺩﺓ ﻤﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜـﺔ
ﻴﺠﺏ ﺃﻥ ﺘﻜﻭﻥ ﻏﻴﺭ ﻤﺘﻤﻴﺯﺓ ﻓﻲ ﻓﻀﺎﺀ v-ﺃﻱ ﺃﻥ ﺘﻜﻭﻥ ﻋﺭﻴﻀﺔ ﺠﺩﺍﹰ. ﻭﻫﺫﺍ ﺍﻟﺤﺩ )ﺍﻟﺤﺼﺭ( ﻭﺍﺤﺩ ﻤﻥ ﺇﻟﺯﺍﻤﺎﺕ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻋﻠﻰ ﻭﺼﻔﻨﺎ ﻟﻸﻨﻅﻤﺔ
ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺒﺎﻟﻤﻔﺎﻫﻴﻡ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ .ﻓﻲ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻜﺎﻷﻤﺭ ) p, xﺍﻟﻤﻭﻗـﻊ ،ﻜﻤﻴـﺔ ﺍﻟﺤﺭﻜﺔ( ﻤﺴﺘﻘﻼﺕ ﻋﻥ ﺒﻌﻀﻬﻤﺎ. ﺒﻴﻨﻤﺎ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ p, xﻤﻜﻤﻠﻴﻥ ﻟﺒﻌﻀﻬﻤﺎ. ﻻﺤﻅ ﺃﻨﻪ ﻤﻥ ﻗﻴﻤﺔ
ﺍﻟﺼﻐﻴﺭﺓ ﻴﺘﻀﺢ ﻟﻨﺎ ﺃﻥ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺘﻔﺸل ﻓـﻲ
h
ﺤﺎﻟﺔ ﺍﻷﻨﻅﻤﺔ ﺍﻟﻤﺠﻬﺭﻴﺔ ﻓﻠﻭ ﺍﻋﺘﺒﺭﻨﺎ ﺃﻥ ﺤﺒﺔ ﻏﺒﺎﺭ ﻤﺜﺎل :ﺨﺫ ﺠﺴﻤﺎﹰ ﻜﺘﻠﺘﻪ
ﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ
Δp
m 1μ g
m = 10 −4 g υ = 10 4 cn / s
ﻭﻤﻭﻗﻌﻪ ﻤﻌﺭﻭﻑ ﻟﺩﻗﺔ ، 1μ mﺍﺤﺴﺏ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻓـﻲ
ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺴﺎﺒﻘﺔ. m2 s2
J = kg.
D h ≥ ⇒ Δp 2 2Δ v
≥ Δp Δv
6.63×10 −34 J.S J. S m ≥Δ p = 0.5× 10 −28 = 5 × 10 −29 k g −6 m s 4D ×1 ×10 m
ﻭﻫﺫﺍ ﻴﻁﺎﺒﻕ ﺴﺭﻋﺔ ﻤﻘﺩﺍﺭﻫﺎ
−29 m 5 ×10 u g m/s = s 1×10 −9 u g
− 20
Δν = 5 ×10
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ﻭﻫﺫﻩ ﺍﻟﻘﻴﻤﺔ ﻤﻬﻤﻠﺔ .ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻟﻸﺠﺴﺎﻡ ﺍﻟﻜﺒﻴﺭﺓ ﻋﻼﻗﺔ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻟﻴﺴﺕ ﺫﺍﺕ ﺃﻫﻤﻴﺔ ﻭﻋﻠﻴﻪ ﻓﺈﻥ )ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ( ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺘﻌﻁﻲ ﻨﻔﺱ ﺍﻟﻘﻴﻤﺔ. ﻗﻴﺎﺱ ﻤﻜﺎﻥ ﺇﻟﻜﺘﺭﻭﻥ )ﻤﻴﻜﺭﻭﺴﻜﻭﺏ ﻫﻴﺯﺒﻨﺭﺝ( )Measwrcmerg position of aneleiton (Hesibdoceg Microswgee
ﺸﻌﺎﻉ ﻤﻥ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻟﻬﺎ ﻜﻤﻴﺔ ﺤﺭﻜﺔ ﻤﺤﺩﺩﺓ ﺠﻴﺩﺍﹰ
px
ﺘﺘﺤﺭﻙ ﻓﻲ
ﺍﺘﺠﺎﻩ t x
ﻭﺍﻟﻐﺭﺽ ﻤﻥ ﺍﻟﻤﻴﻜﺭﻭﺴﻜﻭﺏ ﺍﻟﺸﺎﺸﺔ ﻫﻲ ﺭﺅﻴﺔ ﻤﻭﻗﻊ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻭﺫﻟﻙ ﺒﻤـﺸﺎﻫﺩﺓ ﺍﻟﻀﻭﺀ ﺍﻟﻤﺘﺸﺘﺕ ﻤﻥ ﺍﻹﻟﻜﺘﺭﻭﻥ.
ﺇﻥ ﺍﺴﺘﻁﺎﻡ ﺍﻟﻔﻭﺘﻭﻨﺎﺕ ﺒﺎﻹﻟﻜﺘﺭﻭﻥ ﻴﺭﺘﺩ ﺨﻼل ﺍﻟﻤﻴﻜﺭﻭﺴﻜﻭﺏ .ﻭﺇﻥ ﻗـﺩﺭﺓ ﺘﺤﻠﻴل ﺍﻟﻤﻴﻜﺭﻭﺴﻜﻭﺏ ) bastionﺃﻱ ﺍﻟﺩﻗﺔ ﺍﻟﺘﻲ ﺒﻬﺎ ﻴﻤﻜﻨﻨﺎ ﺘﺤﺩﻴﺩ ﻤﻭﻗﻊ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻤﻥ ﻤﻌﻠﻭﻤﺎﺕ ﻤﻥ ﺍﻟﻀﻭﺀ(.
ﺤﻴﺙ λﻁﻭل ﻤﻭﺠﺔ ﺍﻟﻀﻭﺀ
λ sin ϕ
~ Δν
ﻴﺠﻌل λﺃﻗل ﻤﺎ ﻴﻤﻜﻥ ﻭ ϕﺃﻜﺒﺭ ﻤﺎ ﻴﻤﻜﻥ ﺒﺈﻤﻜﺎﻨﻨﺎ ﺘﺼﻐﻴﺭ
∆v
ﻟﻠﺤﺩ ﺍﻟﻤﺭﻏﻭﺏ
ﻫﺫﺍ ﻴﻤﻜﻨﻨﺎ ﻓﻌﻠﻪ ﻋﻠﻰ ﺤﺴﺎﺏ ﻤﻌﻠﻭﻤﺎﺘﻨﺎ ﻤﻥ ﺩﻗﺔ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻓﻲ ﺍﺘﺠﺎﻩ .νx ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﺘﺨﺒﺭﻨﺎ ﺒﺄﻥ ﺍﻹﺸﺎﺭﺍﺕ ﺍﻟﻤﺴﺠﻠﺔ ﻋﻠﻰ ﺍﻟﺸﺎﺸﺔ ﻫﻲ ﻋﺒﺎﺭﺓ ﻋـﻥ
ﻓﻭﺘﻭﻨﺎﺕ ﺘﺸﺘﺕ ﻤﻥ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﺘﺠﺎﻩ ﺍﻟﻔﻭﺘﻭﻨﺎﺕ ﺒﻌﺩ ﺍﻟﺘﺸﺘﺕ ﻏﻴﺭ ﻤﺤﺩﺩ ﺨـﻼل ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻘﺎﺒﻠﺔ ﻟﻠﻔﺘﺤﺔ ﻭﻋﻠﻴﻪ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﻤﺭﺘﺩﺓ ﻟﻺﻟﻜﺘﺭﻭﻥ ﻏﻴﺭ ﻤﺤﺩﺩﺓ ﺒـ h λ c h hv λ ⇒ ϕ = 2 sin ϕ . ⇒ Δ p x ~ 2 sin φ c v c v
Δp (sin ϕ + sin ϕ ).
ﻭﻋﻠﻴﻪ
Dv λ sin ϕ . ~ 4Dh c sin ϕ
Δ p v Δv ~ 2
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ﻟﺘﺤﺴﻴﻥ ﺩﻗﺔ ﺍﻟﻤﻭﻗﻊ ﻟﻺﻟﻜﺘﺭﻭﻥ ﻴﺠﺏ ﺃﻥ ﻨﺴﺘﺨﺩﻡ ﻀﻭﺀ ﺫﻭ λﺼﻐﻴﺭﺓ ﻟﻜﻥ ﻫﺫﺍ ﻴﻨﺘﺞ ﺨﻁﺄ ﻜﺒﻴﺭ ﺃﻭ ﺘﺸﻭﻴﺵ ﻜﺒﻴﺭ ﻓﻲ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻭﺍﻟﻌﻜﺱ :ﻟﻭ ﺃﺭﺩﻨﺎ ﺃﻥ ﻨﻘﻠل ﺍﻟﺨﻁﺄ ﺒﻘﻴﺎﺴﻨﺎ ﺒﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻴﺠﺏ ﺃﻥ ﻨﺴﺘﺨﺩﻡ ﺇﺸﻌﺎﻉ ﺫﻭ ﻁﻭل ﻤﻭﺠﻲ ﻜﺒﻴﺭ ﻭﻫﺫﺍ
ﺒﺩﻭﺭﻩ ﻴﺅﺩﻱ ﺇﻟﻰ ﺨﻁﺄ ﻜﺒﻴﺭ ﻓﻲ ﺍﻟﻤﻭﻗﻊ.
ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻤﺒﺩﺃ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻫﻭ ﻨﺎﺸﺊ ﻤﻥ ﻁﺭﻴﻘﺔ ﺍﻟﻘﻴـﺎﺱ ﻨﻔـﺴﻬﺎ .ﻭﻋﻠـﻰ
ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﺫﺭﻱ ﻓﺈﻥ ﺍﻟﻘﻴﺎﺱ ﻨﻔﺴﻪ ﻴﻨﺸﺄ ﻋﻨﻪ ﺨﻁﺄ ﻭﺘﺸﻭﻴﺵ ﻤﻠﺤﻭﻅ ﻟﻠﻨﻅﺎﻡ ﻭﺫﻟـﻙ ﻟﻠﺘﻔﺎﻋل ﺒﻴﻥ ﺠﻬﺎﺯ ﺍﻟﻘﻴﺎﺱ ﺍﻟﻜﻤﻴﺔ ﺍﻟﻤﻘﺎﺴﺔ.
ﻭﻤﺒﺩﺃ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻴﻘﺭﺭ ﺃﻨﻨﺎ ﻻ ﻨﺴﺘﻁﻴﻊ ﺘﺤﺩﻴﺩ ﻤﺴﺎﺭ ﺠﺴﻡ ﺒﺩﻗﺔ ﻤﺘﻨﺎﻫﻴﺔ ﻜﻤـﺎ
ﻫﻭ ﺍﻟﺤﺎل ﻓﻲ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ.
ﻤﺜﺎل :ﺒﺎﻓﺘﺭﺍﺽ ﺃﻥ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻓﻲ ﻤﻭﻗﻊ ﺠﺯﺉ ﻫﻴﺩﺭﻭﺠﻴﻥ )ﺍﻟﺫﻱ ﻜﺘﻠﺘﻪ
kg
−27
( 2 ×10
ﻴﻘﺩﺭ ﺒﺤﻭﺍﻟﻲ 10-10 mﺍﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻓﻲ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ؟ ﻭﻜﺫﻟﻙ ﺍﻟﻨـﺴﺒﺔ Δ px px 6.6 ×10 −34 h ≥ Δp x = ≥ 6.6 ×10 − 24 kg − m/s −10 Δx (2D )10
ﻜﻤﻴﺔ ﺤﺭﻜﺔ ﺍﻟﺠﺯﺉ ﺍﻟﺫﻱ ﻴﺘﺤﺭﻙ ﺒﺴﺭﻋﺔ
m s
2000
)ﺍﻟﺴﺭﻋﺔ ﺍﻟﺤﺭﺍﺭﻴﺔ ﻋﻨﺩ
ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺍﻟﻐﺭﻓﺔ( m = 4 ×10 −24 kg. m/s s
Px = mc = 2 ×10 − 27 kg × 2 ×10 3
ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻨﺴﺒﺔ ﺍﻟﻼﺘﻌﻴﻴﻥ: 0.25
6.6 ×10 −24 = = 1.7 4 ×10 − 24
Δp x px
ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻜﻤﻴﺔ ﺤﺭﻜﺔ ﻫﺫﺍ ﺍﻟﺠﺯﺉ ﻻ ﻴﻤﻜﻥ ﺘﺤﺩﻴﺩﻫﺎ ﺒﺩﻗﺔ ﺃﻜﺜﺭ ﻤﻥ ﺍﻟﻘﻴﻤﺔ
ﺃﻷﺼﻠﻴﺔ.
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ﻟﻜﻥ :ﻟﻭ ﺃﺨﺫﻨﺎ ﻜﺘﻠﺔ ،ﻟﻬﺎ ﺴﺭﻋﺔ ،1000 mﻤﻭﻗﻌﻬﺎ ﻤﺤﺩﺩ ﻟﺩﻗﺔ 1. mmﻫـﺫﺍ s
ﺍﻟﺠﺴﻡ ﻤﻘﺩﺍﺭ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻓﻲ ﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ. 6.63×10 −34 6.6 ≥ Δp = ×10 −31 kg − m/s −3 10 m 2D p = 0.005 ×1000 = 50 kg - m/s
ﻨﺴﺒﺔ ﺍﻟﻼﺘﻌﻴﻴﻥ. Δp 6.6 ×10 −31 kg − k/s = = 1.3×10 −32 p
ﻭﻫﺫﺍ ﺭﻗﻡ ﺼﻐﻴﺭ ﺠﺩﺍﹰ ﻭﻻ ﻴﻤﻜﻥ ﻗﻴﺎﺴﻪ ﺒﺄﻱ ﺠﻬﺎﺯ. ﻤﺜﺎل :ﺯﻤﻥ ﺍﻟﺤﻴﺎﺓ ﻟﻠﺤﺎﻟﺔ ﺍﻟﻤﺜﺎﺭﺓ ﻟﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ
s
−8
.10ﻓﺈﻥ ﺍﻗل ﻗﻴﻤـﺔ
ﻟﻼﺘﺤﺩﻴﺩ ﻟﻠﻁﺎﻗﺔ ﻫﻭ؟ ΔE. Δt ≥ h
h 6.6 ×10 −34 = 1.0 × − 26 J = 6.5 ×10 −8 ev. = ∆t 2D ×10 −8
≥∆E
ﻭﻫﺫﺍ ﻴﺴﻤﻰ ﻋﺭﺽ ﺍﻟﻁﺎﻗﺔ ﻟﻠﺤﺎﻟﺔ ﺍﻟﻤﺜﺎﺭﺓ. ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﻁﻭل ﺍﻟﻤﻭﺠﺔ ﻭﺍﻟﺘﺭﺩﺩ ﻓـﻲ ﻤﻭﺠـﺔ ﺍﻟﻤﻭﺠـﺎﺕ wave Guide
ﺘﻌﻁﻲ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ:
c 2
ﺃﻭﺠﺩ ﺴﺭﻋﺔ ﺍﻟﻤﺠﻤﻭﻋﺔ
vg
=λ
v 2 − v°
ﻟﻬﺫﻩ ﺍﻟﻤﻭﺠﺎﺕ. c2 2
2
v 2 − v°
= λ2
2 2 4D 2 (v − v° ) w − w° = = D2 c2 c2 2
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4D2 2D ⇒ k2 = 2 λ λ w w = 2Dv ⇒ v = 2D k=
v2 = k = 2
w2 − w° c2
w2 vD 2
2
⇒ c k = w − w° ⇒ k = 2
2
2
2
w2 − w°
2
c
w ﺒﺎﻟﺘﻔﺎﻀل ﺒﺎﻟﻨﺴﺒﺔ ﻟـ c 2 2kdk = 2 w dw ⇒
dw c 2 k = dk w 2
vg =
w dw c 2 k c 2 w l − w ° = c 1 − °2 = = dk w w w = c 1−
v°
2
v2
.1 mev = ﺤﺯﻤﺔ ﻤﻥ ﺍﻟﻨﻴﻭﺘﺭﻭﻨﺎﺕ ﺒﻁﺎﻗﺔ ﻤﺘﻭﺴﻁﺔ ﻟﻠﻨﻴﺘﺭﻭﻥ ﺍﻟﻭﺍﺤﺩ:ﻤﺜﺎل 10 km ﺍﺤﺴﺏ ﺍﺘﺴﺎﻉ ﺍﻟﺤﺯﻤﺔ ﺒﻌﺩ ﻗﻁﻊ ﻤﺴﺎﻓﺔ،1 cu ﺒﺩﺃﺕ ﺍﻟﺤﺯﻤﺔ ﺒﺎﺘﺴﺎﻉ
.ﻓﻲ ﺍﻟﻔﺭﺍﻏﺎﺕ b b Δy Δy = 1com, ⇒ Δp y ? ∴Δp y =
,
ﺍﻻﺘﺴﺎﻉ
Δp y . Δy ≥ h
6.6 ×10 −34 D = 5.3×10 −33 kgm/sec = 2DΔy 4D ×10 − 2 m
p x = 2m n E = 2 (1.67 ×10 −27 ) (10 6 en) (1.6 × 10 −19 ) = 2.3×10 − 20 ∴tomθ =
kg . m / sec
−33
5.3 ×10 2.3 × 10 − 20
R = x R 5.3× 10 −33 9 = ton θ ⇒ R = v ton θ = 10 m × = 2 ×10 −4 m − 20 x 2.3×10
.2R ﺒﻤﻘﺩﺍﺭ1am ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﺍﻟﺤﺯﻤﺔ ﺴﺘﺯﺩﺍﺩ ﻤﻥ - ٦٣ PDF created with pdfFactory Pro trial version www.pdffactory.com
2 R = 4 ×10 −4 m = 4 ×10 −2 con = 0.4
ﺴﻴﺼﺒﺢ 10.4 nm
ﺱ /ﻜﻡ ﻴﺒﻠﻎ ﻋﺩﻡ ﺍﻟﺘﺤﺩﻴﺩ )ﺍﻟﻼﺘﺤﺩﻴﺩ( ﻓﻲ ﻤﻭﻀﻊ ﻓﻭﺘﻭﻥ ﻁﻭل ﻤﻭﺠﺘﻪ
°
3000 A
ﻜﺎﻨﺕ ﺩﻗﺔ ﺘﺤﺩﻴﺩ ﻁﻭل ﺍﻟﻤﻭﺠﺔ ﻴﺼل ﺇﻟﻰ ﺠﺯﺀ ﻤﻥ ﺍﻟﻤﻠﻴﻭﻥ؟
ﺇﺫﺍ
∆λ = 10 λ
ﺱ /ﺇﺫﺍ ﻜﺎﻨﺕ ﺇﻟﻜﺘﺭﻭﻥ ﻓﻲ ﺤﺎﻟﺔ ﻤﺸﺘﺎﻗﺔ ﺃﻭﻟﻲ ﻤﺩﺓ 10-6 sﺍﺤﺴﺏ ﻋﺩﺩ ﺍﻟﻠﻔﺎﺕ ﺍﻟﺘﻲ
ﺴﻴﻠﻔﻬﺎ ﻓﻲ ﻫﺫﺍ ﺍﻟﻤﺩﺍﺭ.
13.6 2D 2Dh = = ) → hw = (E 2 = 4 w hw ) 2D (1.05 × 10 −34 =T = 1.2 ×10 −15 sec 13.6 ( ) ) (1.6 ×10 −19 4 =T
~
C 10 −8 sec = =T = 8 × 10 6 tw −15 T 1.2 × 10
ﺍﺘﺴﺎﻉ ﺍﻟﺨﻁ ﺍﻟﻁﻴﻔﻲ ﺍﻟﺫﻱ ﻟﻪ ﻁﻭل ﻤﻭﺠﺔ
°
4000 A
ﻫـﻭ
ﻤﺘﻭﺴﻁ ﺯﻤﻥ ﻤﻜﻭﺙ ﺍﻟﻨﻅﺎﻡ ﺍﻟﺫﺭﻱ ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻁﺎﻗﺔ ﺍﻟﻤﻨﺎﻅﺭﺓ. °
°
0.0001A
ﺍﺤـﺴﺏ
°
D = 4000 A , Δλ = 10 −4 A = 10 −14 m. h c ΔD hc ΔD =ΔE =hΔv = 2 D D D
ﻋﺩﻡ ﺍﻟﺘﺤﺩﻴﺩ ﻓﻲ ﺍﻟﺯﻤﻥ ، ∆tﺯﻤﻥ ﻤﻜﻭﺙ ﺍﻟﺫﺭﺓ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ. 4Dh 4h D 2 = C = Δt = Δ E c D ΔD `
) (4) (16) (10 −14 = 2.1× 10 −7 sec. 8 −14 ) (3×10 ) (10 = 0.21μ sec.
=
ﺱ /ﺍﻟﻁﺎﻗﺔ ﺍﻟﻤﺴﺘﺜﺎﺭﺓ ﻟﻬﺎ ﺇﺸﻌﺎﻉ ﻴﺴﺎﻭﻱ ،1.1 evﺍﺤﺴﺏ ﻤﺘﻭﺴﻁ ﻋﻤﺭ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ؟ ﻤﻥ ﻋﻼﻗﺔ ﺍﻟﻼﺘﺤﺩﻴﺩ
h 2D
= Δt . ΔE
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ΔE = DΔv Δt =
6.6 × 10 −34 h = = ..............? 2D ΔE 2D ×1.1ev α∞
F (x) = ∫ dk g (k) e ikx −∞
ﺤﺯﻤﺔ ﻤﻭﺠﻴﺔ ﻤﻌﺭﻓﺔ ﺒـ ﺘﻌﻁﻲ ﺒـg ( k) ﺤﻴﺙ
g (k ) = o =N
k 〈 − k/2 − k/2 〈 k 〈 k/2 4 〈k 2
=o
ﺍﻟﺘﻲ ﻟﻬﺎN ﻗﻴﻤﺔ +∞
∫ dx f ( x )
2
=1
−∞
2N kx sin x 2 +∞ 2 kx 4N ⇒ ∫ dx 2 sin 2 = 2 x −∞
Qf( x ) =
+∞
2 4N 2 sin 2 n z z 2 du . Sin u 2kN du = ∫ k 2u 2 ∫−∞ u 2 = 2D N = 1 k2 1 ⇒ N= 2Dk
ﺍﻟﺘﻲ ﻟﻬﺎN ﻜﻴﻑ ﻫﺫﻩ ﻤﺭﺘﺒﻁﺔ ﺒﺎﺨﺘﻴﺎﺭ +∞
∫ dk g ( u )
2
−∞
+∞
=
1 2D
2 ∫ dk g ( u) = N
u/z
2
−∞
− k/z
⇒N sin
kv 2
∫ dk = N
2
k=
1 2D
1 2Dk
ﺒﻴﻥ ﻨﻘﻁﺘﻴﻥ ﺍﻟﺘـﻲ ﻋﻨـﺩﻫﺎ ﺍﻟﺩﺍﻟـﺔv ﻤﻥ ﺍﻟﻤﻨﺎﺴﺏ ﺍﺨﺘﻴﺎﺭ ﻋﺭﺽ Δu =
uD 2D ⇐ ×=± u k
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ﻭﻫﺫﺍ ﻴﺤﺩﺙ ﻋﻨﺩ.ﺘﺘﻼﺸﻰ
uD ) u = uD u
( = ∴Δ × Δk
ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ .u )(1
2D dv dv dv dv = = = −D2 ) dk 2D d (1/D ) d (1/D dD
C2 2 v2 − v°
= D2
2
,
2
v 2 − v°
= vg
= QD
2
2v dv dv dv c 2 v2 v° 1 ⇒ ∴ 2 = 2 + 2 ⇒ 2 =− 2 2 = dD D 3 C c c D D c2 c2 dv 2 −D2 = (−D 2 ) . ( 3 ) = = c 1 − v ° /v 2 dD D v Dv c c c 2 v2 − v° = = = c 1 v v° (v ) 2 2 2 2 2 v − v ° /v v − v°
)(1
u g = c 1 − v ° /v 2 2
ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﺸﺭﻭﺩﻨﺠﺭ ﻭﺘﻔﺴﻴﺭ ﺍﻻﺤﺘﻤﺎل The Schrödinger wave Eqn. nd the Probability Interpretation
ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺴﻨﻨﺎﻗﺵ ﺒﻌﺽ ﺨﻭﺍﺹ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻟﻠﺠﺴﻡ ﺍﻟﺤﺭ ،ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻲ ﺍﺴﺘﻨﺘﺠﻨﺎﻫﺎ ﻓﻲ ﺍﻟﻔﺼل ﺍﻟﺜﺎﻨﻲ .ﻭﻜﺫﻟﻙ ﺴﻨﺴﺘﻨﺘﺞ ﺘﻔـﺴﻴﺭ ﺍﻻﺤﺘﻤـﺎل
ﻟﻠﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ Ψﻭﻫﺫﺍ ﺒﺩﻭﺭﻩ ﻴﺅﺩﻱ ﺇﻟﻰ ﺘﻌﺭﻴﻑ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ montentumﻓﻲ
ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ،ﻭﻜﺫﻟﻙ ﻟﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﺘﻲ ﺘﺼﻑ ﺠﺴﻡ ﻓﻲ ﺠﻬﺩ ).V (x ﻭﻨﻘﻁﺔ ﺍﻟﺒﺩﺍﻴﺔ ﻫﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ. ).............(1
)∂ ψ (x, t )h 2 ∂ 2 ψ (x, t =− ∂t 2m ∂x 2
ih
ﻭﻫﺫﻩ ﻤﻌﺎﺩﻟﺔ ﺼﺤﻴﺤﺔ ﻟﻭﺼﻑ ﺤﺭﻜﺔ ﺠﺴﻡ ﺤﺭ .ﻭﺤل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻴﻤﻜـﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻴﻪ )ﺒﻌﻜﺱ ﺍﻟﺨﻁﻭﻁ ﺍﻟﺘﻲ ﺃﺩﺕ ﺇﻟﻴﻬﺎ( ..ﻭﻨﺘﺤﺼل ﻋﻠﻰ ﺍﻟﺤل ﺍﻟﻌﺎﻡ: i(px − Et)/h
1
∫ ap ϕ (p) e 2Dh
= )ψ (x, t
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ﻭﺤﻴﺙ ﺃﻥ Eﻟﻠﺠﺴﻡ ﺍﻟﺤﺭ ﻫﻲ p2/2m i(px − (p 2 /2m)t)lh
).............(2
1
∫ ap ϕ (p) e 2Dh
= )ψ (x, t
ﻭﻤﻥ ﺍﻟﺠﺩﻴﺭ ﺒﺎﻟﻤﻼﺤﻅﺔ :ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (١ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺘﻔﺎﻀﻠﻴﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻨﻨﺎ ﺒﻤﺠﺭﺩ ﺃﻥ ﻨﻌﺭﻑ ﺍﻟﻘﻴﻤﺔ ﺍﻷﻭﻟﻴﺔ ﻟـ ) (ψ
) )(ψ ( x, o
ﻋﻨﺩ ﻜل ﺍﻷﺯﻤﻨﺔ ﺍﻷﺨﺭﻯ ﻴﻤﻜﻥ ﻤﻌﺭﻓﺘﻬﺎ.
ﻓـﺈﻥ ﻗﻴﻤـﺔ
)∂ψ (x, t Δt ∂t ∂ψ (x, t) i 2h 2 ∂ψ (x, t) ih ∂ψ = = i2m h ∂xH2 2m xt 2 ∂t ih ∂ 2 ψ Δt ψ ( x, t + Δt) = ψ (x, t) + 2m ∂x 2 )ih ∂ 2 ψ (x, t ψ (x, t + Δt ) = ψ (x, t ) + )Δt .................(3 2m ∂x 2 ψ (x, t + Δt) = ψ (x, t) +
ﻭﺒﺈﻋﻁﺎﺀ ﻗﻴﻤﺔ ) ψ (x, oﺍﻟﺩﺍﻟﺔ ) ϕ (pﻴﻤﻜﻥ ﺇﻴﺠﺎﺩﻫﺎ ﻤﻥ ) (2ﻀﻊ t = oﻓـﻲ
ﺍﻟﻤﻌﺎﺩﻟﺔ ) (2ﺘﻜﺎﻤل ﻓﻭﺭﻴﺭ.
).................(4
ﻴﻤﻜﻥ ﻋﻜﺴﻪ ﻟﻠﺤﺼﻭل ﻋﻠﻰ
ipx/h
1
∫ dp φ (p) e 2xh
= ) ψ (x, o
)ϕ ( p
ﺸﺭﻭﻁ ﺍﻟﻤﻌﺎﻴﺭﺓ: ϕ (P ) = 2Dh ∫ ψ (x, o ) dx e ipx/ h
ﻭﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (2ﻴﻤﻜﻨﻨﺎ ﺤﺴﺎﺏ ) ψ (x, tﻋﻨﺩ ﻜل ﻗﻴﻡ ..t ﻻﺤﻅ ﺃﻨﻪ ﻻ ﻴﻭﺠﺩ "ﻻ ﺘﻌﻴﻴﻥ" ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ: ﻁﺎﻟﻤﺎ onceﻨﻌﺭﻑ ﺍﻟﻘﻴﻡ ﺍﻷﻭﻟﻰ ﻟﻠﺤﺎﻟﺔ )ﻟﻠﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﺒﺔ( ﻻ ﺘﻭﺠﺩ ﻗﻴـﻭﺩ ﻋﻠﻰ ﻗﻴﻡ ) ψ (x, oﻋﻨﺩﺌﺫ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻜﻠﻬﺎ ﻤﻌﺭﻓﺔ ﻋﻨﺩ ﻜل ﺍﻷﺯﻤﻨﺔ ﺍﻟﺘﺎﻟﻴﺔ.. ﺘﻔﺴﻴﺭ ﺍﻻﺤﺘﻤﺎل ﻟﻠﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ
) ψ ( x, t
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ﻭﻨﺤﻥ ﺒﺼﺩﺩ ﺍﻟﺒﺤﺙ ﻋﻥ ﺘﺄﻭﻴل )ﺘﻔﺴﻴﺭ( ﻟﻠﺩﺍﻟﺔ
) ψ ( x, t
ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ ﻓـﻲ
ﺘﻔﻜﻴﺭﻨﺎ. ) (1ﺃﻥ ) ψ (x, tﻫﻲ ﺒﺸﻜل ﻫﺎﻡ ﺩﺍﻟﺔ ﻤﻌﻘﺩﺓ .conglex fetom ) (2ﺍﻟﺩﺍﻟﺔ
) ψ (x, t 1 424 3
ﺘﻜﻭﻥ ﻜﺒﻴﺭﺓ ﺤﻴﺙ ﻴﻔﺘﺭﺽ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﺠﺴﻡ ﻤﻭﺠـﻭﺩ ﻭﺘﻜـﻭﻥ
ﻛﻤﯿﺔﺣﻘﯿﻘﯿﺔ
ﺼﻐﻴﺭﺓ ﻓﻲ ﺃﻤﺎﻜﻥ ﻏﻴﺭ ﻫﺫﺍ ﺍﻟﻤﻜﺎﻥ. ) (3ﻭﻜﺫﻟﻙ ﻤﺼﺎﺤﺏ ﻟﻠﺩﺍﻟﺔ ﻅﺎﻫﺭﺓ ﺍﻻﻨﺒﺴﺎﻁ )) (Spreadingﺍﻟﻼﺘﺤﻴﺯ(. ﻭﻤﺒﺎﺸﺭﺓ ﺒﻌﺩ ﺍﻜﺘﺸﺎﻑ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ )ﺍﻟﺘﻲ ﺘﺒﻌﺙ ﺍﻜﺘﺸﺎﻑ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ
ﺒﻭﺍﺴﻁﺔ ﻫﻴﺯﺒﻨﺭﺝ ﺴﻨﺔ (١٩٢٥ﻗﺎﻡ MaxBornﺒﺩﺭﺍﺴﺔ ﺘﺒﻌﺜﺭ )ﺘﻁﺎﻴﺭ( ﺸﻌﺎﻉ ﻤﻥ
ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺒﻬﺩﻑ ﺃﻭ ﺍﻟﺘﻲ ﺃﺭﺸﺩﺘﻪ ﺇﻟﻰ ﺍﻟﺘﺄﻭﻴل ﺍﻟﺼﺤﻴﺢ ﻟﻠﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ .ﻭﻋﻠﻴـﻪ
ﻓﻘﺩ ﺍﻗﺘﺭﺡ ﺃﻥ.
ﺘﻜﻭﻥ ﻜﺒﻴﺭﺓ ﺤﻴﻨﻤﺎ ﻴﻔﺘﺭﺽ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﺠـﺴﻡ
⇒) ⇐ p (x, t
ﺩﺍﻟـﺔ ﺤﻘﻴﻘﻴـﺔ
ﻭﺘﻔﻠﻁﺤﻬﺎ ﻻ ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺠﺴﻡ ﻤﺘﻔﻠﻁﺢ. ﻭﻜل ﻤﺎ ﺘﻌﻨﻴﻪ ﻫﻭ ﺃﻨﻪ ﻋﻨﺩﻤﺎ ﻴﺘﻐﻴﺭ ﺍﻟﺯﻤﻥ ﻴﻜﻭﻥ ﻤﻥ ﺍﻷﻗل ﺍﺤﺘﻤﺎﻻﹰ ﺃﻥ ﻨﺠـﺩ
ﺍﻟﺠﺴﻡ ﺤﻴﻨﻤﺎ ﻜﺎﻥ ﻋﻨﺩ .t = 0
ﻭﻟﻜﻲ ﻴﻜﻭﻥ ﺍﻟﺘﺄﻭﻴل ﺼﺤﻴﺤﺎﹰ ﻴﻠﺯﻤﻨﺎ: ∞x
∫ p (x, t ) dx =1
)(6
∞
ﺸﺭﻁ ﺍﻟﻤﻌﺎﻴﺭﺓ vormalization
ﻭﺫﻟﻙ ﻷﻥ ﺍﻟﺠﺴﻡ ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ ﻓﻲ ﻤﻜﺎﻥ ﻤﺎ ﻭﺴﻴﺘﻀﺢ ﻟﻨﺎ ﻻﺤﻘﺎﹰ ﺃﻥ ﻴﻜﻔﻲ ﺃﻥ ﻴﻠﺯﻡ ﺃﻥ: ) (7
∞〈
2
∞x
) ∫ dx ψ (x, o
∞
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ـﺔ ـﺔ ﻤﺘﻜﺎﻤﻠـ ـﻭﻥ ﻤﺭﺒﻌـ ـﺏ ﺃﻥ ﺘﻜـ ـﺔ ) ψ (x, oﻴﺠـ ـﻲ ﺃﻥ ﺍﻟﺩﺍﻟـ ـﺫﺍ ﻴﻌﻨـ ﻭﻫـ squaredintegrabliﻭﺒﻨﻁﺎﻕ ﺘﻜﺎﻤل
∞ + ∞←−
ﻴﺠﺏ ﺃﻥ ﺘﺅﻭل ﺇﻟﻰ ﺼﻔﺭ ﺃﺴﺭﻉ ﻤﻥ ﻤﺘﺼﻠﺔ ﻓﻲ x
) ψ (x, t
ﻴﻌﻨﻲ ﺃﻥ ).ψ (x, o ﻭﻫﺫﺍ ﻴﻠـﺯﻡ ﺃﻥ ﺘﻜـﻭﻥ ﺍﻟﺩﺍﻟـﺔ
x −1 / 2
.contineon sin x.
ﺃﻫﻤﻴﺔ ﺍﻷﻁﻭﺍﺭImport on ce of the ases :
ﻴﺒﺩﻭ ﻟﻠﻭﻫﻠﺔ ﺍﻷﻭﻟﻰ ﻜﻭﻥ ﺍﻟﺩﺍﻟﺔ
2
ﺫﺍﺕ ﻤﻌﻨﻰ ﻓﻴﺯﻴـﺎﺌﻲ ،ﻴﺒـﺩﻭ ﺃﻥ
) ψ (x, t
ﺍﻟﻁﻭل ﻟﻴﺱ ﻟﻪ ﺃﻫﻤﻴﺔ .ﻭﻫﺫﺍ ﺨﻁﺄ. ﺤل.
ﻭﺒﻤﺎ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺨﻁﻴﺔ) ،ﻭﺒﺎﻗﺘﺭﺍﻥ ﺃﻥ )ψ (x, t ) = ψ1 (x, t) + ψ 2 (x, t
)(8
ﻟﻭ ﺃﻥ
iθ1
ﺤﻴﺙ ﺃﻥ
)ψ 2 (x, t ), ψ1 (x, t
ﻫﻲ
ﻫﺫﺍ ﻜﺫﻟﻙ ﺤل.
←
ψ 2 (x, t ) = R 2 e iθθ , ψ1 (x, t) = R 1 e R 1 , R 2 , θ1 , θ 2
ﻫﻲ ﺤﻘﻴﻘﺔ ،ﻋﻨﺩﻫﺎ 2
)
) i (θ 2 − θ 1
(R1 +R 2 e
iθ 1
= e
2
) ψ (x, t
) = R 1 + R 2 + 2R 1 cos (θ 1 − θ 2 2
)(9
2
ﻭﻫﺫﺍ ﻴﻅﻬﺭ ﺃﻥ ﺍﻟﻁﻭﺭ ﻤﻬﻡ ،ﺍﻟﻁﻭﺭ ﺍﻟﻜﻠﻲ ﻓﻲ ) ψ (x, tﻴﻤﻜﻥ ﺇﻫﻤﺎﻟـﻪ ،ﺃﻤـﺎ ﺍﻟﻁﻭل ﺍﻟﻨﺴﺒﻲ
) (θ 1 − θ 2
ﺒﻴﻥ ﺩﺍﻟﺘﻴﻥ ﻤﻭﺠﺒﺘﻴﻥ
ψ1
و ψ2
ﺘﻅﻬﺭ ﻓﻲ
2
ψ
.
ﻻﺤﻅ ﺃﻥ ﺍﻟﻁﺭﻑ ﺍﻷﻴﻤﻥ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (9ﻫﻭ ﺍﻟﺫﻱ ﻨـﺭﺍﻩ ﻋﻨـﺩ ﺘﺭﺍﻜـﺏ
ﺍﻟﻤﻭﺠﺎﺕ .ﻭﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ ﺃﻨﻬﺎ ﺍﻟﺨﻁﻭﺓ lineallyﻟﻠﺩﻭﺍل ﺍﻟﻤﻭﺠﺒﺔ ﺍﻟﺘﻲ ﺃﺩﺕ ﻟـ ﻟﺸﻜل ﺍﻟﺘﺩﺍﺨل ﺍﻟﺫﻱ ﻴﻅﻬﺭ ﻤﻥ ﺨﻼل cosineﻓﻲ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ. ﻭﻋﻨﺩﻤﺎ ﻨﺘﺤﺩﺙ ﻋﻥ ﺍﻟﺴﻠﻭﻙ ﺍﻟﻤﻭﺠﻲ ﻟﻺﻟﻜﺘﺭﻭﻨﺎﺕ ،ﺍﻟﻔﻭﺘﻭﻨﺎﺕ ،ﻓﺘﺤﺩﺙ ﻤﻥ
ﺍﻟﺨﻁﻴﺔ...
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Peobalrility current ﺍﺤﺘﻤﺎل ﺍﻟﺘﻴﺎﺭ
.(1) ﺍﻟﻤﺭﺍﻓﻕ ﺍﻟﻤﻌﻘﺩ ﻟﻠﻤﻌﺎﺩﻟﺔcomplex conjugate ﻟﻨﺄﺨﺫ ∂ψ h2 ∂2ψ =− ⇒ 2m ∂t 2 ∂t ∂ψ * (x, t) h 2 ∂ 2 ψ * (x, t) ⇒ − ih =− 2m ∂t ∂t 2
ih
∂ψ ih ∂ 2 ψ = ∂t 2m ∂x 2 ih ∂ 2 ψ * ∂ψ * ⇒ =− 2m ∂x 2 ∂t
ψ 2
644744 8 ∂ ∂ p (x, t) = ( ψ (x, t) ψ x (x, t) ) ∂t ∂t ∂ψ * (x, t) ∂ψ ψ + ψ* = ∂t ∂t * ih ∂ψ * ih ∂ 2 ψ ψ = − 2 2m ∂x 2 2m ∂x 2 1 h 2 ∂ 2ψ* ∂ 2ψ * h ψ ψ − ih 2m ∂x 2 2m ∂x t ∂ ∂ p (x, t) = − ∂t ∂x ∂ ∂ j (x, t) ⇒ t (x, t) = − ∂t ∂x
=
+∞
∂ dx p (x, t) ∂x −∫∞
ﻴﺘﻌـﻭﺽ ﺒـﺎﻟﺘﻐﻴﺭ
x
+∞
= − ∫ dx −∞
∂ j (x, t) = o ∂x
= − [J (− ∞ ) − j (− ∞ ) ]= 0 , j → 0
ﺍﻟﺘﻐﻴﺭ ﻓﻲ ﺍﻟﻜﺜﺎﻓﺔ ﻓﻲ ﻤﻨﻁﻘﺔ ﻓـﻲ:ﺘﺒﻴﻥ
⇐
(12) ﻤﻌﺎﺩﻟﺔ
.ﺍﻟﺼﺎﻓﻲ ﻓﻲ ﺍﻟﻔﻴﺽ ﻓﻲ ﺘﻠﻙ ﺍﻟﻤﻨﻁﻘﺔ ∞
ﺘﺘﻼﺸﻰ ﻋﻨﺩsquare integreble ﺍﻟﺩﻭﺍل ﻤﺘﻜﺎﻤﻠﺔ ﺍﻟﺘﺭﺒﻴﻊ
Q
∞ ∂ ∂ ax p(x, t) dx j (x, t) = − ∫ ∫ ∂t ∂x −∞
= j (a, t) − j (b, t)
(1) ﻭﻗﺎﻨﻭﻥ ﺍﻟﺒﻘﺎﺀ ﺘﻜﻭﻥ ﺼﺤﻴﺤﺔ ﻟـﻭ ﺍﻟﻤﻌﺎﺩﻟـﺔ
(14)
j (x, t) , p(x, t )
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ﻭﺘﻌﺭﻴﻑ ﻟـ
ﺘﺘﻐﻴﺭ ﻟـ
ﻭﻫﺫﻩ ﺍﻟﺘﻌﺭﻴﻔﺎﺕ ﻟـ ) d (x, t ), p (x, tﻭﻗﺎﻨﻭﻥ ﺍﻟﺤﻔﻅ )ﺍﻟﺒﻘﺎﺀ( ﺘﺒﻘـﻰ ﺼـﺤﻴﺤﺔ ﺤﺘﻰ ﻟﻭ ﻏﻴﺭﻨﺎ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺠﺴﻡ ﺍﻟﺤﺭ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ. )(15
ﺃﺒﻌﺎﺩ.
) ∂ ψ (x, t )h 2 ∂ 2 ψ (x, t =− )+ ψ (x, t ∂t 2m ∂x
it
ﻭﻫﺫﻩ ﻤﻬﻤﺔ ﻷﻨﻬﺎ ﺘﻤﺜل ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻟﺠﺴﻡ ﻓﻲ ﺠﻬﺩ ) V(xﻓﻲ ﺜﻼﺜـﺔ ψ (x, y, z, t), ) V (x, y, z ∂2 ∂2 ∂2 + + ⇒∇2 ∂x 2 ∂y 2 ∂z 2
ﻜﺫﻟﻙ ﺘﻌﻤﻴﻡ )(12 ∂ p (r, t ) + ∇ . J (r, t) = 0 ∂t 2
) p (r, t ) = ψ (r, t
)∂ψ (x, t h2 2 =− ) ∇ ψ (r, t ) + v (r ) ψ (r, t ∂t 2m h = ) j (r, t ) ψ * (r, t ) ∇ψ (x, t ) − ∇ψ * (r, t ) ψ (r, t 2im
ih
[
]
ﺍﻟﻘﻴﻡ ﺍﻟﻤﺘﻭﻗﻌﺔ ﻭﻜﻤﻴﺔ ﺤﺭﻜﺔ ﺍﻟﺠﺴﻡ Expectation Values of Canticle nonctn
ﺒﺈﻋﻁﺎﺀ ﻜﺜﺎﻓﺔ ﺍﻻﺤﺘﻤﺎل.P (x,t) . ﻓﺈﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻭﻗﻌﺔ ﻟـ ) f (xﻴﻤﻜﻥ ﺤﺴﺎﺒﻬﺎ ﺒﺸﻜل ﻋﺎﻡ. )(20
) f (x ) = ∫ ax f (x ) p (x, t
) = ∫ ax ψ * (x, t ) f (x ) ψ (x, t
ﻻﺴﺘﻨﺘﺎﺝ ﻤﺅﺜﺭ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻤﻥ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ
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p = mv = m
dx dt
(21)
d d x = m ∫ dx ψ * 9x, t) v ψ (x, t ) dt dt +∞ ∂ψ ∂ψ * ) xψ + ψ * x p (x ) = m ∫ dx ( ∂Z ∂Z −∞
p =m
+∞
p (x ) = m ∫ dx ( −∞
،ﻫﻲ ﺍﻟﺘﻲ ﺘﺘﻐﻴﺭ ﺒﺎﻟﺯﻤﻥ
ψ (x, t )
(22)
∂ψ * ∂ψ ) + ψ (x, t ) + ψ * x ∂t ∂Z
ﻓﻘﻁ.ﺘﺤﺕ ﺍﻟﺘﻜﺎﻤل
.ﻤﻊ ﺍﻟﺯﻤﻥ
x
dx dt
ﻻﺤﻅ ﺃﻨﻪ ﻻ ﻴﻭﺠﺩ
ﻭﻫﺫﺍ ﺍﻟﺘﻐﻴﺭ ﻫﻭ ﺍﻟﺫﻱ ﻴﺅﺩﻱ ﺇﻟﻰ ﺍﻟﺘﻐﻴﺭ ﻓﻲ .ﻭﺒﻤﺎ ﺃﻥ ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺁﺨﺭ ﻤﻌﺎﺩﻟﺔ
∂ψ h ∂ψ =− ∂t 2m ∂x 2 ∂ψ h ∂ 2ψ m =− ∂t 2i ∂x 2 ∂ψ * h ∂ 2 ψ * m = ∂t 2i ∂x 2
ih
2 h +∞ ∂ 2 ψ * * ∂ ψ xψ − ψ x 2 )dx p = ∫( 2i −∞ ∂x 2 ∂x
∂ 2 ψ* ∂ ∂ψ * ∂ψ * ∂ψ * ∂ψ xψ ( xψψ ψ x = − − x 24 x ∂x ∂x ∂x ∂4 ∂3 ∂x 2 1 ∂x * ∂ψ ψ ∂x ∂x
∂ ∂ψ * ∂ ∂ψ ( xψψ− (ψ * ψ) + ψ * ∂x ∂x ∂x ∂x 2 ∂ ∂ψ ∂ψ ∂ ψ − (ψ * x ) + ψ * + ψ*x 2 ∂x ∂x ∂x ∂x =
.ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﺍﻟﺤﺩ ﺍﻟﺫﻱ ﻴﻘﻊ ﺩﺍﺨل ﺍﻟﺘﻜﺎﻤل ﻴﺼﺒﺢ 2 ∂ 2 ψ* ∂ ∂ψ * ∂ ∂ψ * ∂ ψ (xψx − ψ x = ( x ψ) − (ψ * ψ) + ψ * 2 2 ∂x ∂x ∂x ∂x ∂x ∂x ∂ ∂ψ ∂ψ − ( ψ* x ) + ψ* ∂x ∂x ∂x * ∂ ∂ψ ∂ψ ∂ψ = xψ − ψ * ψ − ψ * x + 2ψ * ∂x ∂x ∂x ∂x
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+∞
+∞
h ∂ p = ∫ [ 2i − ∞ ∂x
]+ h ∫ 2ψ * (x, t) ∂ ψ (x, t) dx 2i −∞ ∂x
+∞
p = ∫ dx ψ * (x, t) −∞
h ∂ ψ (x, t) i ∂x
.operator ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻘﺘﺭﺡ ﺃﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻴﻤﻜﻥ ﺃﻥ ﺘﻤﺜل ﺒﺎﻟﻤﺅﺜﺭ p=
h ∂ i ∂x
. ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل،ﻭﻟﻬﺫﺍ p 2 = ∫ dxψ * (x, t) (−h 2
∂2 ) ψ (x, t) ∂x 2
f ( p ) = ∫ dxψ * (x, t) f ( φ ( p)
D ∂ ) ψ (x, t) t ∂x
.ﻭﺒﺸﻜل ﻋﺎﻡ
ﻭﺒﻬﺫﺍ ﺍﻟﺘﻤﺜﻴل ﻟﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻴﻤﻜﻨﻨﺎ ﻤﻨﺎﻗﺸﺔ ﺍﻟﻤﺩﻟﻭل ﺍﻟﻔﻴﺯﻴـﺎﺌﻲ ﻟــ
.ﺍﻟﺘﻲ ﺘﻅﻬﺭ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻓﻲ ﻓﺭﺍﻍ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ 1
∫ 2Dh
ψ(x, t )=
ap φ ( p ) e i [px - (p/2m) t ] / h
.t ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ψ (x ) =
1
∫ 2Dh
ap φ ( p ) e ipn/ h =
ﻭﺫﻟﻙ ﻷﻥt = 0 ﺩﻉ:ﺃﻭﻻﹰ
φ (p )
h ou φ (hD ) e ikv 1D ∫
ﻭﺒﺎﺴﺘﺨﺩﺍﻡ ﻤﻌﻜﻭﺱ ﺘﻜﺎﻤل ﻓﻭﺭﻴﺭ φ ( hD ) = ∫ axψ ( x) e −ikx or ⊗ φ (p) =
1 2Dh
∫
ax ψ (x) e - ipx/ h
⇒ ∫ ap φ * ( p) φ ( p) = ∫ ap φ * ( p) = ∫ ax ψ
1
1 2Dh
∫
ax ψ (x) e - ipx/ h
∫
ap φ * e - ipx/ h Dh4424443 124 ψ*
= ∫ ax ψψ * = 1
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ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺘﺴﻤﻰ ﺒﻨﻅﺭﻴﺔ ﺒﺎﺭﺴﻴﻔﺎل Parsevali
ﻭﻤﻨﻁﻭﻗﻬﺎ :ﻟﻭ ﺃﻥ ﺩﺍﻟﺔ ﻤﻌﺎﻴﺭﺓ ﻟـ :1ﻓﺈﻥ ﺘﺤﻭﻴل ﻓﻭﺭﻴﺭ ﻟﻠﺩﺍﻟﺔ ﻴﻜﻭﻥ ﻤﻌﺎﻴﺭ ﻜﺫﻟﻙ If a function is normalized to 1, 50 is its Fourier Transform
) ﻋﻨﺩﻨﺎ ﺘﻤﺜﻴﻼﻥ ﻟﻠﺩﻭﺍل :ﺘﻤﺜﻴل ،xﺘﻤﺜﻴل .(regasrauney p ﺨﺫ: h d )ψ (x i dx h d 1 ) = ∫ ax ψ * ( x ap φ (p) e ipx/h ∫ i dx 2Dh 1 = ∫ ap φ ( p) p ax ψ * (x) e ipx/ h ∫ 2Dh
) p = ∫ ax ψ * ( x
)p = ∫ ap φ ( p ) p φ * (p
ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺘﻌﻨﻲ ﺃﻥ
ﺍﻟﺤﺭﻜﺔ ﺒـ
2
) φ( p
) φ (p
ﺘﻔﺴﻴﺭ ﺒﺄﻨﻬﺎ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻓﻲ ﻓـﺭﺍﻍ ﻜﻤﻴـﺔ
ﺘﻌﻁﻴﻨﺎ ﻜﺜﺎﻓﺔ ﺍﺤﺘﻤﺎل ﺇﻴﺠﺎﺩ ﺍﻟﺠﺴﻡ ﺍﻟﺫﻱ ﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ .p
ﻓﻲ ﻓﺭﺍﻍ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ xﻟﻬﺎ ﺍﻟﺘﻤﺜﻴل
∂ ∂p
x = ih
ﺴﺘﺠﺩ ﻓﻴﻤﺎ ﺒﻌﺩ ﺃﻥ ﺍﻟﻤﺅﺜﺭﺍﺕ
ogenatrsﺘﻠﻌﺏ ﺩﻭﺭﺍﹰ ﻤﻬﻤﺎﹰ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ. ﻭﺍﻵﻥ ﺴﻨﻭﻀﺢ ﺒﻌﺽ ﺨﺼﺎﺌﺼﻬﺎ. ﻋﻠﻰ ﻋﻜﺱ ﺍﻷﺭﻗﺎﻡ ﺍﻟﻌﺎﺩﻴﺔ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻟﻴﺴﺕ ﺒﻨﺩ ﺩﺍﺌﻤﺎﹰ ،ﻟﻭ ﻋﺭﻓﻨﺎ
[ A B ] = AB − BA 1
∂ )D∂ ψ (x, t x ψ (x, t ) − x i ∂x i ∂x D ∂ψ ∂ψ = (x +ψ ( x, t ) − x ) i ∂x ∂x ) [ p, x ] ψ (x, t) = h ψ (x, t i
[ p, x ] ψ (x, t) = h
ﺤﻴﻨﺌﺫ:
ﻭﻋﻠﻴﻪ ﻨﻘﻭل ﺃﻨﻨﺎ ﻋﻨﺩﻨﺎ ﻋﻼﻗﺔ ﺍﻟﺘﺒﺩﻴل Commutation relatively
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[ p, x ] = h i
normalized ware fumets ﺨﺫ ﺠﺴﻡ ﺍﻟﺫﻱ ﺩﺍﻟﺘﻪ ﺍﻟﻤﻌﺎﻴﺭﺓ:ﻤﺜﺎل ψ (x ) = 2 α α v e − αv v > 0 =
v<0
ﻓﻲ ﺍﻟﻘﻴﻤﺔ؟
p(x )= ψ (x )
2
ﺘﻜﻭﻥ
( ﻷﻱ ﻗﻴﻡ ﻤﻥa)
ν
v2 ,
ﻭﺍﺴﺘﺨﺩﻡ ﺍﻟﻨﺘﻴﺠﺔ ﻟﺤﺴﺎﺏ
p
ap =o ax ⇒=
( ﺍﺤﺴﺏb)
( ﻤﺎ ﻗﻴﻤﺔ ﺍﺤﺘﻤﺎل ﺃﻥ ﺍﻟﺠﺴﻡ ﻴﻭﺠﺩ ﺒﻴﻥc )
v = 1 α ,v= 0 p2 ,
v
φ (p)
( ﺍﺤﺴﺏb)
ﺘﺤﺩﺙ ﻋﻨﺩﻤﺎp (x) ﻗﻴﻤﺔ:ﺃﻭﻻﹰ
a a ( ψ ψ * ) = (2α α2 e −αv * 2α x α e − αv ) ax ax a = (4 α 3 x 2 e − 2αα ) = 0 ax = ( − 2 α e −2αα v 2 + e − 2αα 2v) = 0 = 2 v (1 − 2 α v ) e − 2αα = 0 1 ⇒ 1 = 2 xn ⇒ ν = α
(b)
v =
∞
∫
dxψ ( x) vψ ( x) = *
0
=
∞
∫
2α α v e −αv . v . 2 α e −αx
0
∞
∫
∞
∫
4α 3 v(v 2 e − 2αv ) dx =
0
ax.v (4α 3 v 2 e −2αx )
0
ﻋﻭﺽ ﻋﻥ 1 v = 4α ∞
∫ 0
−t
t e dt = t e n
n
−t
∞
∫ 0
1 1 y = 2α v , dy = 2 αv , dy = 2α ax ⇒ ax = dy , x = y 2 2α ∞
∫
dy y 3 e − y =
0
∞
+n ∫ t 0
n −1
e
− tdt
3i 3 = 4α 2α
∞
= ∫ t n e − t at = Γ (n + 1) = n ! 0
n is aminke Γ (n + 1) = n ( n − 1) (n − 2).........1 = n!
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4i 3 = 2 2 8α α
= ) dy y 2 ( 4 α 3 x 2 e − 2αx
∞
∫
=
2
a
0
=
)ﻭﺍﺼل ﺍﻟﻤﺤﺎﻀﺭﺓ ( ﺍﻷﺨﻴﺭﺓ )ﺍﻟﺴﺅﺍل ﻓﻲ ﺍﻟﻤﻨﺘﺼﻑ(: ﻭﺍﺼل ﺨﻭﺍﺹ ﺍﻟﻤﺅﺜﺭﺍﺕ: ﻅﻬﻭﺭ ﺍﻟﻤﺅﺜﺭﺍﺕ pﺒـ Iﻤﻤﻜﻨﻪ ﺃﻥ ﻴﺅﺩﻱ ﺇﻟﻰ ﺍﻟﺸﺩﺓ ﻓـﻲ ﺃﻥ
p
ﻜﻤﻴـﺔ
ﺤﻘﻴﻘﻴﺔ ﻭﻟﻜﻨﻨﺎ ﻴﻤﻜﻨﻨﺎ ﻤﺭﺍﺠﻌﺔ ﻫﺫﻩ ﺍﻟﺤﻘﻴﻘﺔ ﺃﻥ pﻜﻤﻴﺔ ﺤﻘﻴﻘﻴﺔ. h ∂ψ * h ∂ψ ) − ∫ ax ψ ( − i ∂x i ∂x * ∂ψ ∂ψ )ψ * ax ( ψ + ∂x ∂x ∂ ax ( ψ * ψ) = 0 ∂x
* = ∫ ax ψ
α
p − p
h ∫ i h ∫ = i =
ﺒﺸﺭﻁ ﺃﻥ ﺍﻟﺩﺍﻟﺔ ﺘﺘﻼﺸﻰ ﻋﻨﺩ ∞ ،ﻭﻫﺫﺍ ﺍﻟﺸﺭﻁ ﻤﺘﺤﻘﻕ ﻓﻲ ﺍﻟﺩﻭﺍل ﻓﻲ ﻤﺭﺒﻌﺔ ﺍﻟﺘﻜﺎﻤل ﻭﻟﻜﻥ ﺇﺫﺍ ﻟﻡ ﻴﺘﺤﻘﻕ ﻫﺫﺍ ﺍﻟﺸﺭﻁ ،ﻤﺜﻼﹰ ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﺩﻭﺍل ﺍﻟﺘﻲ ﺘﻐﻴﺭ ﻨﻔـﺴﻬﺎ Periodical )ψ(x )=ψ(x +L
ﻭﻟﻭ ﻗﺼﺭﻨﺎ ﺃﻨﻔﺴﻨﺎ ﻟﻠﻌﻤل ﻓـﻲ
0≤ v < L
ﻋﻨـﺩ
∂t lax
ﻻﺯﺍل ﻴﻜـﻭﻥ ﻤـﺅﺜﺭ
ﻫﻴﺭﻭﻤﺘﻲ hermitical poetry h ∂ ax ) )(ψ * (x) ψ (x ∫ i 0 ∂x L
h h 2 2 ψ(L) − ψ(o) =0 i i
= p − p *
=
ﺍﻟﻤﺅﺜﺭ ﺍﻟﺫﻱ ﻗﻴﻤﺘﻪ ﺍﻟﻤﺘﻭﻗﻌﺔ ﻟﻜل ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺴﻤﻭﺤﺔ ﻟﻠﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴـﺔ ﺘﻜـﻭﻥ ﺤﻘﻴﻘﻴﺔ ﻴﺴﻤﻰ ،lerimti of openerﻭﻋﻠﻴﻪ ﻓﺈﻥ pﻤﺜل xﻫﻭ ﻤﺅﺜﺭ ﻫﻴﺭﻤﺘﻲ. An operator whose expectation value for all admissible wave functims is real is called a hermitical operator
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ﻭﻟﻌﻠﻨﺎ ﻨﺴﺘﻨﺞ ﻫﺫﺍ ﺍﻟﻔﺼل ﺒﻤﻼﺤﻅﺔ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ: )∂ψ (x, t )h 2 ∂ 2 ψ (x, t = − ∂t 2m ∂x 2
ﺒﺎﻟﺘﻌﺭﻴﻑ:
∂ h i ∂x
= p op
= ih
ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻬﺎ ﺒﺎﻟﺼﻭﺭﺓ ﺍﻟﺘﺎﻟﻴﺔ: 2
p op )∂ψ (x, t = ih )ψ (x, t = − 2m ∂t
ﺍﻟﻤﺅﺜﺭ ﻋﻠﻰ ﺍﻟﺠﺎﻨﺏ ﺍﻷﻴﻤﻥ ﻫﻭ ﺍﻟﻁﺎﻗﺔ ﻟﺠﺴﻡ ﺤﺭ .ﻭﻟﻭ ﻋﻤﻤﻨﺎ ﻫﺫﺍ ﺍﻟﺠـﺴﻡ
ﻓﻲ ﺠﻬﺩ ،ﻨﺘﺤﺼل ﻋﻠﻰ:
) ψ (x, t
2 ∂ψ (x, t) p op = ih = )+ V (x ∂t 2m
or )∂ψ (x, t )h 2 ∂ 2 ψ (x, t )+ V (x) ψ (x, t =− 2m ∂x 2 ∂t
= ih or
)∂ψ (x, t )= H ψ (x, t ∂t
ih
ﺤﻴﺙ Hﻫﻭ ﻤﺅﺜﺭ ﺍﻟﻁﺎﻗﺔ .ﻭﻴﺴﻤﻰ ﻋﺎﺩﺓ ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ Hamiltonianﻷﻨـﻪ
ﺼﻭﺭﺓ ﻤﺅﺜﺭﺓ ﻟﻠﺩﺍﻟﺔ ﻫﺎﻤﻠﺘﻭﻨﻴﺎﻥ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ. ﻭﺒﻤﺎ ﺃﻥ pﻫﻭ ﻤﺅﺜﺭ ﻫﻴﺭﻤﻴﺘﻲ
⇐ p2
p2 )+ V (x 2m
=H
ﺠﻬﺩ ﺤﻘﻴﻘﻲ
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ﺱ /ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻌﻁﺎﺓ: N x +a2 2
= )ψ (x
) (aﺍﺤﺴﺏ Nﺍﻟﻼﺯﻤﺔ ﺍﻟﻤﻌﺎﻴﺭﺓ ﺍﻟﺩﺍﻟﺔ . ψ ) (bﺍﺴﺘﺨﺩﺍﻡ ﺍﻟﺩﺍﻟﺔ ) (cﺍﺤﺴﺏ
ψ
ﻟﺤﺴﺎﺏ
vn
ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻓﻲ ﻓﺭﺍﻍ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ.
p2
) (aﺍﺤﺴﺏ ﻟﺤﺴﺎﺏ
،ﻤﺎ ﻗﻴﻡ nﺍﻟﺘﻲ ﺘﺅﺩﻱ ﺇﻟﻰ ﺘﻜﺎﻤل ﻤﺘﺠﻤﻊ.
x1 − x2
=∆ x
p1 − p 2
=Δx
؟
Δx Δp
!)D (2n − 3 dx = 2n −1 2 n ) (a + x 2a !)(2n − 2 2
∞
dx =1 2 (x + a 2 ) 2
∫ 0
∞+
∞
∫
∞−
dx x 1 x = 2 2 + 3 + tu −1 ( ) 2 2 2 ∞a − (x + a ) 2a (x + a ) 2a D 2a 3 ⇒ = N D 2a 3
N2 ∞+
∫
∞−
=
2a 3 1 2 D x +a2
= ∴ψ
)(b a2 (x 2 + a 1 ) 2
ﺘﺘﻘﺎﺭﺏ ⇐
xn
∞+
∫
∞−
2a 2 = D
n
x
ﻓﻘﻁ ﻋﻨﺩﻤﺎ n < 3
n = 0, 1, 2,
ﺍﻟﻤﻜﺎﻤل ﻟﺯﻭﺍﻴﺎ
=0
xn
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x
n
+∞
2a 3 = D
x2 dx (x 2 + a 2 ) 2
∫
−∞
+∞
∫
x 2 ax x 1 = + + tam −1 (x/a) 2 2 2 2 2 (x + a ) 2 (x + a ) 2a −∞ = D / 2a N
φ (p) = = Z = x tiy
a D
2Dh
a h
+∞
∫
n2
,
= 2a 2
ax e −2p x/h
dx e −i px/h p = hk → p/h = k x2 +a2
∫
−∞
ﻟﻠﺘﻜﺎﻤل ﻨﺴﺘﺨﺩﻡ ﻜﻭﻨﺘﻭﺭ ﻨﺼﻑ ﺩﺍﺌﺭﺓ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﺭﻜﺏ = 2 D i R (−ia) 2Di − k ( −ik) D −ka e = = e 2ia a φ (p) = D/h e − pa/h
p>0
= a/h e −pa/ h p2 p
2
= =
2a h
∫
∞
∫
p 2 e − pa/h ap
0
φ * p 2 φ ap
ap =z h p
2
p>0
p2 = z2
2a h 3 ( ) = h a = 2(
p =0 Δ p = 2h/a
∞
∫
h2 a2
z 2 e − z az
0
3
h h ) Γ (3) = 4 ( ) 2 a a Γ (n + 1 ) = n ! Γ (2 + 1) 2! Δp =
p2
Δx =
Δx
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.ﺃﺜﺒﺕ ﺃﻥ ﺍﻟﻤﺅﺜﺭ
ﻟﻭ ﺃﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺘﺤﻘﻕ ﺍﻟﺸﺭﻁ/ﺱ
ψ (D ) = ψ ( − D )
∠=
h d i dθ
ﻤﺅﺜﺭ ﻫﻴﺭﻤﻴﺘﻲ ∠ − ∠ D
∫
∠ =
dθ ψ * ( θ )
−D D
∠
*
=−
∫
dθ ψ
−D
∠ − ∠ = *
*
ﻹﺜﺒﺎﺕ ﻫﺫﺍ ﻴﺠﺏ ﺃﻥ
=0
h d ψ i dθ
h d * ψ i dθ
d d D h * dθ dθ + dθ ( ψ ( θ ) ψ ) (ψ ψ * ) ∫ i −D
=
h i
D
∫
dθ
−D
d ( ψ* ψ ) dθ
[
h * = ψ ( D ) ψ ( D ) − ψ * ( − D ) ψ (D ) i ψ (D ) =ψ ( − D ) = 0
]
∠ = ∠ ∞
3.c.
∫
k/2
dk N 2 = N 2
−∞
∫
dk = N 2 k =
− k/2
1 ⇒ 2D
N=
*
∴ ﺍﻟﻤﺅﺜﺭ
1 2D k
. ﻓﻲ ﺍﻟﺤﺎﻟﺘﻴﻥ ﻤﺘﺴﺎﻭﻴﺔN ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻗﻴﻡ ( ﻴﻤﻜﻥ ﺘﻘﺩﻴﺭﻩ ﺒﺎﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﻨﻘﻁﺘـﻴﻥ ﻋﻨـﺩﻫﺎp) ﻋﺭﺽ ﺍﻟﺩﺍﻟﺔ ﻤﻥ ﺍﻟﺠﺯﺀ : ﻭﻫﺫﺍ ﻴﺤﺩﺙ ﻋﻨﺩ3.d ﺘﺨﺘﻔﻲ sin
.k ﻭﻫﻲ ﻨﺘﻴﺠﺔ ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ
2D kx x = + k =0 2D 2 x = − k
⇒ Δx=
4D k
⇒ Δ x. Δ k =
, 4D . k
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k u Δk = − (− ) = k 2 2
k = 4D
sin
kx 2
ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻭﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔEigenfunctions and Eigenvalues :
ﻟﻭ ﺃﺨﺫﻨﺎ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﺘﻲ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ: )∂ψ (x, t )h 2 ∂ 2 ψ(x, t =− )+ v (x) ψ (x, t ∂t 2m ∂x 2
)(1
ih
ﻭﻟﺤل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻨﻘﺼﻬﺎ ﻟﻤﻌﺎﺩﻟﺘﻴﻥ ﻨﻕ )ψ (x, t) , T(t) u (x )dT (t) h 2 d 2 u (x = − + v (x) u (x) 2 dt 2m dx
)⇒ ih u (x
ﺒﺎﻟﻘﺴﻤﺔ ﻋﻠﻰ )u (x) T (t )2m) (a 2 u (x) / dx 2 ) + V (x)u (x )u (x
(sin 1
dT (t)dt ⇒ ih = )T (t
ﻭﻫﺫﺍ ﻜﻤﻴﺔ ﻴﻤﻜﻥ ﺘﺤﻘﻴﻘﻪ ﻓﻘﻁ ﺍﻓﺘﺭﺍﺽ ﺃﻥ ﻜﻼ ﺍﻟﻁﺭﻓﻴﻥ ﻴﺴﺎﻭﻱ ﺜﺎﺒﺕ ﻭﺍﻟﺫﻱ ﻨﺴﻤﻴﻪ E )dT (E i1 i = ∫ − E dt = − Et h )T (E ih
∫
)d T (t ⇒ )= E T (t dt
⇒ ih
i
− ET i lm T (t) = - Et ⇒ T (t) = Ce t h
ﻭﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ: )h 2 d 2 u (x )+ v (x) u (x) = Eu (u 2m dx 2
−
ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﻐﻴﺭ ﻤﻌﺘﻤﺩﺓ ﻋﻠﻰ ﺍﻟﺯﻤﻥ time indgader S-E :ﻭﻫـﺫﻩ
ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻔﺭﻕ ﻋﻥ ) (1ﻭﺍﻟﺘﻲ ﺘﺼﻑ ﺍﻟﺘﻐﻴﺭ ﺍﻟﺯﻤﻨﻲ ﻟﻠﺩﺍﻟﺔ
)ψ (x, t
ﻤﻌﺎﺩﻟﺔ ﻫﻲ ﺩﺍﻟﺔ
ﺨﺎﺼﺔ .Ligea valu equation .ﻭﻟﻔﻬﻡ ﻤﺎﺫﺍ ﻴﻌﻨﻲ ﻫﺫﺍ ﺍﻟﻜﻼﻡ ﺩﻋﻨﺎ ﻨﻌـﺩﺩ ﻤـﺭﺓ ﺃﺨﺭﻯ ﻟﻤﻔﻬﻭﻡ ﺍﻟﻤﺅﺜﺭﺓ.
ﺒﺸﻜل ﻋﺎﻡ ﻤﺅﺜﺭ ﻤﺎ :ﻴﺅﺜﺭ ﻋﻠﻰ ﺩﺍﻟﺔ ﻭﻴﺄﺨﺫﻫﺎ ﻟﺩﺍﻟﺔ ﺃﺨﺭﻯ ،ﻤﺜﺎل ﻋﻠﻰ ﺫﻟﻙ:
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of (x) = f (x) + x 2
1.
] )of (x) = [f (x
2.
)of (x) = f (3 x 2 + 1
3.
2
3
d f (x) 4. of (x) = dx )5. of (x) = df / dx − 2f (x )6. of (x) = λf (x
ﻜل ﻫﺫﻩ ﺍﻷﻤﺜﻠﺔ ﺘﺅﺜﺭ ﻓﻲ ﺍﻟﺨﺎﺼﻴﺔ ﻫﻲ ﺒﺈﻋﻁﺎﺀ ﺩﺍﻟﺔ ) f (xﺘﻭﺠﺩ ﻗﺎﻋﺩﺓ ﺍﻟﺘﻲ ﺘﺤﺩﺩ ) of (xﻟﻨﺎ. ﻴﻭﺠﺩ ﻨﻭﻉ ﺨﺎﺹ ﻤﻥ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻴﺴﻤﻰ ﺍﻟﻤﺅﺜﺭﺍﺕ ﺍﻟﺨﻁﻴﺔ. ﻭﻫﺫﻩ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻟﻪ ﺍﻟﺨﺎﺼﻴﺔ: ] )∠ [f 1 (x) + f 2 (x) ] = ∠ [f 1 (x) + ∠ f 2 (x
ﻭﺒﻭﺠﻭﺩ cﻜﺭﻗﻡ ﻤﻭﺠﺏ ﺍﺨﺘﻴﺎﺭﻱ. ﻤﻼﺤﻅﺔ :ﻴﻭﺠﺩ ﻤﺅﺜﺭﺍﺕ ﻋﻜﺴﻴﺔ ﺍﻟﺨﻁﻴﺔ ﻭﻟﻬﺎ )∠ cf (x) = c * ∠ f (x
ﺍﻟﺨﻁﻴﺎﻥ.
)∠cf(x) = c * ∠ f(x
ﻭﻋﻠﻴﻪ ﻓﺈﻨﻪ ﻓﻲ ﺍﻟﻘﺎﺌﻤﺔ ﺍﻟﺴﺎﺒﻘﺔ ﻓﺈﻥ ﺍﻟﻤﺅﺜﺭﻴﻥ ﺍﻷﺨﻴﺭﻴﻥ ﻫﻤـﺎ
ﺍﻟﻤﺅﺜﺭ ﺍﻟﺨﻁﻲ ﺴﻴﻨﻘل ﺩﺍﻟﺔ ﻤﺎ ﻟﺩﺍﻟﺔ ﺃﺨﺭﻯ. )df (x )− 2 f (x ax
= )∠ f (x
ﻭﻴﻤﻜﻨﻨﺎ ﺘﺸﺒﻪ ﺍﻟﺩﻭﺍل ﻜﻤﺘﺠﻬﺎﺕ ﻓﻲ ﺜﻼﺜﺔ ﺃﺒﻌﺎﺩ ﻭﺘﺄﺜﻴﺭ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻴﻜﻭﻥ ﺒﻨﻘل )ﺘﺤﻭﻴل( ﺍﻟﻤﺘﺠﻪ ﻟﻤﺘﺠﻪ ﺁﺨﺭ ،ﻭﻴﺄﺨﺫ ﺍﻟﻤﺘﺠﻪ ﻜﻭﺤﺩﺓ ) (amitvedreﻭﻤـﺅﺜﺭ ﻴﻨﻘـل
ﻨﻘﻁﺔ ﻋﻠﻰ unit sphereﻟﻨﻘﻁﺔ ﺃﺨﺭﻯ ﻭﻤﺅﺜﺭ ﻤﺎ ﻴﻤﻜﻥ ﺍﻋﺘﺒـﺎﺭﻩ ﺒﻘﺎﺒـل ﺩﻭﺭﺍﻥ ﺍﻟﻨﻘﻁﺔ ﺤﻭل ﻤﺤﻭﺭ zﺒﺜﻼﺜﻴﻥ . 30 °
ﺜﻼﺙ ﺤﺎﻻﺕ:
A → A′
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B → B′ C → C′
ﻻﺤﻅ ﺃﻥ ﺍﻟﻨﻘﻁﺔ ﻋﻠﻰ ﺍﻟﻘﻁﺏ ﺘﻨﻘل ﻋﻠﻰ ﻨﻔﺴﻬﺎ )ﻭﻜﺫﻟﻙ ﺍﻟﻨﻘﻁﺔ ﺍﻟﺘﻲ ﻋﻠـﻰ
ﺍﻟﻘﻁﺏ ﺍﻟﺠﻨﻭﺒﻲ( ﻭﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺨﺎﺼﺔ ﻟﻤﺜﺎل ﻤﺅﺜﺭ ﻤﺜل ﻤﻌﺎﺩﻟﺔ ) ( 4 – 6ﻭﺍﻟـﺫﻱ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻪ ﻜﻤﺎ ﻴﻠﻲ: )(4-10
)u E (x
H u E (x) = E
ﻤﻨﻁﻭﻕ ﺍﻟﻤﻌﺎﺩﻟﺔ ،H :ﻤﺅﺜﺭ ﺍﻟﻬﺎﻤﻠﺘﻴﻭﻨﻴﺎﻥ ﻴﺅﺜﺭ ﻋﻠﻰ ﻨﻭﻑ ﻤﻌﻴﻥ ﻤﻥ ﺍﻟﺩﻭﺍل
ﺴﻴﻌﻁﻴﻨﺎ ﻤﻥ ﺠﺩﻴﺩ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﻲ ﻴﺅﺜﺭ ﻋﻠﻴﻬﺎ ﻤﻀﺭﻭﺒﺔ ﻓﻲ ﺜﺎﺒﺕ.
ﻫﺫﺍ ﺍﻟﺜﺎﺒﺕ ﻴﺴﻤﻰ eigeulalwﻭﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻴﻌﺘﻤﺩ ﻋﻠﻰ Eﻭﺤل ﺍﻟﻤﻌﺎﺩﻟـﺔ ﻴﻌﺘﻤﺩ ﻋﻠﻰ .E ﻭﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ
)u E (x
ﻟﻠﻤﺅﺜﺭ .H
ﻴﺴﻤﻰ eigeufactionﺍﻟﺫﻱ ﻴﻨﺎﻅﺭ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ E
ﻤﺜﺎل :ﺤل ﻤﺴﺄﻟﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ
)∠ f (x) = λ f (x
ﺤﻴﺙ: )h df (x )− β x f (x i ax
ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ
a ≤x≤a
= )∠ f (x
ﻭﻟﻬﺎ ﺸﺭﻁ ﺍﻷﻁﺭﺍﻑ )ﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻴﺔ( .bouday )f (a) = f ( − a
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:ﺍﻟﺤل ∠ f (x) = λ f (x) h a f (x) − β v f (x) = λ f (x) i ax df (x) i i = [β + t (x) + λ f (x)]= [β x + λ ]+ (x) dx h h df i x = (β + λ) ax f h i x2 h f = (β + λx )+c h 2 x 2 iβ + iλλxh 2 f (x) = Ae
energy eigenvahes ﺘﺴﻤﻰ ﻗﻴﻡ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺨﺎﺼﺔH ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ ﻟﻠﻤﺅﺜﺭ H=
p 2 op + V (x) 2m
the Eigenvalne Poeem for a ﻤﺴﺄﻟﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ ﻟﺠﺴﻡ ﻓﻲ ﺼﻨﺩﻭﻕ
.Particle in box
h 2 a 2 u (x) − + V (x) u (x) = Eu (x) 2m ax 2
(4 – 6) ﺒﺎﻋﺘﺒﺎﺭ
:ﺤﻴﺙ V (x) = ∞ = 0 =∞
x <0 °<x
:(4 – 6 ) ﻤﻥ u (x) = 0 =0
x<0 a<x
.( 4 – 6 ) ﺍﻟﻤﻌﺎﺩﻟﺔV ( x ) = 0 ﺩﺍﺨل ﺍﻟﺼﻨﺩﻭﻕ ﺤﻴﺙ d 2 u (x) 2mE + 2 u (x) = 0 dx 2 h
. ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﺄﺨﺫ ﺍﻟﺸﻜل،E < 0 ﻟﻭﻥ ﺃﻥ - ٨٤ PDF created with pdfFactory Pro trial version www.pdffactory.com
d 2 u (x) − K 2 u (x) = 0 2 dx e − lx
, ,
k2 =
2m E h2
e kx
ﺍﻟﺤل ﺍﻟﻌﺎﻡ ﻫﻭ ﺇﺘﺤﺎﺩ ﺨﻁﻲ ﻤﻥ
k2 =
2mE ,E h2
⇐ ﺒﻘﻴﻡ ﺇﻴﺠﺎﺒﻴﺔ
d 2 u (x) + k 2 u (x) = 0 dx 2
A sin Kx + B con lex ﺍﻟﺤﺎﻟﺔ ﺍﻟﻌﺎﻡ u ( 0 ) = 0 ﻭﻟﻜﻥ ﻟﺯﻭﻡ
:ﻴﺤﺼﺭﻨﺎ ﻻﺨﺘﻴﺎﺭ ﺍﻟﺸﺭﻁ u (x) = Asin k u (a ) = 0 ⇒ ⇐ ka = nD
n = 1, 2, 3.........
.ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻗﻴﻡ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺨﺎﺼﺔ En =
h 2u 2 h 2 D 2 u 2 = 2m 2ma 2
n = 1, 2, 3.........
.ﻭﺒﺈﻤﻜﺎﻨﻨﺎ ﺍﺨﺘﺒﺎﺭ ﺃﻥ ﺍﻟﺤﻠﻭل ﻤﻌﺎﻴﺭﺓ ﻟﻭ ﻜﺎﻨﺕ ﻗﻴﻤﺔ A = 2la u n (x ) =
2 hDx sin a a
.ﻭﺍﻟﺤﻠﻭل ﻟﻬﺎ ﺍﻟﺨﺎﺼﻴﺔ a
∫ dx u 0
a
* n
(x ) u m (x) = ∫ dx 0
m Dx 2 nD χ sin sin a a a
a (n − m) Dx (n + m) Dx 1 ) − cos ( ) cos ( ∫ a a0 a sin (n − m) D sin (n + m) D = − ( n − m )D ( n + m )D
=
= 0 = 1
whem whem
n± m n =m
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a
(x ) u m (x) = smn
)..................(4 − 23
* n
∫ dx u 0
* ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺍﻟﻤﻨﺎﻅﺭﺓ ﻟﻘﻴﻡ ﺨﺎﺼﺔ ﻤﺨﺘﻠﻔﺔ ﺘﻜـﻭﻥ ﻤﺘﻌﺎﻤـﺩﺓ
.ortagoral
* ﻟﻭ ﺃﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻤﻌﺎﻴﺭﺓ .ﺒﺩﻗﺔ ،ﻜﻤﺎ ﻫﻭ ﻅـﺎﻫﺭ ﻤـﻥ ) ،(4-23ﻴـﺴﻤﻰ thonormality covelitionﺸﺭﻁ ﺍﻟﺘﻁﺒﻴﻊ ﻭﺍﻟﺘﻌﺎﻤﺩ.
ﺍﺴﺘﻨﺒﺎﻁ ﺒﻌﺽ ﺍﻟﻤﻌﻠﻭﻤﺎﺕ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﻤﻥ ﺍﻟﺤﻠﻭل ﺍﻟﺨﺎﺼﺔ .eigensalutions -1ﺍﻟﺤﺎﻟﺔ ﺍﻟﺘﻲ ﻟﻬﺎ ﺃﻗل ﻁﺎﻗﺔ ،ﺍﻟﺤﺎﻟﺔ ﺍﻟﺩﻨﻴﺎ ،grourdstateﺘﻭﺼـﻑ ﺒــ )u1 (x
ﻭﺘﻘﺎﺒﻠﻬﺎ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ.
D 2h 2 2ma 2
ﻟﻪ
= E1
ﻻﺤﻅ ﺃﻨﻪ ﻤﻥ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺃﻥ ﺃﻗل ﻁﺎﻗﺔ ﻟﺠﺴﻡ ﺴﺎﻜﻥ atesrﻓﻲ ﻓﺠﻭﺓ، V(x)=0 , p = 0
ﻓﺈﻥ ﻤﺠﻤﻭﻉ ﻁﺎﻗﺔ ﺍﻟﻭﻀﻊ ﻭﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ = 0ﻭﻟﻜﻥ ﻨﺠﺩ ﻓﻲ
ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﺃﻥ ﺃﻗل ﻁﺎﻗﺔ ﻗﻴﻤﺘﻬﺎ ﻟﻴﺴﺕ ﺼﻔﺭ ﺃﻱ ﺘﻭﺠﺩ ﻗﻴﻤﺔ "ﻟﻠﻁﺎﻗﺔ ﺍﻟﺩﻨﻴﺎ" ﻫـﺫﺍ ﻤﺭﺘﺒﻁ ﺒﻤﺒﺩﺃ ﺍﻟﻼﺘﺤﺩﻴﺩ
Δn Δp ≥ D
-2ﺒﻤﺎ ﺃﻥ ﺍﻟﺤﻠﻭل ﺤﻘﻴﻘﻴﺔ ﻓﺈﻥ
ﺘﺨﻴﻠﻴﺔ
h d ⇒ )R (x l dx
ﺃﻥ p = 0
)dx R(x
ﻤﻥ ﻨﺎﺤﻴﺔ ﺃﺨﺭﻯ
ﻟﻤﺎﺫﺍ. p =0
ﻭﺫﻟﻙ ﻷﻥ ﻷﻱ ﺩﺍﻟﺔ ﺤﻘﻴﻘﻴـﺔ ) R(xﻗﻴﻤـﺔ
∫ ﻭﻫﺫﺍ ﻻ ﻴﺘﻔﻕ ﻤﻊ ﺍﻟﻤﻁﻠﻭﺏ
x
= p
p
ﺇﻻ ﻓﻲ ﺤﺎﻟـﺔ
ﻓﺈﻥ p 2 ≠ 0
ﻭﺒﻤﺎ ﺃﻨﻪ ﻓﻲ ﺩﺍﺨل ﺍﻟﺼﻨﺩﻭﻕ
p 2 = 2mE = 2mE n
p2
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-3ﻭﺒﺎﺯﺩﻴﺎﺩ ﻋﺩﺩ ﺍﻟﻌﻘﺩ ﻓﻲ ﺍﻟﺤل ،ﻓﺈﻥ ﻫﺫﺍ ﻴﻘﺎﺒل ﻁﺎﻗﺔ ﺃﻋﻠﻰ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﻓﻲ ﺍﻟﺸﻜل ﺍﻟﺘﺎﻟﻲ ﻫﺫﺍ ﻴﻤﻜﻥ ﻓﻬﻤﻪ ﻭﺫﻟﻙ ﻷﻥ ﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ ﺘﺯﺩﺍﺩ ﺒـ urith caruature
of the solntionsﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻭﻗﻌﺔ ﻟﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ. )d 2 u (x d x2
)dx u * (x
∫
h2 2m
=
u
ﻓﺭﻀﻴﺔ ﺍﻟﻔﻙ: The Expansion Postulate and its physical Interpretation
ﻨﻅﺭﻴﺔ ﻓﻭﺭﻴﺭ ﺘﻨﺹ ﻋﻠﻰ ﺃﻥ ﺃﻱ ﺩﺍﻟﺔ ψ (0 ) and ψ ( a ) = 0
)ψ(x
ﺍﻟﺘﻲ ﺘﺤﻘﻕ ﺍﻟـﺸﺭﻁ ﺍﻟﺤـﺩﻱ
ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻬﺎ. nD x a
ψ ( x ) = ∑ Cu sin
ﻭﺒﻤﺎ ﺃﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ )ﺍﻟﻤﻤﻴﺯﺓ( ﻟــ H
∞
∞
ﺘﺘﻨﺎﺴـﺏ ﻤـﻊ
nDx a
sin
ﻓﺒﺈﻤﻜﺎﻨﻨﺎ ﻜﺘﺎﺒﺔ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟـ ) Un (xﻜﺎﻟﺘﺎﻟﻲ: )ψ (x) = ∑ A n U n (x
ﻭﻤﻥ ﺸﺭﻁ ﺍﻟﺘﻌﺎﻤﺩ ﻭﺍﻟﻤﻌﺎﻴﺭﺓ ﻓﺈﻨﻪ ﺒﺈﻤﻜﺎﻨﻨﺎ ﺇﻴﺠﺎﺩ An؟
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a
∫
a
∫
dx u (x) ψ (x) = * m
0
dx u *m (x)
∑
A n U n (x)
0
=
a
∑
An
∫
dx u *m (x) =
∑
A n Smn
0
a
An =
∫
dx u *m (x) ψ (x)
0
ﺩﺍﻟﺘﻪ ﺍﻟﻤﻭﺠﻴﺔ ﺘﻌﻁﻲ ﺒـ. ﺍﻋﺘﺒﺭ ﺠﺴﻴﻡ ﻓﻲ ﺼﻨﺩﻭﻕ/ﻤﺜﺎل ψ (x) = A ( x / a ) x = A (1 − ) a
0 < x < a/2 a
2
<x < a
a
∫
ψ (x)
dx
2
=1
0
ﻟﻜﻲ ﺘﺤﻘﻕ
A=
12 a
ﺤﻴﺙ
؟En ﺍﺤﺴﺏ ﺍﺤﺘﻤﺎل ﺃﻥ ﻗﻴﺎﺱ ﺍﻟﻁﺎﻗﺔ ﻴﻌﻁﻴﻨﺎ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺨﺎﺼﺔ . ﻓﻲ ﺍﻟﻔﻙAn ﻨﺤﻥ ﻨﺭﻴﺩ ﺤﺴﺎﺏ a
An =
∫
2 nD x sin α a
dx ψ (x)
0
a
An =
∫
2 nDx sin α a
dx ψ (x)
0
= QA = =
An = An = An =
∫
a/2 0
12 a dx A
x a
2 nDx sin + a a
∫
a/2
24 a/2 x nDx + ∫0 dx ( ) sin a a a 24 a 24 a
∫ ∫
nDx x dx ( A 1 − ) sin a a
0
∫
0 a/2
dx (1 −
x nDx ) sin a a
2 nDx sin α a
D/a 0 D/a 0
an
u sin nu (1 − (−1) n ) D
، ﻭﺫﻟﻙ ﻟﻭﺠﻭﺩ ﺍﻟﺤﺩ ﺍﻟﺜﺎﻨﻲ ﺍﻟﺘﻜﺎﻤل ﻴﻤﻜﻥ ﺤﺴﺎﺒﻪ ﺒـﺴﻬﻭﻟﺔ0 ← An ← ﻗﻴﻤﺔN .ﻓﻘﻁ ﻟﻘﻴﻡ ﻓﺭﺩﻴﺔ
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24 1 2 2 ( − 1) n +1 D πn a6 An = 4 4 n odd D n =0 n euen = An
ﻤﺜﺎل /ﻓﻲ ﺤﺎﻟﺔ ﻭﺠﻭﺩ ﺠﺴﻴﻡ ﺩﺍﺨل ﺼﻨﺩﻭﻕ ) 3ﺃﺒﻌﺎﺩ( ﻤﻥ ﺨـﻼل ﺍﻟﺤـل ﺍﻟـﺫﻱ
ﺘﻭﺼﻠﻨﺎ ﺇﻟﻴﻪ ﻓﻲ ﺤﺎﻟﺔ ﺒﻌﺩ ﻭﺍﺤﺩ.
Dhn z c
nD nD ⇒ p = hu = h a a Dhn y = , py , b
= pz
= u a = nD ⇒ u Dhn x a
= ⇒ px
2 2 n 2z 1 2 1 D 2h 2 n x n y 2 2 2 =E = p = ) ( p + py + pz ( + )+ 2 2m 2m x 2m a 2 b 2 c
ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻌﻁﻲ ﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﻁﺎﻗﺔ ﻟﺠﺴﻴﻡ ﻓﻲ ﺼﻨﺩﻭﻕ ﺠﻬﺩ ﻓﻲ ﺜﻼﺜﺔ ﺃﺒﻌﺎﺩ. n x hz c
ﺃﻋﻼﻩ.
sin
n x hy b
sin
n x hx a
ψ = C sin
ﻟﻭ ﻋﻭﻀﻨﺎ ﺒﻬﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻓﻲ ﻤﻌﺎﺩﻟﺔ ﺴﺘﺭﻭﺩﻨﺠﺭ ﻓﻨﺘﺤـﺼل ﻋﻠـﻰ ﻗـﻴﻡ E
ﻓﻲ ﺤﺎﻟﺔ ﻤﻜﺴﺏ ) .(a = b = c D 2h 2 D 2h 2 2 2 2 2 =E = ) ( nx + ny + nz k 2m 2m 2
ﻭﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﺒﺔ ﺍﻟﺘﻲ ﺘﻘﺎﺒﻠﻬﺎ. n x hz c
sin
n x hy b
sin
n x hx a
ψ = C sin
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ﻻﺤﻅ ﺃﻥ ﺍﻟﻁﺎﻗﺔ ﺘﻌﺘﻤﺩ ﻋﻠﻰ .n2ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﻜل ﺍﻟﺤﺎﻻﺕ ﺍﻟﺘﻲ ﺘﻨﺎﻅﺭ ﻜل ﺍﻷﺭﻗﺎﻡ
D,ny ,nz
ﻭﺍﻟﺘﻲ ﺘﻌﻁﻲ ﻨﻔﺱ ﺍﻟﻁﺎﻗﺔ .ﻭﻟﻜﻥ ﻋﻨﺩﻤﺎ ﻨﻐﻴﺭ ﺍﻷﺭﻗـﺎﻡ
nx , n y , n z
ﺒﺩﻭﻥ ﻤﺎ ﻨﻐﻴﺭ ﻗﻴﻤﺔ nﻓﺈﻥ Ψﺘﺘﻐﻴﺭ ﻭﻋﻠﻴﻪ ﻓﺈﻨﻪ ﻴﻭﺠﺩ ﻤﺴﺘﻭﻴﺎﺕ ﻤﻥ ﺍﻟﻁﺎﻗـﺔ ﻟﻬـﺎ ﺍﻟﻌﺩﻴﺩ ﻤﻥ ﺍﻟﺩﻭﺍل ﺍﻟﻤﻭﺠﻴﺔ ﻭﻫﺫﺍ ﻤﺎ ﻴﺴﻤﻰ ﺒﺎﻟﺘﻜﺭﺍﺭ .degeneracy g
nx , n y , n z
ﺍﻟﻁﺎﻗﺔ
1
) (1,1,1,
3 E1
2
) (2 , 1, 1 ) (1, 2 ,1) (1,1, 2
6 E1
3
) (2 , 2 ,1 ) (2 , 1, 2 ) ( 2 , 2 , 2
9 E1
( ) (3 ,1, 1
11 E 1
ﻓﺭﻀﻴﺔ ﺍﻟﻔﻙ: Expansion Postulate and Its physical Interpretation
ﺇﻥ ﻨﻅﺭﻴﺔ ﻓﻭﺭﻴﺭ ﺘﻨﺹ ﻋﻠﻰ ﺃﻥ ﺃﻱ ﺩﺍﻟﺔ ψ (0 ) = ψ ( a ) = 0 ⇐ bowdary canditions
nD x a
)ψ(x
ﺍﻟﺘﻲ ﺘﺤﻘﻕ ﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻴﺔ
ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻬﺎ ﺒﺎﻟﺸﻜل ﺍﻟﺘﺎﻟﻲ.
ψ ( x ) = ∑ Cu sin
ﻭﺒﻤﺎ ﺃﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟـ Hﻟﻠﺒﺌﺭ ﺍﻟﻼﻨﻬﺎﺌﻲ infinite wellﺘﺘﻨﺎﺴﺏ ﻤـﻊ nDx a
sin
ﻓﺒﺈﻤﻜﺎﻨﻨﺎ ﻜﺘﺎﺒﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺴﺎﺒﻘﺔ ﺒﺩﻻﻟﺔ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟـ )Un (x )ψ (x) = ∑ A n U n (x
ﻭﺒﺎﺴﺘﺨﺩﺍﻡ ﻋﻼﻗﺔ ﺍﻟﻤﻌﺎﻴﺭﺓ
a
dx u *m (x) u n (x) = smn ↔ orthonormelnty
∫ 0
ﻴﻤﻜﻨﻨﺎ ﺇﻴﺠﺎﺩ ﻗﻴﻤﺔ .An
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)A n U n (x
∑
a
a
∫
)dx u *m (x
∫
= )dx u (x) ψ (x * m
0
A n Smn
∑
a
∫
= )dx u *m u n (x 0 1 442443
0
∑
An
=
Smn a
)dx u *m (x) ψ (x
∫
= An
0
a
sin = 1, if m = n
∫
)dx u *n (x) ψ (x
,
= ⇒ An
0
ﻭﻜﻤﺎ ﻓﻌﻠﻨﺎ ﻓﻲ ﺤﺎﻟﺔ ﺍﻨﺘﺸﺎﺭ ﺍﻟﺤﺯﻤﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺍﻟﺤﺭﺓ ،ﺒﺈﻤﻜﺎﻨﻨﺎ ﺤﺴﺎﺏ ﺍﻟﺘﻁﻭﺭ
ﺍﻟﺯﻤﻨﻲ ﻟﻬﺫﻩ ﺍﻟﺩﺍﻟﺔ ﺍﻻﺨﺘﻴﺎﺭﻴﺔ
)ψ (x
ﻤﻊ ﺍﻟﺯﻤﻥ .ﻭﺒﻤﺎ ﺃﻥ ﻜل ﻤﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ
ﺘﺘﻁﻠﺏ ﺩﺍﻟﺔ ﺍﻋﺘﻤﺎﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﺨﺎﺼﺔ ﺒﻬﺎ
)U n (x
ﻜﺘﺎﺒﺔ.
A n U n (x) e − iEnt/h
∑
− iEnt/h
)
، ( eﺒﺸﻜل ﻋﺎﻡ ﻴﻤﻜﻨﻨـﺎ
= ) ψ (x, t
ﻭﻟﺘﺄﻭﻴل ﺍﻟﻤﻌﺎﻤﻼﺕ Anﻓﺒﺈﻤﻜﺎﻨﻨﺎ ﺤﺴﺎﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻭﻗﻌﺔ ﻟﻠﻁﺎﻗﺔ ﻓﻲ ﺃﻱ ﺤﺎﻟﺔ ﺍﺨﺘﻴﺎﺭﻴﺔ ) (xﻭﺒﻤﺎ ﺃﻥ ﺩﺍﺨل ﺍﻟﺼﻨﺩﻭﻕ
ﻭﺒﻤﺎ ﺃﻥ
/2m
2
، H = pﻭﺨﺎﺭﺝ ﺍﻟـﺼﻨﺩﻭﻕ
ψ (x) = 0
)H U n (x) = E n U n (x
)A n U n (x
∑
a
dx ψ * (x) H
∫
a
= )dx ψ (x) Hψ (x *
0
0
a
)(a
∫
=
H
∫
)dx ψ * (x) E n U n (x
An
∑
=
0
a
= ∑ En
∫
)dx ψ * (x) A n U n (x
0
= ∑ En An
2
)(B
∑
)A n U n (x
a
)dx ψ * (x
∫
= An
∑
= 1
0
2
An
ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺴﺎﺒﻘﺔ ) (aﻤﻊ ﺸﺭﻁ ﺍﻟﺘﻁﺒﻴﻊ ﻴﻘﺘﺭﺡ ﺃﻥ
∑
=
2
An
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)A n = ∫ dx U *n (x) ψ (x
ﻴﻤﻜﻥ ﺘﺄﻭﻴﻠﻬﺎ ﺒﺎﺤﺘﻤﺎل ﺃﻥ ﻗﻴﺎﺱ ﺍﻟﻁﺎﻗﺔ ﻷﻱ ﺩﺍﻟـﺔ
)ψ (x
ﺍﻟﺨﺎﺼﺔ .En
ﻴﻌﻁﻴﻨـﺎ ﺍﻟﻘﻴﻤـﺔ
ﻤﺜﺎل :ﺨﺫ ﺠﺴﻡ ﻓﻲ ﺼﻨﺩﻭﻕ .ﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﺔ. )ψ (x) = A (x/a x ) = A (1 − a
0 < x < a/2 a/2 < x < a
ﺤﻴﺙ
12 a
=A
ﻭﺫﻟﻙ ﻟﺘﺤﻘﻕ
∫ dxﺍﺤﺴﺏ ﺍﺤﺘﻤـﺎل ﺃﻥ ﺍﻟﻘﻴـﺎﺱ a
ψ (x) = 1 2
0
ﻟﻠﻁﺎﻗﺔ ﻴﻌﻁﻴﻨﺎ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ .Enﻨﺤﻥ ﻨﺭﻴﺩ ﺃﻥ ﻨﺤﺴﺏ Anﻓﻲ ﺍﻟﻔﻙ. 2 nDx sin α a
)ψ (x ↑
a
dx
∫
= An
0
12 x a a x nDx ) sin a a
ﺒﺘﻐﻴﻴﺭ ﺍﻟﻤﺘﻐﻴﺭﺍﺕ
dx (1 −
0 a/2
∫
Dx a , dx = du a D Dx a
24 a/2 x nDx dx ( ) sin + ∫ 0 a a a
ﻟﻠﺘﻜﺎﻤل ﺍﻷﻭل.
=u
ﻟﻠﺘﻜﺎﻤل ﺍﻟﺜﺎﻨﻲ
=D-u
a
u
∫ ∫ D an D sin (nu) + u ) sin nu (1 − (−1) n D
D/2
∫
dn
0
24 a 24 = D
= An
Anﻟﻘﻴﻡ nﻤﺯﺩﻭﺠﺔ ← 0ﻷﻥ ﺍﻟﺤﺩ ﺍﻷﺨﻴﺭ ﻭﻟﻘﻴﻤﺔ nﺍﻟﻔﺭﺩﻴﺔ. 24 1 2. ( − 1 ) n +1 a πn 2
ﻭﻋﻠﻴﻪ.
n odd neven
96 D 4n 4
=
2
An
= An ⇒
=0
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Momentum Eiglnhetion of the Fremantle
ﺍﻟﺘﻤﺎﺜل :Parity ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟﺠﺴﻡ ﺤﺭ ) (sin kx, cos kxﻭﻜﺫﻟﻙ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟﺠﺴﻡ
ﻓﻲ ﺼﻨﺩﻭﻕ ﻤﻤﺘﺩ ﻤﻥ
−a /2 →a /2
)(n = 2, 4, 6, ..............
nDx 2 sin a a
= )U n (x
)(n = 1, 3, 5, ..............
nDx 2 cos a a
= )U n (x
ﻫﺫﻩ ﺍﻟﺩﻭﺍل ﺇﻤﺎ ﺃﻥ ﺘﻜﻭﻥ ﺯﻭﺠﻴﺔ ﺃﻭ ﻓﺭﺩﻴﺔ ﺘﺤﺕ ﺍﻟﺘﻐﻴﻴﺭ
ﺠﺴﻡ ﻓﻲ ﺼﻨﺩﻭﻕ ﻤﺘﻤﺎﺜل ﺤﻭل
x= 0
x →−x
ﻭﻜﺎﻨﺕ ﺍﻟﻘﻴﻡ ﺍﻻﺒﺘﺩﺍﺌﻴـﺔ ﻟﻠﺤﺎﻟـﺔ
ﻟﻭ ﺃﺨـﺫﻨﺎ )ψ (x
ﺩﻭﺍل
ﺯﻭﺠﻴﺔ. ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﻴﻤﺘﺩ ﻋﻠﻰ
)ψ (x
ﻴﺠﺏ ﺃﻥ ﺘﻜﻭﻥ
)(n =1, 3, 5, ..............
2 nDx cos a a
An
∑
= )ψ (x
ﺤﻴﺙ ﺃﻥ ﺍﻟﺠﻤﻊ
ﻋﻨﺩ ﻭﻗﺕ ﺁﺨﺭ ﺘﺼﺒﺢ ﺍﻟﺩﺍﻟﺔ
2 nDx − iEnt/h e cos a a
ψ (x , t) = ∑ An
ﻭﻫﺫﻩ ﺍﻟﺩﺍﻟﺔ ﻻ ﺯﺍﻟﺕ ﺯﻭﺠﻴﺔ ﻓﻲ xﻋﻨﺩ ﺃﻭﻗﺎﺕ ﺃﺨﺭﻯ .ﻭﻨﻔﺱ ﺍﻟﺸﻲﺀ ﻴﺤﺩﺙ
ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻘﻴﻡ ﺍﻷﻭﻟﻰ ﺍﻻﺒﺘﺩﺍﺌﻴﺔ ﺘﻜﻭﻥ ﻓﺭﺩﻴﺔ.
ﻭﻋﻠﻴﻪ :ﻟﻠﺼﻨﺩﻭﻕ ﺍﻟﻤﺘﻤﺎﺜل ﻭﻤﺘﻤﺤﻭﺭ ﺤﻭل x = 0ﻓﺈﻥ ﺍﻟﻔﺭﺩﻴﺔ ﻭﺍﻟﺯﻭﺠﻴـﺔ
ﻫﻲ ﺨﻭﺍﺹ ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ .ﻭﻋﻠﻴﻪ ﺒﺄﻨﻬﺎ )ﺍﻟﺯﻭﺠﻴـﺔ ﻭﺍﻟﻔﺭﺩﻴـﺔ( ﺜﻭﺍﺒـﺕ ﺍﻟﺴﺭﻋﺔ.
ﻴﻌﻜﺱ
ﻓﻠﻭ ﺍﺴﺘﺤﺩﺜﻨﺎ ﻤﺅﺜﺭ Parihyopenator pﻤﺅﺜﺭ ﺍﻟﺘﻤﺎﺜل ﻭﺍﻟﺫﻱ ﻗﺎﻋﺩﺘـﻪ ﺃﻥ x →−x
ﻭﻫﻜﺫﺍ ﻷﻱ ﺩﺍﻟﺔ ﻤﻭﺠﺒﺔ
)ψ (x
)p ψ (x) = ψ ( - x
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ﻭﻟﺩﺍﻟﺔ ﻤﻭﺠﺒﺔ ﺯﻭﺠﻴﺔ. )p ψ ( + ) (x) = ψ ( + ) ( x
ﻭﻟﺩﺍﻟﺔ ﻓﺭﺩﻴﺔ. )p ψ ( − ) (x) = − ψ ( − ) ( x
ﻭﻫﺎﺘﻴﻥ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ﻫﻲ ﻤﻌﺎﺩﻻﺕ ﺫﺍﺕ ﻗﻴﻡ ﺨﺎﺼﺔ ﻭﺍﻟﺫﻱ ﻭﻀﺤﻨﺎﻩ ﻫـﺫﺍ ﺃﻥ
ﺍﻟﺩﻭﺍل ﺍﻟﺯﻭﺠﻴﺔ ﻫﻲ ﺩﻭﺍل ﺨﺎﺼﺔ ﻟـ pﻟﻬﺎ ﻗﻴﻤﺔ ﺨﺎﺼﺔ + 1ﺒﻴﻨﻤﺎ ﺍﻟﺩﻭﺍل ﺍﻟﻔﺭﺩﻴﺔ
ﻫﻲ ﺩﻭﺍل ﺨﺎﺼﺔ ﻟـ pﻟﻬﺎ ﻗﻴﻤﺔ ﺨﺎﺼﺔ -1
ﻭﻓﻲ ﻤﺴﺄﻟﺔ ﺍﻟﺠﺴﻡ ﻓﻲ ﺼﻨﺩﻭﻕ ﻓﺈﻥ ﺍﻟﺩﻭﺍل
nDx nDx ( ) , sin ) a a
ﺩﻭﺍل ﺨﺎﺼﺔ ﻟـ Hﺒل ﻫﻲ ﺩﻭﺍل ﺨﺎﺼﺔ ﻟـ Pﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼـﺔ ﺍﻟﻤﻤﻜﻨﺔ ﻓﻘﻁ.
ﺍﻓﺘﺭﺍﺽ ﺃﻨﻨﺎ ﻋﻨﺩﻨﺎ.
)PU (x) = D u (x
( cos
±1
ﻟﻴﺴﺕ ﻓﻘﻁ
ﻫـﻲ ﺍﻟﻘـﻴﻡ
ﺒﺎﻟﺘﺄﺜﻴﺭ ﺒـ Pﻤﺭﺓ ﺃﺨﺭﻯ.
)P 2 U (x) = P D u (x) = D 2 u ( x
ﻟﻜﻥ
)P 2 U (x) = U (x
⇒ λ 2 =1 ⇒ λ = ± 1
ﻭﻓﺭﺩﻴﺔ.
ﻭﺫﻟﻙ ﻷﻥ ﻻﻨﻌﻜﺎﺴﻴﺔ ﺍﻻﺜﻨﻴﻥ )ﻤﺭﺘﻴﻥ( ﻴﺠﺏ ﺃﻻ ﻴﻐﻴﺭ ﺸﻴﺌﺎﹰ.
ﻭﺃﻱ ﺩﺍﻟﺔ ﺍﺨﺘﻴﺎﺭﻴﺔ ﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺘﻬﺎ ﻜﻤﺠﻤﻭﻉ ﻟـﺩﻭﺍل ﺯﻭﺠﻴـﺔ 1 ] )[ψ (x) + ψ (- x) ] + 1 [ψ (x) − ψ (- x 2 2
= )ψ (x
ﻭﻫﺫﺍ ﻜﻤﺎ ﻋﻤﻠﻨﺎ ﺒﺎﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟـ ⇐ Hﺃﻱ ﺩﺍﻟﺔ ﻴﻤﻜﻥ ﺃﻥ ﺘﻔﻙ ﺒﺩﻻﻟـﺔ
ﺍﻟﺩﻭﺍل ﻟﻠﻤﺅﺜﺭ ﺍﻟﺠﺩﻴﺩ.
ﺇﻥ ﻅﻬﻭﺭ "ﺍﻟﺯﻭﺠﻴﺔ" ﻭ "ﺍﻟﻔﺭﺩﻴﺔ" ﺠﺎﺩ ﻷﻨﻨﺎ ﺭﻜﺯﻨﺎ ﺍﻟﺼﻨﺩﻭﻕ ﺤـﻭل x = 0
ﻭﺇﺫﺍ ﻜﻨﺎ ﻭﻀﻌﻨﺎﻩ ﺒﻴﻥ a , 0ﻻ ﺸﻲﺀ ﺴﻴﺘﻐﻴﺭ ﻭﺴﺘﺒﻘﻰ ﺍﻟﺘﻤﺎﺜل )ﺍﻟﺘﻨـﺎﻅﺭ( ﺘﺤـﺕ ﺍﻻﻨﻌﻜﺎﺱ ﻋﻨﺩ .x = a/2ﺇﻥ ﺍﻟﺩﺭﺱ ﺍﻟﻤﻁﻠﻭﺏ ﺘﻌﻠﻤﻪ ﻫﻨﺎ ﻫﻭ ﺃﻨﻨﺎ ﺒﻭﻀﻊ ﻤﺴﺄﻟﺔ ﻓﻲ
ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻴﺠﺏ ﺃﻥ ﻨﺭﺍﻋﻲ ﺒﺎﻫﺘﻤﺎﻡ ﺘﻤﺎﺜل ﻭﺘﻨـﺎﻅﺭ ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴـﺎﻥ ،ﻭﺍﺨﺘﻴـﺎﺭ - ٩٤PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺍﻻﺤﺩﺍﺜﻴﺎﺕ ﺒﻁﺭﻴﻘﺔ ﺘﻅﻬﺭ ﺍﻟﺘﻤﺎﺜل ﺒﻭﻀﻭﺡ ﻭﺍﻟﺤﻘﻴﻘﺔ ﺍﻟﻬﺎﻤﺔ ﻫﻭ ﺘﻤﺎﺜـل )ﺘﻨـﺎﻅﺭ( ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ .ﻫﺫﺍ ﺭﺒﻤﺎ ﺃﻥ ﻴﺭﻯ ﺒﻭﻀﻭﺡ ﺒﺎﻟﺴﺅﺍل ﺘﺤﺕ ﺃﻱ ﺤﻴﺜﻴﺎﺕ ﺴﺘﺒﻘﻰ ﺍﻟﺩﺍﻟﺔ
ﺍﻟﺯﻭﺠﻴﺔ ﺯﻭﺠﻴﺔ ﻋﻨﺩ ﻜل ﺍﻷﺯﻤﻨﺔ. ﺍﻟﺠﻬﺩ ﺫﻭ ﺍﻟﺒﻌﺩ ﺍﻟﻭﺍﺤﺩ
One – Dimensional Potential
ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺴﻭﻑ ﻨﺤل ﻤﺴﺎﺌل ﺒﺴﻴﻁﺔ ﻟﻠﺤﺭﻜﺔ ﻓﻲ ﺒﻌـﺩ ﻭﺍﺤـﺩ .ﻫـﺫﻩ
ﺍﻟﻤﺴﺎﺌل ﺫﺍﺕ ﺃﻫﻤﻴﺔ ﻭﺫﻟﻙ ﻷﻨﻬـﺎ ﺘﻭﻀـﺢ ﺒﻌـﺽ ﺍﻟﻅـﻭﺍﻫﺭ ﺍﻟﻐﻴـﺭ ﺘﻘﻠﻴﺩﻴـﺔ nonclaesical efpeotsﻭﻜﺫﻟﻙ ﻷﻥ ﺍﻟﻌﺩﻴﺩ ﻤﻥ ﺍﻟﺤـﺎﻻﺕ ﺍﻟﻔﻴﺯﻴﺎﺌﻴـﺔ ﻫـﻲ ﻓـﻲ ﻤﺤﺼﻠﺘﻬﺎ ﻓﻲ ﺒﻌﺩ ﻭﺍﺤﺩ ﺒﺎﻟﺭﻏﻡ ﻤﻥ ﺃﻨﻨﺎ ﻨﻘﻴﺱ ﻓﻲ ﺜﻼﺜﺔ ﺃﺒﻌﺎﺩ.
ﻋﺘﺒﺔ ﺍﻟﺠﻬﺩ )ﻤﻌﺎﻤﻠﻲ ﺍﻟﻨﻔﺎﺫ ﻭﺍﻻﻨﻌﻜﺎﺱ( :The Potential step ﺒﺎﻟﻨﻅﺭ ﺇﻟﻰ ﺸﻜل ﺍﻟﻤﻭﻀﺢ.
x < 0 x>0
V (x) = 0 = V° 0
ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ. )h 2 d 2 u (x − )+ v (x) u (x) = E u (x 2m dx 2
ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻬﺎ: [ E - v (x) ] u (x) = 0
)d 2 u (x 2m + 2 2 dx h
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ﺒﻜﺘﺎﺒﺔ: 2m E =K2 2 h
ﻋﻨﺩ
V (x) = 0 ← x< 0
[
]
2m E − V0 = q 2 2 h
,
ﻓﺈﻥ ﺍﻟﺤل ﺍﻟﻌﺎﻡ. u (x) = e ikz + R e − ikz
ﻭﻫﺫﺍ ﻴﻘﺎﺒل ﻓﻴﺽ ﻴﺘﺤﺭﻙ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﻤﻭﺠﺏ ﻟـ xﻭﻤﻘﺩﺍﺭ ﺍﻟﻔﻴﺽ ﻫﻭ: * ∂ψ ∂ψ * − ψ )(x = ψ ∂x ∂x complex coujngate 644444 4 474444444 8 h iux * iux 2ikx − iux iux = − Rik e ) − ( ike + ikR * e − iux ) (e iux + R * e − iux ) ( e + R e ) ( iue 2im h ) ik (1 − RR * ) + ik ( R * e 2ikx − Re −ikx 2im h =} ) − − ik (1 − RR * ) + ik ( R * e 2ikx − Re − ikx ) [ 2 ik ( 1 − R) = hk (1 − R 2 2im m h 2im
]
[{
]
ﺒﺈﻤﻜﺎﻨﻨﺎ ﺃﻥ ﻨﻨﻅﺭ ﻟـ
e ikx
ﻫﻨﺎﻙ ﺠﻬﺩ ﺒﺈﻤﻜﺎﻨﻨﺎ ﺃﻥ ﻨﺨﺘﺎﺭ
[
ﺒﻔﻴﺽ ﻗﺩﺭﺓ
e ikx
ﻫﺫﺍ ﺍﻟﺠﻬﺩ ﻴﻨﺸﺊ ﺍﻟﻤﻭﺠﺔ ﺍﻟﻤﻨﻌﻜﺴﺔ
=j
hk m
ﻜﻤﻭﺠﺔ ﺴﺎﻗﻁﺔ .ﻭﻟﻭ ﻟـﻡ ﻴﻜـﻥ
ﻜﺤل ﻟﻜل ﻗﻴﻡ xﻭﻋﻠﻴﻪ ﻨﻌﺯﻱ Rﻟﻭﺠﻭﺩ ﺍﻟﺠﻬﺩ R e - ikx
ﺒﻔﻴﺽ ﻤﻨﻌﻜﺱ ﻗﺩﺭﻩ
2
hk R m
ﻟﻘﻴﻡx > 0
ﺒﺎﻤﻜﺎﻨﻨﺎ ﺃﻥ ﻨﻜﺘﺏ ﺍﻟﺤل. U (x) = T e ijx
ﻭﻟﻜﻥ ﻫﺫﺍ ﺍﻟﺤﺩ ﻴﺼﻑ ﻤﻭﺠﺔ ﻗﺎﺩﻤﺔ ﻤﻥ
∞+
ﻓﻲ ﺍﺘﺠﺎﻩ
−x ←
ﻭﻟﻜـﻥ ﻓـﻲ
ﺤﺎﻟﺔ ﻤﻭﺠﺔ ﺴﺎﻗﻁﺔ ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﻓﺈﻥ ﺍﻟﻤﻭﺠﺔ ﺍﻟﻭﺤﻴﺩﺓ ﻋﻠﻰ ﺍﻟﻴﻤﻴﻥ ﻴﻤﻜﻥ ﺃﻥ ﺘﻜـﻭﻥ ﻤﻭﺠﺔ ﻨﺎﻓﺫﺓ transmitted wereﻭﺍﻟﻔﻴﺽ ﺍﻟﻤﻘﺎﺒل ﻟﻬﺎ.
]
[
h T * e - iux ( Tiq e iqx ) − (−iq T * e −iqx ) T e iqx 2im
j
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ﻭﺒﻤﺎ ﺃﻨﻪ ﻻ ﻴﻭﺠﺩ ﺍﻋﺘﻤﺎﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﻓﻲ ﻫﺫﻩ ﺍﻟﻤﺴﺄﻟﺔ ﻓﺈﻥ ﻗـﺎﻨﻭﻥ ﺍﻟﺤﻔـﻅ )ﺍﻟﺒﻨﺎﺀ(
∂ ∂ P (x, t) + j (x, t) = 0 ∂t ∂t
ﻴﻌﻨﻲ ﺃﻥ ) j(xﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ .xﻭﻋﻠﻴﻪ ﻓﺈﻥ ﺍﻟﻔﻴﺽ
ﻋﻠﻰ ﺍﻟﺠﺎﻨﺏ ﺍﻷﻴﺴﺭ ﻴﺠﺏ ﺃﻥ ﻴﺴﺎﻭﻱ ﺍﻟﻔﻴﺽ ﻋﻠﻰ ﺍﻟﺠﺎﻨﺏ ﺍﻷﻴﻤﻥ. 2
hi T m
= )
2
hk (1 − R m
⇒
ﻭﺇﻥ ﺍﺘﺼﺎﻟﻴﺔ contiuwityﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻴﺅﺩﻱ
)U (x) = U (x
e iux + R e - iux = T e 2ikx ⇒ at x = 0 ⇒ 1 + R = T
)(a
ﻭﻜﺫﻟﻙ ﻓﺈﻥ ﻤﻴل ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﺒﺔ ﻤﺘﺼل ﻋﻨﺩ x = 0ﺒﺎﻟﺭﻏﻡ ﻤﻥ ﺃﻥ ﺍﻟﺠﻬـﺩ
ﻏﻴﺭ ﻤﺘﺼل ﻭﻫﺫﺍ ﻴﺘﻀﺢ ﺒﺘﻜﺎﻤل ﺍﻟﻤﻌﺎﺩﻟﺔ ﺇﻟﻰ
∈+
ﺤﻴﺙ
∈
d 2 u (x) 2m ]+ 2 [E − V (x) u (x) = 0 dx 2 h
(
ﻤﻥ
∈−
ﺭﻗﻡ ﻤﻭﺠﺏ ﻓﺈﻥ. ∈
∈
du du d du 2m ( )∈ − ( ) − ∈ = ∫ dx = ∫ dx 2 (V (x) − E ) u (x) = 0 dx dx ∈dx dx − h ∈−
ﻤﻼﺤﻅﺔ ﺍﻟﺠﻬﺩ ﻴﺤﺘﻭﻱ ﻋﻠﻰ ﺤﺩ ﻤﺜل:
ﻟﻠﻤﻌﺎﺩﻟﺔ ﻤﻥ
∈+ a −
ﺇﻟﻰ
2m )V u (a h 2 °ο
∈a +
)V°ο 8 (x − a
ﻴﻌﻁﻴﻨﺎ
∈a +
= )V°ο 8 (x − a) u (x
∫
∈a −
ﻋﻨﺩﺌـﺫ ﻓـﺈﻥ ﺍﻟﺘﻜﺎﻤـل
du du 2m ( ) at ∈ − ( ) a − ∈ = 2 dx dx h
ﺍﺘﺼﺎل ﺍﻟﺘﻔﺎﻀل ﻟﻠﺠﻬﺩ ﻓﻲ ﺍﻟﻤﺴﺄﻟﺔ ﻴﻌﻁﻲ: 2k k+q
=T
,
K −q K +q
= R
ﻭﻤﻥ ﻫﺫﺍ ﻴﻤﻜﻨﻨﺎ ﺤﺴﺎﺏ ﺍﻟﻔﻴﺽ ﺍﻟﻤﻨﻌﻜﺱ ﻭﺍﻟﻨﺎﻓﺫ flected and transmitted feuxes h k k −q 2 ( ) m k+q hq 2k 2 hq hk 4 hq = Transmitte d j = T2 ( = ) m m k +q m (k + q) 2 =
)(C
2
hk R m
= Refleted j
ﻻﺤﻅ ﺍﻵﺘﻲ: - ٩٧PDF created with pdfFactory Pro trial version www.pdffactory.com
-١ﺒﺨﻼﻑ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻭﺍﻟﺘﻲ ﻁﺒﻘﺎﹰ ﻟﻬﺎ :ﺠﺴﻡ ﻴﺴﻴﺭ ﻓﻭﻕ ﺠﻬﺩ ﻓﺈﻨﻪ ﺴﻭﻑ ﻴﺘﺒﺎﻁﺄ ﻭﺫﻟﻙ ﻟﺤﻔﻅ ﺍﻟﻁﺎﻗﺔ .ﻭﻟﻜﻨﻪ ﻻ ﻴﻤﻜﻥ ﺃﻥ ﻴﻨﻌﻜﺱ .ﺃﻤﺎ ﻫﺫﺍ ﻓﻨﻼﺤﻅ ﺃﻥ ﺠـﺯﺀ ﻤﻌﻴﻥ ﻤﻥ ﺍﻟﺠﺴﻴﻤﺎﺕ ﺍﻟﺴﺎﻗﻁﺔ ﺍﻨﻌﻜﺱ .ﻭﻫﺫﺍ ﺒﺎﻟﻁﺒﻊ ﻨﺘﻴﺠـﺔ ﻟﻠﻁﺒﻴﻌـﺔ ﺍﻟﻤﻭﺠﻴـﺔ
ﻟﻠﺠﺴﻴﻤﺎﺕ ﻓﺈﻥ ﺍﻻﻨﻌﻜﺎﺱ ﺍﻟﺠﺯﻱﺀ ﻟﻠﻀﻭﺀ ﺍﻟﺴﺎﻗﻁ ﻋﻨﺩ ﺍﻟﺘﻤـﺎﺱ ﺒـﻴﻥ ﻭﺴـﻁﻴﻥ
ﻓﻅﺎﻫﺭﺓ ﻤﻌﺘﺎﺩﺓ. -٢ﻤﻥ ﺨﻼل ﺍﻟﺘﻌﺒﻴﺭﺍﺕ )ﻤﻌﺎﺩﻟﺔ (Cﻨﻼﺤﻅ ﺃﻥ ﻗﺎﻨﻭﻥ ﺍﻟﺤﻔﻅ )ﺍﻟﺒﻘﺎﺀ( ﻤﺘﺤﻘﻕ ﺃﻱ ﺃﻥ ﺍﻟﻔﻴﺽ ﻋﻠﻰ ﺍﻟﺠﺎﻨﺏ ﺍﻷﻴﺴﺭ = ﺍﻟﻔﻴﺽ ﻋﻠﻰ ﺍﻟﺠﺎﻨﺏ ﺍﻷﻴﻤﻥ. hk hq 4k 2 k −q 2 ( (1 − =) ) m k+q m (k + q) 2 hk h k 4qk 4qk = 2 )m (k + q m (k + q) 2
-٣ﻓﻲ ﺤﺎﻟﺔ
2m 2m E ≈ 2 ( E − Vο ) = q ⇐ E >> Vο 2 h h
ﺍﻟﻤﻨﻌﻜﺱ ﻭﺍﻟﺴﺎﻗﻁ
2
= k
ﻓﺈﻥ ﺍﻟﻨـﺴﺒﺔ ﺒـﻴﻥ ﺍﻟﻔـﻴﺽ
ο← R ( k − q) 2 JR = →0 JT 4kq
ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺘﺘﻔﻕ ﻤﻊ ﺍﻟﺤﺩﺱ ﻭﺍﻟﺫﻱ ﻴﺩل ﻋﻠﻰ ﺃﻨﻪ ﻋﻨﺩ ﺍﻟﻁﺎﻗﺎﺕ ﺍﻟﺤﺎﻟﻴـﺔ
ﻓﺈﻥ ﻭﺠﻭﺩ ﻋﻘﺒﺔ ﺍﻟﺠﻬﺩ ﻤﺎ ﻫﻭ ﺇﻻ ﺘﺸﻭﻴﻪ ﻗﻠﻴل ﻋﻠﻰ ﺍﻨﺘﺸﺎﺭ ﺍﻟﻤﻭﺠﻪ. -٤ﻟﻭ ﺃﻥ
E < Vο
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ
2m )(E − Vο h2
=q
ﺘﺼﺒﺢ ﻜﻤﻴﺔ ﺘﺨﻴﻠﻴﺔ ﻓﻲ ﻫـﺫﻩ
ﺍﻟﺤﺎﻟﺔ ﻓﺈﻥ ﺍﻟﺤﻠﻭل x > 0ﻴﻜﻭﻥ ﻟﻬﺎ. − q x
U (x) = T e i (iq) x = T e
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ: *
k −i q k −i q k −i q k +i q = = R R = =1 k +i q k + i q k + i q k − i q *
2
R
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ﻭﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﺘﻭﺠﺩ ﻤﻭﺠﺔ ﻤﻨﺘﺸﺭﺓ ﻋﻥ ﻴﻤﻴﻥ ﺍﻟﺠﻬﺩ ﻷﻥ
−qx
u = Te
ﻟﻴﺴﺕ
ﻤﻭﺠﺔ ﺭﺍﺤﻠﺔ .ﻭﻫﻜﺫﺍ ،ﻜﻤﺎ ﻓﻲ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻴﻜﻭﻥ ﺍﻨﻌﻜﺎﺱ ﻜﻠﻲ .ﻻﺤﻅ )ﻤﻥ
ﻨﺎﺤﻴﺔ ﺃﺨﺭﻯ( ﺃﻥ.
2k ≠0 k+i q
=T
ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺠﺯﺀ ﻤﻥ ﺍﻟﻤﻭﺠﺔ ﻴﺨﺘﺭﻕ Penetratesﺇﻟﻰ ﺩﺍﺨل ﺍﻟﻤﻨﻁﻘـﺔ ﺍﻟﻤﻤﻨﻭﻋﺔ )ﺍﻟﻤﺤﺭﻤﺔ( ﻭﻅﺎﻫﺭﺓ ﺍﻻﺨﺘﺭﺍﻕ ﻫﺫﻩ ﻤﻥ ﻅﻭﺍﻫﺭ ﺍﻟﻤﻭﺠﺎﺕ ﻭﺴﻨﺭﻯ ﺒﻌـﺩ
ﻗﻠﻴل ﺃﻥ ﻫﺫﻩ ﺴﺘﺴﻤﺢ ﺒـ " "tunnelingﺍﻻﺨﺘﺭﺍﻕ ﻤﻥ ﺨﻼل ﺍﻟﺠﻬﺩ .ﻭﺍﻟـﺫﻱ ﻤـﻥ
ﻨﺎﺤﻴﺔ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺴﻴﻤﻨﻊ ﺒﺎﻟﻜﺎﻤل ﺍﻟﺠﺴﻴﻤﺎﺕ ﻭﻴﻤﻨﻌﻬﺎ ﻤﻥ ﺍﻻﻓﺘـﺭﺍﻕ .ﻭﻓـﻲ
ﺍﻟﺤﻘﻴﻘﺔ ﻓﺈﻨﻪ ﻻ ﻴﻭﺠﺩ ﻓﻴﺽ ﻤﻥ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻟﻠﻴﻤﻴﻥ ﻭﺫﻟﻙ ﻷﻥ ) j (xﺘﺨﺘﻔﻲ ﻟﻠﺤﻠﻭل
ﺍﻟﺤﻘﻴﻘﻴﺔ ﺤﺘﻰ ﻟﻭ ﺍﻟﻤﻌﺎﻤل ﻟﻸﻤﺎﻡ ﺃﺨﺫ ﻜﻤﻴﺔ ﻤﺭﻜﺒﺔ ﻅﺎﻫﺭﺓ ﺍﻻﻨﻌﻜﺎﺱ ﺍﻟﻜﻠﻲ ﺭﻴﺎﻀﻴﺎﹰ
ﺘﻜﺎﻓﺊ ﻟﻠﺫﻱ ﻴﺤﺩﺙ ﻟﻠﻀﻭﺀ ﻋﻨﺩﻤﺎ ﻴﺴﻘﻁ ﻋﻠﻰ ﺴﻁﺢ )ﺘﻤﺎﺱ( ﺒـﻴﻥ ﻭﺴـﻁﻴﺔ ﻟﻬـﺎ ﻤﻌﺎﻤل ﺍﻨﻜﺴﺎﺭ ﻤﺨﺘﻠﻑ ) ( n2 > n1ﺒﺯﺍﻭﻴﺔ ﺃﻜﺒﺭ ﻤﻥ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﺤﺭﺠﺔ .ﻓﺈﻥ ﺍﻟﻀﻭﺀ
ﻴﻌﺎﻨﻲ ﻤﻥ ﺍﻨﻌﻜﺎﺱ ﻜﻠﻲ ﻭﻟﻜﻥ ﺴﻴﻜﻭﻥ ﻫﻨﺎﻙ ﺍﻀﻤﺤﻼل ﻟﻠﻤﺠﺎل ﺍﻟﻜﻬﺭﻭﻤﻴﻐﻨﺎﻁﻴﺴﻲ ﺍﻟﺫﻱ ﻴﺨﺘﺭﻕ ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﻤﺤﺭﻤﺔ.
ﺒﺌﺭ ﺍﻟﺠﻬﺩ :The Potential well ﻨﺄﺨﺫ ﺍﻟﺠﻬﺩ ﺍﻟﻤﻭﻀﻊ ﺒـ x <−a −a < x
V (x) = 0 = − Vο
a<x ) 2m (E + Vο h2
= q2
=o 2mE , h2
= k2
ﺍﻟﺤﻠﻭل:
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⇒ u ( x ) = e ikx + R e −ikx
< −a
⇒ u ( x ) = A e iqx + B e -iqx
⇒ u ( x ) = T e ikx
< a
ﻭﻫﺫﻩ ﺘﻁﺎﺒﻕ ﻟﻔﻴﺽ ﻗﺎﺩﻡ )ﺴﺎﻗﻁ( ﺍﻟﻴﻤﻴﻥ
2
hk R m
ﻭﻓﻴﺽ ﻨﺎﻓﺫ
2
T
hk m
x -a< x x
ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﻭﻓـﻴﺽ ﻤـﻨﻌﻜﺱ ﻤـﻥ
. hk m
ﺃﻤﺎ ﺩﺍﺨل ﺍﻟﺒﺌﺭ ﻓﺈﻨﻪ ﺘﻭﺠﺩ ﻤﻭﺠﺎﺕ ﺫﺍﻫﺒﺔ ﻓﻲ ﻜﻼ ﺍﻻﺘﺠـﺎﻫﻴﻥ ﻭﺫﻟـﻙ ﻷﻥ
ﺍﻻﻨﻌﻜﺎﺱ ﻴﺤﺩﺙ ﻤﻥ ﺍﻟﻁﺭﻓﻴﻥ ﻋﻨﺩ 2
T
hk m
x=±
=) − B 2
2
ﻭﻁﺒﻘﺎﹰ ﻟﻘﺎﻨﻭﻥ ﺤﻔﻅ ﺍﻟﻔﻴﺽ.
hq ( A m
= ) (1− R 2
hk m
ﻭﺒﺘﻁﺎﺒﻕ ﺍﻟﺩﻭﺍل ﺍﻟﻤﻭﺠﻴﺔ ﻭﺘﻔﺎﻀﻼﺘﻬﺎ ﻴﻌﻁﻴﻨﺎ ﺃﺭﺒﻊ ﻤﻌﺎﺩﻻﺕ. e - ika + R e + ika = A e -iqa + B e iqa ) ik ( e - ika − R e ika ) = iq ( A e -iqa − B e iqa A e + iqa + B e - iqa T e ika iq ( Ae iqa − B e - iqa = ik T e ika
ﻭﺒﺤل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻻﺕ (q 2 − k 2 ) sin 2qa 2k q cos 2qa − i (q 2 + k 2 ) sin 2qa 2kq 2k q cos 2qa − i (q 2 + k 2 ) sin 2qa
ﻭﻓﻲ ﺤﺎﻟﺔ
ﻭﻜﺫﻟﻙ ﻋﻨﺩﻤﺎ
E >> Vο
E→ο
R = ie −2ika T = e −2ika
ﻓﺈﻨﻪ ﻻ ﻴﻭﺠﺩ ﻋﻤﻠﻴﺎﹰ ﺍﻨﻌﻜـﺎﺱ ﻭﺫﻟـﻙ ﻷﻥ
ﻓﺈﻥ ﺍﻟﻨﻔﺎﺫﻴﺔ
q 2 − k 2 << 2k q
←ο
ﻓﻲ ﺤﺎﻟﺔ sin 2 ga = 0ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻁﺎﻗﺎﺕ ﻤﻭﺠﺒﺔ ﺍﻟﻤﻌﻁﺎﺓ ﺒـ ﻟﻬﺫﻩ ﺍﻟﻘﻴﻡ ﻤﻥ ﺍﻟﻁﺎﻗﺔ ﻻ ﻴﻭﺠﺩ ﺍﻨﻌﻜﺎﺱ .R = 0ﻭﻫﺫﺍ ﻨﻤﻭﺫﺝ ﻟﻤﺎ ﻴﺤﺩﺙ ﻓﻲ
ﺍﺴﺘﻁﺎﺭﺓ ﺍﻟﻜﺘﺭﻭﻨﺎﺕ ﺫﺍﺕ ﻁﺎﻗﺔ ﻤﻨﺨﻔﻀﺔ ) ( 0.1 eVﺒﻭﺍﺴـﻁﺔ ﻏـﺎﺯﺍﺕ ﻓﺎﻋﻠـﺔ
)ﻨﻴﻭﻥ ،ﺃﺭﺠﻭﻥ( ﻭﺍﻟﺫﻱ ﻴﻭﺠﺩ ﺒﻬﺎ ﻨﻔﺎﺫﻴﺔ ﻋﺎﻟﻴﺔ ﻏﻴﺭ ﻁﺒﻴﻌﻴﺔ .ﻫﺫﻩ ﺍﻟﻅـﺎﻫﺭﺓ ﺘـﻡ
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ﻤﺸﺎﻫﺩﺘﻬﺎ ﺃﻭﻻﹰ ﺒﻭﺍﺴﻁﺔ Townsend , Ramsauerﻭﻭﺼﻔﺕ ﺒﺄﻨﻪ ﺭﻨﻴﻥ ﺍﻟﻨﻔﺎﺫﻴـﺔ .transmission resonance ﻭﻟﺘﻘﺩﻴﻡ ﻤﻨﺎﻗﺸﺔ ﺩﻗﻴﻘﺔ ﻓﺈﻨﻪ ﻤﻥ ﺍﻟﻀﺭﻭﺭﻱ ﺇﺩﺨﺎل )ﻭﻀﻊ ﻓـﻲ ﺍﻟﺤـﺴﺒﺎﻥ(
ﺍﻋﺘﺒﺎﺭﺍﺕ ﺜﻼﺜﺔ ﻭﺒﻠﻐﺔ ﺍﻟﻤﻭﺠﺎﺕ ،ﻓﺈﻥ ﺍﻟﺘﺄﺜﻴﺭ ﺒﺴﺒﺏ ﺍﻟﺘﺩﺍﺨل ﺍﻟﻬﺩﺍﻡ ﺒﻴﻥ ﺍﻟﻤﻭﺠـﺎﺕ ﺍﻟﻤﻨﻌﻜﺴﺔ ﻋﻨﺩ c = - aﻭﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﻤﻨﻌﻜﺴﺔ ﻤﺭﺓ ﻭﺍﺜﻨﻴﻥ ﻭﺜﻼﺜـﺔ ﻋﻨـﺩ x = 0
ﻭﺸﺭﻁ
2D 2D 4a = = ﺍﻟﺭﻨﻴﻥ⇐ 2 qa = nh . q nD n
=λ
ﻭﻫﺫﺍ ﻫﻭ ﺍﻟـﺫﻱ ﻴـﺼﻑ – Fairy
.Perotintenforometeerﻭﻫﺫﺍ ﻫﻭ ﺸﺭﻁ ﻋﻤل ﺠﻬﺎﺯ. ﺍﻟﺤﻠﻭل ﻓﻲ E < 0ﺒﺸﺭﻁ ﺃﻥ
Vο
ﺴﺎﻟﺒﺔ ) ﺃﻱ ﺃﻥ
Vο
ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ
V (x) = ο
= Vο → > ο ο
=
ﺴﺘﻜﻭﻥ )ﻜﻤﺎ ﺴﻨﺭﻯ ﺍﻟﺤﻠﻭل ﻤﺠﺯﺃﺓ .(discreteﺨﺫ 2mE = −K2 h2
ﺍﻟﺤﻠﻭل ﺨﺎﺭﺝ ﺍﻟﺒﺌﺭ ﻭﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻤﻘﻴﺩﺓ ﻋﻨﺩ
∞
x < −a
u ( x ) = C1 e kx
a <x
u ( x ) = C 2 e − kx
x<−a
u ( x ) = e + ikx + Re −ikx
−a<x
u ( x ) = A e iqx + Be −iqx
a<x
u ( x ) = T e iux
ﻭﺒﻤﺎ ﺃﻨﻨﺎ ﻨﺘﻌﺎﻤل ﻤﻊ ﺩﻭﺍل ﺤﻘﻴﻘﻴﺔ ﻓﺈﻨﻪ ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻤﻥ ﺍﻟﻤﻨﺎﺴﺏ ﺃﻥ ﻨﻜﺘﺏ
ﺍﻟﺤﻠﻭل ﺩﺍﺨل ﺍﻟﺒﺌﺭ ﺒﺎﻟﺸﻜل. − a<x
ﻻﺤــﻅ ﺃﻥ
u ( x ) = A cos qx + B sin qx
2m (Vο − E ) > ο h2
ﻭﺘﻔﺎﻀﻼﺘﻬﺎ ﻋﻨﺩ ﺍﻟﺠﻭﺍﺏ
x = ±a
= q2
ﻭﺫﻟــﻙ ﻷﻥ
Vο > E
ﺒﺘﻁــﺎﺒﻕ ﺍﻟﺤﻠــﻭل
ﻴﻌﻁﻴﻨﺎ - ١٠١ -
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C1 e− ka = A cos qa − B sin q a
u C1 e− ka = [A ( sin qa) q ]+ B (cos q a) = q ( A sin qa + B cos q a)
C 2 e − ka = A cos qa + B sin q a − k C 2 e − ka = q (A sin qa − B cos q a)
.K ﻤﻥ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻻﺕ ﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺔ A sin A sin A sin =q A sin
k=q
qa − Bcos qa qa + Bcos qa qa + Bcos qa qa − Bcos qa
(1) (2)
ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺤﻠـﻭل ﺇﻤـﺎA B = 0 ﻭﻫﺎﺘﻴﻥ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ﻤﻌﺎﹰ ﺘﻌﻨﻲ ﺃﻥ
( A = 0 ) x = ( ﺃﻭ ﻓﺭﺩﻴﺔB = 0 ) x = ﺯﻭﺠﻴﺔ
.ﻭﺍﻟﺩﻭﺍل ﺘﻘﺭﻴﺒﺎﹰ ﺸﻜﻠﻬﺎ ﺍﻟﻤﻭﻀﺢ ﻓﻲ ﺍﻟﺸﻜل
)ﺍﻟﺘﻲ ﻟﻴﺱ ﻟﻬﺎ ﻋﻘﺩ( ﺘﻜﻭﻥ ﺯﻭﺠﻴﺔ ﻭﺍﻟـﺸﺭﻭﻁgroudstate ﻭﺍﻟﺤﺎﻟﺔ ﺍﻟﺩﻨﻴﺎ .ﺍﻟﺘﻲ ﺘﺤﺩﺩ ﺍﻟﻁﺎﻗﺔ (1) ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ K = q tam qa
(B = 0 ) ﺤﻠﻭل ﺯﻭﺠﻴﺔ
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ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ )(2 ﺤﻠﻭل ﺯﻭﺠﻴﺔ ) (A = 0
K = q tam qa
ﺩﻋﻨﺎ ﻨﺨﺘﺒﺭ ﻫﺫﻩ ﺍﻟﺤﻠﻭل ﻤﻨﻔﺼﻠﺔ. ) (aﺍﻟﺤﻠﻭل ﺍﻟﺯﻭﺠﻴﺔ: ﺍﺴﺘﺨﺩﻡ ﺍﻟﺭﻤﺯ
2m Vο a 2 h2
)ﺘﺘﻨﺎﺴﺏ ﻤﻊ ﻋﻤﻕ ﺍﻟﺒﺌﺭ(
=λ
) ﺘﺘﻨﺎﺴﺏ ﻤﻊ ﻋﺭﺽ ﺍﻟﺒﺌﺭ(
y= q a
ﺇﻥ ﻨﻘﺎﻁ ﺍﻟﺘﻘﺎﻁﻊ ﺘﺤﺩﺩ ﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔ .ﻭﻫﺫﻩ ﺘﺘﻤﺜل ﻓﺌﺔ ﻤﺘﻘﻁﻌﺔ discete ser
ﻭﻜﻠﻤﺎ ﻜﺒﺭﺕ ﻗﻴﻤﺔ
λ
ﻓﺈﻥ ﺍﻟﻤﻨﺤﻨﻴﺎﺕ ﻟﻠﻤﻌﺎﺩﻟﺔ
λ − y2 y
ﺘﻤﻀﻲ.
ﺃﻱ :ﻋﻨﺩﻤﺎ ﻴﺯﺩﺍﺩ ﻋﻤﻕ ﺍﻟﺠﻬﺩ ﺃﻭ ﻴﺴﺘﻌﺭﺽ ﻓﺈﻨﻪ ﻴﻭﺠﺩ ﺍﻟﻤﺯﻴﺩ ﻤﻥ ﺍﻟﺤﺎﻻﺕ
ﺍﻟﻤﺭﺒﻭﻁﺔ )ﺍﻟﻤﻘﻴﺩﺓ( lundstatesﻭﺍﻟﺸﻜل ﻴﺒﻴﻥ ﻜﺫﻟﻙ ﺃﻨﻪ ﻻ ﻴﻬﻡ ﻜﻡ ﺘﻜﻭﻥ ﻗﻴﻤﺔ
λ
ﺼﻐﻴﺭﺓ ﻓﺈﻨﻪ ﺴﻴﻜﻭﻥ ﺩﺍﺌﻤﺎﹰ ﻋﻠﻰ ﺍﻷﻗل ﺤﺎﻟﺔ ﻤﺭﺒﻭﻁﺔ ﻭﺍﺤﺩﺓ .ﻭﻫﺫﺍ ﺃﻤﺭ ﻴﻤﻴﺯ ﺍﻟﺠﻬﺩ
ﺍﻟﺠﺎﺫﺏ ttractiuepotﻓﻲ ﺒﻌﺩ ﻭﺍﺤﺩ ﻭﻟﻴﺱ ﺼﺤﻴﺤﺎﹰ ﻓﻲ ﺜﻼﺜﺔ ﺃﺒﻌﺎﺩ ﻭﺍﻟﺫﻱ ﻴـﺴﻠﻙ
ﺴﻠﻭﻙ ﺍﻟﺤﻠﻭل ﺍﻟﻔﺭﺩﻴﺔ )ﺴﺘﻨﺎﻓﺱ ﻓﻴﻤﺎ ﺒﻌﺩ( .ﻭﻜﻠﻤﺎ ﻜﺒﺭﺕ
ﺘﺅﻭل ﻭﺘﺼﺒﺢ ﻤﺘﺴﺎﻭﻴﺔ ﺍﻟﻤﺴﺎﻓﺔ ﻓﻲ , n = 0, 1, 2, ..........
1 )D 2
λ
qa = y = (n +
λ
ﻓﺈﻥ ﺍﻟﻘـﻴﻡ ﺍﻟﺨﺎﺼـﺔ
ﻭﺒﻨﻘﺎﻁ ﺍﻟﺘﻘﺎﻁﻊ ﺍﻟﻤﻌﻁﺎﺓ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ.
ﻭﻫﺫﺍ ﻫﻭ ﺸﺭﻁ ﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼـﺔ ﻟﻠﺤﻠـﻭل
ﺍﻟﺯﻭﺠﻴﺔ ﻟﺼﻨﺩﻭﻕ ﻤﺎ ﻻ ﻨﻬﺎﻴﺔ ﻤﺘﻤﺭﻜﺯ ﺤﻭل ﺍﻟﻨﻘﻁﺔ ﺍﻷﺼل.
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Figure 5-11. Location of discrete eigenvalues for even solutions in square well. The rising curves represent tan y; the falling curves are λ − y 2 / y for different values of λ
. ﻫﻨﺎ ﺴﻴﻜﻭﻥ ﺸﺭﻁ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ:( ﺍﻟﺤﻠﻭل ﺍﻟﻔﺭﺩﻴﺔb) λ − y2 = − cos y y
ﻓﺈﻥ ﺍﻟﺭﺴﻤﺔ ﺍﻟﺘﺎﻟﻴﺔ ﺘﺒﺩﻭ ﻤﺜل ﺍﻟﺭﺴﻤﺔ ﺃﻋﻼﻩ ﻭﻟﻜـﻥ ﻓﺈﻨﻬﺎ ﺘﺴﻠﻙ ﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘـﺔ y=nD
,
λ
ﻭﻜﻠﻤﺎ ﻜﺒﺭﺕ
n = 1, 2, 3, .................
D 2
D − cory = ( + y ) 2
ﻭﺒﻤﺎ ﺃﻥ
ﻤﺘﺯﺤﺯﺤﺔ ﺒﻤﻘﺩﺍﺭtan ﻤﻨﺤﻨﻴﺎﺕ
ﺴﺘﺴﺘﺒﺩل ﺒـ
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1 y=(n + ) 2
ﻭﻟﻜﻥ ﻤﻌﺎﺩﻟﺔ
Figure 5-12. Location of discrete eigenvalues for odd solutions in square well. The rising curves represent – cot y; the falling curves are λ − y 2 / y for different values of λ Note that thee is no eigenvalue for λ < (π / 2) 2 λ−
D2 4
>ο
.ﻭﻋﻠﻰ ﻋﻜﺱ ﺍﻟﺤﻠﻭل ﺍﻟﺯﻭﺠﻴﺔ ﻓﺈﻨﻪ ﺴﻴﻜﻭﻥ ﺘﻘﺎﻁﻊ ﻟـﻭﻥ ﺃﻥ
λ>
D2 = 2.40 4
ﻭﻻﺤﻅ ﺃﻥ ﺘﻘﺎﻁﻊ ﻴﺤﺩﺙ ﻋﻨﺩﻤﺎ،ﺍﻨﻅﺭ ﺇﻟﻰ ﺍﻟﺸﻜل ﺃﻋﻼﻩ
ﻭﻋﻠﻴـﻪx = 0 ﻤﻥ ﺍﻟﺸﻜل ﻨﻼﺤﻅ ﺃﻥ ﺍﻟﺤﻠﻭل ﺍﻟﻔﺭﺩﻴﺔ ﺘﺨﺘﻔﻲ )ﺘﺘﻼﺸﻰ( ﻋﻨﺩ
ﻓﺈﻥ ﻤﺴﺄﻟﺔ ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﻘﻴﺩﺓ ﻟﻠﺤﻠﻭل ﺍﻟﻔﺭﺩﻴﺔ ﺴﺘﻜﻭﻥ ﻫﻲ ﻨﻔﺴﻬﺎ ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ ﻟﻬﺎ ﺒﺌﺭ .ﺍﻟﺠﻬﺩ ﻫﻭ ﺍﻟﻤﻭﻀﺢ ﻓﻲ ﺍﻟﺸﻜل ﺍﻟﺘﺎﻟﻲ
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ﻭﺫﻟﻙ ﻷﻥ ﺍﻟﺸﺭﻁ )ﻋﻨﺩ ﻫﺫﺍ ﺍﻟﺠﻬﺩ(
U (ο ) = 0
ﺴﻴﻜﻭﻥ ﺼﺤﻴﺢ.
ﻭﻤﻥ ﺍﻟﺤﺴﺎﺒﺎﺕ ﺍﻟﺘﻔﺼﻴﻠﻴﺔ ﻓﺈﻨﻪ ﻴﻤﻜﻨﻨﺎ ﺘﺄﻜﻴﺩ ﺍﻟﻔﻬﻡ ﺍﻟﻌﺎﻡ ﻟﻠﺴﺒﺏ ﻓﻲ ﺍﻟﺤﺼﻭل
ﻋﻠﻰ ﻗﻴﻡ ﺨﺎﺼﺔ )ﻏﻴﺭ ﻤﺘﺼﻠﺔ( .ﻫﺫﻩ ﺍﻟﻘﻴﻡ ﺘﻅﻬﺭ ﻭﺫﻟﻙ ﻷﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﺒﺔ ﻴﺠـﺏ ﺃﻥ ﺘﺘﻼﺸﻰ ﻋﻨﺩ ∞ .
ﺤﺎﺠﺯ ﺍﻟﺠﻬﺩ )ﺍﻻﺨﺘﺭﺍﻕ ﺍﻟﻨﻔﻘﻲ( The Potential Barrier
ﺨﺫ ﺍﻟﺠﻬﺩ ﻜﺎﻟﺘﺎﻟﻲ: x <−a −a < x
V (x) = 0 = − Vο =o
a<x
ﻭﻟﻌﻨﺎ ﻨﺤﺼﺭ ﺍﻟﻤﻨﺎﻗﺸﺔ ﻟﻠﻁﺎﻗﺎﺕ ﺒﺤﻴﺙ
E < Vο
ﺘﺴﻤﺢ ﺒﺎﺨﺘﺭﺍﻕ ﺍﻟﺤﺎﺠﺯ ﻁﺒﻘﺎﹰ ﻟﻠﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ.
ﺃﻱ ﺃﻨﻬﺎ ﺍﻟﻁﺎﻗـﺎﺕ ﺍﻟﺘـﻲ ﻻ
ﺩﺍﺨل ﺍﻟﺤﺎﺠﺯ )ﺍﻟﺤﻠﻭل( ﻟﻠﻤﻌﺎﺩﻟﺔ. d 2 u (x) 2m + 2 ( E − Vο ) u (x) = 0 h 42 dx 2 1 4 43 4 )d 2 u (x − K 2 u (x) = 0 2 dx 2m ) K 2 = 2 ( Vο - E h
ﺍﻟﺤل ﺍﻟﻌﺎﻡ. x
U ( x ) = A e − kx + Be kx
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ﺃﻱ ﺃﻨﻪ ﺘﻭﺠﺩ ﻨﻔﺎﺫﻴﺔ ﺒﺎﻟﺭﻏﻡ ﻤﻥ ﺃﻥ ﺍﻟﻁﺎﻗﺔ ﺃﻗل ﻤﻥ ﻗﻤـﺔ ﺍﻟﺤـﺎﺠﺯ
) (E < Vο
ﻭﻫﺫﻩ ﻅﺎﻫﺭﺓ ﻤﻭﺠﻴﺔ ﻭﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻨﻔﺱ ﺍﻟﻅﺎﻫﺭﺓ ﺘﻼﺤﻅ ﻤﻥ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻫﺫﻩ ﻫﻲ ﻅﺎﻫﺭﺓ ﺍﻻﺨﺘﺭﺍﻕ ﺍﻟﻨﻔﻘﻲ ﻟﺠﺴﻴﻡ ﺨﻼل ﺤﺎﺠﺯ ﻋﺎﺩﺓ ﻤﺎ ﺘﻼﺤﻅﻪ.
ﺇﻥ ﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﺔ ﻻ ﺘﺘﻼﺸﻰ ﺩﺍﺨل ﺍﻟﺤﺎﺠﺯ ﻜﻤﺎ ﻴﻌﻨﻲ ﺍﺤﺘﻤﺎل ﻭﺠﻭﺩ ﺍﻟﺠـﺴﻡ ﺒﻁﺎﻗﺔ ﺴﺎﻟﺒﺔ .ﻭﻟﻜﻥ ﻜﻴﻑ ﻴﻤﻜﻥ ﺃﻥ ﻟﻬﺫﺍ ﻤﻌﻨﻰ؟ ﻟﻭ ﻨﻅﺭﻨﺎ ﻟﻤﺒﺩﺃ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻟﺘﺴﻌﻔﻨﺎ ﻤﻥ ﺍﻟﺘﻨﺎﻗﺽ ﻤﻊ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ. ﺇﻥ ﺘﺠﺭﺒﺔ ﺩﺭﺍﺴﺔ ﺍﻟﺠﺴﻡ ﺩﺍﺨل ﺤﺎﺠﺯ ﺍﻟﺠﻬﺩ ﻴﺠﺏ ﺃﻥ ﺘﺤﻴﺯﻩ ﺒﺩﻗﺔ ﻗﺩﺭﻫﺎ Δ x << 2a
ﻭﻫﺫﻩ ﺍﻟﺘﺠﺭﺒﺔ )ﺍﻟﻘﻴﺎﺴﺎﺕ( ﺴﺘﻨﻘل ﻟﻜﻤﻴﺔ ﺍﻟﺤﺭﻜـﺔ ﻟﻠﺠـﺴﻴﻡ ﺒﻌـﺩﻡ ﺘﺤﺩﻴـﺩ
ﺘﻜﺭﺭﻫﺎ.
h 2a
>> Δ p
ﻭﻫﺫﺍ ﻴﻘﺎﺒﻠﻪ ﺍﻨﺘﻘﺎل ﻟﻠﻁﺎﻗﺔ ﻗـﺩﺭﻩ
h2 8ma
>> Δ E
ﻭﻟﻤـﺸﺎﻫﺩﺓ
ﻁﺎﻗﺔ ﺤﺭﻜﺔ ﺴﺎﻟﺒﺔ ،ﻓﺈﻥ ﻫﺫﺍ ﺍﻟﻼﺘﻌﻴﻴﻥ ﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ ﺃﻗل ﺒﻜﺜﻴـﺭ ﻤـﻥ
ﺒﺤﻴﺙ ﺃﻥ
h2 8ma 2
>> >> E
E − Vο
h2 k2 2m
ﻭﺒﺸﻜل ﻋﺎﻡ ﻓﺈﻥ ﺍﻟﺤﻭﺍﺠﺯ ﺍﻟﺘﻲ ﺘﺤﺩﺙ ﻓﻲ ﺍﻟﻅﻭﺍﻫﺭ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﺤﻘﻴﻘﻴـﺔ ﻻ
ﺘﻜﻭﻥ ﻤﺭﺒﻌﺔ ،ﻭﻟﻤﻨﺎﻗﺸﺔ ﺒﻌﺽ ﺍﻟﺘﻁﺒﻴﻘﺎﺕ ﻴﺠﺏ ﺃﻭﻻﹰ ﺃﻥ ﻨﺴﺘﻨﺘﺞ ﺘﻌﺒﻴﺭﺍﺕ ﺘﻘﺭﻴﺒﻴـﺔ ﻟﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ
2
T
ﺨﻼل ﺤﻭﺍﺠﺯ ﺠﻬﺩ ﻏﻴﺭ ﻤﻨﺘﻅﻤﺔ ﻭﺍﻟﻁﺭﻴﻘﺔ ﻟﻌﻤل ﻫﺫﺍ ﻤـﻊ
ﺍﻋﺘﺒﺎﺭ ﺃﻨﻪ ﻻ ﺘﻭﺠﺩ ﺤﻠﻭل ﺩﻗﻴﻘﺔ ﻟﻤﻌﻅﻡ ﺃﺸﻜﺎل ﺍﻟﺠﻬﺩ ﻫﻭ ﻋـﻥ ﻁﺭﻴﻘـﺔ ﺘﻘﺭﻴـﺏ
) Wentzel- kreners – Birllouin (WKBﺒﻤﻼﺤﻅﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ 2
− 4ka e
4ku ≈ 2 2 K +u
2
T
ﻭﺍﻟﺘﻲ ﺘﺘﻜﻭﻥ ﻤﻥ ﺤﺩﻴﻥ ،ﻓﺈﻥ ﺍﻟﺤﺩ ﺍﻟﺜﺎﻨﻲ ﻫﻭ ﺍﻟﻤﻬﻴﻤﻥ ﻭﺍﻟﻐﺎﻟـﺏ .ﻭﺒﻜﺘﺎﺒـﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻤﻊ ﺍﻟﺸﻜل .itsholdbe4
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)2 (ka) (ua = 2u (2a) + 2lm 123 (ka) 2 + (ua) 2
2
lm T
ﺘﺤﺕ ﻜل ﺍﻟﺤﻴﺜﻴﺎﺕ ﻓﺈﻥ ﺍﻟﺤﺩ ﺍﻷﻭل ﻫﻭ ﺍﻟﻐﺎﻟﺏ )ﺍﻟﻤﻬﻴﻤﻥ( ﻋﻠﻰ ﺍﻟﺤﺩ ﺍﻟﺜﺎﻨﻲ
ﻷﻱ ﻗﻴﻤﺔ Kaﻭﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺘﻲ ﺴﻨﺘﺒﻌﻬﺎ ﻫﻲ ﻤﻌﺎﻟﺠﺔ ﺤﺎﺠﺯ ﻤﻨﺤﻨﻰ ﻜﺤﻭﺍﺠﺯ ﻤﺭﺒﻌﺔ ﻤﺘﺠﺎﻭﺭﺓ uxtapositionﻭﺒﻤﺎ ﺃﻥ ﻤﻌﺎﻤﻼﺕ ﺍﻟﻨﻔﺎﺫﻴﺔ ﻤﻀﺭﻭﺒﺔ ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ ﺼﻐﻴﺭﺓ
ﻭﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ ﻤﻌﻅﻡ ﺍﻟﻔﻴﺽ ﻤﻨﻌﻜﺱ ،ﻓﺈﻥ ﺍﻟﻨﻔﺎﺫﻴﺔ ﻀﺭﺏ ﻜـل ﺸـﺭﻴﺤﺔ ﺘﻜـﻭﻥ ﻤﺴﺘﻘﻠﺔ ﻓﺈﻥ ﻫﺫﺍ ﻗﻠﻴل ﺍﻻﺤﺘﻤﺎل.
∑ lm T > ~ − 2∑ Δx < k 2
2m ] [V (x) − E h2
ﺤﻴﺙ
∆x
ﻫﻭ ﺍﻟﻌﺭﺽ
u
~
dx
∫
2
lm T
~−2
barier
ﻫﻭ ﻤﺘﻭﺴﻁ ﻗﻴﻤﺔ kﻟﺫﻟﻙ ﺍﻟﺤﺎﺠﺯ.
ﻻﺤﻅ ﺃﻥ ﻫﺫﺍ ﺍﻟﺘﻌﺒﻴﺭ ﻓﻴﻪ ﺘﻘﺭﻴﺏ ﻜﺒﻴﺭ ﻟﺴﺒﺒﻴﻥ ﺨﺎﺼﺔ ﻋﻨﺩ ﻨﻘﻁﺔ ﺍﻻﻨﻌﻁﺎﻑ
ﺤﻴﺙ ﺍﻟﻁﺎﻗﺔ ﻭﺍﻟﺠﻬﺩ ﻴﺘﺴﺎﻭﻴﺎﻥ .ﻭﻤﻥ ﺍﻟﻤﻬﻡ ﺃﻥ ﻴﻜﻭﻥ
)V (x
ﻫﻲ ﺩﺍﻟﺔ ﻤﺘﻐﻴﺭﺓ ﺒﺒﻁﺊ
ﻓﻲ xﻭﺇﻻ ﻓﺈﻥ ﺘﻘﺭﻴﺏ ﺍﻟﺠﻬﺩ ﺍﻟﻤﻨﺤﻨﻰ ﻤﺭﺒﻌﺎﺕ ﻤﺘﺭﺍﺼﺔ ﻴﻜﻭﻥ ﻤﻤﻜﻨﻨﺎﹰ ﻟﻭ ﻜﺎﻨـﺕ ﻫﺫﻩ ﺍﻟﻤﺭﺒﻌﺎﺕ ﻀﻴﻘﺔ ﺠﺩﺍﹰ ﻭﻋﻨﺩﻫﺎ ﻜﺫﻟﻙ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺠﻴﺩ .ﻭﻟﻜﻥ ﻤﻥ ﺍﻟﻤﻨﺎﺴﺏ ﻜﺘﺎﺒﺔ. )dx (2m/h 2 ) 2 (V (x) − E
−2 ∫ ~e
2
T
4ku ~
2
T
ﻫﻭ ﺘﻘﺭﻴﺏ ﻏﻴﺭ
ﺤﻴﺙ ﺃﻥ ﺍﻟﺘﻜﺎﻤل ﻴﻜﻭﻥ ﻋﻠﻰ ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﺘﻲ ﻓﻴﻬﺎ ﻴﻜـﻭﻥ
ﺍﻟﺠﺫﺭ ﺍﻟﺘﺭﺒﻴﻌﻲ ﺤﻘﻴﻘﻲ. ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﻨﻔﻘﻲ Tunneling Phenomena
ﺘﻌﺘﺒﺭ ﻫﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﺸﺎﺌﻌﺔ ﻓﻲ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺫﺭﻴﺔ ﻭﺍﻟﻨﻭﻭﻴﺔ ﻭﺴـﻭﻑ ﻨﻨـﺎﻗﺵ ﺃﻤﺜﻠﺔ. ﺍﻻﻨﺒﻌﺎﺙ )ﺍﻻﻟﻜﺘﺭﻭﻨﻲ( ﺍﻟﺒﺎﺭﺩ .Cold Emission - ١٠٨PDF created with pdfFactory Pro trial version www.pdffactory.com
ﺨﺫ ﺍﻟﻜﺘﺭﻭﻨﺎﺕ ﻓﻲ ﻤﻌﺩﻥ .ﺇﻥ ﻫﺫﻩ ﺍﻻﻟﻜﺘﺭﻭﻨﺎﺕ ﻤﻘﻴﺩﺓ ﺩﺍﺨل ﺍﻟﻤﻌﺩﻥ ﺒﻭﺍﺴﻁﺔ ﺠﻬﺩ ﻭﺍﻟﺫﻱ ﻜﺘﻘﺭﻴﺏ ﺃﻭﻟﻲ ،ﻴﻤﻜﻥ ﻭﺼﻔﻪ ﺒﺼﻨﺩﻭﻕ ﺫﻭ ﻋﻤﻕ ﻤﺤﺩﺩ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ
ﺒﺎﻟﺸﻜل.
ﻻﺤﻅ ﺃﻥ ﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﻁﺎﻗﺔ ﻟﻼﻟﻜﺘﺭﻭﻨﺎﺕ ﻗﺭﻴﺒﺔ ﺠﺩﺍﹰ ﻤﻥ ﺒﻌﻀﻬﺎ ﻭﺫﻟـﻙ ﻷﻥ
ﺍﻟﺼﻨﺩﻭﻕ ﻋﺭﻴﺽ ﺠﺩﺍﹰ.
ﻤﻥ ﺨﻭﺍﺽ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ )ﻭﺍﻟﺘﻲ ﺘﻭﺼﻑ ﺒﻤﺒﺩﺃ ﺒﺎﻟﻭﻟﻲ( ﺃﻨﻬﺎ ﺘﺘﺭﺘـﺏ ﻓـﻲ ﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﻁﺎﻗﺔ ﺒﺤﻴﺙ ﺃﻨﻪ ﻻ ﻴﻭﺠﺩ ﺃﻜﺜﺭ ﻤﻥ ﺍﻟﻜﺘﺭﻭﻨﻴﻥ ﻴﻤﻜﻥ ﺃﻥ ﻴﺸﻐﻠﻬﺎ ﻤﺴﺘﻭﻯ
ﻭﺍﺤﺩ ﻤﻥ ﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﻁﺎﻗﺔ .ﻭﻫﻜﺫﺍ ﻓﺄﻗل ﺤﺎﻟﺔ ﻟﻠﻁﺎﻗﺔ ﻓﻲ ﺍﻟﻤﻌﺩﻥ ،ﻜل ﺍﻟﻤﺴﺘﻭﻴﺎﺕ ﺇﻟﻰ ﺤﺩ ﻤﺴﺘﻭﻯ ﻁﺎﻗﺔ ﻴﺴﻤﻰ ﻤﺴﺘﻭﻯ ﻁﺎﻗﺔ ﻓﻴﺭﺴﻲ )ﻭﺍﻟﺘﻲ ﺘﻌﺘﻤـﺩ ﻋﻠـﻰ ﻜﺜﺎﻓـﺔ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ( ﺘﻜﻭﻥ ﻤﻌﺒﺄﺓ .ﻭﻋﻨﺩ ﺩﺭﺠﺔ ﺤﺭﺍﺭﺓ ﺃﻋﻠﻰ ﻤﻥ ﺍﻟﺼﻔﺭ ﺍﻟﻤﻁﻠـﻕ ﻓـﺈﻥ
ﺒﻌﺽ ﻫﺫﻩ ﺍﻻﻟﻜﺘﺭﻭﻨﺎﺕ ﺘﻜﻭﻥ ﺤﺭﺍﺭﻴﺔ ﻤﻬﻴﺠﺔ ﻟﻤﺴﺘﻭﻴﺎﺕ .ﺍﻟﻔﺭﻕ ﺒـﻴﻥ ﻤـﺴﺘﻭﻯ
ﻁﺎﻗﺔ ﻫﻭ ﺍﻟﺫﻱ ﻴﻠﺯﻡ ﻻﺤﻀﺎﺭ ﺍﻟﻜﺘﺭﻭﻥ ﻟﻠﺨﺎﺭﺝ ،ﺩﺍﻟﺔ ﺍﻟﺸﻐل .ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻴﻤﻜـﻥ ﺇﺯﺍﻟﺘﻬﺎ ﺒﺈﻋﻁﺎﺌﻬﺎ ﺒﺎﻟﻔﻭﺘﻭﻨﺎﺕ )ﻜﻤﺎ ﻓﻲ ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﻜﻬﺭﻭﻀﻭﺌﻲ( ﻭﺇﻤﺎ ﺒﺎﻟﺤﺭﺍﺭﺓ .ﻜﺫﻟﻙ
ﻴﻤﻜﻨﻨﺎ ﺇﺯﺍﻟﺘﻬﺎ ﺒﺎﻟﺘﺄﺜﻴﺭ ﻋﻠﻴﻬﺎ ﺒﻤﺠﺎل ﻜﻬﺭﺒﺎﺌﻲ ﺨﺎﺭﺠﻲ .ﺇﻥ ﺍﻻﻨﺒﻌﺎﺙ ﺍﻹﻟﻜﺘﺭﻭﻨـﻲ
ﺍﻟﺒﺎﺭﺩ ﻴﺤﺩﺙ ﻨﺘﻴﺠﺔ ﺘﺄﺜﻴﺭ ﻤﺠﺎل ﻜﻬﺭﺒﺎﺌﻲ ﺨﺎﺭﺠﻲ ﻭﺍﻟﺫﻱ ﺒﺩﻭﺭﻩ ﻴﻐﻴﺭ ﺍﻟﺠﻬﺩ ﺍﻟﺫﻱ
ﻴﺭﻯ ﺒﺎﻟﻜﺘﺭﻭﻨﺎﺕ ﺃﻱ ﻴﺘﻐﻴﺭ ﺍﻟﺠﻬﺩ
ω −eε x ←W
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻓﺈﻥ ﻫﺫﺍ ﻫﻭ ﺍﻟﺠﻬﺩ
ﺍﻟﺫﻱ ﺘﺭﺍﻩ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻭﻫﺫﺍ ﻴﻜـﺎﻓﺊ ﺍﻟﻔـﺭﻕ }) {V (x) − Eﺍﻟـﺸﻜل (bﻓﻠـﻭ ﺃﻥ ﺍﻻﻟﻜﺘﺭﻭﻥ ﻋﻨﺩ ﺍﻟﻘﻤﺔ ﻟﻠﺒﺤﺭ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻴﺎﺕ ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻓﺈﻥ ﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ. - ١٠٩ -
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a
2m ) ( V−eε x h2
2
]οw/eε
2
3
3
)(A + Bx = 3 B 2
(ω − eεεx 3 − eε 2
2
1
2
2m = 2 h
dx
∫
−2
0
1
2
T ~e
)dx (A + Bx
2
1
∫
2m dx 2 ( w − eεεx h
W/eε
∫ ο
3
2m 2w 2 h 3eε 4 2mw w ) ( 3 h2 eε
−
−2
=
⇒ T =e =e 2
ﻭﻫﺫﺍ ﻫﻭ ﻗﺎﻨﻭﻥ Fowler – Nordheim ﻭﻤﻥ ﺘﻁﺒﻴﻘﺎﺕ ﺍﻻﻨﺒﻌﺎﺙ ﺍﻟﺒﺎﺭﺩ ﻫﻭ Scanning Tunneling Micros cepe
STMﻭﻫﺫﻩ ﺍﻷﺩﺍﺓ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺤﺴﺎﺴﻴﺔ ﺍﻟﻌﺎﻟﻴﺔ ﺠﺩﺍﹰ ﻟﻠﻤﺴﺎﻓﺔ ﻓﻲ ﺍﻻﻨﺒﻌﺎﺙ ﺍﻟﺒﺎﺭﺩ.
ﺇﻥ ﺘﻴﺎﺭ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺍﻟﻤﺤﺙ ﺒﺠﻬﺩ ﻜﻬﺭﺒﺎﺌﻲ ﺨﺎﺭﺠﻲ ﻟﻠﻔﺭﻕ ﻓﻲ ﺍﻟﺠﻬـﺩ ﺴـﻁﺢ ﺍﻟﻤﻌﺩﻥ ﻭﻨﻬﺎﻴﺔ ﺇﺒﺭﺓ ﺤﺎﺩﺓ ﻓﻭﻕ ﺴﻁﺢ ﻴﺘﻐﻴﺭ ﺃﺴﻴﺎﹰ ﺒﺎﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﺍﻹﺒـﺭﺓ ﻭﺍﻟـﺴﻁﺢ
ﻭﻫﺫﺍ ﻴﺴﻤﺢ ﺒﺎﻟﺤﺼﻭل ﻋﻠﻰ ﻤﻌﻠﻭﻤﺎﺕ ﻤﻨﻔﺼﻠﺔ ﻋﻥ ﺘﻭﺒﻭﺠﺭﺍﻓﻴﺔ ﺍﻟﺴﻁﺢ
ﻭﻫﺫﺍ ﻗﺩ ﺘﻡ ﺍﺴﺘﺨﺩﺍﻡ STMﻟﺩﺭﺍﺴﺔ ﺃﺴﻁﺢ ﺍﻟﻤﻌﺎﺩﻥ ﻭﺒﻌﺽ ﺍﻟﻤﻭﺼﻼﺕ ﻭﻗﺩ
ﺘﻡ ﺤﺩﻴﺜﺎﹰ ﻨﻘل ﺫﺭﺍﺕ ﻭﺠﺯﻴﺌﺎﺕ ﻋﻠﻰ ﺍﻷﺴﻁﺢ ﻭﺫﻟﻙ ﺒﺘﺨﺯﻴﻨﻬﺎ.
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ﺍﻟﺘﺄﺜﻴﺭ ﺍﻟﻨﻔﻘﻲ ﻓﻲ ﺍﻟﻤﻭﺍﺩ ﻓﺎﺌﻘﺔ ﺍﻟﺘﻭﺼﻴل .Tunneling in superconductors ﺍﻟﻤﺘﺫﺒﺫﺏ ﺍﻟﺘﻭﺍﻓﻘﻲ ﺍﻟﺒﺴﻴﻁ
The Harmonic Oscillator
ﻓﻲ ﻫﺫﺍ ﺍﻟﻤﺜﺎل ﺴﻨﺠﺩ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ﺍﻟﻤﻁﻠﻭﺏ ﺤﻠﻬﺎ ﺘﺨﺘﻠـﻑ ﻋـﻥ
ﺍﻷﻤﺜﻠﺔ ﺍﻟﺴﺎﺒﻘﺔ ﻭﺃﺤﺩ ﺃﺴﺒﺎﺏ ﻤﻨﺎﻗﺸﺔ ﻫﺫﻩ ﺍﻟﻤﺴﺄﻟﺔ ﻫﻭ ﺘﻌﻠﻡ ﺤل ﻤﺜل ﻫﺫﻩ ﺍﻟﻤﺴﺎﺌل. ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ ﺍﻟﺘﻘﻠﻴﺩﻱ ﻟﻪ ﺍﻟﺸﻜل. p2 1 2 + kx 2m 2 = k. E = p.E
=H
)(1
ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ. h 2 d 2 u (x) 1 + k x h 2 (x) E.u 2 2m dx 2
)(2
) (3ﺒﺘﻌﺭﻴﻑ ﺘﺭﺩﺩ ﺍﻟﻤﺘﺫﺒﺫﺏ ﻭﻜﺘﺎﺒﺔ ﺍﻟﻁﺎﻗﺔ ﻋﻠﻰ ﺍﻟﺸﻜل 2E hω
)(4
) (5ﻭﺘﻐﻴﻴﺭ ﺍﻟﻤﺘﻐﻴﺭﺍﺕ
)(6
mω x h
k m
= k = mω 2 ⇐ ω
=∈
= y
d 2 u ( x) m2ω 2 2 2mE − x u + 2 u=0 2 2 h h dx 2 d u ( x) mω mω 2 2 E − x u =0 h 12 { h3 hω dx 2 2 ← y 2 2 1 d u d u + (∈ − y 2 ) u = 0 ⇒ 2 + (∈− y 2 ) u = 0 2 mω dx dy h
ﻜل ﺍﻟﻜﻤﻴﺎﺕ ﻫﻨﺎ ﺒﺩﻭﻥ ﻭﺤﺩﺍﺕ. ﺤﻠﻭل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻲ ﺍﻟﺩﻭﺍل ﺍﻟﺫﺍﺌﺒﺔ. 2
)H n ( y
− y3
e
2n n D
= un
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ﺤﻴﺙ
ﺘﻌﺭﻑ ﺒﻜﺜﻴﺭﺍﺕ ﻤﺤﺩﻭﺩﺓ.
)H n ( y
H 0 (y) = 1 H1 (y) = 2 y H 2 (y) = 4 y 2 − 2 H 3 (y) = 8 y 2
ﺍﻟﻨﻅﺭ ﻟﻠﺤﻠﻭل ﻋﻨﺩ ﻤﺎ ﻻ ﻨﻬﺎﻴﺔ )ﺃﻭﻻﹰ( ﺜﻡ ﻋﻨﺩ ﺍﻟﺼﻔﺭ. ﻭﻷﻱ ﻗﻴﻤﺔ ﺨﺎﺼﺔ )ﺫﺍﺘﻴﺔ( ،E. eijenvalneﻋﻨﺩﻤﺎ
∞ → y2
ﻓﺈﻥ ﺍﻟﺤﺩ ﺍﻟـﺫﻱ
ﻴﺘﻀﻤﻥ Eﻴﻌﺒﺭ ﻤﻬﻤﻭل .ﻭﻟﻬﺫﺍ ﺍﻟﺴﺒﺏ ﻴﺘﻁﻠﺏ ﺃﻥ ) asymptotically u (yﺘﺤﻘـﻕ ﺍﻟﻤﻌﺎﺩﻟﺔ.
− y 2 u ο (y) = 0
)(7
ﺒﺎﻟﻀﺭﺏ ﻓﻲ
du ο dy
2
)d 2 u ο (y d y2
ﻴﻤﻜﻨﻨﺎ ﻤﻥ ﻜﺘﺎﺒﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺒﺎﻟﺸﻜل. d d duο 2 2 ) − y2 (u 0 ) = 0 ( dy dy dy
)(8
ﺃﻭ ﻤﺎ ﻴﻜﺎﻓﺅﻫﺎ. d d uο 2 2 2 ( ) − y 2u 0 = − 2 y2u 0 dy dy
)(9
ﻭﻟﻭ ﺃﻫﻤﻠﻨﺎ ﺍﻟﺤﺩ ﻋﻠﻰ ﺍﻟﺠﺎﻨﺏ ﺍﻷﻴﻤﻥ ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻓﺈﻨﻪ ﺴﻴﺴﺭ ﺍﻟﺤل ﻭﺒﺘﻜﺎﻤل
ﺍﻟﻤﻌﺎﺩﻟﺔ.
du d d uο 2 2 2 ) − y2u0 = 0⇒ ( ο )2 − y2u0 = C ( dy dy dy 2
ﺒﻤﺎ ﺃﻥ ﻜﻼﹰ ﻤﻥ
), uο ( y
d uο dy
1
) = ( C + y 2 uο 2
d uο dy
ﻴﺠﺏ ﺃﻥ ﺘﺘﻼﺸﻰ ﻋﻥ ∞ ،ﻓﺈﻥ ﻫﺫﺍ ﻴﺤـﺘﻡ ﺃﻥ
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= ± ∫ y dy
)(10
d uο uο
∫
= ± yuο
d uο dy
ﺍﻟﺤﻠﻭل ﺍﻟﻤﺘﺒﺩﻟﺔ ﻋﻨﺩ ∞ ،ﻫﻲ. 2
ﺒﺈﻤﻜﺎﻨﻨﺎ ﺍﻵﻥ ﺍﺨﺘﺒﺎﺭ ﺃﻥ
2
−y2
U ο (y) = e
= 2ye − y
2y U ο2
2 d 2 − y2 (y e ) ~ − 4 y 3 e − y dy
=
ﻜﻤﻴﺔ ﻤﻬﻤﻠﺔ ﻤﻘﺎﺭﻨﺔ ﺒـ d ) (y 2 U 2ο dy
ﻭﺫﻟﻙ ﺒﺎﻋﺘﺒﺎﺭ ﻗﻴﻤﺔ yﻜﺒﻴﺭﺓ. ﺍﻵﻥ ﺴﻨﺴﺘﺤﺩﺙ ﺩﺍﻟﺔ ﺠﺩﻴﺩﺓ
)h (y
ﺒﺤﻴﺙ ﺃﻥ 2
).......... ........(1 2
2
U ( y ) = h (y) e − y
ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ﺘﺼﺒﺢ. )dh (y )d 2 h (y − 2y + (∈ − 1) h (y) = 0 2 dy dy
).............(13
ﻭﺸﻜل ﺍﻟﻤﻌﺎﺩﻟﺔ ﻻ ﻴﺒﺩﻭ ﺒﺴﻴﻁﺎﹰ ،ﻭﻟﻜﻨﻨﺎ ﻨﺎﻗﺸﻨﺎ ﺍﻟﺴﻠﻭﻙ ﻋﻨﺩ ﻤـﺎ ﻻ ﻨﻬﺎﻴـﺔ، ﻭﺍﻵﻥ ﻨﺭﻴﺩ ﺃﻥ ﻨﻨﻅﺭ ﻟﻠﺴﻠﻭﻙ ﻋﻨﺩ .y = 0ﺩﻋﻨﺎ ﻨﺤﺎﻭل ﻜﺘﺎﺒﺔ ﺍﻟﺤل ﻋﻠـﻰ ﺸـﻜل ﻤﻔﻜﻭﻙ ﺴﻠﺴﻠﺔ ﺍﻟﻘﻭﺓ.Power series expansion .
∞
h (y ) = ∑ a m y m
)..................(14
m =0
a m ym = 0
∞
∑
m =0
m−1
)+ (∈ −1
∞
− 2y∑ m a m y m =0 14 4244 3
m-2
∞
m (m - 1) a m y
∑
m =0
∞ m ∑ m am y m =0
ﺒﺄﺨﺫ ﻤﻌﺎﻤﻼﺕ
ym
ﻓﻘﻁ. (m + 2) (m + 1) a m+ 2 − 2 (m + 1) a m + (∈ −1) a m = 0
).......... ..... (15
(m + 2) (m + 1) a m+ 2 = ( 2m -∈ +1) a m
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ﻭﻫﺫﻩ ﻫﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻻﺴﺘﻁﺭﺍﺩ ﻟﻠﺤﺼﻭل ﻋﻠﻰ
am
ﻭﻟﻘﻴﻡ ﺍﺨﺘﻴﺎﺭﻴﺔ ∈ ،ﻨﺠﺩ ﺃﻥ ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ mﻜﺒﻴﺭﺓ 2 am m
ﺇﺫﺍ ﻋﺭﻓﻨﺎ
, a0
1
.a
) (m > N
~ am + 2
ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺤﻠﻭل ﺘﻘﺭﻴﺒﺎﹰ )ﺒﺎﺨﺘﻴﺎﺭ ﺍﻟﺤﻠﻭل ﺍﻟﺯﻭﺠﻴﺔ( .ﻭﻫﺫﺍ )h ( y ) = (atolyromialiny
2 22 23 + a N y N + y N+2 + yN+4 + y N + 6 + .......... N ) N (N + 2 ) N (N + 2 ) (N + 4
ﻭﺒﺎﻟﺘﺒﺴﻴﻁ N N −1 +1 N 2 2 2 2 2 2 N (y ) (y ) (y ) + + + .......... a N y 2 ( −1 )! N N 2 ! ) ( N − 1 !) ( !)( + 1 2 2 2
ﺒﺎﺨﺘﻴﺎﺭ
N = 2k
)ﻟﻠﻤﻼﺌﻤﺔ( ﻓﺈﻥ ﺍﻟﻤﺘﺴﻠﺴﻠﺔ ﺘﺄﺨﺫ ﺍﻟﺸﻜل.
(y 2 ) k − 1 (y 2 ) k (y 2 ) k + 1 y 2 (k − 1 )! + + + .......... !k ! )(k + 1 ! ) (k − 1
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2 (y 2 ) k −1 (y 2 ) 2 = y 2 (k − 1) ! e y − {1 + y 2 + + .................. + (k − 2)! !2
ﻭﻫﺫﺍ ﺸﻜل ﻤﺘﻌﺩﺩ ﺍﻟﺤﺩﻭﺩ +ﺜﺎﺒﺕ x
y2 ey
ﻭﺒﺎﻟﺘﻌﻭﻴﺽ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ
2
−y2
ﻓﺈﻨﻨﺎ ﺴﻨﺘﺤﺼل ﻋﻠﻰ ﺤل ﻻ ﻴﺘﻼﺸﻰ ﻋﻨﺩ
4 (y ) = h (y ) e ∞
ﻟﻭ ﺃﻥ ﻤﻌﺎﺩﻟﺔ ﺍﻻﺴﺘﻁﺭﺍﺩ ﺘﻨﺘﻬﻲ ،ﺃﻱ ﺃﻥ ﻟﻭ ﺃﻥ.
ﻭﺍﻟﺤل ﺍﻟﻤﻘﺒﻭل ﻴﻤﻜﻥ ﺃﻥ ﻴﻭﺠﺩ
( N + 1 ) ( N + 2 ) a n + 2 = 2N − ∈+ 1 a n = 0 1424 3 =0
⇒ ∈ = 2N + 1 2E = 2N + 1 hw
)(17
ﻭﺍﻟﻨﺘﺎﺌﺞ ﻫﻲ: -1ﺃﻨﻪ ﻴﻭﺠﺩ ﻗﻴﻡ ﺫﺍﺘﻴﺔ ﻤﻨﻔﺼﻠﺔ )ﻏﻴﺭ ﻤﺘﺼﻠﺔ( ﻤﺘﺴﺎﻭﻴﺔ ﺍﻟﻤﺴﺎﻓﺎﺕ ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟـﺔ
) (4ﻭﺍﻟﻤﻌﺎﺩﻟﺔ ) (17ﻴﻤﻜﻨﻨﺎ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ. 1 1 hw (2m + 1) = (n + ) h w 2 2
)(18
=E
ﻭﻫﺫﻩ ﻗﺭﻴﺒﺔ ﻤﻥ ﻤﻌﺎﺩﻟﺔ Planckﻟﻺﺸﻌﺎﻉ. 1 E = (n + ) hw 2
-2ﻤﺎ ﻋﺩﺍ ﺜﻭﺍﺒﺕ ﺍﻟﻤﻌﺎﻴﺭﺓ ،ﻓﺈﻥ ﻤﺘﻌﺩﺩ ﺍﻟﺤﺩﻭﺩ ﻫﻭ ﻤﺘﻌﺩﺩ ﺍﻟﺤﺩﻭﺩ
ﺍﻟﻬﻴﺭﻤﻴﺘﻲ)Hn (y
ﻭﺍﻟﺫﻱ ﻴﻤﻜﻥ ﺇﻴﺠﺎﺩ ﺨﻭﺍﺼﻪ ﺒﺴﻬﻭﻟﺔ ﻓﻲ ﺃﻱ ﻜﺘﺎﺏ ﻓﻲ ﺍﻟﻔﻴﺯﻴﺎﺀ ﺍﻟﺭﻴﺎﻀﻴﺔ. ﻭﻟﻭ ﻗﺼﺭﻨﺎ ﺃﻨﻔﺴﻨﺎ ﻟﻠﺨﻭﺍﺹ ﺍﻟﺘﺎﻟﻴﺔ ﻤﻨﻬﺎ: )Hn (y
ﺘﺤﻘﻕ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ. ﺘﺤﺕ ﺸﺭﻁ
∈ = 2 n +1
+ 2nH n (y) = 0
)dH n (y dy
− 2y
) d 2 H n (y 2
dy
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ecussionedations ﻭﻫﺫﻩ ﺘﺤﻘﻕ ﺍﻟﻌﻼﻗﺎﺕ ﺍﻻﺴﺘﻁﺭﺍﺩﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ H n +1 − 2y H n + 2n H n −1 = 0 H n +1 +
∑ n
dH n dy
− 2y H n + 2n H n = 0
ﻭﻜﺫﻟﻙ
2 Zn H n (y) = e 2zy − z n!
H n (y) = (−1) n e y
2
ﻭ
d n − y2 e dy n
ﻭﺍﻟﺘﻁﺒﻴﻊ ﻟﻜﺜﻴﺭﺍﺕ ﺤﺩﻭﺩ ﻫﻴﺭﻤﻴﺕ ﻴﻜﻭﻥ ﺒﺤﻴﺙ ∞
∫
2
dy e − y H n (y) 2 = d n n! D
−∞
.ﻫﺫﻩ ﻗﺎﺌﻤﺔ ﺒﺒﻌﺽ ﻋﺩﻴﺩﺍﺕ ﺤﺩﻭﺩ ﻫﻴﺭﻤﻴﺕ H 0 (y) = 1 H1 (y) = 2 y H 2 (y) = 4 y 2 − 2 H 3 (y) = 8 y 2 − 12 y H 4 (y) = 16 y 4 − 48 y 2 + 12 H 5 (y) = 32 y 5 − 160 y 3 + 129 y
(ﻭﻤﻌﺎﺩﻻﺕ ﺍﻟﻘﻴﻡ ﺍﻟﺫﺍﺘﻴﺔ )ﺍﻟﺨﺎﺼﺔ d2u n dx
2
=
mk 2 2mEn x un − un 2 h h2 و
d2u* e mk 2mEn * u e = 2 x 2 u* e − 2 dx h h2
ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ،ﺒﺎﻟﻤﻌﺎﺩﻟﺘﻴﻥ ﺍﻟﺴﺎﺒﻘﺘﻴﻥ ﻋﻠﻰ ﺍﻟﺘﻭﺍﻟﻲ
un
ﻭ
u*e
ﻭﻋﻨﺩ ﻀﺭﺏ .ﺘﻁﺭﺡ ﻤﻥ ﺍﻷﻭﻟﻰ
* 2m d * dun du x ux − u x = 2 (E x − E n ) u *x u n h dx dx dx
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ﻭﺒﺘﻜﺎﻤل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻋﻠﻰ xﻤﻥ
∞−
∞+
←
ﻓﺈﻥ ﺍﻟﺠﺎﻨـﺏ ﺍﻷﻴـﺴﺭ
ﻴﺘﻼﺸﻰ ،ﺒﻤﺎ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺨﺎﺼﺔ ﻭﺘﻔﺎﻀﻼﺘﻬﺎ ﺘﺘﻼﺸﻰ ﻋﻨﺩ dx u *e (x) u n (x) = 0
∞
∫
) − En
x
∞ x= ±
ﻭﻫﻜﺫﺍ.
(E
∞−
ﻭﺍﻟﺫﻱ ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺍﻟﺘﻲ ﻟﻬﺎ -3ﺍﻟﻤﻌﺎﺩﻟﺔ
1 ) E = h w (n + 2
Ee ± En
ﻫﻲ ﻤﺘﻌﺎﻤﺩﺓ.
ﺘﻭﻀﺢ ﺃﻥ ﺃﻗل ﺤﺎﻟﺔ ﻟﻬﺎ ﻁﺎﻗﺔ ﻭﻫﺫﻩ ﻫﻲ ﻁﺎﻗﺔ ﻨﻘﻁـﺔ
ﺍﻟﺼﻔﺭ .Zero – point evergy ﺇﻥ ﻭﺠﻭﺩﻫﺎ ﻫﻭ ﺘﺄﺜﻴﺭ ﻜﻤﻲ ﺒﺤﺕ ،ﻭﻴﻤﻜﻥ ﺘﺄﻭﻴﻠﻪ ﺒﺩﻻﻟﺔ ﻤﺒﺩﺃ ﺍﻟﻼﺘﻌﻴﻴﻥ .ﺇﻨﻬﺎ ﻁﺎﻗﺔ ﻨﻘﻁﺔ ﺍﻟﺼﻔﺭ ﺍﻟﻤﺴﺌﻭﻟﺔ ﻋﻥ ﺍﻟﺤﻘﻴﻘﺔ ﺃﻥ ﺍﻟﻬﻴﻠﻴﻭﻡ ﻻ ﻴﺘﺠﻤـﺩ ﻋﻨـﺩ ﺩﺭﺠـﺎﺕ ﺍﻟﺤﺭﺍﺭﺓ ﺍﻟﻘﻠﻴﻠﺔ ،ﻭﻟﻜﻨﻪ ﻴﺒﻘﻰ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺴﺎﺌﻠﺔ ﺤﺘﻰ ﺩﺭﺠﺎﺕ ﺤـﺭﺍﺭﺓ ﻤﻘـﺩﺍﺭﻫﺎ 10-3kﻭﺫﻟﻙ ﻋﻨﺩ ﺍﻟﻀﻐﻁ ﺍﻟﻌﺎﺩﻱ .ﺍﻟﺘﺭﺩﺩ ﻟﻬﺎ ﻴﻜﻭﻥ ﻜﺒﻴﺭﺍﹰ ﻟﻠﺫﺭﺍﺕ ﺍﻟﺨﻔﻴﻔﺔ .ﻭﺍﻟﺫﻱ
ﻫﻭ ﺍﻟﺴﺒﺏ ﻓﻲ ﺃﻥ ﺍﻟﻨﻴﺘﺭﻭﺠﻴﻥ ﻻ ﻴﻅﻬﺭ ﻫﺫﺍ ﺍﻟﺘﺄﺜﻴﺭ ﻤﺜل ﺍﻟﻬﻴﻠﻴﻭﻡ .ﺇﻨﻪ ﻜﺫﻟﻙ ﻴﻌﺘﻤﺩ
ﻋﻠﻰ ﺍﻟﺨﻭﺍﺹ ﺍﻟﺘﻔﻀﻴﻠﻴﺔ ﻟﻠﻘﻭﻯ ﺍﻟﻔﺎﻋﻠﺔ ﺒﻴﻥ ﺍﻟﺠﺯﻴﺌﺎﺕ ﻭﺍﻟﺘﻲ ﺘﺸﺭﺡ ﺍﻟﺴﺒﺏ ﻓﻲ ﺃﻥ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﻴﺘﺠﻤﺩ.
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ﺍﻟﺒﻨﺎﺀ ﺍﻟﻌﺎﻡ ﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ )ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻤﻭﺠﻴﺔ(: ﺴﻨﺘﻌﺭﺽ ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﻟﺩﺭﺍﺴﺔ ﺍﻟﻤﺒﺎﺩﺉ ﺍﻷﺴﺎﺴﻴﺔ ﺍﻟﺘﻲ ﺫﻜﺭﺕ ﺁﻨﻔﺎﹰ ﻓـﻲ
ﺴﻴﺎﻕ ﺤل ﺒﻌﺽ ﺍﻟﻤﺴﺎﺌل ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﻤﻥ ﻫﺫﻩ ﻨﻅﺭﻴﺔ ﺍﻟﻤﻔﻜﻭﻙ ﻭﻤﻌﻨﺎﻫﺎ ﺍﻟﻔﻴﺯﻴـﺎﺌﻲ ﻓﻲ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻤﺘﺠﻬﺎﺕ ﺍﻟﺤﺎﻟﺔ ،ﺍﻟﺘﻜﺭﺍﺭ ﻭﺍﻟﻨﻬﺎﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ .ﻭﺴﻨﻘﺩﻡ ﺘﺭﻤﻴﺯ ﺩﻴـﺭﺍﻙ
ﺒﺎﺨﺘﺼﺎﺭ.
ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻭﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔ: "ﻤﺅﺜﺭ ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ" ﺇﻥ ﺤﺎﻟﺔ ﺃﻱ ﻨﻅﺎﻡ ﻓﻴﺯﻴﺎﺌﻲ ﻴﻭﺼﻑ ﺒﺎﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻭﺍﻟﺘﻲ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﻜل
ﺍﻟﻤﻌﻠﻭﻤﺎﺕ ﻋﻥ ﻫﺫﺍ ﺍﻟﻨﻅﺎﻡ .ﺇﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ،ﻭﺘﻁﻭﺭﻫﺎ ﻤـﻊ
ﺍﻟﺯﻤﻥ ﻴﻌﻁﻲ ﺒـ:
)(1
∂ )ψ (x, t) = H ψ (x, t xt
ih
ﺇﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺁﺜﺭﻨﺎ ﻋﻠﻴﺎ ﺒﻤﺅﺜﺭ ،Hﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ ﻭﺍﻟـﺫﻱ ﻴﻠﻌـﺏ ﺩﻭﺭﺍﹰ
ﻜﺒﻴﺭﺍﹰ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ .ﺍﻟﻤﺅﺜﺭ Hﻟﻨﻅﺎﻡ ﻤﻥ ﺠﺴﻡ ﻭﺍﺤﺩ ﻓﻲ ﺠﻬﺩ Vﻟﻪ ﺍﻟﺸﻜل. )+ V (x
p sp2 2m
=H
ﻭﺇﺫﺍ ﻜﺎﻨﺕ ) V (xﻟﻴﺱ ﻟﻬﺎ ﺍﻋﺘﻤﺎﺩ ﻭﺍﻀﺢ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﻓﺈﻥ ﺍﻟﻤﻌﺎﺩﻟـﺔ )(١
ﻴﻤﻜﻥ ﺤﻠﻬﺎ ﺒـ:
ψ (x, t) = u E (x) e − Et/h
ﺤﻴﺙ: )H u E (x) = E u E (x
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ﺇﻥ ﺤﻠﻭل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ
)u E (x
ﻫﻲ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟﻠﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ ﻭ Eﻫـﻲ ﺍﻟﻘـﻴﻡ
ﺍﻟﺨﺎﺼﺔ ﻭﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺴﻨﺫﻜﺭ ﻋﻠﻰ ﺨﺎﺼﻴﺘﻥ ﻤﻬﻤﺘﻴﻥ ﻟﻠﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟـ .H -1ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺍﻟﻤﻁﺎﺒﻘﺔ ﻟﻘﻴﻡ ﺨﺎﺼﺔ ﻤﺨﺘﻠﻔﺔ )ﺃﻱ ﻗﻴﻡ ﻤﺨﺘﻠﻔﺔ ﻟـ (Eﺘﻜـﻭﻥ
ﻫﺫﻩ ﺍﻟﺩﻭﺍل ﻤﺘﻌﺎﻤﺩﺓ ﺃﻱ:
E ≠ E′
(x) = 0
E′
dx u *E (x) u
∫
-2ﺇﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺘﻜﻭﻥ ﻓﺌﺔ ﺍﻟﻬﺎﻤﻠﺔ – ﺃﻱ ﺃﻥ – ﺩﺍﻟﺔ ﺍﺨﺘﻴﺎﺭﻴﺔ
ﻤﺭﺒﻌﺔ ﺍﻟﺘﻜﺎﻤل ﻟﻜﻲ.
∞
)(6
<
)dx ψ * (x) ψ(x
)ψ (x
ﺘﻜـﻭﻥ
∫
ﻴﻤﻜﻥ ﺃﻥ ﺘﻤﺩ )ﺃﻭ ﺘﻔﻙ( expandedﺒﺩﻻﻟﺔ ﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔ ﻟﻠﻬﺎﻤﺘﻠﻭﻨﻴﺎﻥ. )C E u E (x
)(7
∑
= )ψ (x
E
ﺇﻥ ﻁﻴﻑ Hﻴﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻥ ﻤﻨﻔﺼل )ﻤﺠﺯﺃ( ﻜﻤﺎ ﻓـﻲ ﺤﺎﻟـﺔ ﺍﻟﻤﺘﺫﺒـﺫﺏ
ﺍﻟﺘﻭﺍﻓﻘﻲ .ﻭﻟﻭ ﺃﻥ ﺍﻟﺠﻬﺩ
)0 ← V (x
ﺘﻜﻭﻥ ﻤﻨﻔﺼﻠﺔ ﻭﻜﺫﻟﻙ ﻤﺘﺼﻠﺔ.
ﻋﻨﺩﻤﺎ ، ∞ ← xﻓﺈﻥ ﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔ ﻴﻤﻜـﻥ ﺃﻥ
ﻭﺇﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (7ﻴﻤﻜﻥ ﺘﻌﻭﻴﻀﻬﺎ ﺒـ: )dp C (p) u p (x
)(8
∫
C n u n (x) +
∑
= )ψ (x
n
ﺇﻥ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻴﻤﻜﻥ ﺃﻥ ﺘﻀﺭﺏ ﺒﺜﻭﺍﺒﺕ ﺒﺤﻴـﺙ ﺃﻥ ﺘـﺼﺒﺢ ﻤﻌـﺎﻴﺭﺓ ﻭﺸﺭﻭﻁ ﺍﻟﻤﻌﺎﻴﺭﺓ ﺍﻟﻤﺘﻌﺎﻤﺩﺓ. dx u *m (x) u n (x) = δ mn
∫
)dx u *p (x) u p (x) = δ (1 − 2 dx u *m (x) u p (x) = 0
∫
∫
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ﺇﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ
)ψ (x, t
ﻴﻤﻜﻥ ﺘﺤﺩﻴﺩﻫﺎ ﺒﻤﻼﺤﻅﺔ ﺃﻥ ﻜل ﺩﺍﻟﺔ ﺨﺎﺼﺔ ﻟﻬـﺎ
ﺍﻋﺘﻤﺎﺩ ﺴﺘﺤﺼل ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﻴﻌﻁﻲ ﺒـ: u E (x, t ) = u E (x) e −iEt/h ⇒ ψ (x, t) = ∑ C E u E (x) e −iEt/h E
2 t/2m h
dp C(p) u (x) e −ip
∫
ψ (x, t) = ∑ C E u E (x) e −iEt/h E
ﻤﺸﺎﻫﺩﺍﺕ ﺃﺨﺭﻯO These Obseruobeas :
ﺇﻥ ﺍﻟﻁﺎﻗﺔ ﻫﻲ ﻭﺍﺤﺩﺓ ﻤﻥ ﺍﻟﻤﺸﺎﻫﺩﺍﺕ ﻟﻨﻅﺎﻡ ﻤﺸﺎﻫﺩﺍﺕ ﺃﺨﺭﻯ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺴﻨﻨﺎﻗﺵ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺯﺍﻭﻴﺔ )ﺍﻟﻔﺼل ﺍﻟﺘﺎﻟﻲ(. ﻤﻌﺎﺩﻟﺔ :ﻤﺅﺜﺭ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ. h d )u (x) = p u p (x i dx p
= )p op u p (x
ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺘﻜـﻭﻥ ﺍﺘـﺼﺎل aconitum
) ∞ < (− ∞ < p
ﻭﺍﻟـﺩﻭﺍل
ﺍﻟﺨﺎﺼﺔ ﺘﺄﺨﺫ ﺍﻟﺸﻜل. 1 e ipx/ h 2D h
= )u p (x
ﻭﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺘﻜﻭﻥ ﻓﺌﺔ ﻋﻴﺎﺭﻴﺔ ﻤﺘﻌﺎﻤﺩﺓ. ∞+
) = δ (p1 − p 2
)dx u *p (x) u p (x 2
1
∫
∞−
ﺇﻥ ﻤﺅﺜﺭ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻤﺜل ﺍﻟﻬﺎﻤﻠﻴﺘﻭﻨﻴﺎﻥ ﻟﻪ ﻗﻴﻡ ﺨﺎﺼﺔ ﺤﻘﻴﻘﻴﺔ. ﺇﻥ ﺍﻟﻤﺅﺜﺭﺍﺕ ،ﺍﻟﺘﻲ ﻟﻬﺎ ﻗﻴﻡ ﺨﺎﺼﺔ ﺤﻘﻴﻘﻴﺔ ،ﺘﺴﻤﻰ ﻤﺅﺜﺭﺍﺕ ﻫﻴﺭﻤﻴﺘﺎﻥ .ﻭﻷﻥ ﻜل ﺍﻟﻤﺅﺜﺭﺍﺕ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺘﺸﺘﺭﻁ ﻓﻲ ﻫﺫﻩ ﺍﻟﺼﻔﺔ ،ﻓﻠﻬﺫﺍ ﺍﻟﺴﺒﺏ ﺘﻭﺼﻑ ﺒﻤـﺅﺜﺭﺍﺕ
ﻫﻴﺭﻤﻴﺘﻴﺔ.
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ﻨﻅﺭﻴﺔ ﺍﻟﻤﻔﻜﻭﻙ ﻭﺘﻤﺎﺜﻠﻬﺎ ﻤﻊ ﺍﻟﻤﺘﺠﻬﺎﺕ: ﻟﻤﺅﺜﺭ ﺍﺨﺘﻴﺎﺭﻱ ﻭﺍﻟﺫﻱ ﻴﺭﻤﺯ ﻟﻪ ﺒـ ،Aﺴﻴﻜﻭﻥ ﻫﻨﺎﻙ ﺩﻭﺍل ﺨﺎﺼﺔ ﺘﻁﺎﺒﻕ
ﻟﻘﻴﻡ ﺨﺎﺼﺔ ﺤﻘﻴﻘﻴﺔ .a
)A u a (x) = a u a (x
ﻭﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺘﺸﻜل ﻓﺌﺔ ﻤﺘﻌﺎﻤﺩﺓ ﻭﻴﻤﻜﻥ ﺃﻥ ﺘﺼﺒﺢ ﻤﻌﺎﻴﺭﺓ ﺒﺤﻴﺙ: dx u *a1 (x) u a2 (x) = δa 1 , a 2
ﻭﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ
)u a (x
∫
ﺘﻜﻭﻥ ﻓﺌﺔ ﻜﺎﻤﻠﺔ ﻭﺍﻟﺘﻲ ﻴﻤﻜﻥ ﺃﻥ ﺘﺤﺩﺩ ﺒﻪ )C a u a (x
∑
ﻷﻥ )u a (x
= )ψ(x
a
ﻤﻥ ﺸﺭﻁ ﺍﻟﻤﻌﺎﻴﺭﺓ ﻭﺍﻟﺘﻌﺎﻤﺩ: )C a = ∫ dx u *a (x) ψ (x
ﺘﻔﺴﻴﺭ ﻤﻌﺎﻤﻼﺕ ﺍﻟﻔﻙ: ﺃﻥ ﺘﻔﺴﻴﺭ :Ca ﻟﻭ ﻫﻨﺎﻙ ﻤﺅﺜﺭ ﺘﻡ ﻗﻴﺎﺴﻪ ﻟﺘﺠﻤﻊ ﻤﻥ ﺍﻷﻨﻅﻤﺔ ﻜل ﻤﻨﻬﺎ ﻴﻭﺼـﻑ ﺒﺩﺍﻟـﺔ )(x
ﻭﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻤﻌﺎﻴﺭﺓ ﻟﻠﻭﺤﺩﺓ ﺒﺤﻴﺙ.
(x) ψ (x) = 1
*
∫ dx ψ
ﻋﻨﺩﺌﺫ: -1ﻨﺘﻴﺠﺔ ﺃﻱ ﻗﻴﺎﺱ ﻤﻌﻁﻰ ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻥ ﻓﻘﻁ ﻭﺍﺤﺩﺓ ﻤﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ .a -2ﺍﺤﺘﻤﺎل ﺃﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ aﺃﻥ ﺘﻭﺠﺩ ﻫﻭ
2
Ca
-3ﻭﻜﻤﻘﺘﻀﻰ ﻟﻬﺫﺍ ﺍﻟﺘﺄﻭﻴل ﻓﺈﻥ ﺍﺤﺘﻤﺎل ﺃﻥ ﻗﻴﻤﺔ ﺍﻟﻤﺅﺜﺭ Aﻟﻨﻅﺎﻡ ﻴﻜﻭﻥ ﻟﻬﺎ ﻭﺍﺤﺩ ﻤﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ ﻫﻭ .unity - ١٢١PDF created with pdfFactory Pro trial version www.pdffactory.com
∑
Ca
2
=1
a
:ﻭﻫﺫﺍ ﻴﻨﺘﺞ ﻤﻥ 1=
∫
dx ψ * (x) ψ (x) =
∫
dx ∑ C *a u *a (x) ψ (x) a * = ∑ Ca C a a
:ﻭﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻌﻁﻰ ﺍﻟﺘﺎﻟﻲ ∑
C a C *a =
∑ ∫
dx u *a (x) ψ (x)
a
a
= ⇒
∑
∫ ∫
∫
dy u a (y) ψ * (y)
dx dy ψ * (y) ψ (x) ∑ u a (y) u *a (x) = 1 a
u a (y) u (x) = δ (x − y) * a
a
ﺇﻥ ﻫﺫﻩ ﺍﻟﺨﺎﺼﻴﺔ ﻟﻠﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﺘﻭﺼﻑ ﺒـ ﺍﻟﻌﻼﻗﺔ ﺍﻟﻜﺎﻤﻠﺔ ﻭﻫﻲ ﻤﻁﺎﺒﻘﺔ
.ﻟﻤﻨﻁﻭﻕ ﻨﻅﺭﻴﺔ ﺍﻟﻤﻔﻜﻭﻙ
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ﺍﻟﺘﻨﺎﻅﺭ ﻤﻊ ﻓﻀﺎﺀ ﺍﻟﻤﺘﺠﻪ
The Vector Space Analogy
ﺍﻟﻤﺅﺜﺭﺍﺕ ﻭﺍﻟﻜﻤﻴﺎﺕ ﺍﻟﻤﺸﺎﻫﺩﺓ :operators and observables ﻓﻲ ﻓﺭﺍﻍ )ﻓﻀﺎﺀ( ﺍﻟﻤﺘﺠﻬﺎﺕ ،ﺍﻟﻤﺅﺜﺭ ﻴﺤﻭل transformsﻤﺘﺠﻪ ﺇﻟﻰ ﻤﺘﺠـﻪ
ﺁﺨﺭ ﺇﻥ ﺍﻟﻤﺅﺜﺭﺍﺕ ﺍﻟﺨﻁﻴﺔ ﻟﻬﺎ ﺍﻟﺨﺎﺼﻴﺔ.
)A (α1 ψ1 (x) + α 2 ψ 2 (x) ) = α 1 A ψ1 (x) + α 2 A ψ 2 (x
ﻭﻟﻭ ﻜﺎﻨﺕ ﻫﺫﻩ ﺍﻟﻤﺅﺜﺭﺍﺕ ﺘﻤﺜل ﻜﻤﻴﺎﺕ ﻤﺸﺎﻫﺩﺓ ،ﻓﺈﻨﻬـﺎ ﻴﺠـﺏ ﺃﻥ ﺘﻜـﻭﻥ
ﻫﻴﺭﻤﻴﺘﻴﺔ .ﻭﻟﻠﻤﺅﺜﺭ ﺍﻟﻬﻴﺭﻤﻴﺘﻲ. =0
ﺃﻱ ﺃﻨﻪ ﻟﻠﺤﺎﻟﺔ
*
A − A
*
A = A
ψ
)dx ψ * (x) Aψ (x) = ∫ dx (A ψ (x) ) ψ (x *
)(a
∫
ﻤﻥ ﻫﺫﺍ ﻴﻨﺘﺞ ﺃﻥ ﻟﻠﻤﺅﺜﺭﺍﺕ ﺍﻟﻬﻴﺭﻤﻴﺘﻴﺔ dx φ * Aψ = ∫ dx (A φ ) ψ *
)(b
ﻭﻷﻱ ﺯﻭﺝ ﻤﻥ ﺍﻟﺩﻭﺍل
φ
,
∫
ψ
ﻭﻹﺜﺒﺎﺕ ﻫﺫﺍ ،ﺨﺫ. )ψ (x) = φ (x) + λ ψ (x
ﺤﻴﺙ
λ
ﻫﻭ ﺜﺎﺒﺕ ﻤﺭﻜﺏ ،ﻋﻨﺩﺌﺫ ﻴﻨﺘﺞ ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (aﺃﻥ: )dx ψ * (x) A ψ = ∫ dx (φ * + λ * ψ * ) A (φ + λ ψ dx ψ * Aψ
∫
2
dx φ * A φ + λ
∫
∫
=
+ λ ∫ dx φ * Adx φ * Aψ + λ * ∫ ψ * Aφ
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ﺍﻟﻤﺭﺍﻓﻕ ﻟﻬﺫﺍ ﻫﻭ ] ) dx (AψA* ψ = ∫ dx [A (φ + λψ) ] [ (φ + λ ψ *
dx (AψA* ψ + λ * ∫ dx (AψA* φ + λ ∫ dx (AφA* ψ
∫
2
∫
= ∫ dx (AφA* φ + λ
ﻭﺍﻟﻤﺅﺜﺭ ﻫﻴﺭﻤﻴﺕ ،ﺍﻟﺠﻭﺍﻨﺏ ﺍﻟﻴﺴﺭﻯ ﻤﻥ ﺍﻟﻤﻌـﺎﺩﻟﺘﻴﻥ ) (d) , (cﻭﺍﻟﺤـﺩﻴﻥ
ﺍﻷﻭﻟﻴﻥ ﻤﺘﺴﺎﻭﻴﻴﻥ .ﻭﺒﻤﺎ ﺃﻥ
λ* , λ
ﻤﻌﺎﻤﻼﺕ ﻤﺴﺘﻘﻠﺔ )ﻫﻲ ﻤﺭﻜﺒﺔ( ﺒﺈﻤﻜﺎﻨﻨﺎ ﺘﺴﺎﻭﻱ
ﻤﻌﺎﻤﻼﺘﻬﻡ ﺒﺎﻨﻔﺼﺎل ،ﻭﻴﻨﺘﺞ ﻟﻨﺎ ﺍﻟﻤﻌﺎﺩﻟﺔ
ﻭﻟﻭ ﺃﻥ ﻤﺅﺜﺭ Aﻏﻴﺭ ﻫﻴﺭﻤﻴﺘﻲ ،ﻓﺒﺈﻤﻜﺎﻨﻨـﺎ ﺘﻌﺭﻴـﻑ ﺍﻟﻤـﺅﺜﺭ ﺍﻟﻤﺭﺍﻓـﻕ
ﺍﻟﻬﻴﺭﻤﻴﺘﻲ hermitian conjugatioput A+
dx (AφA* ψ = ∫ dxφ * A + ψ
⊗⊗
∫
ﻭﺫﻟﻙ ﻷﻱ ﺯﻭﺝ ﻤﻥ ﺍﻟﺩﻭﺍل ﺍﻟﻤﻭﺠﻴﺔ. ⊗
ﻻﺤﻅ ﺃﻥ ﻟﻠﻤﺅﺜﺭ ﺍﻟﻬﻴﺭﻤﻴﺘﻲ
⊗
ﺍﻟﻤﺅﺜﺭﺍﺕ ﺍﻟﺘﺒﺎﺩﻟﻴﺔ [ A , B ] = 0
⊗
ﻟﻠﻤﺅﺜﺭﺍﺕ ﺍﻟﻬﻴﺭﻤﻴﺘﻴﺔ
A = A+
( AB ) + = B + A +
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ﺍﻟﻨﻬﺎﻴﺔ ﺍﻟﻜﻼﺴﻴﻜﻴﺔ )ﺍﻟﺘﻘﻠﻴﺩﻴﺔ( ﻟﻠﻨﻅﺭﻴﺔ ﺍﻟﻜﻤﻴﺔ ﻭﺍﻻﻋﺘﻤﺎﺩ ﺍﻟﺯﻤﻨﻲ Time Dependence The classical limit of Quantum theory
ﻤﻥ ﺍﻷﺴﺌﻠﺔ ﺍﻟﻤﻬﻤﺔ ﺍﻟﺘﻲ ﻨﺤﻥ ﺒﺼﺩﺩﻫﺎ ﺍﻵﻥ ﻫﻲ ﺍﻟﻨﻬﺎﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻟﻠﻨﻅﺭﻴـﺔ
ﺍﻟﻜﻤﻴﺔ .ﻭﻟﻤﻨﺎﻗﺸﺔ ﻫﺫﻩ ﺍﻟﻤﺴﺄﻟﺔ ،ﻓﺈﻨﻨﺎ ﻨﺒﺩﺃ ﺒﺩﺭﺍﺴﺔ ﺍﻟﺘﻁﻭﺭ ﺍﻟﺯﻤﻨﻲ ﻟﻠﻘﻴﻡ ﺍﻟﻤﺘﻭﻗﻌـﺔ
ﻟﻠﻤﺅﺜﺭﺍﺕ ﺍﻟﻤﺘﻐﻴﺭﺓ ﻤﻊ ﺍﻟﺯﻤﻥ .ﻤﻨﻬﺎ ﻴﻤﻜﻥ ﺃﻥ ﺘﺘﻐﻴـﺭ ﻤـﻊ ﺍﻟـﺯﻤﻥ ﻭﺫﻟـﻙ ﻷﻥ
ﺍﻟﻤﺅﺜﺭﺍﺕ ﻟﻬﺎ ﺍﻋﺘﻤﺎﺩ ﻭﺍﻀﺢ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ،ﻋﻠﻰ ﺴﺒﻴل ﺍﻟﻤﺜﺎل ،ﺍﻟﻤـﺅﺜﺭ ، α + pt / m
ﻭﺇﻨﻪ ﻴﺘﻐﻴﺭ ﻤﻊ ﺍﻟﺯﻤﻥ ،ﻭﺫﻟﻙ ﻷﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻭﻗﻌﺔ ﺘﺅﺨﺫ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺩﺍﻟـﺔ ﺍﻟﻤﻭﺠﻴـﺔ ﺍﻟﺘﻲ ﺒﻨﻔﺴﻬﺎ ﺘﺘﻐﻴﺭ ﻤﻊ ﺍﻟﺯﻤﻥ .ﻟﻭ ﻜﺘﺒﻨﺎ.
A t = ∫ ψ * (x, t) A ψ (x , t) dx
ﻋﻨﺩﺌﺫ: d ∂A )A t = ∫ ψ * (x, t ψ (x , t) dx dt ∂t )∂ψ * (x, t )∂ψ(x, t A ψ (x , t ) dx + ∫ ψ * (x , t ) A di ∫+ ∂t ∂t ∂A = ∂t
*
1 1 + ∫ H ψ (x, t) A ψ (x, t) dx + ∫ ψ * (x, t) A H ψ (x, t) ih ih t ∂A i i = + ∫ ψ * (x, t) H A ψ (x, t) dx + − ∫ ψ * (x, t) AHψ (x, t)dx ∂t t h h t
] [H , A
i h
+ t
∂A ∂t
=
ﻨﻼﺤﻅ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻤﺅﺜﺭ Aﻏﻴﺭ ﻤﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﺒﻭﻀﻭﺡ ﺤﻴﻨﺌﺫ = 0 dA dA = =0 dt dt d i ⇒ = A t [H , A dt h ⇒
t
]
ﻭﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻤﺅﺜﺭ ﺘﺒﺎﺩﻟﻲ ﻤﻊ ،Hﻓﺈﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻭﻗﻌﺔ ﺘﻜﻭﻥ ﺩﺍﺌﻤﺎﹰ ﺜﺎﺒﺕ.
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ﻭﻋﻠﻴﻪ ﻓﺒﺈﻤﻜﺎﻨﻨﺎ ﺃﻥ ﻨﻘﻭل ﺃﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻘﻴﺎﺴﻴﺔ ﺍﻟﻤﺸﺎﻫﺩﺓ ﻫﻲ ﺜﺎﺒﺕ ﻟﻠﺤﺭﻜـﺔ. ﻭﻟﻭ ﺃﻥ ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ ﻫﻭ ﻭﺍﺤﺩ ﻤﻥ ﺍﻟﻔﺌﺎﺕ ﺍﻟﻜﺎﻤﻠﺔ ﻟﻠﻤﺸﺎﻫﺩﺍﺕ ﺍﻟﻤﺘﺒﺎﺩﻟﺔ ،ﻋﻨﺩﺌﺫ ﻜل ﺍﻟﻤﺅﺜﺭﺍﺕ ﺘﻜﻭﻥ ﺜﺎﺒﺕ ﻓﻲ ﺍﻟﺤﺭﻜﺔ. ﺨﺫ: A=x A=p
)( 1 )(2
ﺃﻭﻻﹰ: ] [H , x
d i = x dt h
i p2 2m + v (x) , χ h χ
ﺘﺘﺒﺎﺩل ﻤﻊ ﺃﻱ ﺩﺍﻟﺔ ﻓﻲ
χ
=
،ﺃﻱ ﻤﺜﻼﹰ ).v (x [ V (x) , χ ] = 0
?=] χ ] = p [ p , χ ] + [ p , χ ]p
, χ
2
,
2
[p [p
⇒
h =2 p i )(a
p p = 2/ m m
d i/ 2/ χ = . t/ dt h/ i/
⇒
ﺜﺎﻨﻴﺎﹰ: d i ] p = [H , p dt h i p2 + v (x) , p h 2m
=
Pﺘﺘﺒﺎﺩل ﻤﻊ ﺃﻱ ﺩﺍﻟﺔ ﻓﻲ p
] )[ p , V (x
i h
=−
] [ V (x) , p
i h
=
⇒
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[ p , V (x) ]
ﻟﺤﺴﺎﺏ
?
h d [ V (x) ψ (x) ]− h V (x) d ψ (x) i dx i dx h dψ ( x) h dV(x) h dψ (x) = V (x) + ψ (x) − V (x) i dx i dx i dx h dV(x) . ψ (x ) = i dx
pV (x) ψ (x) − V (x) P ψ (x) =
⇒ [ p , V (x)]=
h dV(x) i dx
:ﻭﻋﻠﻴﻪ d i h/ dV (x) p =− . dt h/ i dx
=− t
dV (x) dx
(b) t
( b ) , ( a ) ﺒﺎﺘﺤﺎﺩ d p d χ = ⇒m χ = p dt m dt d p dt ⇒m
= t
dV (x) d d d2 .m χ = m 2 χ t =− dx dt dt dt
dV (x) d2 χ t =− 2 dx dt
t
ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﺒﺩﻭ ﻤﺸﺎﺒﻬﺔ ﻜﺜﻴﺭﺍﹰ ﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺤﺭﻜﺔ ﻟﺠﺴﻡ ﺘﻘﻠﻴﺩﻱ ﻴﺘﺤﺭﻙ ﻓـﻲ
V (x) ﺠﻬﺩ
m χ= χ
d2χ dV (x) =− 2 dx dt
ﻭﺍﻟﺸﻲﺀ ﺍﻟﻭﺤﻴﺩ ﺍﻟﺫﻱ ﻴﻤﻨﻌﻨﺎ ﺃﻥ ﻨﻌﻤل ﺍﻟﺘﻁﺎﺒﻕ :ﻫﻭ dV d =/ V( χ dx dχ
)
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ﺘﻤﺕ ﺍﻟﺤﻴﺜﻴﺎﺕ ﻭﺍﻟﺘﻲ ﻋﻨﺩﻫﺎ ﺍﻟﻤﺘﻁﺎﺒﻘﺔ ﺍﻟﺴﺎﺒﻘﺔ ﺘـﺼﺒﺢ ﺘﻘﺭﻴﺒـﺎﹰ ﻤﺘـﺴﺎﻭﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻫﻲ ﻓﻲ ﺍﻷﺴﺎﺱ ﺘﻘﻠﻴﺩﻴﺔ ،ﻜﻤﺎ ﺘﻡ ﻤﻼﺤﻅﺘﻪ ﺃﻭﻻﹰ ﺒﻭﺍﺴـﻁﺔ hrenfenﻫـﺫﺍ
ﻴﻠﺯﻡ ﺃﻥ ﺩﺍﻟﺔ ﺍﻟﺠﻬﺩ ﺘﺘﻐﻴﺭ ﺒﺒﻁﻲﺀ ﻤﻊ ﺍﻟﻤﺘﻐﻴﺭ.
)dV (x dχ
)
F′′ ( χ
ﻭﻟﻭ ﺃﻥ ﺍﻟﻼﺘﻌﻴﻴﻥ
) )F′ ( χ )+ (χ − χ
2
⇒ F (x) = F ( χ )+ (x − χ
2
)
2
(
= χ− χ
F (x) = −
) (Δ χﺼﻐﻴﺭﺍﹰ ﻓﺈﻥ ﺍﻟﺤﺩﻭﺩ ﺍﻟﻌﻠﻴﺎ ﻤـﻥ ﺍﻟﻔـﻙ
2
ﻴﻤﻜﻥ ﺇﻫﻤﺎﻟﻬﺎ. )
F (χ ) ≅ F ( χ )+ (χ − χ )F′ ( x
)
≅ F( χ
ﻭﻟﻤﺠﺎﻻﺕ ﻋﻴﻨﻴﺔ macroscopic helveﻓـﺈﻥ ﺍﻟﻤﻌﺎﺩﻟـﺔ ﺍﻷﺨﻴـﺭﺓ ﺘﻌﺘﺒـﺭ
ﺼﺤﻴﺤﺔ ﻭﻫﺫﺍ ﻴﺠﻌﻠﻨﺎ ﺃﻥ ﻨﺤﺴﺏ ﻭﻨﻀﻴﻑ ﻤﺩﺍﺭ ﺍﻹﻟﻜﺘﺭﻭﻥ ﺃﻭ ﺍﻟﺒﺭﻭﺘﻭﻥ ﻓﻲ ﺘﻌﺠل ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ. ﺱ /ﺇﺫﺍ ﻜﺎﻥ B , Aﻤﺅﺜﺭﻴﻥ ﻫﻴﺭﻤﻴﺘﻴﻴﻥ ،ﺃﺜﺒﺕ ﺃﻥ: -1ﺍﻟﻤﺅﺜﺭ ABﻴﻜﻭﻥ ﻓﻘﻁ ﻫﻴﺭﻤﻴﺘﻲ ﻟﻭ ﺃﻥ B , AﻴﺘﺒﺎﺩﻟﻴﺎﻥAB = BA ،
-2ﺍﻟﻤﺅﺜﺭ (A + B)nﻫﻭ ﻜﺫﻟﻙ ﻫﻴﺭﻤﻴﺘﻲ. ﺍﻟﺤل/ -1ﻟﻤﺅﺜﺭﻴﻥ ﻫﻴﺭﻤﻴﺘﻴﻴﻥ: (AB) + = B + A + = BA
ﻭﻫﺫﺍ ﻴﺴﺎﻭﻱ ABﻓﻘﻁ ﻟﻭ ﻜﺎﻥ AB = BA
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-2
)[ (A + B) ] = (A + B
(A + B)+ ................(A + B)+ )= (A + B)(A + B)..................... (A + B n )= (A + B +
ﺱ /ﺃﺜﺒﺕ ﺃﻥ
) i (A − A+
, A + A+
n +
ﻴﻜﻭﻨﻭ ﻫﻴﺭﻤﻴﻴﺘﺎﻥ ﻷﻱ ﻤﺅﺜﺭ ﻤﺜل
A A+
ﺍﻟﺤل/ = A + + (A + ) + = A + + A
)
( A + A+ )+
(
) = (− i ) A + − (A + ) + = (−i) (A + − A
]) [i ( A - A
+
+
) =i (A − A+ ( A A + ) + = (A + ) + A + = A A +
ﺱ /ﺃﺜﺒﺕ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻨﺕ Hﻫﻭ ﻤﺅﺜﺭ ﻫﻴﺭﻤﻴﺘﻲ ،ﻓﺈﻥ ﺍﻟﻤﺅﺜﺭ ﺍﻟﻬﻴﺭﻤﻴﺘﻲ ﺍﻟﻤﺭﺍﻓﻕ ﻟـ e iH
)ﻭﺍﻟﺫﻱ ﻴﻌﺭﻑ ﺒـ
ﺍﻟﻤﺅﺜﺭ e - iH
∑ ( ﻫﻭ ∞
!in Hn / n
n =0
ﺍﻟﺤل/
) (
+
+
∞ in n (−i) n ∑ = (e ) = ∑ H Hn !n ! 0 n (−i) n n − iH ∑= H =e !n +
ﺱ /ﺨﺫ ﺍﻟﻤﺅﺜﺭ ﺍﻟﻬﻴﺭﻤﻴﺘﻲ Hﻭﺍﻟﺫﻱ ﻟﻪ ﺍﻟﺨﺎﺼﻴﺔ
iH
H 4 =1
ﺃﻭﺠﺩ ﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔ )ﺍﻟﺫﺍﺘﻴﺔ( ﻟﻠﻤﺅﺜﺭ H؟ ﺃﻭﺠﺩ ﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔ )ﺍﻟﺫﺍﺘﻴﺔ( ﻟﻠﻤﺅﺜﺭ Hﻟﻭ ﺃﻥ Hﻏﻴﺭ ﻤﻘﻴـﺩ ﺒـﺄﻥ ﻴﻜـﻭﻥ
ﻫﻴﺭﻤﻴﺘﻲ؟
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ﺍﻟﺤل /ﻜﻭﻥ
H 4 =1
ﻤﻌﻨﺎﻩ
(H 4 − 1) ψ = 0
ﻭﻫﺫﺍ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻪ: ﻭﻋﻠﻴﻪ ﻓﺈﻥ
ﻟﻜل ﻗﻴﻡ
ψ
( H + 1) ( H − 1) ( H + i) ( H − i) ψ = 0
(H −i)ψ=0
ﻭﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ
Hψ − iψ = 0 ⇒ Hψ = i ψ
ﺃﻱ ﺃﻥ ﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔ = i
ﺃﻭ
( H − i ) =/ 0
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻨﻀﻊ ⇐
ﺃﻤﺎ
(H −i)ψ = φ
( H + 1 ) ( H − 1 ) ( H + i) φ = 0 ⇒
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ
(H +i)φ=0
⇐
ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ ﻫﻲ – i
ﺃﻭ ﺃﻥ
(H +i)φ=x ⇒ ( H + 1) ( H − 1 ) x = 0
ﺃﻤﺎ
(H −1 ) x = 0
⇐ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ
( −1) ................+ 1
ﻭﺒﻬﺫﻩ ﺍﻟﻁﺭﻴﻘﺔ ﻭﻀﺤﻨﺎ ﺃﻥ ﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔ ﺘﻜﻭﻥ ﺇﺫﺍ ﻜﺎﻥ
± 1, ± i
H = H+
ﻓﺈﻥ ﺍﻟﻘﻴﻡ ﺍﻟﺨﺎﺼﺔ ﻴﺠﺏ ﺃﻥ ﺘﻜﻭﻥ ﺤﻘﻴﻘﻴﺔ ﻭﻋﻠﻴﻪ ﻓﺎﻟﻘﻴﻡ ﺍﻟﺼﺤﻴﺤﺔ ﻫﻲ
±1
ﻓﻘﻁ.
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ﺱ /ﻭﻀﺢ ﺃﻨﻪ ﺇﺫﺍ ﻜﺎﻥ Aﻫﻭ ﻤﺅﺜﺭ ﻫﻴﺭﻤﻴﺘﻲ ،ﻓﺈﻥ ﻗﺒل ﺃﻨﻪ ﻟﻜل ﻤﺅﺜﺭ ﻫﻴﺭﻤﻴﺘﻲ
= e −iH
ﻫﻭ unitaryﻟﻘﺩ ﺃﺜﺒﺕ ﻤﻥ
e iA
) (e
iH +
ﻭﻋﻠﻴﻪ: , e iA = 1
− iA
(e ) (e )= e iA +
iA
ﻭﻴﻤﻜﻥ ﺘﻭﻀﻴﺢ ﻫﺫﺍ ﻜﺫﻟﻙ ﻤﻥ ﺨﻼل ﻤﻌﺎﻤﻼﺕ ﻟﻬﺎ ﻗﻭﻯ ﻤﺨﺘﻠﻔﺔ ﻓﻲ (iλi 2 2 (iλi 3 3 (iλi 2 2 (iλi 3 3 H + H + ..........) (1 − iλλ + H − )H + .......... !2 !3 !2 !3
ﺱ /ﺍﺴﺘﺨﺩﻡ ﻋﻼﻗﺎﺕ ﺍﻟﺘﺒﺩﻴل ﺒﻴﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﻲ ﺘﺼﻑ ﺍﻋﺘﻤﺎﺩ ﻜـل ﻤـﻥ
p
ﻭ
χ
ﻭﺍﻟﻤﻭﻗﻊ p
χ
λ
(1 + iλλ +
ﻤﺤـﺼﻭل ﻋﻠـﻰ
ﻋﻠـﻰ ﺍﻟـﺯﻤﻥ ،ﻤﻌﻁـﻰ
ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ. 2
(a ) H = p
1 )∈ + m ( ω12 χ 2 + ω 2χ + 2m 2 p2 1 (b ) H = + m ω 2 χ 2 − A2 2m 2 χ
ﺍﻟﺤل/ i p = χ {[ ] p, χ h m h/i
d i i p2 χ = [ H , χ ]= , dt h h 2m p m
]
=
[
dp i i = [H, p ]= − p, m ω12 χ 2 /2 + ω 2 χ dt h h i h χ = − , 2 mω12 + ω 2 = − mω12 χ - ω 2 h i 2 2 ω d x 1 dp = = − ω12 χ − 2 dt m dt m
ﻭﺍﻟﺤل ﻟﻠﻤﻌﺎﺩﻟﺔ ﻫﻭ )p(0 sin ω1 t mω1
+ x (0 ) cos ω1 t +
ω2 2 1
mω
χ =−
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dχ p = dt m
(b)
dp i x2 A 2A = − p , m ω2 − 2 = − mω 2 χ − 3 dt h 2 x x
. ﻭﻁﺎﻗﺘﻪ ﺘﻭﺼﻑ ﺒﻤﺅﺜﺭ ﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ،ﺇﻟﻜﺘﺭﻭﻥ ﻴﺘﺫﺒﺫﺏ ﻓﻲ ﻤﺠﺎل ﻜﻬﺭﺒﺎﺌﻲ p2 H= − ( e E ο cos wt ) χ 2m dH dp dx , , dt dt dt
ﺍﺤﺴﺏ /ﺍﻟﺤل
dx i i p2 i p = [ H, χ ]= , x = [p , x ]= p h 2m h m m dt h dp i i p2 = [ H, p ]= − p , − (eE ο cos ω t )χ h 2m dt h i [p , − (eE ο cos ω t )χ ]= − i , h eE ο cos ω t = − eE ο cos ω t h h i ∂H dH = = eE ο sin ω t ∂t dt =−
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ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺜﻼﺜﺔ ﺃﺒﻌﺎﺩ: The Schrödinger Equation in three Dimensions ﺍﻟﺠﻬﺩ ﺍﻟﻤﺭﻜﺯﻱThe Central Potential :
ﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل ﺴﻭﻑ ﻨﻨﺎﻗﺵ ﺤﺎﻟﺔ ﻤﻬﻤﺔ ﻭﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻓﻴﻬﺎ ﺍﻟﺠﻬﺩ ﻴﻌﺘﻤﺩ ﻋﻠﻰ
)V (x, y, t
ﻓﻘﻁ .ﻭﻟﻨﻅﺎﻡ ﻴﺘﻜﻭﻥ ﻤﻥ ﺠﺴﻴﻤﻴﻥ ،ﺤﻴﺙ ﺍﻟﺠﻬﺩ ﻴﻌﺘﻤﺩ
r = x2 + y 2 + Z 2
ﻋﻠﻰ ﺍﻟﻤﺴﺎﻓﺔ ﺍﻟﻔﺎﺼﻠﺔ ﺒﻴﻨﻬﻤﺎ ﻓﻘﻁ ،ﻓﺈﻥ ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ ﻟﻪ ﺍﻟﺸﻜل. ) + V ( r1 − r2
)(1
p 22 2m 2
+
p12 2m1
=H
ﻭﺒﺎﻟﺘﺤﻠﻴل ﺍﻟﻌﺎﺩﻱ ﻟﻤﺭﻜﺯ ﺍﻟﻜﺘﻠﺔ ﻭﺍﻹﺤﺩﺍﺜﻲ ﺍﻟﻨﺴﺒﻲ. m1 r1 + m 2 r 2
)( 2
m1 + m 2
=R
r = r1 − r 2
ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﻜﻠﻴﺔ ﻭﺍﻟﻨﺴﺒﻴﺔ p = p1 + p 2 m 2 p1 − m1 p 2
)( 3
m1 + m 2
=p
ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﻥ ﻴﺄﺨﺫ ﺍﻟﺸﻜل. 2
) + V( r
)( 4
ﻫﻨﺎ
M
p
2M
2
+
p
2M
=H
ﻫﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻜﻠﻴﺔ ﻟﻠﻨﻅﺎﻡ M = m1 + m 2
)(5
ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﺨﺘﺯﻟﺔ )ﺍﻟﻤﺨﻔﻔﺔ(
μ
)(6
m1 m 2 M
=μ
ﺒﺈﻤﻜﺎﻨﻨﺎ ﺃﻥ ﻨﺘﺄﻜﺩ ﻤﻥ ﺃﻥ: - ١٣٣PDF created with pdfFactory Pro trial version www.pdffactory.com
]hi sij
, Rj
i
, rj
i
]hi sij
) (7
[p [p
ﻭﺒﻤﺎ ﺃﻥ ﺍﻟﺠﻬﺩ ﻻ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﺇﺤﺩﺍﺜﻲ ﻤﺭﻜﺯ ﺍﻟﻜﺘﻠﺔ ، Rﻓﺈﻥ ﻤـﺅﺜﺭ ﻜﻤﻴـﺔ
ﺍﻟﺤﺭﻜﺔ ﺍﻟﻜﻠﻴﺔ Pﺘﺒﺎﺩﻟﻲ ﻤﻊ Hﻭﺒﺈﻤﻜﺎﻨﻨﺎ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟﻜل ﻤﻥP
ﻭ Hﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟـ Pﻟﻬﺎ ﺍﻟﺸﻜل. 1
e i P, R / h
)(8
2
3
= ) U (P , R
)( 2 D h
ﻭﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺨﺎﺼﺔ ﻟـ Hﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻬﺎ ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ. )ψ(R ,r) =U(P,R )uE (r
)(9
ﺤﻴﺙ: p2 ) + V (r ) u E (r ) = E u E (r 2μ
)(10 M =w
,
P2 2M
E = E tot −
ﻫﻲ ﺍﻟﻁﺎﻗﺔ ﺍﻟﺩﺍﺨﻠﻴﺔ ،ﺒﻌﺩ ﻁﺭﺡ ﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ ﻟﻠﺠﺴﻴﻤﻴﻥ. ﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺔ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ )(1
h2 2 − ) Δ r + V (r ) u E (r ) = E u E (r 2μ
ﻻﺤﻅ ﺃﻥ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﺘﺠﺎﻩ ﻨﺘﺎﺌﺞ ﻋﺩﻡ ﺍﻟﺘﻐﻴﻴﺭ ﺒﺎﻟﺩﻭﺭﺍﻥ
r
ﺃﻱ ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ
θ
ﺃﻭ
φ
Consequences of Rotational Invariance
ﻓﻲ ﻫﺫﺍ ﺍﻟﺠﺯﺀ ﺴﻨﻭﻀﺢ ﺃﻥ ) (11ﻴﻤﻜﻥ ﺃﻥ ﺘﻔﺼل ﺒﺤﻴـﺙ ﺃﻥ ﺍﻹﺤـﺩﺍﺜﻲ
ﺍﻟﺨﻁﻲ rﻴﻅﻬﺭ ﻓﻲ ﻤﻌﺎﺩﻟﺔ ﻁﺎﻗﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ.
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ﻓﻲ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ،ﺍﻟﻔﺼل ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﺍﺴﺘﺨﺩﺍﻡ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺯﺍﻭﻴـﺔ ﺒﺄﺨﺫ: L =r×p
) (12
)
2
(
) ( ) ( 1 1 = (r . p ) + L r r
⇒
L = r×p . r×p = r p − r −p 2
2
2
2
2
2
)(13
2
1 2 L r2
p2
= p 2r +
ﻭﻟﺠﻬﺩ ﻤﺭﻜﺯﻱ ) Vﺘﻌﺘﻤﺩ ﻋﻠﻰ rﻓﻘﻁ( ﺍﻟﻘﻭﺓ ﺘﻜﻭﻥ ﻨﺼﻑ ﻗﻁﺭﻴﺔ radial
ﻭﻻ ﺘﺒﺫل ﻋﺯﻡ ﻋﻠﻰ ﺍﻟﻨﻅﺎﻡ .ﻭﻫﻜﺫﺍ ﻓﺈﻥ Lﺜﺎﺒـﺕ ﻟﻠﺤﺭﻜـﺔ ،ﻭ L2ﻓﻘـﻁ ﻭﻟﻬـﺫﺍ ﻓﺎﻟﻤﻌﺎﺩﻟﺔ.
1 2 )P + V (r 2μ
=E
ﺘﺘﻀﻤﻥ ﺍﻹﺤﺩﺍﺜﻴﺎﺕ ﺍﻟﺨﻁﻴﺔ ﻓﻘﻁ .ﻭﻨﻔﺱ ﺍﻷﻤﺭ ﺼﺤﻴﺢ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ. ﻭﻓﻲ ﻫﺫﺍ ﺍﻟﻔﺼل: -1ﺴﻨﺤﺩﺩ ﻤﺅﺜﺭﺍﺕ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﻤﻥ ﺍﺤﺘﻴﺎﺝ ﺃﻥ ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﻥ ﻏﻴﺭ ﻤﺘﻐﻴﺭ ﺘﺤـﺕ
ﺍﻟﺩﻭﺭﺍﻥ.
-2ﺍﺸﺘﻘﺎﻕ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺨﻁﻴﺔ. ﻋﺩﻡ ﺍﻟﺘﻐﻴﺭ ﺘﺤﺕ ﺍﻟﺩﻭﺭﺍﻥ ﺤﻭل ﻤﺤﻭل ) Zﻹﺜﺒﺎﺕ ﺃﻥ
∆2
ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺩﻭﺭﺍﻥ(
Invariance under Rotations about the z-axis
ﺨﺫ ﺤﺎﻟﺔ ﺨﺎﺼﺔ ﻟﺩﻭﺭﺍﻥ ﺨﻼل ﺯﺍﻭﻴﺔ )(14
θ
ﺤﻭل ﻤﺤﻭﺭ Z
χ ′ = χ cos θ − y sin θ y′ = χ sin θ − y cos θ
ﺒﺈﻤﻜﺎﻨﻨﺎ ﺃﻥ ﻨﺭﻯ ﺃﻥ
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r ′ = ( x′ 2 + y′ 2 + Z ′ 2 ) 2
1
2
= ( χ 2 + y2 + Z 2 )
2
1
2
=r
2
∂ ∂ ∂ ∂ ∂ ∂ + sin θ = cos θ − sin θ + + cos θ ∂ y ∂x ∂ y ∂x ∂ x′ ∂ y′ ∂ ∂ = + ∂x ∂y 2
2
2
ﻻ ﻴﺘﺄﺜﺭ ﺒﺎﻟﺩﻭﺭﺍﻥ ﻭﺒﻤﺎ ﺃﻥ ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﻥ ﻟﻪ ﺨﺎﺼـﺔ ﻋـﺩﻡH ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ
ﻭﻟﺘﺤﺩﻴﺩ ﺍﻟﻤﺅﺜﺭﺍﺕ ﺍﻟﺘﻲ ﺘﺘﺒﺎﺩل ﻤﻊ، ﻓﻨﺤﻥ ﻨﺘﻭﻗﻊ ﻗﺎﻨﻭﻥ ﺍﻟﺤﻔﻅInvariace ﺍﻟﺘﻐﻴﻴﺭ .ﻓﻘﻁ
θ
ﻭﺍﺤﺘﻔﻅ ﺒﺎﻟﺤﺩﻭﺩ ﻟـZ ﺨﺫ ﺩﻭﺭﺍﻥ ﻤﺘﻨﺎﻫﻲ ﻓﻲ ﺍﻟﺼﻐﺭ ﺤﻭل ﻤﺤﻭﺭ،H χ′ = χ − θ y y′ = y + x θ
(15)
:ﻨﺤﻥ ﻨﻠﺯﻡ ﺃﻥ H U E (x − θ y , y + θ x , Z ) = E U E (x − θ y , y + θ x , Z)
f ( x + Δ ) = f (x) +
af (Δ x) ax
ﻭﻁﺭﺤﻨﺎ ﻤﻨﻪ
θ
(16)
ﻭﻟﻭ ﻓﻜﻜﻨﺎ ﻫﺫﺍ ﻟﻠﺤﺩ ﺍﻷﻭل ﻓﻲ
H U E (x , y , Z ) = E U E (x , y , Z)
(17)
:ﻨﺘﺤﺼل ﻋﻠﻰ ∂ ∂ ∂ ∂ H x − y U E (x, t, z ) = E x − y U E (x, y, z ) ∂x ∂x ∂y ∂y
(18)
.ﺍﻟﺠﺎﻨﺏ ﺍﻷﻴﻤﻥ ﻓﻲ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻪ ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ ∂ ∂ ∂ ∂ x − y E U E ( x, y, z) = x − y ∂x ∂x ∂y ∂y
H U E ( x, y, z )
.H ﺘﺒﺎﺩﻟﻲ ﻤﻊ
∂ ∂ x − y ∂x ∂y
:ﻓﺈﻥ h ∂ ∂ L z = x − y i ∂y ∂x
(19)
= x Py − y Px
L =r
ﺃﻱ ﺃﻥ ﺍﻟﻤﺅﺜﺭ , P
ﻟﻭ ﻋﺭﻓﻨﺎ
(20)
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(H L z − L z H) U E (x, y, z) = 0
ﺒﻤﺎ ﺃﻥ
) U E (r
ﺘﻜﻭﻥ ﻓﺌﺔ ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﻋﻼﻗﺔ ﺍﻟﻤﺅﺜﺭ.
[H , L ] = 0
)(21
ﺘﻜﻭﻥ ﺼﺤﻴﺤﺔ.
Lz
z
ﻫﻭ ﺍﻟﻤﺭﻜﺒﺔ
z−
ﻟﻠﻤﺅﺜﺭ. × P
)(22
L =r
ﻭﻫﻭ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺯﺍﻭﻴﺔ .ﻭﻟﻭ ﺃﺨﺫﻨﺎ ﺍﻟﺩﻭﺭﺍﻥ ﺤﻭل y, xﺒﺈﻤﻜﺎﻨﻨﺎ ﺃﻥ ﻨﺠﺩ
ﺍﻟﻌﻼﻗﺎﺕ.
[H , L ] = 0
[H , L ] = 0 x
)(23
y
ﻭﻫﻜﺫﺍ ،ﻓﺈﻥ ﻤﺅﺜﺭﺍﺕ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﺜﻼﺜﺔ ﻤﺘﺒﺎﺩﻟﺔ ﻤـﻊ Hﺃﻱ ﺃﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺯﺍﻭﻴﺔ ﺜﺎﺒﺕ ﻓﻲ ﺍﻟﺤﺭﻜﺔ .ﻭﻫﺫﺍ ﻴﻭﺍﺯﻱ ﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻤﻥ ﺤﻴـﺙ ﺃﻥ ﺍﻟﻘﻭﻯ ﺍﻟﻤﺭﻜﺯﻴﺔ ﻴﻨﺘﺞ ﻋﻨﻬﺎ ﺤﻔﻅ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺯﺍﻭﻴﺔ. ﻋﻼﻗﺎﺕ ﺍﻟﺘﺒﺩﻴل ﻟﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺯﺍﻭﻴﺔ.
)(24
− [yp z , xp z ]
] ]
]
[ ] = [yp
, L y = yp z − zp y , zp x − xp z x
[ ] + [ zp
, zp y − zp y , zp x , xp z
y
z
[L
= y [ Pz , z ] Px + x [z , p z ]p y h (yp x − xp x ) = ih L z i
=
ﻭﺒﺎﻟﻤﺜل: )(25
, L z = ih L x
y
[L
]
)(26
, L x ] = ih L y
z
[L
ﻋﻼﻗﺎﺕ ﺍﻟﺘﺒﺩﻴل ) (24ﻭ ) (26ﻴﻤﻜﻥ ﺘﻠﺨﻴﺼﻬﺎ ﺒﺎﻟﻘﺎﻨﻭﻥ - ١٣٧PDF created with pdfFactory Pro trial version www.pdffactory.com
L× L = i h L
ﻭﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﻤﺘﻭﻗﻌﺔ ﻋﻠﻰ ﻋﻼﻗﺎﺕ ﺍﻟﺘﺒﺩﻴل ﻫﺫﻩ ﻫﻲ ﺃﻨﻪ ﺘﻭﺠﺩ ﻓﻘـﻁ ﻤﺭﻜﺒـﺔ
ﻭﺍﺤﺩﺓ ﻟـ
L
ﻴﻤﻜﻥ ﺍﺨﺘﻴﺎﺭﻫﺎ ﻟﺘﺒﺎﺩل ﻤﻊ Hﻟﺘﻜﻭﻥ ﻓﺌﺔ ﻤﻥ ﺍﻟﻜﻤﻴﺎﺕ ﺍﻟﻤﺸﺎﻫﺩﺓ.
ﺸﺭﺡ ﻤﻔﺼل ﻟﻠﺨﻁﻭﺓ:
]
]
[
]
− zp y , zp x − xp z
[
z
[ L , L ] = [yp
= [yp z , zp x ]− zp y , zp x − [ yp z , xp z ] + zp y , xp z
y
x
ﻭﻟﺘﻭﻀﻴﺢ ﻫﺫﺍ ،ﻨﻔﺘﺭﺽ ﺃﻨﻪ ﻴﻭﺠﺩ ﻟﺩﻴﻨﺎ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺨﺎﺼﺔ ﻟﻠﻤﺅﺜﺭ Lxﻭﺍﻟﺫﻱ ﻟﻪ
ﻗﻴﻤﺔ ﺨﺎﺼﺔ ﻭﺍﻟﺘﻲ ﻫﻲ ﺘﻜﻭﻥ ﺩﺍﻟﺔ ﺨﺎﺼﺔ ﻟﻠﻤﺅﺜﺭ Lyﺒﻘﻴﻤﺔ ﺨﺎﺼﺔ .L2
ﻭﻫﻜﺫﺍ ﺒﺎﺴﺘﻁﺎﻋﺘﻨﺎ ﺍﺨﺘﻴﺎﺭ ﻓﺌﺔ ﻜﺎﻤﻠﺔ ﻤـﻥ ﺍﻟﻜﻤﻴـﺎﺕ ﺍﻟﻤـﺸﺎﻫﺩﺓ ﺍﻟﻤﺘﺒﺎﺩﻟـﺔ ﻟﻠﻤﺅﺜﺭﺍﺕ
L2 , H L z
ﻓﺼل ﺍﻟﻤﺘﻐﻴﺭﺍﺕ ﻓﻲ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ Separation of Variables is the Schrödinger Equation
ﻓﻲ ﺍﻟﻔﺼل ﺍﻟﻘﺎﺩﻡ ) (11ﺴﻨﺤﺩﺩ ﺍﻟﻘﻴﻡ ﻭﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟﻠﻤﺅﺜﺭﺍﺕ
, Lz
، Lﻫﻨﺎ
2
ﺴﻨﻭﻀﺢ ﺃﻥ ﺍﺴﺘﺨﺩﺍﻤﻬﺎ ﻴﺒﺴﻁ ﻜﺜﻴﺭﺍﹰ ﺤل ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ .ﻭﻫﺫﺍ ﻴﻨﺘﺞ ﻤﻥ ﺍﻟﻌﻼﻗﺔ ﺍﻟﻤﺸﺘﻘﺔ ﺍﻟﺘﺎﻟﻴﺔ:
] ) ( ) = [(r × p ) ] + [ (r × p ) ] + [ (r × p 2
2
z
2
y
∂ ∂ y − z ∂y ∂z ∂ ∂ z − x ∂z ∂x
2 ∂2 ∂2 + z 2 + 2 ∂x ∂y
2
x
2
L = r×p
∂ ∂ = − h 2 y − z ∂y ∂z ∂ ∂ −h2 z − x ∂z ∂x
∂ ∂ ∂ ∂ − h 2 x − y x − y ∂z ∂x ∂y ∂y ∂2 ∂2 ∂2 ∂2 = − h 2 x 2 2 + 2 + y 2 2 + 2 ∂y ∂z ∂z ∂y
∂2 ∂2 ∂2 ∂ ∂ ∂ − 2 yz − 2 xy − 2 zx − 2x − 2 y − 2z ∂y∂z ∂z∂x ∂x ∂y ∂z ∂x∂y
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(r − p) = − h
∂ ∂ ∂ ∂ ∂ ∂ x + y + z + y + z x ∂z ∂y ∂z ∂x ∂y ∂x ∂2 ∂2 ∂2 ∂2 ∂2 + 2yz = − h 2 (x 2 2 + y 2 2 + z 2 2 + 2xy ∂y∂z ∂x∂y ∂z ∂y ∂x
2
2
+ 2zx
=h2 { y
∂ ∂ y ∂ z ∂ z
∂2 ∂ ∂ ∂ +x +y +z ) ∂z∂x ∂x ∂y ∂z
∂ ∂ ∂ − z − y z ∂z ∂y ∂y
∂ ∂ ∂ − z z ∂x ∂x ∂x
+x
∂ ∂y
∂ + z ∂y
∂ z ∂y
∂ ∂ ∂ ∂ ∂ + x x − x z x ∂ z ∂ z ∂z ∂x ∂z ∂ ∂ ∂ ∂ ∂ ∂ y x + y y − y −x ∂ x ∂ z ∂x ∂y ∂y ∂z
+z
∂ x ∂y
∂ y ∂z
(28)
∂2 ∂2 ∂ ∂ ∂2 + + = y y 2 − y z − z y ∂ z ∂ y ∂ y ∂ y∂ z ∂ z ∂z ∂2 ∂2 ∂2 ∂2 ∂ ∂ z z z x x + z z + + − + − z 2 2 ∂z∂ x ∂x ∂ x ∂ z ∂ z ∂y ∂x ∂2 ∂2 ∂2 ∂ x + x x 2 + x x − + y 2 ∂ y∂ z ∂ z ∂z ∂y ∂2 ∂ ∂2 y + + − y x y ∂ x ∂ y ∂ y ∂x∂z 2 ∂2 ∂2 2 ∂2 ∂2 ∂2 2 ∂ x z = y2 2 + + + + + 2 2 ∂x∂z ∂ y2 ∂ x2 ∂ z ∂z ∂ y ∂2 ∂2 ∂2 − 2yz − 2xy − 2z x ∂ x ∂ y ∂ z ∂ y ∂ z ∂ y ∂ ∂ ∂ − 2y − 2z − 2x ∂y ∂z ∂x
.( ﻴﻌﻁﻴﻨﺎr . p ) ﻭ 2
2
→2
(
→ → + r . p = − h2 x2 + y2 + z2
) ∂∂x
2
2
+
∂2 ∂2 + ∂ y2 ∂ z2
2
ﻭﻤﺠﻤﻭﻉ
2 ∂ ∂ ∂ + h x (29) +y +z ∂y ∂ z ∂x
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2
→ → → → → → + r . p = r . p + ih r . p
)(30
→2
ﻻﺤﻅ ﺃﻨﻨﺎ ﻨﺘﻌﺎﻤل ﻤﻊ ﻤﺅﺜﺭﺍﺕ ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﺍﻟﻤﺤﺎﻓﻅﺔ ﻋﻠﻰ ﺘﺭﺘﻴﺏ ﺍﻟﺤـﺩﻭﺩ ﻤﻬﻡ ﺠﺩﺍﹰ ﻤﻥ ).(30 → → → → + r . p − ih r . p 2
)(31
2
1 p = 2 r 2
ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺘﺨﺘﻠﻑ ﻤﻥ ﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻭﺫﻟﻙ ﻷﻥ 2
ﻭﺒﺎﻋﺘﺒﺎﺭ
∂ ∂ 1 r − h 2 r ∂r ∂r
)ﻻﺤﻅ ﻫﻨﺎ ﺃﻨﻨﺎ ﻋﻭﻀﻨﺎ ﻋﻥ
1 −h 2 r
∂ ∂r
2
p . r
1 p = 2 r 2
( p = − i hﻓﺈﻥ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺘﺄﺨﺫ ﺍﻟﺸﻜل →
u E (r) + V (r) u E (r) = Eu E (r) #
)(33
2
→
→
ﻻ ﻴﺘﺒﺎﺩﻻ.
∂ ∂ ∂ 1 1 r r + − 2 2 ∂r ∂r r ∂r h r
2
h2 1 − 2μ r 2
ﻭﻟﻭ ﺃﻨﻨﺎ ﺃﺨﺫﻨﺎ ﺍﻟﺤﺴﺎﺒﺎﺕ ﺒﺎﻻﺤﺩﺍﺜﻴﺎﺕ ﺍﻟﻜﺭﻭﻴﺔ )ﺍﻟﻤﻭﻀﺤﺔ ﻓﻲ ﺍﻟﺸﻜل( ﻭﻫﺫﺍ
ﻫﻭ ﺍﻟﻁﺒﻴﻌﻲ ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ،ﺤﻴﻨﺌﺫ ﻓﺈﻥ ﺍﻟﻤﺅﺜﺭ ﺍﻟﻭﺤﻴﺩ ﺍﻟﺫﻱ ﻴﺤﺘﻭﻱ ﻋﻠﻰ ﺍﻟﺯﻭﺍﻴﺎ ﺍﻟﻜﺭﻭﻴﺔ
y,θ
ﻫﻭ
2
.ﻭﻟﻬﺫﺍ ﻓﻠﻭ ﺃﺨﺫﻨﺎ ﺍﻟﺩﻭﺍل ﻟﻬﺎ ﺍﻟﺸﻜل. )(34
)u E (r) = Yλ (θ , φ ) R Eλ (r
ﺤﻴﺙ ) Yλ (θ , φ ) = λ Yλ (θ , φ
)(35
ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺨﺎﺼﺔ ﻟﻠﻤﺅﺜﺭ
2
2
،ﻋﻨﺩﺌﺫ ﻓﺈﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺴﺘﻨﻔﺼل ﺇﻟـﻰ
) (35ﻭﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻘﻁﺭﻴﺔ )ﺍﻟﺨﻁﻴﺔ( ﻻﺤﻘﺎﹰ ).(39
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ﺍﻟﺸﻜل ﺃﻋﻼﻩ ﻴﺒﻴﻥ ﺍﻹﺤﺩﺍﺜﻴﺎﺕ ﺍﻟﻜﺭﻭﻴﺔ ،ﻭﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻹﺤﺩﺍﺜﻴﺎﺕ ﺍﻟﻜﺎﺭﺘﻴﺯﻴﺔ
) (x,y,zﻭﺍﻟﻜﺭﻭﻴﺔ ).(r, o, y
ﺇﻥ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﻤﺘﺒﻌﺔ ﻫﻨﺎ ﻻ ﺘﺨﺘﻠﻑ ﻋﻥ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﻌﺎﺩﻴﺔ ﺍﻟﻤﺘﻌﺎﺭﻑ ﻋﻠﻴﻬﺎ ﻓﻲ ﻓﺼل ﺍﻟﻤﺘﻐﻴﺭﺍﺕ .ﻭﻟﻜﻥ ﻫﻨﺎ ﻨﺅﻜﺩ ﻋﻠﻰ ﺩﻭﺭ ﺍﻟﺘﻤﺎﺜل symmetryﻓﻲ ﺘﺤﺩﻴﺩ ﺍﻟﻔﺌـﺔ ﺍﻟﻭﺍﺼﻠﺔ ﻟﻠﻤﺅﺜﺭﺍﺕ ﺍﻟﻤﺘﺒﺎﺩﻟﺔ )ﺍﻟﺘﺒﺎﺩﻟﻴﺔ( .commuting epertes ﻨﺭﻯ ﺃﻥ λﻟﻬﺎ ﺃﺒﻌﺎﺩ ، hﻭﻨﻜﺘﺒﻬﺎ ﺒﺎﻟﺸﻜل. 2
λ = l (l + 1) h 2
)(36
ﻓﻲ ﺍﻟﻔﺼل ﺍﻟﻘﺎﺩﻡ ﺴﻨﺜﺒﺕ ﺃﻥ
l = 0, 1, 2, 3, .............
ﻭﻨﺴﺘﺨﺩﻡ ﻫﺫﻩ ﺍﻟﺤﻘﻴﻘـﺔ ﻓـﻲ
ﺍﻷﺠﺯﺍﺀ ﺍﻟﺘﺎﻟﻴﺔ .ﺴﻨﻜﺘﺏ ﺍﻟﺩﻭﺍل ﺍﻟﺨﺎﺼﺔ ﻟﻜﻤﻴﺔ ﺍﻟﺘﺤﺭﻙ ﺍﻟﺯﺍﻭﻱ ﻋﻠـﻰ ﺍﻟـﺼﻭﺭﺓ ) ، Y (θ , φﺤﻴﺙ lm
l
ﺘﺩل ﻋﻠﻰ )(37
) Ylm (θ , φ
ﻫﻲ ﻜﺫﻟﻙ ﺩﺍﻟﺔ ﺨﺎﺼﺔ ﻟﻠﻤﺅﺜﺭ
) Ylm (θ , φ ) = mh Ylm (θ , φ
ﻭﺴﻨﺠﺩ ﻜﺫﻟﻙ ﺃﻥ mﻫﻭ ﺭﻗﻡ ﺼﺤﻴﺢ
2
ﻭ
z
.
z
−l ≤ m ≤ l
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ﻭﺒﻤﺎ ﺃﻥ
Ylm
ﻫﻲ ﺩﻭﺍل ﺨﺎﺼﺔ ﻟﻤﺅﺜﺭﺍﺕ ﻫﻴﺭﻤﻴﺘﻴﺔ ،ﻓﺈﻨﻨﺎ ﻨﺘﻭﻗﻊ ﺃﻥ
Ylm
ﺍﻟﺘﻲ
ﺘﻨﺎﻅﺭ ﻗﻴﻡ ﺨﺎﺼﺔ ﻤﺨﺘﻠﻔﺔ ﺘﻜﻭﻥ ﻤﺘﻌﺎﻤﺩﺓ .ﻭﺴﻨﺜﺒﺕ ﻓﻲ ﺍﻟﻔﺼل ﺍﻟﻘـﺎﺩﻡ ﺃﻥ ﻋﻨـﺩﻤﺎ ﻴﻜﻭﻥ ﻫﻨﺎﻙ ﻤﻌﺎﻴﺭﺓ ﺤﻘﻴﻘﻴﺔ ،ﻭﺍﻟﺘﻜﺎﻤل ﻋﻠﻰ ﻤﺩﻯ ﺍﻟﻤﺘﻐﻴﺭﺍﺕ . φ ,θ ) (θ , φ (θ , φ ) = Sl1l2 S m1m2
)(38
d Ω Yl m (θ , φ ) Yl
∫
*
2m2
1 1
2D
D
= ∫ sin θ d θ ∫ dφ Yl m (θ , φ ) Yl *
2m 2
1 1
ο
ο
ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻨﺼﻑ ﻗﹸﻁﺭﻴﺔ The Radial Equation
ﻋﻨﺩﻤﺎ ﻨﻌﻭﺽ ) (35ﻭ ) (36ﻓﻲ ) (33ﻭ ) ،(34ﻓﺈﻨﻨﺎ ﻨﺘﺤﺼل ﻋﻠﻰ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ. h 2 1 d d l ( l + 1) ) R (r) + V (r) R Elm = R Elm (r r − 2μ r dr dr r 2 Elm
) (39
ﻤﻌﻁﺎﻩ
−
ﻻﺤﻅ ﺃﻨﻪ ﻻ ﻴﻭﺠﺩ ﺍﻋﺘﻤﺎﺩ ﻋﻠﻰ mﻓﻲ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ .ﻭﻋﻠﻴﻪ ،ﻓﺈﻨـﻪ ﻟﻘﻴﻤـﺔ l
ﺴﻴﻜﻭﻥ ﻫﻨﺎﻙ
)( 2l + 1
ﻤﻥ ﺍﻟﺘﻜﺭﺍﺭ ،deglnemacyﻭﺫﻟﻙ ﻷﻥ ﻜل ﻗﻴﻡ m
ﺍﻟﻤﻤﻜﻨﺔ ﺴﻴﻜﻭﻥ ﻟﻬﺎ ﻨﻔﺱ ﺍﻟﻁﺎﻗﺔ .ﻭﻟﻬﺫﺍ ﺤﺫﻓﻨﺎ mﻤﻥ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴـﺔ
ﻭﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺘﻬﺎ ﻋﻠﻰ ﺍﻟﺼﻴﻐﺔ.
d2 2 d 2μ l (l + 1) h 2 2 μE 2 + R nl (r ) − 2 V (r ) + R nl (r ) + 2 R nl = 0 2 r dr 2μ r h h dr
)(40
ﺤﻴﺙ ﺍﺴﺘﺒﺩﻟﻨﺎ ﺍﻟﺭﻤﺯ Eﺒﺎﻟﺭﻤﺯ nﻓﻲ ﺍﻟﺩﺍﻟـﺔ ﺍﻟﺨﺎﺼـﺔ
)(r
ne
. Rﻭﺴـﻨﻘﻭﻡ
ﺒﺎﺨﺘﺒﺎﺭ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻟﻠﻌﺩﻴﺩ ﻤﻥ ﺃﺸﻜﺎل ﺍﻟﺠﻬﺩ ﻤﻘﻴﺩﺓ ﺒﺎﻟﺸﺭﻁ ﺃﻨﻬـﺎ ﺘـﺼﻴﺭ ﺇﻟـﻰ ﺍﻟﺼﻔﺭ ﻋﻨﺩ
ﺃﺴﺭﻉ ﻤﻥ
∞
1 r
ﻤﺎ ﻋﺩﺍ ﻟﻠﺤﺎﻟﺔ ﺍﻟﻤﻬﻤـﺔ ﺍﻟﺨﺎﺼـﺔ ﻟﺠﻬـﺩ ﻜﻭﻟـﻭﻡ
Coulomb potentialﻭﺍﻟﺫﻱ ﺴﻴﻨﺎﻗﺵ ﻓﻲ ﺍﻟﻔﺼل ) .(12ﻭﺴـﻨﻔﺘﺭﺽ ﻜـﺫﻟﻙ ﺃﻥ ﺍﻟﺠﻬﺩ ﻟﻴﺱ ﻤﻔﺭﺩ .Singular ﻜﻔﺭﺩﻴﺔ
1 r
ﻋﻨﺩ ﻨﻘﻁﺔ ﺍﻷﺼل،
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)(41
lim r 2 V (r) = 0 r →ο
ﻭﺇﻨﻪ ﻤﻥ ﺍﻟﻤﻨﺎﺴﺏ ﺃﺤﻴﺎﻨﺎﹰ ﺃﻥ ﻨﺴﺘﺤﺩﺙ ﺍﻟﺩﺍﻟﺔ. )(42
ﻭﺒﻤﺎ ﺃﻥ:
)(43
)u nl (r) = r R nl (r
d 2 2 d u re (r ) d u 1 du 2 u 1 du 2 + = − 2 + + − + r dr r dr r r dr r r r dr dr 2 1 du 1 d 2 u 1 du 2u 2 du = 3 u− 2 + − − + r r dr r dr 2 r 2 dr r 3 r 2 dr 1 d2u = r dr 2
∴ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻴﻤﻜﻨﻨﺎ ﻜﺘﺎﺒﺘﻬﺎ ﺍﻵﻥ )ﻤﻊ ﻀﺭﺏ ﻁﺭﻓﻴﻬﺎ ﻓﻲ (r ) (44
u nl (r ) = 0
d 2 u re (r) 2μ l (l + 1 ) h 2 ( ) + E − V r − dr 2 h2 2μ r 2
ﻭﻫﺫﻩ ﺘﺒﺩﻭ ﻤﺸﺎﺒﻬﺔ ﻟﻠﻤﻌﺎﺩﻟﺔ ﻓﻲ ﺒﻌﺩ ﻭﺍﺤﺩ ﻤﺎ ﻋﺩﺍ
) (aﺃﻥ ﺍﻟﺠﻬﺩ .barrier
) V (r
l (l + 1)h 2 2μ r 2
ﺘﻐﻴﺭ ﺒﺈﻀﺎﻓﺔ ﺍﻟﺤﺎﺠﺯ ﺍﻟﻤﺭﻜﺯ ﺍﻟﻤﻨﹶﻔﱢﺭ repulsive centrifugal
V (r ) → V (r ) +
ﺍﻟﺸﻜل :ﺍﻟﺠﻬﺩ ﺍﻟﻔﻌﺎل ﺍﻟﻤﺅﺜﺭ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴـﺔ ﻟــ
) u = r R (r
ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ ﺍﻟﺠﻬﺩ ﺍﻟﺤﻘﻴﻘﻲ ﻤﺭﺒﻊ.
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) (bﺍﻟﺘﻌﺭﻴﻑ ﻟﻠﺩﺍﻟﺔ ﺃﻥ:
) u nl (r
ﻭﻤﺤﺩﻭﺩﻴﺔ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻋﻨﺩ ﻨﻘﻁﺔ ﺍﻷﺼل ﻴﻠﺯﻡ
u nl (0 ) = 0
ﻤﻤﺎ ﻴﺠﻌﻠﻬﺎ ﻤﺜل ﻤﺴﺄﻟﺔ ﺍﻟﺒﻌﺩ ﺍﻟﻭﺍﺤﺩ ﺍﻟﺫﻱ ﻟﻪ
∞V=r
ﻓﻲ ﺍﻟﺠﺎﻨﺏ ﺍﻷﻴﺴﺭ.
ﺃﻭﻻﹰ :ﻨﻌﺘﺒﺭ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ ﻗﺭﻴﺒﺎﹰ ﻤﻥ ﺍﻷﺼل ،Origin
ﻋﻨﺩﻤﺎ
ﻭﺒﺈﺴﻘﺎﻁ u nl
ﻓﺈﻥ ﺍﻟﺤﺩﻭﺩ ﺍﻟﻤﻬﻴﻤﻨﺔ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ.
r→ο
)d 2 u l (l + 1 − 2 u~ο dr 2 r
) (47
ﻭﺒﺎﻟﺘﻌﻭﻴﺽ Ansatz
u (r ) ~ r s
ﺴﻨﺠﺩ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﻭﻥ ﺼﺤﻴﺤﺔ ﺒﺸﺭﻁ ﺃﻥ: s ( s −1) − l ( l + 1 ) = 0
ﺃﻱ ﺃﻥ
s = l +1
ﺃﻭ
s =−l
ﻭﺍﻟﺤل ﺍﻟﺫﻱ ﻴﺤﻘﻕ ﺍﻟﺸﺭﻁ u (ο) = 0ﺃﻱ ﺍﻟﺤل ﺍﻟﺫﻱ ﻴﺴﻠﻙ ﻤﺜـل ﺍﻟﺤل ﺍﻟﻤﻨﺘﻅﻡ regular sohtimﺍﻟﺤل ﺍﻟﺫﻱ ﻴﺴﻠﻙ ﻤﺜل
r -l
r l +1
ﻴـﺴﻤﻰ
ﻫﻭ ﺍﻟﺫﻱ ﺍﻟﻐﻴﺭ ﻤﻨـﺘﻅﻡ
irregnlarﻭﺒﺩﻻﻟﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ ) R (rﻟﺩﻴﻨﺎ ﺍﻟﺤل ﺍﻟﻤﻨﺘﻅﻡ ﻴﺴﻠﻙ ﻤﺜل
ﻭﺍﻟﺤل ﺍﻟﻐﻴﺭ ﻤﻨﺘﻅﻡ ﻴﺴﻠﻙ ﻤﺜل
ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ rﻜﺒﻴﺭﺓ ﺒﺈﻤﻜﺎﻨﻨﺎ ﺇﻫﻤﺎل ﺤﺩﻭﺩ ﺍﻟﺠﻬﺩ ﻭﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﺼﺒﺢ. ) (50
ﻭﺸﺭﻁ ﻤﺭﺒﻊ ﺍﻟﺘﻜﺎﻤل )(51
2
d 2 u 2μ E + 2 u =0 dr 2 h
) r 2 dr ∫ d Ω R nl (r ) Ylm ( θ , ϕ 2
) r 2 dr R nl (r
∞
∫
∞
∫ ο
2
) 1 = ∫ d r ψ (r 2
=
ο
ﺃﻱ: ) (52
2
) dr u nl (r
∞
∫ ο
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ﻟﻜﻲ ﺘﺘﻼﺸﻰ ﺍﻟﺩﺍﻟﺔ ﻋﻨﺩ * ﻓﻲ ﺤﺎﻟﺔ
∞
E < 0
)(53
ﺍﻟﺤل ﺍﻟﻤﻘﺎﺭﺏ
d 2u +α 2 u = 0 2 dr
)(54
* ﻓﻲ ﺤﺎﻟﺔ
2μ E = −α2 h2
u (r ) ~ e − α r
E > 0
ﻟﺩﻴﻨﺎ ﺍﻟﺤﻠﻭل ﻭﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻤﻌﺎﻴﺭﺓ ﻓﻲ ﺩﺍﺨل ﺼﻨﺩﻭﻕ. )(55
ﺍﻟﺤل ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ ﻤﻥ
e ikr
ﻭ
− ikr
γ
2μ E =k2 2 h d 2u −u2 =0 2 dr
ﻜﺒﻴﺭﺓ ﺒﺤﻴﺙ ﺃﻥ ) V(rﻤﻬﻤﻠﺔ ﻴﻜﻭﻥ ﻫﻭ ﺍﺘﺤﺎﺩ ﺨﻁـﻲ
.e
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ﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ
The Hydrogen Atom
ﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﺃﺒﺴﻁ ﺫﺭﺓ ﻭﺫﻟﻙ ﻷﻨﻬﺎ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﺇﻟﻜﺘﺭﻭﻥ ﻭﺍﺤﺩ .ﻭﻟﻬﺫﺍ
ﺍﻟﺴﺒﺏ ﺘﺼﺒﺢ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻤﻌﺎﺩﻟﺔ ﺠﺴﻡ ﻭﺍﺤﺩ ﻭﺫﻟﻙ ﺒﻌﺩ ﻓﺼل ﺤﺭﻜﺔ ﻤﺭﻜﺯ ﺍﻟﻜﺘﻠﺔ ﻭﺴﻭﻑ ﻨﺘﻌﺎﻤل ﻤﻊ ﺫﺭﺍﺕ ﻤﺸﺎﺒﻬﺔ ﻟﻠﻬﻴﺩﺭﻭﺠﻴﻥ ،ﺃﻱ ﺃﻥ ﺍﻟﺫﺭﺍﺕ ﺍﻟﺘﻲ ﺘﺤﺘﻭﻱ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﺃﻜﺜﺭ ﻤـﻥ ﺒﺭﻭﺘـﻭﻥ
ﻋﻠﻰ ﺇﻟﻜﺘﺭﻭﻥ ﻭﺍﺤﺩ ﻓﻘﻁ ،ﻭﻟﻜﻥ ﺴﻨﺴﻤﺢ ﻭﺍﺤﺩ .ﻭﺍﻟﺠﻬﺩ ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﻴﻜﻭﻥ
Ze2 r
)(1
V(r ) = −
ﻭﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ ﺘﻜﻭﻥ d2 2 d 2μ Ze 2 l (l + 1) h 2 2 + R + 2 E + − R =0 r dr r 2μ r 2 h dr
)(2
ﻭﺴﻨﺭﻜﺯ ﺍﻟﺩﺭﺍﺴﺔ ﻋﻠﻰ ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﻘﻴﺩﺓ ،ﺃﻱ ﺤﻠﻭل
E < 0
ﺃﻥ ﻨﻌﻤل ﺘﻐﻴﻴﺭ ﻓﻲ ﺍﻟﻤﺘﻐﻴﺭﺍﺕ.
2
)(3
1
r
ﺇﻨﻪ ﻤﻥ ﺍﻟﻤﻨﺎﺴﺏ
8µ E δ = 2 h
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﺘﺼﺒﺢ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (2ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ ﺍﻟﺘﺎﻟﻴﺔ: )d 2 R 2 dR l ( l + 1 λ 1 + − R+ − R = 0 2 2 δ dδ dδ δ δ 4
)(4
ﻭﻗﺩ ﺍﺴﺘﺤﺩﺜﻨﺎ ﺍﻟﻤﻌﺎﻤل )ﺒﺩﻭﻥ ﻭﺤﺩﺓ( )(5
2
1
μ C2 = Zα 2 E
ﺍﻟﻤﻌﺎﺩﻟﺔ ) (4ﻤﻥ ﺍﻟﺴﻬل ﺤﻠﻬﺎ .ﻭﺫﻟﻙ
2
1
1 137
μ 2 E
Z e2 =λ h
= ، αﻭﺍﻟﻁﺎﻗﺔ ﻤﻌﺒﺭ ﻋﻨﻬﺎ ﺒﻭﺤﺩﺍﺕ
ﺍﻟﻜﺘﻠﺔ ﻋﻨﺩ ﺍﻟﺴﻜﻭﻥ . rest – mass
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ﻁﻴﻑ ﺍﻟﻁﺎﻗﺔ
The Energy Spectrum
ﺴﻨﺤﺎﻭل ﺤل ﻤﻌﺎﺩﻟﺔ ) (4ﺒﻁﺭﻴﻘﺔ ﻤﻌﺭﻭﻓﺔ ﻟﺩﻴﻨﺎ ﺍﻵﻥ .ﺃﻭﻻﹰ ﺴـﻨﺠﺩ ﺴـﻠﻭﻙ
ﺍﻟﺩﺍﻟﺔ ﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ δﻜﺒﻴﺭﺓ .ﻭﻋﻨﺩﻤﺎ ﺘﻜﻭﻥ
δ
ﻜﺒﻴﺭﺓ ﻓﺎﻟﺤﺩﻭﺩ ﺍﻟﻤﺘﺒﻘﻴﺔ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ.
d2R 1 − R=0 4 dδ 2
) (6
ﻭﺍﻟﺤل ﺍﻟﻤﻘﺒﻭل ﻋﻨﺩ ﻤﺎ ﻻ ﻨﻬﺎﻴﺔ 2
+δ
⇐ R ~e
ﻋﻨﺩ 2
−δ
∞ → ∞ ← R ⇐δ
R~e
ﻭﻜﻤﺎ ﻋﺎﻟﺠﻨﺎ ﺍﻟﻤﺘﺫﺒﺫﺏ ﺍﻟﺘﻭﺍﻓﻘﻲ ) G (δ
) (7
δ 2
−
R= e
ﻭﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ )(4
ﺴﻨﺘﺤﺼل ﻋﻠﻰ ﻤﻌﺎﺩﻟﺔ ﻟـ ):G (s d 2 G 2 dG λ − 1 l (l + ) G =0 − 1 − + − dδ 2 δ dδ δ δ 2
)(8
ﻭﺍﻵﻥ ﻨﻜﺘﺏ ﻤﻔﻜﻭﻙ ﺍﻟﻘﻭﺓ ﻟـ ) G (δﻭﺍﻟﺘﻲ ﺘﺄﺨﺫ ﺍﻟﺸﻜل: )(9
δn
∞
n
∑a
n =ο
G (δ ) = δ l
ﻭﺒﺘﻌﻭﻴﺽ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (aﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀـﻠﻴﺔ ،ﺴـﻨﺠﺩ ﺍﻟﻌﻼﻗـﺔ ﺒـﻴﻥ
ﺍﻟﻤﻌﺎﻤﻼﺕ ﺍﻟﻤﺨﺘﻠﻔﺔ
an
ﻭﻤﻌﺎﺩﻟﺔ ﺍﻻﺴﺘﻁﺭﺍﺩ ﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻴﻬﺎ ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟـﺔ
ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ﺍﻟﺨﺎﻀﻌﺔ ﻟﻠﻌﻼﻗﺔ
)(10
∞
H (δ ) = ∑ a n δ n n =ο
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ﻭﺍﻟﺘﻲ ﺘﻜﻭﻥ d 2 H 2 l + 2 d H λ −1− l H=0 + + − 1 δ dδ 2 δ dδ
)(11
ﻭﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟـﺔ ﻴﻤﻜﻨﻨـﺎ ﺍﻟﺤـﺼﻭل ﻋﻠﻴﻬـﺎ ﺒـﺴﻬﻭﻟﺔ ﻭﺫﻟـﻙ ﺒﺘﻌـﻭﻴﺽ ) G (δ ) = δ l H ( p
ﻓﻲ ﻤﻌﺎﺩﻟﺔ )(8
ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ: ) (12
2l + 2 δ n − 2 + n a n δ n −1 − 1 + (λ − 1 − l ) a n δ n −1 = 0 δ
]
}
+ (2 l + 2 ) a n + 1 + ( λ − 1 − l − n ) a n δ n −1 = 0
∑ n ( n − 1 ) a ∞
n
n=0
∑ {(n + 1)[n a ∞
n +1
n=0
ﻭﺒﻤﺎ ﺃﻥ ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﺘﻼﺸﻰ ﺤﺩ ﺒﺤﺩ ،ﻭﺴﻨﺘﺤﺼل ﻋﻠﻰ ﻤﻌﺎﺩﻟﺔ ﺍﻻﺴﺘﻁﺭﺍﺩ. n + l +1− λ ) ( n + 1 )(n + 2l + 2
)(13
=
a n +1 an
ﻭﻟﻘﻴﻡ ﻜﺒﻴﺭﺓ ﻟﻘﻴﻤﺔ nﻓﺈﻥ ﺍﻟﻨﺴﺒﺔ ﺘﺼﺒﺢ ) (14
1 n
≈
a n +1 an
ﻭﻜﻤﺎ ﻓﻌﻠﻨﺎ ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻤﺘﺫﺒﺫﺏ ﺍﻟﺘﻭﺍﻓﻘﻲ ،ﻓﺈﻨﻨﺎ ﻟﻥ ﻨﺘﺤﺼل ﻋﻠﻰ ﺴﻠﻭﻙ ﻤﻘﺒﻭل
ﻟﻠﺩﺍﻟﺔ ) R ( pﻋﻨﺩ
∞
ﺇﻻ ﺇﺫﺍ ﺍﻨﺘﻬﺕ ﺍﻟﺴﻠﺴﻠﺔ ) (aﺃﻱ ﺃﻨﻪ ﻟﻘﻴﻤﺔ ﻤﻌﻁﺎﺓ ﻟـ
l
ﻭ
n = nr
ﻴﺠﺏ ﺃﻥ ﻨﺘﺤﺼل ﻋﻠﻰ. ﺩﻋﻨﺎ ﺍﻵﻥ ﻨﺴﺘﺤﺩﺙ ﺍﻟﺭﻗﻡ ﺍﻟﻜﻤﻲ ﺍﻷﺴﺎﺴﻲ Nﺍﻟﻤﻌﺭﻑ ﺒـ )(16
ﻭﻴﻨﺘﺞ ﻤﻥ ﺍﻟﺤﻘﻴﻘﺔ ﺃﻥ
n = nr + l + 1
⇐ nr ≥ 0
1. n ≥ l +1 2. n is au int eger
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) (Zα 1 E = − µ c2 2 n2 E = hw
2
)(14
ﻭﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﺸﺒﻴﻬﺔ ﺒﻨﻤﻭﺫﺝ ﺒﻭﺭ .ﻻﺤﻅ ﺃﻥ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﺨﺘﺯﻟﺔ
µ
ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ .ﺇﻥ ﻭﺠﻭﺩ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﺨﺘﺯﻟﺔ. )(18
mµ m+ M
ﺍﻟﺘﻲ ﺘﻅﻬﺭ
=µ
ﺤﻴﺙ mﻜﺘﻠﺔ ﺍﻹﻟﻜﺘﺭﻭﻥ ،ﻭ Mﻜﺘﻠﺔ ﺍﻟﻨﻭﺍﺓ ،ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺘﺭﺩﺩﺍﺕ. )(19
Ei − Ej mC 2 /2h (Zα )2 12 − 12 = m h n ∫ n ∫ 1+ M
= ωiy
ﻭﺍﻟﺫﻱ ﻴﺨﺘﻠﻑ ﻗﻠﻴﻼﹰ ﻻﺨﺘﻼﻑ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﺍﻟﺸﺒﻴﻬﺔ ﻋﻠـﻰ ﺍﻷﺨـﺹ
ﺍﻟﻔﺭﻕ ﺒﻴﻥ ﻁﻴﻑ ﺫﺭﺓ ﺍﻟﻬﻴﺩﺭﻭﺠﻴﻥ ﻭﺫﺭﺓ ﺍﻟﺩﻴﺘﻭﻴﺭﻭﻡ.
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ﻅﺎﻫﺭﺓ ﺘﺤﻠل ﺍﻟﻁﻴﻑ
The Degeneracy of the spectrum
ﺍﻵﻥ ﺴﻨﻨﺎﻗﺵ ﺘﺤﻠل ﻁﻴﻑ ﺍﻟﻁﺎﻗﺔ ﻟﻘﻴﻤﺔ ، λ =1ﺍﻟﺤﺎﻟﺔ ﺍﻟﺩﻨﻴﺎ ،ﻴﺠﺏ ﺃﻥ ﻴﻜـﻭﻥ )ﻤـﻥ ﺍﻟﻤﻌﺎﺩﻟـﺔ = o, l = 0
ﺃﺨﺫﻨﺎ
r
λ = nr + l +1
ﻭ
. nﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻨﻪ ﺘﻭﺠﺩ ﺤﺎﻟﺔ ﻭﺤﻴﺩﺓ ﻓﺭﻴﺩﺓ ﻟﻠﺤﺎﻟﺔ ﺍﻟﺩﻨﻴﺎ ground staleﺇﺫﺍ
⇐ λ=2
ﻴﻜﻭﻥ ﻋﻨﺩﻨﺎ ﺍﺤﺘﻤﺎﻟﻴﻴﻥ
(i ) n r =1, l = 0 (ii ) n r = 0 , l =1
ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ
=1, l = 0
r
(i ) nﻓﺈﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) ،(13ﺘﻜﺘﺏ ﺒﺎﻟﺸﻜل
n + l + 1− nr − l −1 n + l +1 − λ = )(n + 1)(n + 2l + 2 ) (n + 1)(n + 2 l + 2 n − nr
) (20
=
) (n + 1)(n + 2 l + 2
a n +1 an
=
ﻭﻫﺫﺍ ﻴﺒﻴﻥ ﺃﻥ: a −1 −1 = ⇒ a1 = − ο (ο + ο + 2) 1 × 2 2
=
a1 aο
∞
⇒ H (δ ) = ∑ a n δ n = a ο δ ο + a 1 δ1 n=ο
δ
)(21
a = a ο + − ο 2 δ = a ο 1 − 2
) H (δ
ﺒﻴﻨﻤﺎ ﺍﻟﺘﻭﺯﻴﻊ ﺍﻟﺯﺍﻭﻱ ﻤﺘﻤﺎﺜل ﻜﺭﻭﻴﺎﹰ ﻟﻨﺄﺨﺫ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ(i i ) n r = 0, l = 1 .
ﻫﻨﺎ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ ﺜﺎﺒﺘﺔ H (δ ) = a ο
ﻟﻜﻥ ﺍﻟﺠﺯﺀ ﺍﻟﺯﺍﻭﻱ ﻤﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻴﺘﻀﻤﻥ ) . Y (ο , ϕ 1m
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ﺍﻟﺘﺤﻠل ﻴﻜﻭﻥ ) (2 l +1ﻭﻋﻠﻴﻪ ﻴﻭﺠﺩ ﺜﻼﺙ ﻤﻥ ﻤﺜل ﻫﺫﻩ ﺍﻟﺤﺎﻻﺕ. ﺍﻟﺘﺤﻠل ﺍﻟﻜﻠﻲ ﻟﻠﺤﺎﻟﺔ
ﺍﻻﺤﺘﻤﺎﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ:
λ =n=2
ﻴﻜﻭﻥ
3 +1= 4 = 2 2
ﻓﻲ ﺤﺎﻟﺔ
ﻴﻭﺠﺩ ﻟـﺩﻴﻨﺎ
λ =3
n r = 2, l = 0
ﻫﻨﺎ ﻴﻭﺠﺩ ﺤﺎﻟﺔ ﻭﺍﺤﺩﺓ ﻭﺍﻟﺘﻲ ﻟﻬﺎ
1 6
=−
a2 a1
ﻭ
= −1
a1 aο
ﻭﻋﻠﻴـﻪ ﻓـﺈﻥ ﺍﻟﺩﺍﻟـﺔ
ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ ﺘﺘﻀﻤﻥ. 1 H (δ ) = a ο 1 − δ + δ 2 6
)(22 nr = 1 , l = 1
ﺒﻴﻨﻤﺎ
ﻟﻬﺎ ﺜﻼﺙ ﺤﺎﻻﺕ ،ﺒـ
nr = 0 , l = 2
ﻟﻬﺎ
ﻭﻫﻜﺫﺍ ﻴﻭﺠﺩ λ =n=3
)5 = (2l +1
δ 1 − 4
ο
ﺤﺎﻻﺕ ﺒـ
1+ 3 + 5 = 9 = 32
، H (δ ) = a H (δ ) = a ο
ﺘﺤﻠل ﻟﻠﺤﺎﻻﺕ ﺍﻟﺘﻲ ﻟﻬﺎ ﻗﻴﻤﺔ ﺨﺎﺼـﺔ )ﺫﺍﺘﻴـﺔ(
ﻭﺒﺸﻜل ﻋﺎﻡ ،ﺍﻟﺘﺤﻠل ﻓﻲ ﺤﺎﻟﺔ
ﻴﻜﻭﻥ
λ=n
1 + 3 + 5 + ... + [2 (n − 1) + 1]= n 2
ﻭﻤﺴﺒﻘﺎﹰ ﻨﺤﻥ ﺘﻭﻗﻌﻨﺎ ﺘﺤﻠل ) (2l +1ﻟﻠﺠﻬـﺩ ﺍﻟﻨـﺼﻑ ﻗﻁـﺭﻱ ،ﻭﺫﻟـﻙ ﻷﻥ
ﺍﻟﻬﺎﻤﻠﺘﻭﻨﻴﺎﻥ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻱ ﻻ ﻴﻌﺘﻤﺩ ﻋﻠﻰ
z
،ﻭﻟﻜﻥ ﻴﻌﺘﻤﺩ ﻋﻠﻰ
.
2
ﻫﻨﺎ ﻴﻭﺠﺩ ﺘﺤﻠل ﺇﻀﺎﻓﻲ .ﻭﺍﻟﺘﺤﻠل ﺍﻟﺨﺎﺹ ﺼﻔﺔ ﻤﻤﻴﺯﺓ ﻟﻠﺠﻬﺩ . 1 r
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ﺍﻟﺩﻭﺍل ﺍﻟﺫﺍﺘﻴﺔ ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ The Radial Eigen functions
ﺒﺎﻟﺭﺠﻭﻉ ﻟﻠﺩﺍﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ .ﻟﻭ ﻋﻭﻀﻨﺎ
λ=n
ﺒﺤﻴﺙ:
ﻓﻲ ﻋﻼﻗـﺔ ﺍﻻﺴـﺘﻁﺭﺍﺩ )(13
k + l +1− n a (k + 1)(k + 2l + 2 ) k
) (26
= a k +1
ﺴﻨﺠﺩ ﺃﻥ: )n − (k + l + 1 ) n − (k + l . )(k + 1)(k + 2l + 2) k (k + 2l +1 )n − ( l + 1 a ......... 1. ( 2l + 2) ο
) (27
k +1
)a k + 1 = (− 1
ﺒﻤﺴﺎﻋﺩﺓ ﻫﺫﻩ ﺍﻟﻌﻼﻗﺔ ﺒﺈﻤﻜﺎﻨﻨﺎ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﻤﻔﻜﻭﻙ ﺴﻠﺴﻠﺔ ﺍﻟﻘﻭﺱ ﻟﻠﺩﺍﻟـﺔ ) . H (δﻭﻤﺎ ﻴﻜﺎﻓﺊ ﻫﺫﺍ.ﻨﺠﺩ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻟـ ) H (δﺘﻜﻭﻥ ﻟﻤﻌﻘﺩﺓ ﺍﻟﺤﺩﻭﺩ ) (δ
)(28
= ) H (δ
)( 2 l + 1 n − l −1
ﻭﺒﻌﺩﻤﺎ ﻨﹸﺤﻭل ﺒﺎﻟﻌﻜﺱ ﻟﻠﻭﺼﻭل ﻟﻺﺤـﺩﺍﺜﻴﺎﺕ ﺍﻟﻨـﺼﻑ ﻗﻁﺭﻴـﺔ rﻭﺒﻌـﺩ ﺍﻟﻤﻌﺎﻴﺭﺓ ﻨﺠﺩ ﺃﻥ ﺍﻟﺩﻭﺍل ﺍﻟﻨﺼﻑ ﻗﻁﺭﻴﺔ ﺍﻷﻭﻟﻰ ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻬﺎ ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ 12 − 30
h = a ο µ c α
Rnl (r ): zr aο
− 2 aο e zr
−
2
e
1 − z r 2a ο
2 2 z r 2 ( z r ) − 3 aο + 1− e 3 aο 27 a ο2 zr
2
2
3
2 2 z r 2 ( z r ) − 3 aο + 1− e 3 aο 27 a ο2 zr
3
3
2
z R10 (r ) = 2 aο
z R20 (r ) = 2 2 aο
z 3 2 a ο
1
3
= ) R21 (r
z R30 (r ) = 2 3 aο
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ﺭﺴﻡ ﻜﺜﺎﻓﺔ ﺍﻻﺤﺘﻤﺎل ﺍﻟﻨﺼﻑ ﻗﻁﺭﻱ
ﻟﻭﺠﻭﺩ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻋﻨﺩ ﻤﺴﺎﻓﺔ r
) P (r
ﻤﻥ ﻨﻘﻁﺔ ﺍﻷﺼل ﻴﻤﻜﻥ ﺇﻴﺠﺎﺩﻩ ﻤﻥ ﻫﺫﻩ ﺍﻟﺩﻭﺍل ﺍﻟﻤﻭﺠﻴﺔ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﺒﺎﻟـﺸﻜل
) (12-2ﻴﺠﺏ ﺃﻥ ﻨﺘﺫﻜﺭ ﺃﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻬﺎ ﺠﺯﺀ ﺯﺍﻭﻱ ،ﺍﻟﺫﻱ ﻤﺭﺒﻌﻪ ﺍﻟﻤﻁﻠﻕ ﻴﻜﻭﻥ
2
) Pem (cosθ
ﺍﻟﺭﺴﻭﻤﺎﺕ ﻟﺩﻭﺍل ﻟﻴﺠﻨﺩﺭ ﺍﻟﻤﺸﺎﺭﻜﺔ
) Pem (cosθ
2
ﻤﻭﻀﺤﺔ ﺒﺎﻟﺭﺴﻡ
).(12-3 ﻭﺒﻤﻌﺭﻓﺔ ﺍﻟﺩﻭﺍل ﺍﻟﻤﻭﺠﻴﺔ ﺒﺈﻤﻜﺎﻨﻨﺎ ﺤﺴﺎﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻭﻗﻌﺔ. ∞
]) = ∫ dr r 2 + k [ Rnl (r
2
k
r
ο
ﻭﻫﺫﻩ ﺒﻌﺽ ﺍﻟﻘﻴﻡ ﺍﻟﻤﺘﻭﻗﻌﺔ ﺍﻟﻤﻔﻴﺩﺓ. 2 2 a n = ο 2 5 n 2 + 1 − 3 l (l + 1) 2z ) (12 − 36 z = aο n 2 2 z = 1 2 3 aο n l + 2
]
)− l (l + 1
]
2
[
[3n 2z aο
= r r2 1 r 1 r2
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