© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﺍﻟﺒﺎﺏ ﺍﻟﺜﺎﻨﻲ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﻤﻭﺠﻴ ﺔ ﻭﺘﻁﺒﻴﻘﺎﺘﻬﺎ Applications of the Schrödinger’s Wave Equation
ﺍﻟﺼﻔﺤﺔ 1ﻓﺭﻭﺽ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ 2ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺒﻌﺩ ﻭﺍﺤﺩ 3ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻏﻴﺭ ﺍ ﻟﺯ ﻤ ﻨ ﻴ ﺔ 4ﻜ ﺜ ﺎ ﻓ ﺔ ﺍﻟﺘﻴﺎﺭ 5ﺘﻁﺒﻴﻘﺎﺕ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ 5.1ﺩﺭﺍﺴﺔ ﺤﺭﻜﺔ ﺠ ﺴ ﻴ ﻡ ﺤﺭ 5.2ﺩ ﺭ ﺍ ﺴ ﺔ ﺤﺭﻜﺔ ﺠ ﺴ ﻴ ﻡ ﺩﺍﺨل ﺼ ﻨ ﺩ ﻭ ﻕ )ﺫﻱ ﺒﻌﺩﻴﻥ( 5.3ﺍ ﻟ ﺠ ﻬ ﺩ ﺍ ﻟﺩﺭﺠﻲ 6ﺘﻁﺒﻴﻕ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭﻓﻰ ﺜﻼﺜﺔ ﺃﺒﻌﺎﺩ 7ﻤﺴﺎﺌل ﻤ ﻨ ﻭ ﻋ ﺔ ﻤﻠﺤﻕ 2.Aﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ﺍ ﻟ ﺒ ﺴ ﻴ ﻁ ﺔ
1
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﻤﻥ ﺩﺭﺍﺴﺘﻨﺎ ﺒﺎﻟﺒﺎﺏ ﺍﻷﻭل ﻋﺭﻓﻨﺎ ﺃﻥ ﻗﻭﺍﻨﻴﻥ ﻨ ﻴ ﻭﺘ ﻥ ﻭﺼﻔﺕ ﻤﺴﺎﺭ ﻭﺤﺭﻜﺔ ﺍﻟ ﺠ ﺴ ﻡ ﺍ ﻟ ﺘ ﻘ ﻠ ﻴﺩﻱ ،ﻭ ﻤﻌﺎﺩﻻﺕ ﻤﺎﻜﺴﻭﻴل ﻭﺼﻔﺕ ﺤﺭﻜﺔ ﺍﻟﻤﻭﺠﺎﺕ ﺍﻟﺘ ﻘﻠﻴ ﺩﻴ ﺔ ﺍﻟﻜﻬﺭﻭﻤﻐﻨﺎﻁﻴﺴﻴﺔ .ﻭﺍﻵﻥ ﻨﺤﻥ ﻨﺤﺘﺎﺝ ﺇﻟﻰ ﻭﺼﻔﺔ ﺭﻴﺎﻀﻴﺔ ﺘﺄﺨﺫ ﻓﻲ ﺍﻻﻋﺘﺒﺎﺭ ﺍﻟﺼﻔﺔ ﺍﻻﺯﺩﻭﺍﺠﻴﺔ ﻟﻠﻤ ﺎ ﺩ ﺓ .ﻫﺫ ﻩ ﺍﻟﻭﺼﻔﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﻴﺠﺏ ﺃﻥ ﺘﺄﺨﺫ ﻓﻲ ﺍﻻﻋﺘﺒﺎﺭ ﻭﺼﻑ ﺍﻟ ﺩﺍﻟ ﺔ ﺍﻟﻤ ﻭﺠﻴ ﺔ ψ ﻜﻤﺎ ﻋﺭﻓﻨﺎﻫﺎ ﺒﺎﻟ ﺒﺎﺏ ﺍﻷﻭل. ﻓﻲ ﻫﺫﺍ ﺍﻟﺒﺎﺏ ﺴﻨﺘﻨﺎﻭل ﺍﻟﻭﺼﻔﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﻭﺍﻟﺘﻲ ﻗ ﺩﻤﻬ ﺎ ﺸﺭﻭﺩﻨﺠﺭ ) (1926ﻭﺴﻤﻴﺕ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍ ﻟ ﻤﻭﺠﻴ ﺔ ﺃﻭ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻤﻊ ﺘﻁﺒ ﻴﻘﺎﺘ ﻬ ﺎ ﺍﻟﺒ ﺴﻴ ﻁ ﺔ .ﻭﻫﻨﺎﻙ ﻭﺼﻔﺔ ﺃﺨﺭﻯ ﺴﻤﻴﺕ ﺒﻁﺭﻴﻘﺔ ﺍﻟﻤﺼﻔﻭﻓﺎﺕ ﺍﻟﻤﻴﻜﺎﻨﻴﻜﻴﺔ ﻭﺒﺩﺃﻫﺎ ﻫﻴﺯﻨﺒﺭﺝ ﻭ ﺒﻭﺭﻥ ﻓﻲ ﻨﻔ ﺱ ﺍﻟﻌﺎﻡ ﻭ ﺴﻭﻑ ﻨ ﺘ ﻨ ﺎ ﻭﻟﻬ ﺎ ﻓﻲ ﺒﺎﺏ ﺁﺨﺭ .ﻭ ﺒﺎﻟﻁﺒﻊ ﻓﻨﺤﻥ ﻫﻨﺎ ﺴﻭﻑ ﻨﺘﻌﺎﻤل ﻤﻊ ﺍﻟ ﻨﻅﺎﻡ ﺍ ﻟﻤﺠﻬﺭﻱ ﻓﻘﻁ .ﻭﺘﺒﻌﺎ ﻟﻠﻨﻅﺎﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻲ ﺍ ﻟﻤﺠﻬﺭﻱ ﻴﻭﺠﺩ ﻫﻨﺎﻙ ﻜﻤﻴﺎﺕ ) ﻗﻴﻡ( ﻓﻴﺯﻴﺎﺌﻴﺔ ﻭﻫﻲ ﺍ ﻟ ﻘ ﻴ ﻡ ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﻗ ﻴ ﺎ ﺴ ﻬ ﺎ ﺃﻭ ﺤ ﺴ ﺎ ﺒ ﻬ ﺎ ﻤﺜل ﻁ ﺎ ﻗ ﺔ ﺍﻟﺤﺭﻜﺔ ،T ﻁ ﺎ ﻗ ﺔ ﺍﻟﻭﻀﻊ V ﺃﻭ ﺍﻟﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ ... E ﺇ ﻟﺦ.
-1ﻓﺭﻭﺽ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﹸﺘﺒﻨﻰ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻋﻠﻲ ﺜﻼﺜ ﺔ ﻓﺭﻭﺽ ﺃﺴ ﺎ ﺴ ﻴ ﺔ
ﺍﻟﻔﺭﺽ
ﺍﻷﻭل:
ﻭ ﻫﻲ :
” ﺃﻱ ﻨﻅﺎﻡ ﻓﻴﺯﻴﺎﺌﻲ ﻴﻭﺼﻑ ﺒ ﺩ ﺍ ﻟ ﺔ ﻤ ﻭ ﺠ ﻴ ﻪ
."ψ
ﻫﺫ ﻩ ﺍﻟ ﺩﺍﻟ ﺔ ﻜﻤﺎ ﻋﺭ ﻓﻨﺎﻫﺎ ﻤﻥ ﻗﺒل ﻋﻠﻰ ﺃﻨ ﻬ ﺎ ﺍﻓﺘﺭﺍﻀﻴﺔ) ﺤ ﻘ ﻴ ﻘ ﻴ ﺔ ﺃﻭ ﻤﺭﻜﺒﺔ( ﻭﻻ ﻴﻤﻜﻥ ﻗ ﻴ ﺎ ﺴ ﻬ ﺎ ﻋﻤﻠﻴﺎﹰ ﻭﻟﻜﻨ ﻬﺎ ﺘﻌﺘﺒﺭ ﻜ ﻤ ﻘ ﻴ ﺎ ﺱ .ﺍﻟﻤﻘﺩﺍﺭ | ) |ψ (r , t ﻴﻌﻁﻴﻨﺎ ﻜﺜﺎ ﻓﺔ ﺍﻻﺤﺘﻤﺎل ﻭﺘﻤﺜل ﺍﺤﺘﻤﺎل ﺘﻭﺍﺠﺩ ﺍﻟ ﺠ ﺴﻴ ﻡ ﻋﻨﺩ ﺍﻟﻭﻀﻊ " "r ﻋﻨﺩ ﺍﻟﻠﺤﻅﺔ " "t ﻭﻫﻨﺎ ﻴﻅﻬﺭ ﺴﺅﺍل :ﻤﺎ ﻫﻲ ﺍﻟﺸﺭﻭﻁ ﺍﻟﻭﺍﺠﺏ ﺘﻭﻓﺭﻫﺎ ﻓﻲ ﺍﻟ ﺩﺍﻟ ﺔ ψ ؟ ﻭﺍﻟﺠﻭﺍﺏ ﻫﻭ :ﺇﺫﺍ ﻋﺭﻓﺕ ﺍ ﻟ ﺩ ﺍ ﻟ ﺔ ψ ﻓﻲ ﻤﺩﻯ ﻤﺤﺩﺩ ،ﻤﺜﻼﹰ ﺒﻴﻥ ﺍﻹﺤﺩﺍﺜﻴﺎﺕ ) ، (a , bﻓﺈﻥ ﺍ ﻟ ﺩ ﺍ ﻟ ﺔ ﻴﺠﺏ ﺃﻥ ﺘﻜﻭﻥ: ﺃ -ﻭ ﺤ ﻴ ﺩ ﺓ ﺍ ﻟ ﻘ ﻴ ﻤ ﺔ ) ( Single valued ﺏ -ﻤﺤﺩ ﺩﺓ ) ( Finite and bounded ﺕ ) -ﻭﺃﻴﻀ ﺎﻤ ﺸ ﺘ ﻘ ﺘ ﻬ ﺎ ﺍﻷﻭﻟﻰ( ﻤﺘﺼﻠﺔ ﻭﻤ ﺴ ﺘ ﻤ ﺭ ﺓ ) .( Continous everywhere ﺃﻱ ﻭﺼﻑ ﺃﻭ ﺸﺭﻁ ﺃﺨﺭ ﻴﻌﺘﺒﺭ ﻤﻜﻤ ﹰﻼ ) ﻟﻴ ﺱ ﻀﺭﻭﺭﻴﹰﺎ( ﻟﻠ ﺸﺭﻭﻁ ﺍ ﻟ ﺴ ﺎ ﺒ ﻘ ﺔ .ﻤﺜﻼﹰ ،ﺘﻌﺘﺒﺭ ﺍﻟ ﺩﺍﻟ ﺔ ﻤﺘﻌﺎﻤﺩﺓ ﺃﻭ ﻤﻌﻴﺭﺓ ﺇﺫﺍ ﺤﻘﻘﺕ ﺍﻟﺸﺭﻁ: 2
if m = n
)(1
if m ≠ n
⎧1 ⎩0
* ⎨ = ∫ ψm (r )ψ n (r ) d τ = δmn , δmn
all space
ﺤﻴﺙ ﺇﻥ ﺍﻟﺩﺍﻟﺔ * ψ ﻫﻲ ﺍﻟﻤﺭﺍﻓﻕ ﺍﻟﻤﺭﻜﺏ ﻟﻠ ﺩﺍﻟ ﺔ ψ ﻭﺍﻟ ﺩﺍﻟ ﺔ δ ﺘﻌﺭﻑ ﺒﻜﺭﻭﻨﻜﺭ ﺩﻟﺘﺎ .ﺤﻴﺙ ﺇﻥ ﻫﻭ ﺸﺭﻁ ﺍﻟﻤﻌﺎﻴﺭﺓ ﻭ δ = 0ﻫﻭ ﺸﺭﻁ ﺍﻟﺘﻌﺎﻤﺩ. ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ ﻤﻌﻴﺭﺓ ﻓﺈﻥ ﺍﻟﺼﻴﻐﺔ ﺍ ﻟ ﺘ ﺎ ﻟ ﻴ ﺔ: ij
ij
b
≤ b} = ∫ ψn* (x ) ψ n (x ) dx a b
) |2 dx |)= ∫ |ψ n ( x a
2
Prob. {a ≤ x
=1
δ ij
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
. {a ≤ x
ﺘﻌﻁﻲ ﺍ ﺤ ﺘ ﻤ ﺎ ﻟ ﻴ ﺔ ﻭﺠﻭﺩ ﺍ ﻟ ﺠ ﺴ ﻴ ﻡ ﻓﻲ ﺍﻟﻤﺩﻯ ﺍﻟﻤﺤﺩﺩ ) ≤ b} ≡ (a,b
ﻤﺜﺎل :ﻭﻀﺢ ،ﻟﻤﺎﺫﺍ ﻻ ﺘ ﺤ ﻘﻕ ﺍﻷﺸﻜﺎل ﺍﻵﺘ ﻴ ﺔ ﺸﺭﻭﻁ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ؟ ( x ) x
x
) x Ψ( x
Ψ
( x ) x
x
a
)(a
Ψ
x
)(c
)(b
ﺍﻟﺤل :ﺍﻷﺸﻜﺎل ﺍ ﻟ ﺴ ﺎ ﺒ ﻘ ﺔ ﻻ ﺘ ﺤ ﻘﻕ ) (aﺍﻟﻤﻴل ﺍ)ﻟ ﻤ ﺸ ﺘ ﻘ ﺔ ﺍﻷﻭﻟﻰ( ﻟ ﻠ ﺩ ﺍ ﻟ ﺔ ) Ψ ( xﻏﻴﺭ ﻤﺘﺼل ﻋﻨﺩ ﺍﻟﻨﻘﻁﺔ . x = a ) ( bﺍﻟﺩﺍﻟﺔ ) Ψ ( xﻏﻴﺭ ﻤﺘﺼﻠﺔ. ) (cﺍ ﻟ ﺩ ﺍ ﻟ ﺔ ) Ψ ( xﻤﺘﻌﺩﺩﺓ ﺍ ﻟ ﻘ ﻴ ﻡ ﺤﻴﺙ ﺇﻥ ﻟﻜل ﻗ ﻴ ﻤ ﺔ xﻴﻭﺠﺩ ﻋﺩﺩ ﻻ ﻨ ﻬ ﺎ ﺌ ﻲ ﻤﻥ ﺍﻟﺩﻭﺍل. ﺷﺮوط ﻣﻴﻜﺎﻧﻴﻜﺎ اﻟﻜﻢ ﻟﻸﺴﺒﺎﺏ ﺍ ﻟ ﺘ ﺎ ﻟ ﻴ ﺔ :
ﻤﺜﺎل :ﻓﻲ ﺍﻟﻤﺩﻯ ﺍﻟﻤﺤﺩﺩ ﺒﻴﻥ ﻗ ﻭﺴﻴ ﻥ ،ﻭﻀﺢ ﻟﻤﺎﺫﺍ ﻻ ﺘ ﺤ ﻘﻕ ﺍﻟﺩﻭﺍل ﺍﻟﻤ ﻭﺠﻴ ﺔ ﺍﻵﺘ ﻴ ﺔ ﺸﺭﻭﻁ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ؟ = e − x ( − ∞ , 0 ) , − | x | ( − ∞ , ∞ ) , ψ 2 = e
) (b
ψ 3
) (c
ψ 1
(0,5) .
1
−4
x
=
) (a
ﺍﻟﺤل: ∞ → ψ 1ﻭﺘﺼﺒﺢ ﻏﻴﺭ ﻤﺤﺩﺩﺓ.
) (aﻋﻨﺩﻤﺎ ∞ x → −ﻓﺈﻥ ﺍﻟ ﺩﺍﻟ ﺔ ) ( bﺍﻟﻤﻴل ﻟ ﻠ ﺩ ﺍ ﻟ ﺔ ψ ﻏﻴﺭ ﻤﺘﺼل ﻋﻨﺩ ﺍ ﻟ ﻨ ﻘ ﻁ ﺔ . x = 0 ) (cﻋﻨﺩﻤﺎ x → 4ﻓﺈﻥ ﺍ ﻟ ﺩ ﺍ ﻟ ﺔ ∞ → ψ ﻭﺘﺼﺒﺢ ﻏﻴﺭ ﻤﺤﺩﺩﺓ. 2
3
ﻤﺜﺎل :ﻟﻤﻭﺠﺔ ﻋﻴﺎﺭﻴﺔ )ﻤﺼﺎﺤﺒﺔ ﻟ ﺠ ﺴ ﻴ ﻡ( ﻤﻌﺭﻓﺔ ﻓﻲ ﺍﻟﻤﺩﻯ )ψ ( x) = csin ( bx ﺤﻴﺙ ، b = π / Lﺍﺤﺴﺏ: ﺃ -ﺜﺎﺒﺕ ﺍﻟﻌﻴﺎﺭﻴﺔ ، c ﺏ -ﺍﺤﺘﻤﺎﻟﻴﺔ ﻭﺠﻭﺩ ﺍﻟ ﺠ ﺴﻴ ﻡ ﻓﻲ ﺍﻟﻤﺩﻯ ، 0 → 0.5 L ﺕ -ﺍﺤﺘﻤﺎﻟﻴﺔ ﻭﺠﻭﺩ ﺍ ﻟ ﺠ ﺴ ﻴ ﻡ ﻓﻲ ﺍﻟﻤﺩﻯ . 0.25 L → 0.75L
) ( 0, Lﺒﺎﻟﻌﻼﻗﺔ:
3
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
:ﺍﻟﺤل :ﺍﻟﺘﻜﺎﻤل L
L
0
0
ﺜﺎﺒﺕ ﺍﻟﻌﻴﺎﺭﻴﺔ ﻴﺤﺴﺏ ﻤﻥ -ﺃ
I = ∫ψ n* ( x) ψ n ( x) dx = c2 ∫ sin ( bx)sin ( bx) dx L
sin(2 bx ) c2 = c ∫ sin (bx ) dx = x − 2 ( 2b ) 0 0 2
L
2
= L
=
c
2
2
( L )
:ﺃﻥ 2 L c ( ) =1 2
⇒ c= :ﺍﻟﺘﻌﺭﻴﻑ
b
Prob. {a ≤ x
b
≤ b } = ∫ψ *n ( x ) ψ n ( x )dx = ∫ a
" ﻨﺠﺩ I = 1 " ﺒﺎﺴﺘﺨﺩﺍﻡ ﺸﺭﻁ ﺍﻟﻌﻴﺎﺭﻴﺔﻭ 2
L
ﺴﺘﺨﺩﻡ ﻟﺤﺴﺎﺏ ﺍﺤﺘﻤﺎﻟﻴﺔ ﻭﺠﻭﺩ ﺍﻟﺠﺴﻴﻡ ﻓﻲ ﻤﺩﻯ ﻤﺤﺩﺩ ﻨ 2 L
a
⎛ nπ ⎞ 2 ⎛ nπ ⎞ x⎟ sin ⎜ x ⎟ dx ⎝ L ⎠ L ⎝ L ⎠
sin ⎜
b
nπ ⎞ x 1 2n π x = ∫ sin ⎛⎜ − x ⎟ dx = sin( ) La L 2π n L a ⎝ L ⎠
2
b
2
:ﻨﺤﺼل ﻋﻠﻰ ، 0 → 0.5 L Prob.{0 ≤ x
L / 2
2 nπ x − sin( ) L 2π n L 0 x
≤ L / 2} = =
ﻭ ﻓﻲ ﺍﻟﻤﺩﻯ-ﺏ
1
1
(for all n )
2
:ﻨﺤﺼل ﻋﻠﻰ ، 0.25 L → 0.75L Prob. {0.25 L≤ x ≤ 0.75 L} =
x L
−
1 2π
sin(
2π x L
ﻭ ﻓﻲ ﺍﻟﻤﺩﻯ-ﺕ
0.75 L
) 0.25 L
= 0.818 ψ
ﺍﺤﺴﺏﺍﺤﺘﻤﺎﻟﻴﺔ ﻭﺠﻭﺩ،ﻟﻠﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ﺍﻟﻤﺭﺴﻭﻤﺔ :ﻤﺜﺎل . X = {2,4} ﺠﺴﻴﻡ ﻓﻲ ﺍﻟﻤﺩﻯ
3 2 1
2
4
6
X
4
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﺍﻟﺤل :ﺒﺤﺴﺎﺏﺍﺤﺘﻤﺎﻟﻴﺔ ﻭﺠﻭﺩ ﺠﺴﻴﻡ ﻓﻲ ﺍﻟﻤﺩﻯ
}: X = {2,4
= 9 + 4 = 13
4
∑ψ
2 i
= I
i =2
ﻭﻤﻘﺎﺭﻨﺘﻬﺎ ﺍﺤﺘﻤﺎﻟﻴﺔ ﻭﺠﻭﺩ ﺠﺴﻴﻡ ﻓﻲ ﺍﻟﻤﺩﻯ
}: X = {0,5
= 1 + 1+ 4 + 9 + 1 = 16,
5
∑ψ
2 i
= II
i =1
ﻨﺠﺩ
ﺃﻥ: 4
13 16
=
∑ψ
2 i
i =2 5
∑ψ
2 i
=
I II
= }Prob. {2 ≤ X ≤ 4
i =0
ﺍﻟﻔﺭﺽ ﺍﻟﺜﺎﻨﻰ" :ﻴﻭﺠﺩ
ﻤﺅﺜﺭ ) ( Operator ﻟﻜل ﻜﻤﻴﺔ ﻓﻴﺯﻴﺎﺌﻴﺔ ﻴﻤﻜﻥ ﻗﻴﺎﺴﻬﺎ".
ﻭﺍﻟﻤﺅﺜﺭ ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻥ ﻋﻤﻠﻴﺔ ﺠﻤﻊ ) (+ﺃﻭ ﻁﺭﺡ )– ( ﺃﻭ ﺘﻔﺎﻀل ....ﻟﺘﻌﺭﻴﻑ ﺍﻟﻤﺅﺜﺭ ﺴﻭﻑ ﻨﻀﻊ ﺃﻋﻼﻩ ﺍﻹﺸﺎﺭﺓ "^" .ﻭﺍﺨﺘﻴﺎﺭ ﺍﻟﻤﺅﺜﺭ ﻴﻔﻀل ﺃﻥ ﻴﺤﻘﻕ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻘﻴﻤ ﺔ ﺍﻟﻤﻤﻴﺯﺓ ﻭﻫﻲ ﺒﺎﻟﺼﻭﺭﺓ: operator eigenfunction
eigenfunction
)(2
= a
ϕ
ˆ A
ϕ
eigenvalue
ﺤﻴﺙ ﺇﻥ ﺍﻟﺩﺍﻟﺔ ϕ ﻫﻲ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻤﻴﺯﺓ ) ) ( Eigenfunctionﻭﻴﺠﺏ ﺃﻥ ﺘﻅﻬﺭ ﺒﻜﻼ ﺍﻟﻁﺭﻓﻴﻥ( ﻭ aﻫﻲ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻤﻴﺯﺓ ) .( Eigenvalueﻭﺘﹸﻌﺭﻑ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻤﻴﺯﺓ ﺒﺄﻨﻬﺎ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﻲ ﺘﺤﻘﻕ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻘﻴﻡ ﺍﻟﻤﻤﻴﺯﺓ ،ﻭﻻ ﻴﻭﺠﺩ ﻋﻠﻴﻬﺎ ﺃﻱ ﺸﺭﻭﻁ ﺃﺨﺭﻯ ﺇﻻ ﺇﺫﺍ ﺍﺴﺘﺨﺩﻤﺕ ﻟﻭﺼﻑ ﺍﻟﻨﻅﺎﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻲ ،ﻓﻴﺠﺏ ﺃﻥ ﻴﻨﻁﺒﻕ ﻋﻠﻴﻬﺎ ﺍﻟﺸﺭﻭﻁ ﺍﻟﺜﻼﺜﺔ ﺍﻟﺴﺎﺒﻘﺔ . ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺅﺜﺭ ﻴﻔﻀل )ﻭ ﻟﻴﺱ ﺸﺭﻁﹰﺎ ﻀﺭﻭﺭﻴﹰﺎ( ﺃﻥ ﻴﻜﻭﻥ ﻟ ﻪ ﺨﺼﺎﺌﺹ ﺨﻁﻴﺔ ،ﺒﻤﻌﻨﻰ: )(3
ˆ ( x ) + Ag ˆ ( x ), Aˆ ⎡⎣ f ( x ) + g ( x ) ⎤⎦ = Af ˆ ⎡ kf ( x )⎤ = kAf )ˆ ( x A ⎣ ⎦ ﺍﺨﺘﻴﺎﺭﻴﺘﺎﻥ.
ﺤﻴﺙ k ﻤﻘﺩﺍﺭ ﺜﺎﺒﺕ ﻭ ) g (xﻭ ) f (xﻫﻤﺎ ﺩﺍﻟﺘﺎﻥ ﺍﻟﺠﺩﻭل ﺍﻟﺘﺎﻟﻲ ﻴﺤﺘﻭﻱ ﻋﻠﻰ ﺒﻌﺽ ﺍﻟﻜﻤﻴﺎﺕﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﻭﺍﻟﻤﺅﺜﺭﺍﺕ ﺍﻟﻤﻨﺎﻅﺭﺓ ﻟﻬﺎ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ. ﺭﻤﺯﻫﺎ
ﺍﻟﻜﻤﻴﺔﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺇﺤﺩﺍﺜﻲ ﺍﻟﻤﻭﻀﻊ
x
5
ﺍﻟﻤﺅﺜﺭ ﺍﻟﻤﻨﺎﻅﺭ =x
ˆ x
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﻤﺭﻜﺒﺔ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺨﻁﻴﺔ
∂ ∂ x ∇ pˆ = −i = −i
p x
ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺨﻁﻴﺔ ﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﻤﺩﺍﺭﻴﺔ ﻁﺎﻗﺔ ﺍﻟﻭﻀﻊ ﺍﻟﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ
p
= p 2 / 2m
∇2
K
pˆ x
2 ˆ =− K 2m
L
ˆLˆ = rˆ × p
V
Vˆ =V
∂ ∂t
E
=i
ˆ E
d 2
ﻤﺜﺎل :ﻫل ﺍﻟﻤﺅﺜﺭ dxﻭﺍﻟﺩﺍﻟﺔ ) sin (axﻴﻜﻭﻨﺎﻥ ﻤﻌﺎﺩﻟﺔ ﻗﻴﻡ ﻤﻤﻴﺯﺓ؟ 2
d 2
ﺍﻟﺤل :ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﻤﺅﺜﺭ ﺍﻟﺘﻔﺎﻀﻠﻲ
dx 2
ﻋﻠﻰ ﺍﻟﺩﺍﻟﺔ
) sin ( ax
ﻨﺠﺩ
ﺃﻥ: 2
d
⎦⎤) ⎡⎣sin (ax )⎤⎦ = −a 2 ⎡⎣sin ( ax dx 2
d 2
ﻭﻤﻨﻬ ﺎ ﻨﺴﺘﻨﺘﺞ ﺃﻥ ﺍﻟﻤﺅﺜﺭ dxﻭﺍﻟﺩﺍﻟﺔ ) sin (axﻴﻜﻭﻨﺎﻥ ﻤﻌﺎﺩﻟﺔ ﻗﻴﻡ ﻤﻤﻴﺯﺓ ،ﻭﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻤﻴﺯﺓ ﻟﻬﺎ ﻫﻲ 2
. −a 2
ﻤﺜﺎل :ﺃﻱ ﻤﻥ ﺍﻟﻤﺅﺜﺭﺍﺕ ﺍﻵﺘﻴﺔ ﺘﻌﺘﺒﺭ ﻤﺅﺜﺭﺍﺕ ﺨﻁﻴﺔ؟ d ) ( dx
) (e
) (d
) ((b) sin
)(c
)(
ﺍﻟﺤل: ≠
+ (ψ 2 ) (b ) sin(ψ 1 + ψ 2 ) ≠ sin(ψ 1 ) + sin(ψ 2 ) ) (ψ 1
≠ e (ψ 1 ) + e (ψ 2 )
) (ψ 1 + ψ 2
) (a
e
) (c
) (ψ 1 +ψ 2
d d d = ) (ψ 1 + ψ 2 (ψ 1 ) + (ψ 2 ) dx dx dx
ﺇﺫﹰﺍ ﺍﻟﻤﺅﺜﺭ ) (d ﻫﻭ ﺍﻟﻤﺅﺜﺭ ﺍﻟﺨﻁﻲ ﺍﻟﻭﺤﻴﺩ ﻓﻲ ﻤﺜﺎل :ﺍﺤﺴﺏ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻤﻴﺯﺓ ﻟﻠﻤﺅﺜﺭ
− 2 x
ﻫﺫﻩ ﺍﻟﻤﺠﻤﻭﻋﺔ.
d dx
=
ˆ A
6
)(d
)(a
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﺍﻟﺤل :ﺒﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﻴﺯﺓ Aˆψ = aψ ﻨﺠﺩ
ﺃﻥ:
⎛ d ⎞ ⎜ dx − 2 x ⎟ψ = aψ ⎝ ⎠
ﻭﺘﺨﺘﺼﺭ ﺇﻟﻰ ﺍﻟﺼﻭﺭﺓ d ψ dx
= ( a + 2x )ψ
ﻭﺒﺈﺠﺭﺍﺀ
ﺍﻟﺘﻜﺎﻤل ﻴﻨﺘﺞ: d ψ
∫ ψ = ∫ ( a + 2x )dx 2
ﺤﻴﺙ cﻫﻭ ﺜﺎﺒﺕ
ψ
ﺍﻟﺘﻜﺎﻤل. ˆ x ≡ d x D dx
ﻤﺜﺎل :ﺍﺤﺴﺏ ﺍﻟﺘﻌﺒﻴﺭ ﺍﻟﻨﻬﺎﺌﻲ ﻟﻠﻤﺅﺜﺭ ﺍﻟﺤل :ﺒﺎﻟﺘﺄﺜﻴﺭ
= c e a x +x
⇒
d
ﻋﻠﻰ ﺩﺍﻟﺔ ﺍﺨﺘﻴﺎﺭﻴﺔ ψ ﺒﺎﻟﻤﺅﺜﺭ x dx dψ ⎞ dx ⎛ d x +ψ = ⎜ x +1⎟ψ dx ⎠ dx ⎝ dx
ﻨﺠﺩ: = ) xψ
(
d dx
⎞ ⎠
= x⎟ψ
⎛d ⎜ dx ⎝
ﻭﻫﺫﺍ ﻴﻌﻨﻰ ﺃﻥ
)
xˆD+ 1
( )
≡ ˆDx
ﻤﺜﺎل :ﺍﺤﺴﺏ ﺍﻟﺘﻌﺒﻴﺭ ﺍﻟﻨﻬﺎﺌﻲ ﻟﻠﻤﺅﺜﺭ ) ˆ ، ( Dˆ + xﺤﻴﺙ 2
d dx
(
= ˆ. D
ﺍﻟﺤل:
) ˆ= ( Dˆ + xˆ )( Dˆ + x
( Dˆ + xˆ ) 2
= Dˆ 2 + xˆ Dˆ + Dˆ xˆ + xˆ 2 = Dˆ 2 + xˆ Dˆ + xˆ Dˆ + 1 + xˆ 2
xˆ Dˆ +1
= Dˆ 2 + 2 xˆ Dˆ + xˆ 2 + 1
ﻫﺫﺍ ﻭﻗﺩﺍﺴﺘﺨﺩﻤﻨﺎ ﻨﺘﻴﺠﺔ
ﺍﻟﻔﺭﺽ ﺍﻟﺜﺎﻟﺙ
ﺍﻟﻤﺜﺎل ﺍﻟﺴﺎﺒﻕ .
" :ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻭﻗﻌﺔ ) ﺍﻟﻤﺘﻭﺴﻁﺔ ( ﻟﻠﻘﻴ ﻡ ﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﻗﻴﺎﺴﻬﺎ ﺘﻌﺭﻑ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ: ψ Aˆ ψ d τ ∫ = ∫ ψ ψ d τ *
)(4
*
ˆ≡ A
7
ˆ A
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﻭﺘﻠﻌﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻭﺴﻁﺔ ﺩﻭﺭﹰﺍ ﺭﺌﻴﺴﻴ ﺎﹰ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﺤﻴﺙ ﺇﻨﻨﺎ ﻨﺘﻌﺎﻤل ﻤﻊ ﺃﻋﺩﺍﺩ ﻻ ﻨﻬﺎﺌﻴﺔ ﻤﻥ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻭﻻ ﻨﺴﺘﻁﻴﻊ ﺍﻟﺘﻌﺎﻤل ﻤﻊﻜل ﺠﺴﻴﻡ ﻋﻠﻰ ﺤﺩﺓ. ﻤﺜﺎل :ﺍﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﻁﺎﻗﺔﺍﻟﻤﺘﻭﺴﻁﺔ ﻟﻠﺩﺍﻟﺔ: 5 > |ϕ 19 5
ﻤﻊ ﺍﻋﺘﺒﺎﺭ
>+
2 3 |ϕ 3 > + |ϕ 19 19 4
1 4 |ϕ1 > + |ϕ 19 19 2
>+
=>
a5
a4
a3
a2
a1
|ψ
ﺃﻥ:
= En ϕ n , = δ mn
En = nε o
ﺍﻟﺤل :ﺒﻔﺭﺽ ﺃﻥ ﻤﻌﺎﻤﻼﺕ ﺍﻟﻤﺭﻜﺒﺎﺕ ϕ ﻫﻲ ، aﻓﻴﺠﺏ ﺃﻥ ﻨ ﺘﺄﻜﺩ ﺃﻭﻻﹰ n
15 19
=
5
+
3
+
2
n
4
+
+
1
19 19 19 19 19
H ϕn ϕ m ϕ n
ﻤﻥ ﻋﻴﺎﺭﻴﺔ ﺍﻟﺩﺍﻟﺔ ﻜﺎﻟﺘﺎﻟﻲ:
= ∑ an2 => < ψ |ψ > = ∑ an2 < ϕn | ϕn
ﻏﻴﺭ ﻋﻴﺎﺭﻴﺔ.
ﺤﻴﺙ ﺇﻥ ﺍﻟﻤﺠﻤﻭﻉ ﻻ ﻴﺘﺴﺎﻭﻯ ﺒﺎﻟﻭﺍﺤﺩ ﻟﺫﻟﻙ ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ ﺜﺎﻨﻴﺎ :ﺍﺤﺴﺏ ﺍﺤﺘﻤﺎل ﺘﻭﺍﺠﺩ ﺍﻟﺠﺴﻴﻡ ﻓﻲ ﻜل ﻤﺴ ﺘﻭﻯ ﻋﻠﻰ ﺤﺩ ﺓ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﺼﻭﺭﺓ: > < ϕi |ψ > <ψ |ψ
ﻟﺘﻌﻁﻴﻨﺎ
=
Pi
ﺍﻵﺘﻲ: .
5 15
= P5
,
3 15
ﺜﺎﻟﺜﺎ :ﻗﻴﻤﺔ ﺍﻟﻁﺎﻗﺔ ﺍﻟﻤﺘﻭﺴﻁﺔ ε o.
52 15
= ) (5ε o
5 15
= P4
,
2 15
=
P3
,
4 15
=
| a1 |2 1/19 1 = P1 = = , 15/19 15/19 15
P2
ﺘﺤﺴﺏ ﻜﺎﻵﺘﻲ: (4ε o ) +
3 15
(3ε o ) +
2 15
(2εo ) +
4 15
(εo ) +
1 15
= < E > = ∑P E n
n
n
ﻭ ﺒﻁﺭﻴﻘﺔ ﺃﺨﺭﻯ: ⎞ 19 ⎛ 1 8 6 12 25 52 + + + + ⎟εo = ε o ⎜ ⎠ 15 ⎝ 19 19 19 19 19 15
=
> < ϕn | H | ϕ n 15/19
∑a
2 n
= ><ψ | H |ψ = E ><ψ |ψ
2.2ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺒﻌﺩ ﻭﺍﺤﺩ ﻋﻠﻤﻨﺎ ﻤﻥ ﺍﻟﻔﺭﻀﻴﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﺃﻥ ﻟﻜل ﻜﻤﻴﺔ ﻓﻴﺯﻴﺎﺌﻴﺔ ﻴﻤﻜﻥ ﻗﻴﺎﺴﻬﺎ ،ﻜﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ ﻭﺍﻟﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ ، E ﻴﻜﻭﻥ ﻫﻨﺎﻙ ﻤﺅﺜﺭﺍﹰ ،ﻭﺤﺘﻰ ﺍﻵﻥ ﻟﻡ ﻨﻌﺭﻑ ﻜﻴﻔﻴﺔ ﺤﺴﺎﺏ ﻫﺫﺍ ﺍﻟﻤﺅﺜﺭ .ﻭﻟﺤﺴﺎﺏ ﺍﻟﻤﺅﺜﺭﺍﺕ ﻨﺘﺒﻊ ﺍﻟﺨﻁﻭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ ) ﻟﻠﺴﻬﻭﻟﺔ ﺴﻨﺄﺨﺫ ﺍﻟﺤﺭﻜﺔ ﻓﻲ ﺍﺘﺠﺎﻩ ﻭﺍﺤﺩ ﻓﻘﻁ ﻭﻫﻭ ﺍﺘﺠﺎﻩ :( x T
8
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
- 1ﻨﺒﺩﺃ ﺒﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻭﺠﺔ ﺍﻟﻤﺴﺘﻭﻴﺔ ﺍﻟﻤﺼﺎﺤﺒﺔ ﻟﻠﺠﺴﻴﻡ ﺍﻟﺤﺭ
ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ:
ψ = ei (kx −wt ) = ei ( p x −Et )/ =e ip x / e−iEt / = f (x ) f (t )
)(1
x
x
ﺤﻴﺙ ﻁﺎﻗﺔ ﺍﻟﺠﺴﻴﻡ ﻫﻲ E = ω ﻭ p = k ﻫﻲ ﻜﻤﻴﺔ ﺤﺭﻜﺘﻪ - 2ﻟﻠﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻤﺅﺜﺭ ˆ ، E ﻨﻔﺎﻀل ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺯﻤﻥ ،ﻓﻨﺤﺼل ﻋﻠﻰ: ﺍﻟﺨﻁﻴﺔ.
x
∂ ψ = E ψ ∂t
)(2
⇒
i
i ∂ψ = − Eψ ∂t
ﻭﻫﻲ ﻤﻌﺎﺩﻟﺔ ﻤﻤﻴﺯﺓ ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ Eˆψ = E ψ ﻭﻟﺫﻟﻙ ∂ ∂t
) (3
ﻓ ﺈ ﻥ:
ˆ E
=i
- 3ﻟﻠﺤﺼﻭل ﻋﻠﻰ ﺍﻟﻤﺅﺜﺭ ، pˆ xﻓﺈﻨﻨﺎ ﻨﻔﺎﻀل ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ − Et ) /
x
∂ψ i = p e i ( p ∂ x
x
i p x ψ
)(4
∂ ψ ∂ x
= −i
xﻟﻨﺠﺩ:
= ⇒
pˆ x ψ
∂ ﺤﻴﺙ ﺤﺼﻠﻨﺎ ﻋﻠﻰ ﻤﻌﺎﺩﻟﺔ ﻤﻤﻴﺯﺓ ﻋﻠﻰ ﺍﻟﺼﻭﺭﺓ ، pˆ xψ = p xψ ﻭ ﺘﻌﻁﻴﻨﺎ ﺍﻟﻤﺅﺜﺭ ∂ x 2 2 ∂ 2 ˆx= p ˆ x ˆp x = − .p ﻭﻤﻨﻬﺎ ﻨﺠﺩ ∂ x 2 -4 )(5
ﻨﻘﺎﺭﻥ ﺍﻵﻥ ﺒﻴﻥ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻭﺍﻟﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺤﻴﺙ 2
ﻭﺘﺼﺒﺢ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺍﻟﺸﻜل ﺍﻟﻌﺎﻡ ﻫﻲ ﻫﺎﻤﻴﻠﺘﻭﻨﻲ( )(6
ﺇﻥ:
∂2 ∂2 ∂2 =− ∇, ∇ = 2+ 2+ 2 2m ∂x ∂y ∂z 2
2
∂ ∂t
ˆE= i
ˆψ = E
pˆ x2 ˆ = K 2m
Hˆ ψ
ﺤﻴﺙ
2 2 ⎛ ⎞ ∂ ˆH= ⎜ − ˆ⎟, V + 2 ⎝ 2m ∂ x ⎠
ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭﺍﻟﻤﺴﺘﻘﺭﺓ )ﻏﻴﺭ ﺍﻟﺯﻤﻨﻴﺔ
-3 ﺒﺩ ﺀﺍﹰ ﺒﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ:
9
= −i
pˆ x
(
ˆ H
ﻴﺩﻋﻰ ﻫ ﺎﻤﻴﻠﺘﻭﻨﻴﺎﻥ )ﻤﺅﺜﺭ
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
⎛ 2 ∂2 ⎞ ˆ ⎟ Ψ ( x , t ) = i ∂ Ψ( x , t ) V + ⎜− 2 ∂t ⎝ 2m ∂ x ⎠
)(1
ﻨﺠﺩ ﺃﻨﻬﺎ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﻭﻨﺤﻥ ﻓﻲ ﺩﺭﺍﺴﺘﻨﺎ ﻫﺫﻩ ﻨﺒﺤﺙ ﻋﻥ ﺩﻭﺍل ﻭ ﻗﻴﻡ ﻤﻤﻴﺯﺓ ﻤﺴﺘﻘﺭﺓ )ﺃﻱ ﺃﻨﻬﻥ ﻻ ﻴﻌﺘﻤﺩﻥ ﻋﻠﻰ ﺍﻟﺯﻤﻥ( .ﻟﻬﺫﺍ ﺴﻭﻑ ﻨﺤﺎﻭل ﻓﺼل ﻤﻌﺎﺩﻻﺕ ﺍﻟﺯﻤﻥ ﻋﻥ ﻤﻌﺎﺩﻻﺕ ﺍﻟﻤﻜﺎﻥ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺇﺤﺩﻯ ﺍﻟﻁﺭﻕ ﻭﻫﻲ ﻁﺭﻴﻘﺔ ﻓﺼل ﺍﻟﻤﺘﻐﻴﺭﺍﺕ ،ﻭﺍﻟﻁﺭﻴﻘﺔ ﻜﺎﻟﺘﺎﻟﻲ: - 1ﻨﻔﺘﺭﺽ ﺍﻨﻔﺼﺎل ﺍﻟﺩﺍﻟﺔ ) Ψ ( x, t ﺇﻟﻰ: Ψ ( x , t) = ψ ( x )ψ ( t) ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﻭ ) ψ ( xﻫﻲ ﺩﺍﻟﺔ
)(2
ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺤﻴﺙ ) ψ (t ﻫﻲ ﺩﺍﻟﺔ - 2ﻓﺎﻀل ﺍﻟﺩﺍﻟﺔ ) (2ﻤﺭ ﺓ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺯﻤﻥ ﻭﻤﺭﺘﻴﻥ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻜﺎﻥ ،ﻓﻨﺤﺼل ﻋﻠﻰ: ∂ )∂ψ (t )Ψ ( x, t) = ψ ( x ∂t ∂t 2 2 ∂ ) ∂ ψ ( x ( x , t ) ψ ( t ) Ψ = ∂ x 2 ∂x 2
)(3a )(3b -3
ﺍﻟﻤﻜﺎﻥ.
ﺒﺎﻟﺘﻌﻭﻴﺽ ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ) (3aﻭ) (3bﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﻭﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ
) ψ ( x ) ψ (t ﻨﺠﺩ:
) ⎧⎪ 2 ∂ 2ψ ( x ⎫ )⎫⎪ 1 ⎧ ∂ψ (t ˆ = ⎬ +V ⎨− ⎨i ⎬ ∂ ψ ( x) ⎪ 2 m ∂ x 2 ψ ( ) t t ⎩ ⎭ ⎭⎪ ⎩ 1
)(4
- 4ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (4ﻨﺠﺩ ﺃﻥ ﺍﻟﻁﺭﻑ ﺍﻷﻴﺴﺭ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﻤﺘﻐﻴﺭ " " xﺒﻴﻨﻤﺎ ﺍﻟﻁﺭﻑ ﺍﻷﻴﻤﻥ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﻤﺘﻐﻴﺭ " "t ﻓﻘﻁ .ﻤﻥ ﻨﻅﺭﻴﺎﺕ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ﺘﻌﻠﻤﻨﺎ ﺃﻥ ﺫﻟﻙ ﻤﻤﻜﻥ ﻓﻘﻁ ﻓﻲ ﺤﺎﻟﺔ ﻜﻭﻥﻜل ﻁﺭﻑ ﻴﺴﺎﻭﻯ ﻜﻤﻴﺔ ﺜﺎﺒﺘﺔ ،ﺴﻭﻑ ﻨﻔﺘﺭﻀﻬﺎ " ،"cﻭﻟﻬﺫﺍ )(5a )(5b
)∂ψ (t )= cψ (t ∂t
i
ˆ ) ∂ 2ψ ( x − ) +V ψ ( x ) = cψ (x 2m ∂x 2 2
- 5ﺘﻜﺎﻤل ﺍﻟﻤﻌﺎﺩﻟﺔ ) (5aﻴﻌﻁﻰ ﺍﻟﺩﺍﻟﺔ ﺍﻟﺯﻤﻨﻴﺔ )(16
ψ (t ) = e − ict / ﻭﺫﻟﻙﻟﻠﺴﻬﻭﻟﺔ . ﻫﻭ ﺍﻟﻭﺤﺩﺓ
ﺒﺎﻓﺘﺭﺍﺽ ﺃﻥ ﺜﺎﺒﺕ ﺍﻟﺘﻜﺎﻤل - 6ﻟﺤﺴﺎﺏ ﺍﻟﺜﺎﺒﺕ cﻨﻘﺎﺭﻥ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ) (6ﻭ) (5ﻓﻨﺠﺩ ﺃﻥ ﺍﻟﺜﺎﺒﺕ cﻤﺎ ﻫﻭ ﺇﻻ ﺍﻟﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ E ﻟﻠﺠﺴﻴﻡ ﻭﻨﺤﻥ ﻨﻌﻠﻡ ﺃﻨﻬ ﺎ ﻜﻤﻴﺔ ﻤﺤﻔﻭﻅﺔ ) ﺜﺎﺒﺘﺔ(. - 7ﺃﺨﻴﺭﹰﺍ ﻨﺤﺼل ﻋﻠﻰ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﺘﻲ ﻻ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﻭﻫﻲ:
10
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
) ∇ x2ψ (x ) +V (x )ψ ( x ) = E ψ ( x )(2.17
[ E − V ( x ) ]ψ ( x ) = 0
2m 2
2
2m
−
) d 2ψ (x
+
dx 2
⇒
ﻭﻫﻲ ﺘﻤﺜل ﺤﺭﻜﺔ ﺠﺴﻴﻡ ﻴﺘﺤﺭﻙ ﺒﻁﺎﻗﺔ ﻜﻠﻴﺔ E ﻭﻁﺎﻗﺔ ﻭﻀﻊ ) ﻁﺎﻗﺔ ﻜﺎﻤﻨﺔ( V ﻭﻫﻡ ﻻ ﻴﻌﺘﻤﺩﻭﻥ ﻋﻠﻰ ﺍﻟﺯﻤﻥ ﻭﻟﺫﻟﻙ ﻴﻘﺎل ﻋﻨﻬﺎ ﺒﺄﻨﻬﺎ ﺘﻤﺜل ﺤﺎﻟ ﺔ ﻤﺴﺘﻘﺭﺓ .
-4ﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ * ψ ﺘﻜﺘﺒﺎﻥ ﺒﺎﻟﺼﻭﺭﺓ:
ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻟﻠﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﻴﺔ ψ ﻭﺍﻟﻤﺭﺍﻓﻕ ﺍﻟﻤﺭﻜﺏ ﻟﻬﺎ
∂ψ 2m ∂t 2 * ∂ψ − ∇2ψ * = −i 2m ∂t ∇ 2ψ = i
)(1 )(2
2
−
ﺒﻀﺭﺏ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﺒﺎﻟﺩﺍﻟﺔ * ،ψ ﻭﺍﻟﻤﻌﺎﺩﻟﺔ ) (2ﻤﻥ ﺍﻟﻴﺴﺎﺭ ﺒﺎﻟﺩﺍﻟﺔ ،ψ ﺜﻡ ⎞ * ∂ψ ⎟ +ψ ⎠ ∂t
)(3
ﺇﺫﺍ ﺃﺨﺫﻨﺎ ﻓﻲ
∂
2
ψ ⎛ * ψ * ∇ ψ −ψ∇ ψ *) = i ⎜ψ ( ∂t 2m ⎝ 2
2
ﺒﺎﻟﻁﺭﺡ ﻴﻨﺘﺞ:
−
ﺍﻻﻋﺘﺒﺎﺭ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥﺍﻟﺘﺎﻟﻴﺘﻴﻥ :
)(ψ * ∇ ψ −ψ∇ ψ *) = ∇.(ψ * ∇ψ − ψ∇ψ * 2
)(4
2
∂ψ ∂ * ∂ψ +ψ = (ψ *ψ ) ∂t ∂t ∂t
)(5
*ψ
ﻭﺍﻋﺘﻤﺩﻨﺎ ﺍﻟﺭﻤﺯﻴﻥ ﺍﻟﺘﺎﻟﻴﻴﻥ:
ﻜﺜﺎﻓﺔ
ﺍﻟﺘﻴﺎﺭ: i
) ، J = − 2m (ψ * ∇ψ −ψ∇ψ * ﻭﺍﻟﻜﺜﺎﻓﺔ
ﺍﻻﺤﺘﻤﺎ ﻟﻴﺔ: ρ = ψ *ψ
،
ﻓﺈﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (3ﺘﺄﺨﺫ ﺍﻟﺼﻭﺭﺓ )(6
∂ ρ =0 ∂t
∇ ⋅ J +
ﻭﻫﻲ ﺘﺸﺒﻪ ﻤﻌﺎﺩﻟﺔ ﺍﻻﺴﺘﻤﺭﺍﺭﻴﺔ ) ﺍﻻﺘﺼﺎل( ﻓﻲ ﺩﻴﻨﺎﻤﻴﻜﺎ ﺍﻟﻤﻭ ﺍﺌﻊ ﻭﺍﻟﺘﻲ ﺘﻌﺒﺭ ﻫﻨﺎ ﻋﻥ ﻗﺎﻨﻭﻥ ﻤﺜﺎل :ﺍﺤﺴﺏ ﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﻟﻠﺩﺍﻟﺔ
ikx
ψ ( x ) = A eﺤﻴﺙ 11
Aﺜﺎﺒﺕ.
ﺤﻔﻅﺍﻟﺸﺤﻨﺔ .
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
:ﺍﻟﺤل J
=−
=
i
2m
(ψ * ∇ψ −ψ ∇ψ *) =
⎛ Im ⎜ A * e−ikx m ⎝
m
∂ A eikx ∂x
ﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﻫﻲ ﺍﺜﺒﺕ ﺃﻥ،ﺜﻭﺍﺒﺕ A , B ﺤﻴﺙ
⎛ ⎝
I m ⎜ψ *
∂ ψ ∂x
⎞ ⎟ ⎠
⎞ k 2 2 ⎟ = m | A| = v | A| ⎠
ψ ( x ) = A e
ikx
+ B e −ikx ﻟﻠﺩﺍﻟﺔ: ﻤﻨﺯﻟﻲ . J = v (| A |
2
12
ﻭﺍﺠﺏ −|B| ) 2
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
-5ﺘﻁﺒﻴﻘﺎﺕ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺒﻌﺩ ﻭﺍﺤﺩ 5.I
ﺩﺭﺍ ﺴﺔ ﺤﺭﻜﺔ ﺠﺴﻴﻡ ﺤﺭ )ﺒ ﻤﻌﻨﻰ ﺃﻥ
( V = 0
ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻟﺠﺴﻴﻡ ﺤﺭ ،ﻓﻲ ﺒﻌﺩ ﻭﺍﺤﺩ ،ﺘﻜﺘﺏ ﻋﻠﻰ ) ∂ 2ψ ( x ) = E ψ ( x ∂x 2 = − k 2ψ ,
d 2ψ dx 2
ﺍﻟﺼﻭﺭﺓ: 2 − 2m
⇒ 2mE ﻭ 2
= . k 2ﻨﻼﺤﻅ ﻫﻨﺎ
ﻭﺤﻠﻬﺎ ﻫﻭ ψ ( x ) = A e± i k x / ﺤﻴﺙ ﺇﻥ Aﻫﻭ ﺜﺎﺒﺕ ﺍﻟﺘﻜﺎﻤل ﻗﻴﻤ ﺎﺤﻘﻴﻘﻴﺔ ﺤﻴﺙ ﺇﻨ ﻬﺎ ﻤﺭﺘﺒﻁﺔ ﺒﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ ﻓﻘﻁ. ﻭﻜﻤﺜﺎل ﺘﻭﻀﻴﺤﻲ ﺩﻋﻭﻨﺎ ﻨﺴﺘﺭﺠﻊ ﺍﻟﻤﺴﺄﻟﺔ ﺍﻟﺭﺌﻴﺴﻴﺔ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﺃﻻ ﻭﻫﻲ ﺩﺭﺍﺴﺔ ﺤﺭﻜﺔ ﺠﺴﻴﻡ ﺩﺍﺨل ﺼﻨﺩﻭﻕ ﻤﻐﻠﻕ ﺘﻤﺎﻤﺎﹰ. ﺃﻥ k
ﺘﺄﺨﺫ
-5.IIﺩﺭﺍﺴﺔ ﺤﺭﻜﺔ ﺠﺴﻴﻡ ﺩﺍﺨل ﺼﻨﺩﻭﻕ ﻤﻐﻠﻕ ﺘﻤﺎﻤﺎﹰ ﻓﻲ ﻫﺫﺍ ﺍﻟﻤﺜﺎل ﺴﻭﻑ ﻨﺩﺭﺱ ﺤﺭﻜﺔ ﺠﺴﻴﻡ ﻓﻲ ﻤﺠﺎل )(1
≤L
0 ≤ x
ﺠﻬﺩ ﻴﻭﺼﻑ ﺒﺎﻟﺘﺎﻟﻲ:
⎧ 0,
⎨ = ) V (x
⎩∞, otherwise
ﻭﻫﺫﺍ ﺍﻟﻤﺜﺎل ﻴﻨﻁﺒﻕ ﻋﻤﻠﻴ ﺎﹰ ﻋﻠﻰ ﺤﺭﻜﺔ ﺇﻟﻜﺘﺭﻭﻥ ﺤﺭ ﺒﺩﺍﺨل ﻗﻁﻌﺔ ﻤﻌﺩﻨﻴﺔ ،ﻟﻬﺎ ﻁﺎﻗﺔ ﺘﻭﺘﺭ ﺴﻁﺤﻲ ﻴﻔﻭﻕﺍﻟﻁﺎﻗ ـﺔ ﺍﻟﺤﺭﻜﻴﺔ ﻟﻺﻟﻜﺘﺭﻭﻥ ،ﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﺍﻹﻟﻜﺘﺭﻭﻥ ﻴﺘﺤﺭﻙ ﺩﺍﺨل ﺍﻟﻤﻌﺩﻥ ﻭﻻ ﻴﺘﻤﻜﻥ ﻤﻥ ﺍﻟﻬﺭﻭﺏ ﻤﻨ ـ ﻪ ،ﺇﻻ ﺒﺈﻋﻁﺎﺌ ـﻪ ﻁﺎﻗﺔ ﺨﺎﺭﺠﻴﺔ .ﻭﻟﺤل ﻫﺫﻩﺍﻟﻤﺴﺄﻟﺔ ﻨﻼﺤﻅ ﻤﻥ ﺸﻜل ) (6ﺃﻥ ﺤﺭﻜﺔ ﺍﻟﺠﺴﻴﻡ ﺤﺭﺓ ﻭﻟﻜﻨﻬ ﺎﻤﻘﻴ ـﺩﺓ ﻓ ـﻲ ﺍﻟﻤ ـﺩﻯ 0 ≤ x ≤ Lﺤﻴﺙ ﺇﻥ ﺍﻟﺠﻬﺩ ∞ = V ﺨﺎﺭﺝ ﻫﺫﺍ ﺍﻟﻤﺩﻯ ﻫﻭ ﺍﻟﻤﺴﺌﻭل ﻋﻥ ﺍﻨﻌﻜﺎﺱﺍﻟﺠﺴﻴﻡ ﻤﻥ ﺤﺎﺌل ﺍﻟﺠﻬﺩ. ∞=V
ﺸﻜل) (6ﺠﺴﻴﻡ ﺩﺍﺨل ﺼﻨﺩﻭﻕ ﻤﻐﻠﻕ ﺘﻤﺎﻤﺎﹰ III
x
ﻫﻭ ﺍﺘﺒﺎﻉ ﺍﻟﺘﺎﻟﻰ:
L
V=0
∞=V
I
II
0
ﻭﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﻤﺜﻠﻰ ﻟﺤل ﻫﺫﻩ ﺍﻟﻨﻭﻋﻴﺔ ﻤﻥ ﺍﻟﻤﺴﺎﺌل ﺃﻭﻻ :ﻜﺘﺎﺒﺔ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﺘﻲ ﻴﻤﻜﻥ ﺃﻥ ﻴﺘﻭﺍﺠﺩ ﻓﻴﻬﺎ ﺍﻟﺠﺴﻴﻡ. ﻤﻥ ﺍﻟﺸﻜل ) (6ﻨﺠﺩ ﺃﻥ ﺍﻟﺠﺴﻴﻡ ﻴﺘﻭﺍﺠﺩ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ ) ( I ﻓﻘﻁ ،ﻭﻟﺫﻟﻙ ﻓﺈﻥ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺘﺄﺨﺫ ﺍ ﻟﺼﻭﺭﺓ: 13
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
⎫ ⎪ ⎪ ⎬ d 2ψ I ⎪ 2 ⇒ = − ψ k , I ⎭⎪ dx 2
= E ψ I
)(2
2 d 2ψ I − 2m dx 2
k = 2mE ﻭ mﻫﻭ ﻜﺘﻠﺔ ﺍﻟﺠﺴﻴﻡ ﻭ E ﻫﻲ ﺍﻟﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ .ﺍﻟﻤﻨﻁﻘﺘﺎﻥ ( II), ( III) ﻏﻴﺭ ﻤﺴﻤﻭﺡ ﺤﻴﺙ ﻟﻠﺠﺴﻴﻡ ﺒﺎﻻﻨﺘﻘﺎل ﺃﻭ ﺒﺎﻟﺘﻭﺍﺠﺩ ﻓﻴﻬﻤﺎ ﻨﺘﻴﺠﺔ ﺍﻟﺠﻬﺩ ∞ = V ﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﺩﺍﻟﺔ ﺍﻟﻤﻭﺠﺔ ﺘﺘﻼﺸﻰ ﻋﻨﺩ x = 0ﻭ ، x = Lﺒﻤﻌﻨﻰ ﺃﻥ: 2
2
.ψ ( xII= 0) = ψ ( xIII= L) = 0
ﺜﺎﻨﻴﺎً :ﻨﻘﺘﺭﺡ ﺤﻼﹰ ﻟﻠﻤﻌﺎﺩﻟﺔ ) ﺍﻟﻤﻌﺎﺩﻻﺕ( ﺍﻟﻤﻌﻁﺎﺓ ﻓﻲ ﺍﻟﻔﻘﺭﺓ ﺍﻷﻭﻟﻰ .ﻨﻼﺤﻅ ﻫﻨﺎ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ) (2ﺘﻤﺜل ﻤﻌﺎﺩﻟﺔ ﺤﺭﻜﺔ ﺘﻭﺍﻓﻘﻴﺔ ﺒﺴﻴﻁﺔ ﻭﺤﻠﻬﺎ ﺇﻤﺎ: )(3
) ψ I ( x ) = A sin(kx ) + B cos(kx
ﺃﻭ + b e −ikx
)(4
ψ I (x ) = a e ikx
ﺍﻟﻌﻴﺎﺭﻴﺔ.
ﺤﻴﺙ i = −1ﻭ a, b, A, Bﺜﻭﺍﺒﺕ ﺘﻌﻴﻥ ﺒﻭﺍﺴﻁﺔ ﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻭﺩﻴﺔ ﺃﻭ ﺍﻟﺸﺭﻭﻁ ﺜﺎﻟﺜﺎ :ﺍﺨﺘﺭ ﺃﺤﺩ ﺍﻟﺤﻠﻭل ﻭﻟﻴﻜﻥ ﺍﻟﺤل ﺍﻟﻤﻌﺭﻑ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ ).(3 ﺭﺍﺒﻌﺎً :ﺍﺴﺘﺨﺩﻡ ﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻭﺩﻴﺔ ﻟﻠﺘﺨﻠﺹ ﻤﻥ ﺍﻟﺜﻭﺍﺒﺕ ﻏﻴﺭ ﺍﻟﻤﻨﻁﻘﻴﺔ .ﻤﺜﻼﹰ ﺍﻟﺸﺭﻁ: ψ I (0) = 0
)(5
ﻴﺘﻴﺢ ﻟﻨﺎ ﺍﻟﺘﺨﻠﺹ ﻤﻥ ﺍﻟﺜﺎﺒﺕ Bﻭﺘﺅﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ
) (3ﺇﻟﻰ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻤﻴﺯﺓ:
) ψ I ( x ) = A sin( kx
)(6
ﺨﺎﻤﺴﺎً :ﻨﺒﺩﺃ ﻓﻲ ﺤﺴﺎﺏ ﺍﻟﺜﻭﺍﺒﺕ ﺃ -ﻟﺤﺴﺎﺏ ﺍﻟﺜﺎﺒﺕ k ﺍﺴﺘﺨﺩﻡ ﺍﻟﺸﺭﻁ ﺍﻟﺤﺩﻱ A, k
)(7
= 1, 2, 3,
،ψ I ( x = L ) = 0ﺒﻤﻌﻨﻰ:
⎫ = L) = 0 n π , = ⎬ ⇒ kn L ⎭ Asin( kL) = 0
ψ I ( x
n
∴
ﺍﻟﻜﻤﻲ.
ﺤﻴﺙ nﻴﻌﺭﻑ ﺒﺄﻨﻪ ﺍﻟﻌﺩﺩ ﺏ -ﻟﺤﺴﺎﺏ ﺍﻟﺜﺎﺒﺕ Aﺍﺴﺘﺨﺩ ﻡ ﺸﺭﻁ ﺍﻟﻌﻴﺎﺭﻴﺔ ﻜﺎﻟﺘﺎﻟﻲ: ⎫ 1 = ⎪ ∫ 0 ⎪ ⎪ L ⇒ A = 2 / L 2 2 ⎪⎬∴ A ∫ sin ( kn x) dx = 1 0 ⎪ L ⎪ 2 ⎭ L
| ψ I |2 dx
)(8
ﻭﺘﺼﺒﺢ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻤﻴﺯﺓ ﻋﻠﻰ
ﺍﻟﺼﻭﺭﺓ: 14
∵
∵
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
n = 1, 2, 3,
)(9
),x
nπ L
(sin
2
=ψ I ( )x
L
ﻤﻼﺤﻅﺎﺕ: E nﻭﻫﻲ:
-1ﺒﻤﻌﻠﻭﻤﻴﺔ k ﻨﺴﺘﻁﻴﻊ ﺃﻥ ﻨﺤﺴﺏ ﺍﻟﻁﺎﻗﺔ ﺍﻟﻜﻠﻴﺔ n
)(10
⎞ ⎛ 2π 2 ⎜= n = n2 E1 , ⎟ 2 ⎠ 2 mL ⎝
n= 1, 2, 3,
2
2
) ( k n 2m
2
=
p
2m
=
En
E 1 2
2
ﻭﻫﻲ ﻗﻴﻡ ﻤﻜﻤﺎﺓ )ﻏﻴﺭ ﻤﺘﺼﻠﺔ( .ﻭﺘﻌﺘﺒﺭﺍﻟﻘﻴﻤﺔ E = π ﻫﻲ ﻁﺎﻗﺔ ﺍﻟﻤﺴﺘﻭﻯ 2mL -2ﺍﻟﻘﻴﻤﺔ n = 0ﺃﻫﻤﻠﺕ ﻷﻨﻬﺎ ﺘﻌﻁﻰ ﺤﻼﹰ ﺼﻔﺭﻴﺎﹰ ﻟﻠﻁﺎﻗﺔ ﻭﻤﻌﻨﺎﻩ ﺃﻥﺍﻟﺠﺴﻴﻡ ﻻ ﻴﺘﺤﺭﻙ .ﻭﺃﻴﻀﹰﺎ ﺘﻌﻁﻲ ﺤﻼﹰ ﺼ ﻔﺭﻴﹰﺎ ﻟﻠﺩﺍﻟﺔ ﻭﻤﺭﺒﻌﻬﺎ ،ﺃﻱ ، ψ (0 < x < L ) = 0ﻭﻫﻭ ﺤل ﻏﻴﺭ ﻓﻴﺯﻴﺎﺌﻲ ﻷﻨﻨﺎ ﻨﻌﺭﻑ ﺃﻥ ﺍﻟﺠﺴﻴﻡ ﻤﻭﺠﻭﺩ ﻭﻟﻪ ﻁﺎﻗﺔ ﺤﺭﻜﺔ. - 3ﺍﻟﻘﻴﻡﺍﻟﺴﺎﻟﺒﺔ ﺘﻌﻁﻰ ﻨﻤﻁﹰﺎ ﻤﺘﻤﺎﺜﻼﹰ ﻟﻠﻘﻴﻡ ﺍﻟﻤﻭﺠﺒﺔ. - 4ﻗﻴﻡ ﺍﻟﻁﺎﻗﺔ ﺘﺘﻨﺎﺴﺏ ﻤﻊ . n - 5ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﻤﺴﺘﻭﻴﺎﺕ ﺍﻟﻁﺎﻗﺔ " " ∆ E ﺘﺯﺩﺍﺩ ﻤﻊ ﺯﻴﺎﺩﺓ nﺘﺒﻌﺎﹰ ﻟﻠﻌﻼﻗﺔ: ﺍﻷﺭﻀﻲ ﻟﻠﺠﺴﻴﻡ.
1
2
2
I
2
∆ E = En +1 − En = ( n + 1) 2 E1 − n 2 E1 = (2 n + 1) E1
) ﺍﻨﻅﺭ ﺍﻟﺸﻜل ﻋﻨﺩ ﺤﺴﺎﺏ ﻤﺘﻭﺴﻁ ﺍﻹﺯﺍﺤﺔ > < xﻨﺠﺩ ﺃﻥ: ( 7 −a
-6 )(11
L
⎞ 2 ⎛ L 2
=⎟ ⎜ L⎝ 4 ⎠ 2
= x) dx
2L
∫ x sin ( k L 2
n
0
L
= = ∫ x | ψ I | dx 2
x
0
ﻭﻴﻔﺴﺭ ﻫﺫﺍ ﻓﻴﺯﻴﺎﺌﻴﺎﹰ ﺒﺄﻥ ﻜﺜﺎﻓﺔ ﺍﻻﺤﺘﻤﺎل ) ﻭﺃﻴﻀﹰﺎ ﺍﻟﺩﺍﻟﺔ( ﻤﺘﻤﺎﺜﻠﺘﺎﻥ ﺤﻭل ﻤﺭﻜﺯ ﺍﻟﺼﻨﺩﻭﻕ ،ﺃﻱ ﻋﻨﺩ . L2ﻭﻟﻬﺫﺍ ﻓﺈﻥ ﺍﺤﺘﻤﺎﻟﻴﺔ ﻭﺠﻭﺩ ﺍﻟﺠﺴﻴﻡ ﻓﻲ ﺍﻟﻨﺼﻑ ﺍﻷﻴﻤﻥ ﻤﻥ ﺍﻟﺼﻨﺩﻭﻕ ﻴﻜﻭﻥ ﻤﺴﺎﻭﻴ ﺎﹰ ﻻﺤﺘﻤﺎﻟﻴﺔ ﻭﺠﻭﺩﻩ ﻓﻲ ﺍﻟﻨﺼﻑ ﺍﻷﻴﺴﺭ ) .ﺍﻨﻅﺭ ﺍﻟﺸﻜل ( 7 − b
15
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
)(a )(b ﺍﻟﺸﻜل )(6
ﺸﻜل ) (7ﻟﺠﻬﺩ
- aﺍﻟﻁﺎﻗﺎﺕﺍﻟﻤﺴﻤﻭﺡ ﺒﻬﺎ
-7ﻤﺘﻭﺴﻁ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺨﻁﻴﺔ
- bﺍﻟﺩﻭﺍل ﺍﻟﻤﺴﻤﻭﺡ ﺒﻬﺎ
ˆ pﺘﺤﺴﺏ ﻜﺎﻟﺘﺎﻟﻲ:
L L ∂ ⎛ ⎞ = ∫ ψ * Iˆpψ Idx = L2 ∫ sin( k nx) ⎜ − i sin( k nx) ⎟ dx = 0 ∂ x 0 ⎝ ⎠
)(12
ˆp
0
ﻭﻴﻔﺴﺭ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (12ﻓﻴﺯﻴﺎﺌﻴﺎﹰ ﺒﺄﻥ ﺍﻨﻌﺩﺍﻡ ﻤﺘﻭﺴﻁ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺨﻁﻴﺔ ﻻ ﺘﻌﻨﻲ ﺃﻥ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻻ ﺘﺘﺤﺭﻙ، ﻭﻟﻜﻥ ﺘﻌﻨﻲ ﺃﻥ ﻋﺩﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ ﺍﻟﺘﻲ ﺘﺘﺤﺭﻙ ﻟﻠﺸﻤﺎل ﻴﻜﻭﻥ ﻤﺴﺎﻭﻴ ﺎﹰ ﻟﻌﺩﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ ﺍﻟﺘﻲ ﺘﺘﺤﺭﻙ ﻟﻠﻴﻤﻴﻥ .ﻭﺫﻟﻙ ﻷﻥ ﻜﻤﻴﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻟﺨﻁﻴﺔ ﻫﻲ ﻜﻤﻴ ﺔ ﻤﺘﺠﻬﺔ ﻭﻟﻴﺴﺕ ﻗﻴﺎﺴﻴﺔ . ﻤﺜﺎل :ﺍﺩﺭﺱ ﺤﺭﻜﺔ ﺠﺴﻴﻡ ﻓﻲ ﻤﺠﺎل ﺠﻬﺩ
a
ﻤﺘﻤﺎﺜل ﻜﻤﺎ ﺒﺎﻟﺸﻜل ﻭﻴﻭﺼﻑ ﺒﺎﻟﺘﺎﻟﻲ:
⎪⎧0, ⎪⎩∞,
x
⎨ = ) V (x
Vx
ﻭﺃﺜﺒﺕ ﺃﻥ
∞
⎧ A sin( nπ x ) n is even 2a ⎪⎪ ⎨ = ) ψ I ( x ⎪ nπ ⎪⎩ Bcos( 2a x) nis odd 1 a
= A= B
∞
n2 ,
π 2 2 2ma 2
I
III
=E
II
x
a
16
0
-a
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﺍﻟﺤل:
ﻨﺘﻴﺠﺔ ﻟﻘﻴﻡ ﺍﻟﺠﻬﺩ ﺍﻟﻼﻨﻬﺎﺌﻴ ﺔ ﺨﺎﺭﺝ ﺍﻟﻤﻨﻁﻘﺔ ) ، (Iﻟﺫﻟﻙ ﻴﻨﻌﺩﻡ ﺘﻭﺍﺠﺩ ﺍﻟﺠﺴﻴﻡ ﺒﺎﻟﻤﻨﻁﻘﺘﻴﻥ )، (II ﺒﺎﻟﻤﻨﻁﻘﺔ ) (Iﻨﺠﺩ ﺃﻥ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻏﻴﺭ ﺍﻟﺯﻤﻨﻴﺔ ﺘﺄﺨﺫ ﺍﻟﺸﻜل: )(1
= −k ψ I , 2
ﺤﻴﺙ
2mE 2
=
2
). (III
2 d 2ψ I − = E ψ I 2m dx 2 d 2ψ I ⇒ dx 2
. k ﺍﻟﺤل ﺍﻟﻌﺎﻡ ﻟﻠﻤﻌﺎﺩﻟﺔ
) (1ﻫﻭ:
) ψ I (x ) = A sin(kx ) + B cos(kx
)(2
ﺤﻴﺙ Aﻭ Bﺜﻭﺍﺒﺕ ﺘﻌﻴﻥ ﻤﻥ ﺨﻼل ﺍﻟﺸﺭﻭﻁﺍﻟﻔﻴﺯﻴﺎﺌﻴﺔ .ﻤﻥ ﺸﺭﻭﻁ ﺍﻨﻌﺩﺍﻡ ﺍﻟﺩﺍﻟﺔ ﻨﺠﺩ ﺃﻥ: )(3
⇒ A cos(ka ) + B sin(ka ) = 0,
)(4
⇒ A cos(ka ) − B sin( ka ) = 0
ﺒﺠﻤﻊ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ) (3ﻭ ) (4ﻴﻨﺘﺞ
ﻋﻨﺩ ﺍﻟﺤﺩﻭﺩ ) (aﻭ ) (−a
ψ I (a ) = 0
ψ I ( −a ) = 0
ﺃﻥ:
2 Acos( ka )=0 )(5
ka0 =A = ) (0 or cos
ﺒﻁﺭﺡ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ) (3ﻭ ) (4ﻴﻨﺘﺞ
⇒
ﺃﻥ:
2 Bsin( ka )=0 )(6
= B0 or sin( ) =ka0
⇒
ﻭﻨﺤﻥ ﻫﻨﺎ ﻻ ﻨﺒﻐﻲ ﺃﻥ ﻨﺴﺎﻭﻱ ﺍﻟﺜﻭﺍﺒﺕ Aﻭ Bﺒﺎﻟﺼﻔﺭ ،ﺤﻴﺙ ﺃﻨﻨﺎ ﺴﻨﺤﺼل ﻋﻠﻰ ﺩﺍﻟﺔ ) ψ I ( xﻏﻴﺭ ﺫﺍﺕ ﻓﺎﺌﺩﺓ ﻓﻴﺯﻴﺎﺌﻴﺎﹰ .ﻭﺃﻴﻀﹰﺎ ﻨﺤﻥ ﻻ ﻨﻭﺩ ﻭﻀﻊ ﺍﻟﺩﻭﺍل ) sin(kaﻭ ) cos( kaﻤﺴﺎﻭﻴﺔ ﻟﻠﺼﻔﺭ ﻟﺒﻌﺽ ﻗﻴﻡ k ﻭ . E ﻭﻟﻬﺫﺍ ﻨﺠﺩ ﺃﻨﻪ ﻴﻭﺠﺩ ﻨﻭﻋﺎﻥ ﻤﻥ ﺍﻟﺤﻠﻭل ﻭﻫﻤﺎ: ≠0 B
⇒ sin( ka ) = 0, ≠A 0 ⇒ cos( ka )=0
)(i )(ii
ﻓﻲ ﺍﻟﻨﻭﻉ ﺍﻷﻭل ﻨﺠﺩ
=0 A =0 B
ﺃﻥ: ,
π 2
⇒ ka = π , 2π , 3π , = n
17
sin( ka ) = 0
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﺤﻴﺙ
n
ﻋﺩﺩ ﺼﺤﻴﺢ ﺯﻭﺠﻲ .ﻓﻲ ﺍﻟﻨﻭﻉ ﺍﻟﺜﺎﻨﻲ ﻨﺠﺩ π , 2
, = n
π 2
,5
π 2
,3
ﺃﻥ:
π 2
= ⇒ ka
cos( ka ) = 0
nﻋﺩﺩ ﺼﺤﻴﺢ ﻓﺭﺩﻱ.
ﺤﻴﺙ ﻭﻤﻨﻪ ﻨﺼل ﺇﻟﻰ ﺍﻟﺨﻼﺼﺔ ﺃﻥ:
n is even
n is odd
= n 2E1
π 2 2 2
2ma
2
=n
k n2 2
2m
) ⎧ A sin( nπ x 2a ⎪⎪ ⎨ = ) ψ I ( x ⎪ nπ B )x (cos ⎩⎪ 2a =
En
⇒
,
2mE n 2
=
2 n
k
E 1
1 a
= =B
A
-5.IIIﺍﻟﺠﻬﺩ ﺍﻟﺩﺭﺠﻲ ﺩﺭﺍﺴﺔ ﺤﺯﻤﺔ ﻤﺘﺠﺎﻨﺴﺔ ﻤﻥ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺘﺘﺤﺭﻙ ﻓﻲ ﺍﺘﺠﺎﻩ ﺤﺎﺠﺯ ﺠﻬﺩ ﻋﻠﻰ
ﺍﻟﺼﻭﺭﺓ:
⎧ 0, x < 0 ⎨ = )V ( x ⎩Vo , x ≥ 0
ﻓﻲ ﻜﻠﺘﺎ ﺍﻟﺤﺎﻟﺘﻴﻥ ﻭ ﻫﻤﺎ ﺃﻥ ﺘﻜﻭﻥ ﻁﺎﻗﺔ ﺍﻟﺠﺴﻴﻡ ﺍﻟﻜﻠﻴﺔ E ﺃﻜﺒﺭ ﻤﻥ V ﺃﻭ ﺃﺼﻐﺭ ﻤﻥ V ﺤﻴﺙ V ﻫﻭ ﺍﺭﺘﻔﺎﻉ ﺤﺎﺠﺯ ﺍﻟﺠﻬﺩ ) ﺍﻨﻅﺭ ﺍﻟﺸﻜل .(2. III .1ﻭﻤﻨﻬﺎ ﺴﻨﻭﺠﺩ ﻤﻌﺎﺩﻻﺕ ﺍ -ﺍﻟﺘﻴﺎﺭ ﺍﻟﺴﺎﻗﻁ ﺏ -ﺍﻟﺘﻴﺎﺭ ﺍﻟﻤﻨﻌﻜﺱ ﺕ -ﺍﻟﺘﻴﺎﺭ ﺍﻟﻨﺎﻓﺫ ﺡ -ﻤﻌﺎﻤل ﺍﻻﻨﻌﻜﺎﺱ ﺩ -ﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ. o
18
o
o
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
)V(x
)V(x
Vo E
Vo
E
x
x
0
0
)(a
ﺸﻜل
)(a
2. III .1
)(b
ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ
E > V o
) ( b
ﺍﻟﺤﺎﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ
E < V o
ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻜﻼﺴﻴﻜﻴﺔ :ﻗﺒل ﺃﻥ ﻨﺒﺩﺃ ﻓﻲ ﺤلﺍﻟﻤﺴﺄﻟﺔ ﻤﻥ ﻭﺠﻬﺔ ﻨﻅﺭ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ،ﺩﻋﻭﻨﺎ ﻨﺴﺘﻌﺭﺽ ﻤﺜﺎ ﹰﻻ ﺒﺴﻴﻁﺎﹰ ﻤﻥ ﺍﻟﺤﻴﺎﺓ ﻟﻨﺘﺼﻭﺭ ﺍﻟﺤل ﺍﻟﻜﻼﺴﻴﻜﻲ .ﻨﺤﻥ ﻓﻲ ﺸﺒﺎﺒﻨﺎ ﻭﺼﺤﺘﻨﺎ ﻻ ﺘﻌﻴﺭﻨﺎ ﺍﻫﺘﻤﺎﻤﺎ ،ﻭﻻ ﺘﻭﻗﻔﻨﺎ ،ﺍﺭﺘﻔﺎﻉ ﺩﺭﺠﺎﺕ ﺍﻟﺴﻼﻟﻡ ،ﺴﻨﺭﻤﺯ ﻟﻬﺎ ﺒﺎﻟﺭﻤﺯ ، V ﻷﻥ ﻁﺎﻗﺎﺘﻨﺎ ،ﺴﻨﺭﻤﺯ ﻟﻬﺎ ﺒﺎﻟﺭﻤﺯ ، E ﺃﻋﻠﻰ ﻜﺜﻴﺭﺍﹰ ،ﺒﻤﻌﻨﻰ ﺃﻥ . E > V ﻭﻟﻜﻥ ﻋﻨﺩ ﺍﻟﻜﺒﺭ ،ﻤﺭﺤﻠﺔ ﺍﻟﻭﻫﻥ ،ﻨﻔﻜﺭﻜﺜﻴﺭﹰﺍ ﻓﻲ ﺍﺭﺘﻔﺎﻉ ﺩﺭﺠﺎﺕ ﺍﻟﺴﻼﻟﻡ ،ﻭﻏﺎﻟﺒﺎﹰ ﻤﺎ ﻨﺭﺠﻊ ﻭﺫﻟﻙ ﻷﻥ ﻁﺎﻗﺎﺘﻨﺎ ﺃﺼﺒﺤﺕ ﺃﻗل ﻜﺜﻴﺭﹰﺍ ﻤﻥ ﺍﺭﺘﻔﺎﻉ ﺩﺭﺠﺎﺕ ﺍﻟﺴﻼﻟﻡ ،ﺒﻤﻌﻨﻰ ﺃﻥ . E < V ﺍﻵﻥ ﺍﻟﻨﻅﺭﺓ ﺴﻭﻑ ﺘﺨﺘﻠﻑ ﺘﻤﺎﻤﺎ ﻤﻥ ﻭﺠﻬﺔ ﻨﻅﺭ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻭﺫﻟﻙ ﻨﺎﺘﺞ ﻤﻥ ﺍﻟﺨﻭﺍﺹ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﺠﺴﻴﻤﺎﺕ. o
o
o
ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ E > V ﻤﺭﺤﻠﺔ ﺍﻟﺸﺒﺎﺏ o
ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﻴﺴﺭﻯ ﻤﻥﺍﻟﻤﺴﺘﻘﻴﻡ = E ψ I )(1
2mE 2
ﻭﺍﻟﺤل ﺍﻟﻌﺎﻡ ﻟﻬﺎ
k
2 2 d ψ I
2m dx 2
= −k ψ I , 2
−
2
d ψ I dx 2
⇒
ﻫﻭ: − ikx
)(2
=
2
x = 0ﺘﻌﺭﻑ:
e
Reflected wave
+B
ikx
e
ψ I ( x) = A
Incident wave
ﺍﻟﺩﺍﻟﺔ e ikxﻫﻲ ﺩﺍﻟﺔ ﻤﻭﺠﻴﺔ ﻤﺴﺘﻭﻴﺔ ﺘﺘﺤﺭﻙ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﻤﻭﺠﺏ ﻟﻠﻤﺤﻭﺭ ﺍﻟﺴﻴﻨﻲ ﻭﺘﺴﻤﻰ ﺩﺍﻟﺔ ﺴﺎﻗﻁﺔ ) e −ikx ، (Incident waveﻫﻲ ﺩﺍﻟﺔ ﻤﻭﺠﻴﺔ ﻤﺴﺘﻭﻴﺔ ﺘﺘﺤﺭﻙ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﺴﺎﻟﺏ ﻟﻠﻤﺤﻭﺭ ﺍﻟﺴﻴﻨﻲ ﻭﺘﺴﻤﻰ 19
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﺩﺍﻟﺔ ﻤﻨﻌﻜﺴﺔ ) A . (Reflected waveﻫﻲ ﺴﻌﺔ ﺍﻟﻤﻭﺠﺔ ﺍﻟﺴﺎﻗﻁﺔ ﻭ Bﻫﻲ ﺴﻌﺔ ﺍﻟﻤﻭﺠ ﺔ ﺍﻟﻤﻨﻌﻜﺴﺔ ﻤﻥ ﺤﺎﺌﻁ ﺍﻟﺠﻬﺩ. ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﻴﻤﻨﻰ ﻤﻥ ﺍﻟ ﺤﺎﺌل ) ( x = 0ﺘﻌﺭﻑ ﺒﺎﻟﺘﺎﻟﻲ: 2
)(3
) ( E −V
o
ﻭﺤﻠﻬﺎ ﺍﻟﻌﺎﻡ
2 d ψ II
+V oψ II = E ψ II 2m 2
=
α II2
2m dx 2
= −α ψ , 2
−
2
d ψ II 2
dx
⇒
ﻫﻭ:
+ D e −i α x
)(4
iαx
ψ II ( x ) = C e
e i α xﻫﻲ ﺩﺍﻟﺔ ﻤﻭﺠﻴﺔ ﻤﺴﺘﻭﻴﺔ ﺘﺘﺤﺭﻙ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﻤﻭﺠﺏ ﻟﻠﻤﺤﻭﺭ ﺍﻟﺴﻴﻨﻲ ﻭﺘﺴﻤﻰ ﻤﻭﺠ ﺔ ﻨﺎﻓﺫﺓ ) (Transmitted waveﻭﺤﻴﺙ ﺇﻨﻪ ﻻ ﻴﻭﺠﺩ ﺤﺎﺌﻁ ﺠﻬﺩ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟ ﻴﻤﻨﻰ ﻟﻜﻲ ﺘﺭﺘﺩ ﻤﻨ ﻪ ﺍﻷﺸﻌﺔ ،ﻓﻠﻬﺫﺍ ﻨﻀﻊ . D = 0ﻭﺍﻵﻥ ﻋﻨﺩ ﺍﻟﺤﺩ x = 0ﻨﺴﺘﻁﻴﻊ ﺃﻥ ﻨﻁﺒﻕ ﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻭﺩ ﻴﺔ ﻟﻠﺩﺍﻟﺔ: )∵ ψ ( x I= 0) = ψ ( xII= 0
)(5
∴
A+ B = C
ﻭﻟﻤﺸﺘﻘﺘﻬﺎ :
)= 0) = ψ ' ( x II= 0 ∴ ik ( A − B ) = i α C I
)(6
ﺤل ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ
∵ψ ' ( x
) (5ﻭ) (6ﻴﻌﻁﻲ:
⎛ k − α ⎞ A, ⎟ ⎠ ⎝ k +α
⎛ 2k ⎞ A ⎟ ⎠ ⎝ k + α
⎜ =C
)(7
⎜ =B
ﺒﺎﻹﻤﻜﺎﻥ ﺘﻌﺭﻴﻑ q ﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﺍﻟﺴﺎﻗﻁ = ، v | A |2 = m | A |2 :
1
2 q | B |2 | | B = ﻭﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻤﻨﻌﻜﺱ = 1 m α = v 2 | C |2 ﻭﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻨﺎﻓﺫ = | C |2 m ﺤﻴﺙ . vi = k i / mﻭﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ:
، v
ﻤﻌﺎﻤل ﺍﻻﻨﻌﻜﺎﺱ
) ( R
=
ﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻤﻨﻌﻜﺱ ﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﺍﻟﺴﺎﻗﻁ
20
2
= ⎞ ⎛ k − α ⎟ ⎜ k + α ⎝ ⎠
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ
) (T
=
ﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻨﺎﻓﺫ ﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﺍﻟﺴﺎﻗﻁ
=
4k α 2
)( k + α
ﺍﻟﺠﺴﻴﻤ ﺎﺕ.
- 1ﻤﻥ ﺍﻟﻌﻼﻗﺘﻴﻥﺍﻟﺴﺎﺒﻘﺘﻴﻥ ﻨﺠﺩ ﺃﻥ T + R = 1ﻭﻫﻭ ﻗﺎﻨﻭﻥ ﺤﻔﻅ - 2ﺍﻟﻔﻴﺯﻴﺎﺀﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﺘﻤﻨﻊ ﺃﻱ ﺍﻨﻌﻜﺎﺱ ﻋﻨﺩﻤﺎ E >V ﻭﻫﺫﺍ ﻏﻴﺭ ﻤﻨﻁﺒﻕ ﻓﻲ ﻗﻭﺍﻨﻴﻥ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻭﺫﻟﻙ ﻨﺎﺘﺞ ﻤﻥ ﺍﻟﺨﻭﺍﺹ ﺍﻟﻤﻭﺠﻴﺔ ﻟﻠﺠﺴﻴﻤ ﺎﺕ. o
ﺍﻟﺤﺎﻟﺔﺍﻟﺜﺎﻨﻴﺔ E < V ﻤﺭﺤﻠﺔ ﺍﻟﻭﻫﻥ o
ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﻴﺴﺭﻯ ﻤﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ x = 0ﻟﻥ ﺘﺘﻐﻴﺭ ،ﻭﺒﺎﻟﺘﺎﻟﻲ ﺴﻭﻑ ﻨﺴﺘﺨﺩﻡ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ) (1ﻭ) .(2ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﻴﻤﻨﻰ ﻤﻥﺍﻟﻤﺴﺘﻘﻴﻡ x = 0ﻫﻲ: 2
)(8
ﻭﺤﻠﻬﺎ ﺍﻟﻌﺎﻡ
)− E o
(V
+V oψ II = E ψ II 2m 2
=β 2
II
2 d ψ II
2m dx 2
=βψ , 2
−
d 2ψ II dx 2
⇒
ﻫﻭ:
)(9
+ D e β x
ψ II (x ) = C e − β x
β x eﻫﻲ ﺩﺍﻟﺔ ﻤﻭﺠﻴﺔ ﺘﺯﺍﻴﺩﻴﺔ ﻓﻲ ﺍﻟﻤﺩﻯ }∞ {0,ﺒﻤﻌﻨﻰ ﺃﻥ ∞ = e limﻭﺒﺎﻟﺘﺎﻟﻲ ﻻ ﺘﺤﻘﻕ ﺸﺭﻭﻁ ﺍﻟﺩﺍﻟﺔ ∞→ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ،ﻟﻬﺫﺍ ﻨﻀﻊ . D = 0ﺍﻵﻥ ﻋﻨﺩ ﺍﻟﺤﺩ x = 0ﻨﺴﺘﻁﻴﻊ ﺃﻥ ﻨﻁﺒﻕ ﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻭﺩﻴﺔ ﻟﻠﺩﺍﻟﺔ: β x
x
)(10
) = 0 ) = ψ ( x II= 0 ∴ A+ B = C I
∵ψ ( x
ﻭﻟﻤﺸﺘﻘﺘﻬﺎ : )(11
ﺒﺤل ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ
) = 0) = ψ ' (x II= 0 ∴ k ( A − B ) = − β C I
∵ψ ' ( x
) (10ﻭ) (11ﻨﺤﺼل ﻋﻠﻰ:
⎞ ⎛ ik + β ⎞ ⎛ k − i β A = ⎟ ⎜ k + i β ⎟ ,A ik β − ⎝ ⎠ ⎝ ⎠ ⎞ ⎛ 2ik ⎞ ⎛ 2k A ⎜= = ⎟ ⎜ k + i β ⎟ A ik β − ⎝ ⎠ ⎝ ⎠
⎜ =B
)(12
ﻤﺜﺎل :ﺍﺴﺘﺨﺩﻡ
ﺍﻟﺘﺤﻭﻴﻼﺕﺍﻟﻘﻁﺒﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ:
21
C
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
β
= r cos δ ,
k
= r sin δ ,
⎜ δ = tan −1
= k 2 + β 2 ,
r
⎛ β ⎞ = V o − E ⎟ E ⎠ ⎝k
)(13
ﻟﺘﺒﺴﻴﻁ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺤل :ﺍﻟ ﺜﺎﺒﺕ Bﻴﺒﺴﻁ ﻜﺎﻟﺘﺎﻟﻲ: )(12 A
⎞δ
⎟
⎠δ
⎞⎛ ik + β ⎞⎛ k − i β ⎛ cos δ− i sin = = A A ⎜ cos δ+ i sin ⎟ ⎟⎜ k + i β ⎝ ⎠⎝ ik − β ⎝ ⎠
⎜ =B
− i δ
A
)(14
i δ
re
re
=
= e −2i δ A
ﺍﻟﺜﺎﺒﺕ
C ﻴﺒﺴﻁ ﻜﺎﻟﺘﺎﻟﻲ:
⎞ ⎛ 2ik ⎞ ⎛ 2k ⎛ 2k ⎞ ⎜= A = A = − + 1 1 ⎟ ⎟⎜ k + i β ⎜ k + i β ⎟A ⎠⎝ ik − β ⎝ ⎠ ⎝ ⎠ ⎞ ⎛ k − i β ⎜= + 1⎟ A ⎠ ⎝ k + i β
)(15
C
= (e −2i δ + 1) A ﺍﻟﻤﻼﺤﻅﺎﺕ:
ﻭﻟﻨﺎ ﻫﻨﺎ ﺒﻌﺽ - 1ﺍﻟﺸﻌﺎﻉ ﺍﻟﺴﺎﻗﻁ ﻭ ﺍﻟﻤﻨﻌﻜﺱ ﻟﻬﻤﺎ ﻨﻔﺱﺍﻟﺸﺩﺓ ،ﺒﻤﻌﻨﻰ ﺃﻥ: ⎛ k − i β ⎞ ⎛ k + i β ⎞ 2 |⎟ × ⎜ k − i β ⎟ | A k i β + ⎝ ⎝ ⎠ ⎠
⎜ =| B|2 = B* B
= | A |2
ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺠﻤﻴﻊ ﺍﻟﺠﺴﻴﻤﺎﺕ ﺍﻟﺴﺎﻗﻁﺔ ﺒﻁﺎﻗﺔ E < V ﺴﻭﻑ ﺘﻨﻌﻜﺱ ﻜﻠﻴﺎﹰ ﻋﻨﺩﻤﺎ ﺘﺼل ﺇﻟﻰ o
ﺍﻟﺤﺎﺌل ،ﻭﻴﺘ ﺴﺎﻭﻯ ﻤﻌﺎﻤل ﺍﻻﻨﻌﻜﺎﺱ ﺒﺎﻟﻭﺤﺩﺓ ) ﺒﻤﻌﻨﻲ ﺃﻥ
=1
| B |2
| .( R = | Aﻭﺒﺎﻟﺘﺎﻟﻲ ﺴﻭﻑ ﻴﻨﻌﺩﻡ
2
ﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ ،ﺃﻱ ﺃﻥ - 2ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍ ﻟﻤﻌﺎﺩﻻﺕ ) (13ﻭ) (14ﻴﻤﻜﻨﻨﺎ ﻭﻀﻊ ﺍﻟﺩﻭﺍل ﺍﻟﻤﻭﺠﻴﺔ ﺒﺎﻟﺼﻭﺭﺓ: ". "T = 0
+ δ ),
)(16
β x
)(17
ψ I ( x ) = 2A e − i δ cos( kx
) e−
−i δ
(
ψ II ( x) = 2 A cos δ e
- 3ﻜﻼﺴﻴﻜﻴﹰﺎ ﺘﻌﺘﺒﺭ ﺍﻟﻤﻨﻁﻘﺔ ) ( II ﻤﻨﻁﻘﺔ ﻏﻴﺭ ﻤﺴﻤﻭﺡ ﺘﻭﺍﺠﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ ﺒﻬ ﺎ ،ﻭﺫﻟﻙ ﻷﻥ ﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ ) ( T = E − V ﺴﻭﻑ ﺘﺼﺒﺢ ﻜﻤﻴﺔ ﺴﺎﻟﺒﺔ ﻨﻅﺭﹰﺍ ﻷﻥ . E < V o - 4ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (17ﻨﺴﺘﻨﺘﺞ ﺃﻥ: o
22
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
) ( II ﻤﻨﻌﺩﻤﺔ ﻷﻥ ﺍﻟﺩﺍﻟ ﺔ ﺤﻘﻴﻘﻴﺔ .
ﺃ -ﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ ﺏ -ﺍﺤﺘﻤﺎﻟﻴﺔ ﻭﺠﻭﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ
) ( II ﺘﻌﻁﻰ ﺒﺎﻟﻤﻌﺎﺩﻟﺔ:
2
= ( 2 A cos δ ) e −2 β x )(15 ﻭﻫﺫﻩ ﺘﻌﻁﻰ ﻗﻴﻤﺎﹰ ﻤﻘﺒﻭﻟﺔ ﻋﻨﺩ x = 0ﻭ ﺘﻘل ﺘﺩﺭﻴﺠﻴﺎ ) ﺃﺴﻴ ﺎﹰ( ﻤﻊ ﺯﻴﺎﺩﺓﺍﻟﻤﺴﺎﻓﺔ .ﺍﻨﻅﺭ ﺍﻟﺸﻜل . b 2. III .1 - 5ﺘﻨﻌﺩﻡ ﺍﻟﺩﺍﻟﺔ ) (x
| P II ( x ) =| ψ II
2
ψ ﺘﻤﺎﻤﺎﹰ ﻋﻨﺩﻤﺎ ∞ → ، V
II
o
ﻭ ﺘﺄﺨﺫ ﺍﻟﺩﺍﻟﺔ ) ψ I (xﺒﺎﻟﺸﻜل:
ψ I ( x ) = A sin kx
ﺤﻴﺙ ﺇﻨ ﻬﺎ ﻴﺠﺏ ﺃﻥ ﺘﻨﻌﺩﻡ ﻋﻨﺩﻤﺎ - 6ﺒﺎﺴﺘﻁﺎﻋﺘﻨﺎﺍﺴﺘﻨﺘﺎﺝ ﺍﻟ ﻤﻌﺎﺩﻟﺔ ) (12ﻤﺒﺎﺸﺭ ﺓﹰ ﻭﺫﻟﻙ ﺒﺘﻐﻴﻴﺭ i α → −β ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ).(7 . x = 0
-5. IVﺤﺎﺠﺯﺍﻟﺠﻬﺩ ﺍﻟﻤﺴﺘﻁﻴل
)( Rectangular potential barrier
ﺩﺭﺍﺴﺔ ﺤﺯﻤﺔ ﻤﺘﺠﺎﻨﺴﺔ ﻤﻥ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﺘﺘﺤﺭﻙ ﻓﻲ ﺍﺘﺠﺎﻩ ﺤﺎﺠﺯ ﺠﻬﺩ ﻋﻠﻰ
ﺍﻟﺼﻭﺭﺓ:
⎧V o , 0 < x < a
⎨ = ) V (x
⎩ 0, a ≤ x ≤ 0
ﻓﻲ ﻜﻠﺘﺎ ﺍﻟﺤﺎﻟﺘﻴﻥ ﻭﻫﻤﺎ ﺃﻥ ﺘﻜﻭﻥ ﻁﺎﻗﺔ ﺍﻟﺠﺴﻴﻡ ﺍﻟﻜﻠﻴﺔ E ﺃﻜﺒﺭ ﻤﻥ V ﺃﻭ ﺃﺼﻐﺭ ﻤﻥ V ﺤﻴﺙ V ﻫﻭ ﺍﺭﺘﻔﺎﻉ ﺤﺎﺠﺯ ﺍﻟﺠﻬﺩ ) ﺍﻨﻅﺭ ﺍﻟﺸﻜل .(2. III .1ﻭﻤﻨﻬﺎ ﺴﻨﻭﺠﺩ ﻤﻌﺎﺩﻻﺕ ﺍ -ﺍﻟﺘﻴﺎﺭ ﺍﻟﺴﺎﻗﻁ ﺏ -ﺍﻟﺘﻴﺎﺭ ﺍﻟﻤﻨﻌﻜﺱ ﺕ -ﺍﻟﺘﻴﺎﺭ ﺍﻟﻨﺎﻓﺫ ﺡ -ﻤﻌﺎﻤل ﺍﻻﻨﻌﻜﺎﺱ ﺩ -ﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ. o
23
o
o
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
)V (x
)V (x
V o
V o E
II
III
x
III II
I
a
x
0
)(a
ﺸﻜل
2. III .1
)(a
ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ
ﺍﻟﺤﺎﻟﺔ ﺍﻷﻭﻟﻰ
E > V o
)(1
ﺤﻴﺙ
2mE 2
) ، −∞ ≤ x ≤ a ، ( I ﺘﻌﻁﻲ ﻜﺎﻷﺘﻲ: d 2ψ I 2
. k ﻭﺍﻟﺤل ﺍﻟﻌﺎﻡ ﻟﻠﻤﻌﺎﺩﻟﺔ
− ikx
)(2
) ( b
ﺍﻟﺤﺎﻟﺔ ﺍﻟﺜﺎﻨﻴﺔ
E < V o
E
= − k 2ψ I , =
a
0
)(b
ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﻼﺯﻤﻨﻴﺔ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ
2
E
I
+B
e
Reflected wave
dx
) (1ﻫﻭ:
ψ I ( x) = A
ikx
e
Incident wave
− e ikx
e ikxﻫﻲ ﺩ ﺍﻟﺔ ﻤﻭﺠﻴﺔ ﻤﺴﺘﻭﻴﺔ ﺘﺘﺤﺭﻙ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﻤﻭﺠﺏ ﻟﻠﻤﺤﻭﺭ ﺍﻟﺴﻴﻨﻲ ﻭﺘﺴﻤﻰ ﺩﺍﻟﺔ ﺴﺎﻗﻁﺔ، ﻫﻲ ﺩﺍﻟﺔ ﻤﻭﺠﻴﺔ ﻤﺴﺘﻭﻴﺔ ﺘﺘﺤﺭﻙ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﺴﺎﻟﺏ ﻟﻠﻤﺤﻭﺭ ﺍﻟﺴﻴﻨﻲ ﻭﺘﺴﻤﻰ ﺩﺍﻟﺔ ﻤﻨﻌﻜﺴﺔ A .ﻫﻲ ﺴﻌﺔ ﺍﻟﻤﻭﺠﺔ ﺍﻟﺴﺎﻗﻁﺔ ﻭ Bﻫﻲ ﺴﻌﺔ ﺍﻟﻤﻭﺠﺔ ﺍﻟﻤﻨﻌﻜﺴﺔ ﻤﻥ ﺤﺎﺌﻁ ﺍﻟﺠﻬﺩ. ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻏﻴﺭ ﺍﻟﺯﻤﻨﻴﺔ ﻓﻲ ﻤﻨﻁﻘﺔ ﺤﺎﺠﺯ ﺍﻟﺠﻬﺩ ) ، 0 ≤ x ≤ a ، ( II ﺘﻌﺭﻑ ﺒﺎﻟﺘﺎﻟﻲ:
)(3
ﺤﻴﺙ )(4
= α ψ II 2
)− E
(V o
2m 2
=
+V oψ II = E ψ II 2
d ψ II dx 2
2
−α x
e
2m dx 2
⇒
ﻟﻠﻤﻌﺎﺩﻟﺔ . α ﻭﺍﻟﺤل ﺍﻟﻌﺎﻡ +D
2 2 d ψ II
αx
) (3ﻫﻭ:
ψ II (x ) = C e
24
−
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﺍﻟﺤﺩ C eﻫﻲ ﺩﺍﻟﺔ ﺃُﺴﻴﺔ ﺘﺯﺍﻴﺩﻴﺔ ،ﺘﻤﺜل ﺍﻀﻁﺭﺍﺒﹰﺎ ﻤﻨﻌﻜﺴﹰﺎ ﻏﻴﺭ ﻤﺘﺫﺒﺫﺏ ﻴﺘﺤﺭﻙ ﺨﻼل ﺍﻟﺤﺎﺌل ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﺴﺎﻟﺏ ﻟﻠﻤﺘﻐﻴﺭ . xﺍﻟﺤﺩ De −α xﻴﻌﺒﺭ ﺩﺍﻟﺔ ﺃُﺴﻴﺔ ﺘﻨﺎﻗﺼﻴﺔ ،ﺘﻤﺜل ﺍﻀﻁﺭﺍﺒﹰﺎ ﻏﻴﺭ ﻤﺘﺭﺩﺩ ﻴﺘﺤﺭﻙ ﺨﻼل ﺍﻟﺤﺎﺌل ﻤﻊ ﺯﻴﺎﺩﺓ ﺍﻟﻤﺘﻐﻴﺭ . xﻭﻨﻅﺭﹰﺍ ﻟﻘﻴﻡ xﺍﻟﻤﺤﺩﺩﺓ ﺒﻴﻥ 0ﻭ aﻓﺈﻨﻪ ﻟﻥ ﻴﺘﻡ ﺇﻫﻤﺎل ﺃﻱ ﻤﻥ ﺍﻟﺤﺩﻭﺩ. α x
ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﻏﻴﺭ ﺍﻟﺯﻤﻨﻴﺔ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ )، a ≤ x ≤ ∞ ، (III )(5
= − k 2ψ III ,
ﻭﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ) (5ﺍﻟﻌﺎﻡ
ﻫﻭ:
− ikx
)(6
e
Reflected wave
+ H
ﺘﹸﻌﻁﻰ ﻜﺎﻷﺘﻲ:
d 2ψ III 2
dx
ψ III ( )x = G
ikx
e
Transmitted wave
e ikxﻫﻲ ﺩﺍﻟﺔ ﻤﻭﺠﻴﺔ ﻤﺴﺘﻭﻴﺔ ﺘﺘﺤﺭﻙ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﻤﻭﺠﺏ ﻟﻠﻤﺤﻭﺭ ﺍﻟﺴﻴﻨ ﻲ ،ﻭﺘﺴﻤﻰ ﻤﻭﺠﺔ ﻨﺎﻓﺫﺓ ) e −ikx . (Transmitted waveﻫﻲ ﺩﺍﻟﺔ ﻤﻭﺠﻴﺔ ﻤﺴﺘﻭﻴﺔ ﺘﺘﺤﺭﻙ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﺴﺎﻟﺏ ﻟﻠﻤﺤﻭﺭ ﺍﻟﺴﻴﻨﻲ ﻭﺘﺴﻤﻰ ﺩﺍﻟﺔ ﻤﺭﺘﺩﺓ G .ﻫﻲ ﺴﻌﺔ ﺍﻟﻤﻭﺠﺔ ﺍﻟﺴﺎﻗﻁﺔ ﻭ H ﻫﻲ ﺴﻌﺔ ﺍﻟﻤﻭﺠﺔ ﺍﻟﻤﺭﺘﺩﺓ ﻤﻥ ∞ .ﻭﺤﻴﺙ ﺇﻨﻪ ﻻﺘﻭﺠﺩ ﻤﻭﺠﺔ ﻤﺭﺘﺩﺓ ﻤﻥ ∞ ﻓﺈﻨﻨﺎ ﻨﺴﺘﻁﻴﻊ ﻭﻀﻊ . H = 0 ﻟﺫﻟﻙ ﻨﺠﺩ ﺃﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭلﺍﻟﻤﻘﺒﻭﻟﺔ ﻫﻲ: + B e − ikx −α x α x ψ II ( x ) = C e + D e ψ III ( x ) = G eikx ψ I ( x ) = A e ikx
)(4
ﺍﻵﻥ :ﻋﻨﺩ ﺍﻟﺤﺩ x = 0ﻨﺴﺘﻁﻴﻊ ﺃﻥ ﻨﻁﺒﻕ ﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻭﺩﻴﺔ )(5
ﻟﻠﺩﻭﺍل:
)∵ ψ ( x I= 0) = ψ ( xII= 0
∴
A+ B = C
ﻭﻟﻤﺸﺘﻘﺎﺘﻬﺎ : )(6
)= 0) = ψ ' ( x II= 0 ∴ ik ( A − B ) = α C − α D I
ﻋﻨﺩ ﺍﻟﺤﺩ x = aﻨﺴﺘﻁﻴﻊ ﺃﻥ ﻨﻁﺒﻕ ﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻭﺩﻴﺔ
∵ψ ' ( x
ﻟﻠﺩﻭﺍل:
)∵ ψ ( x II= a) = ψ ( xIII= a
)(7
+ De −α a = Ge ika
ﻭﻟﻤﺸﺘﻘﺎﺘﻬﺎ :
25
αa
Ce
∴
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
) = a ∵ ψ ' ( x =II a) = ψ ' ( x III
)(8
∴ α Ce αa − α De −α a = ikGe ika
ﺤﺼﻠﻨﺎ ﻫﻨﺎ ﻋﻠﻰ ﺃﺭﺒﻊ ﻤﻌﺎﺩﻻﺕ ﻭﻟﻜﻥ ﻟﺨﻤﺴﺔ ﻤﺠﺎﻫﻴل .ﺒﺤل ﺍﻟﻤﻌﺎﺩﻻﺕ ) (8 ،7 ،6 ، 5ﻟﻜل ﻤﻥ Aﻭ Bﺒﺩﻻﻟﺔ Gﻨﺤﺼل ﻋﻠﻰ: ⎤ ⎦
ika )a⎥ Ge
)(9
⎡ ⎣
⎛ α − k ⎞ sinh(α ⎜ ⎟ ⎠ 2 ⎝ k α i α k ika =B − ⎛⎜ + ⎞⎟ sinh(α ) a Ge ⎠ 2 ⎝ k α i
= ⎢cosh(α )a+ A
ﺒﺎﻹﻤﻜﺎﻥ ﺘﻌﺭﻴﻑ ﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﺍﻟﺴﺎﻗﻁ = ، v | A |2ﻭﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ ﺍﻟﻤﻨﻌﻜﺱ = ، v | B |2ﻭﻜﺜﺎﻓﺔ ﺍﻟﺘﻴﺎﺭ 2 ﺍﻟﻨﺎﻓﺫ = | v | Gﺤﻴﺙ . v = k / mﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﻤﻌﺎﻤل ﺍﻻﻨﻌﻜﺎﺱ ) ( Rﻴﺄﺨﺫ ﺍﻟﺼﻭﺭﺓ: 1
2
i
1
i
2
⎞ 1 ⎛ α k ) + ⎟ sinh 2 (α a ⎜ ⎠ 4 ⎝ k α
= ⎞ ⎛ B⎞ ⎛ B = ⎟ * ⎜⎟ ⎜ 2 * ⎠ A A ⎝ A⎠ ⎝ A ⎞ 1 ⎛ α k 2 ) cosh (α a ) + ⎜ − ⎟ sinh 2 (α a ⎠ 4 ⎝ k α *
BB
=
R
2
)(10
V o
2
) sinh (α a
-1
⎡ 4 E(V − E) ⎤ = ⎢1 + 2 o 2 ⎥ ⎦ ) V o sinh (α a 2 ⎣ ) sinh (α a
ﻭﺫﻟﻙ ﺒﺎﺴﺘﺨﺩﺍﻡ
*
)− E
4 E(Vo
V o2
− E )
4 E (Vo
=
1+
ﻗﻴﻡ α ﻭ . k
ﻭﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ ) (T ﻴﺄﺨﺫ −1
)(11
ﺍﻟﺸﻜل ﻜﺎﻟﺘﺎﻟﻲ:
2 ⎞ V o ⎛ ⎞ * ⎛G ⎞⎛G 2 = T a = = + 1 sinh ( α ) ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ *A A* ⎝ A ⎠ ⎝ A 4 E(Vo − E) ⎠ *
ﻭﻫﻲ:
GG
ﻭﻨﻘﻑ ﻫﻨﺎ ﻟﺴﺭﺩ ﺒﻌﺽ ﺍﻟﺘﻌﻠﻴﻘﺎﺕ - 1ﻤﻥ ﺍﻟﻌﻼﻗﺘﻴﻥ ) (10ﻭ ) (11ﻨﺠﺩ ﺃﻥ T + R = 1ﻭﻫﻭ ﻗﺎﻨﻭﻥ ﺤﻔﻅ ﺍﻟﺠﺴﻴﻤﺎﺕ. - 2ﻟﻠﻤﻘﺎﺭﻨ ﺔ ﺒﺎﻟﻨﻅﺭﻴﺔ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ﻨﻀﻊ = 0ﻟﻨﺠﺩ ﺃﻥ α ﻭ k ﺘﺅﻭﻻﻥ ﺇﻟﻰ ∞ .ﻭﻤﻨﻬﻤﺎ ﻴﺅﻭل R → 1ﻭ ،T → 0ﻜﻤﺎ ﻫﻭ ﻤﺘﻭﻗﻊ ﻤﻥ ﺘﺼﺭﻑ ﺍﻟﺠﺴﻴﻤﺎﺕ ﺍﻟﺘﻘﻠﻴﺩﻴﺔ. - 3ﻤﻥ ﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ ﻨﺠﺩ ﺃﻥ T = 0ﻋﻨﺩﻤﺎ . E = 0ﻭﻤﻊ ﺍﺯﺩﻴﺎﺩ ﻁﺎﻗﺔ ﺍﻟﺤﺭﻜﺔ ﺍﻻﺒﺘﺩﺍﺌﻴﺔ ﻟﻠﺠﺴﻴﻡ ﺒﺤﻴﺙ ﺇﻥ 0 < E
o
o
26
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﻟﻠﺠﺴﻴﻤﺎﺕ .ﻭﻫﺫﻩ ﺍﻟﻅﺎﻫﺭﺓ ﺘﺴﻤﻰ ﺒﺎ ﻟﺘﺄﺜﻴﺭ ﺍﻟﻨﻔﻘﻲ ) . ( Tunneling effectﻭﻴﻭﺠﺩ ﻅﻭﺍﻫﺭ ﻓﻴﺯﻴﺎﺌﻴﺔ ﻜﺜﻴﺭﺓ ﻟﻡ ﺘﻔﺴﺭ ﺇﻻ ﺒﺎﻟﺘﺄﺜﻴﺭﺍﻟﻨﻔﻘﻲ ﻤﻨﻬﺎ: ﺍﻨﺒﻌﺎﺙ ﺍﻹﻟﻜﺘﺭﻭﻨﺎﺕ ﻤﻥ ﺴﻁﺢ ﺍﻟﻤﻌﺎﺩﻥ ﺍﻟﺒﺎﺭﺩﺓ ﻨﺘﻴﺠﺔ ﺘﺄﺜﻴﺭ ﻤﺠﺎل ﺨﺎﺭﺠﻲ. ﺍﻟﺩﺍﻴﻭﺩ )ﺍﻟﺜﻨﺎﺌﻲ( ﺍﻟﻨﻔﻘﻲ ) ( Tunnel diodesﻓﻲ ﺍﻟﺩﻭﺍﺌﺭ ﺍﻟﻜﻬﺭﺒﺎﺌﻴﺔ. ﺍﻟﺘﻭﺼﻴل ﺍﻟﻜ ﻬﺭﺒﻲ ﻟﻠﻤﻭﺍﺩ ﺍﻟﻌﺎﺯﻟﺔ. ﺍﻟﺘﺄﺜﻴﺭﺍﻟﻨﻔﻘﻲ ﻓﻲ ﺍﻷﻨﻭﻴﺔ ﻭﻤﻨﻬﺎ ﺍﻨﺤﻼل ﺠﺴﻴﻤﺎﺕ α ﻤﻥ ﺍﻟﻤﻭﺍﺩ ﺍﻟﻤﺸﻌﺔ. ﺍﻟﺘﺄﺜﻴﺭﺍﻟﻨﻔﻘﻲ ﻓﻲ ﻓﻴﺯﻴﺎﺀ ﺍﻟﺠﻭﺍﻤﺩ. ﺍﻟﺘﺄﺜﻴﺭﺍﻟﻨﻔﻘﻲ ﻓﻲ ﺍﻟﻤﻭﺍﺩ ﻓﺎﺌﻘﺔ ﺍﻟﺘﻭﺼﻴل. - 4ﺇﺫﺍ ﺯﺍﺩﺕ ﺍﻟﻁﺎﻗﺔ ، E ﻨﺠﺩ ﺃﻥ ﺍﻟﻜﻤﻴﺔ ) (V − E ﺘﻘل ،ﻭ ﺘﻘل ﻤﻌﻬﺎ . α ﻭﻤﻊ ﺜﺒ ﻭﺕ ﻋﺭﺽ ﺍﻟﺠﻬﺩ a ﻨﺠﺩ ﺃﻥ ) sinh (α aﺘﻘل ﺃﺴﺭﻉ ﻤﻥ ) . (V − E ﻟﺫﻟﻙ ﻓﺈﻥ T ﺘﺯﺩﺍﺩ ﻤﻊ ﺯﻴﺎﺩﺓ . E 2 - 5ﺇﺫﺍ ﺯﺍﺩ ﻋﺭﺽ ﺍﻟﺠﻬﺩ aﻓﺈﻥ ) sinh (α aﺘﺯﺩﺍﺩ ﺒﺴﺭﻋﺔ ﻭ ﻟﺫﻟﻙ ﺘﻘل ﻤﻌﻬﺎ T ﺒﺴﺭﻋﺔ. o
2
o
ﻤﺜﺎل :ﺍﺴﺘﻨﺘﺞ ﻤﻌﺎﺩﻟﺔ ﺘﻘﺭﻴﺒﻴﺔ ﻟﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ ﻭﺫﻟﻙ ﻋﻨﺩ ﺘﺤﻘﻕ ﺍﻟﺸﺭﻁ ﺍﻟﺤل :ﻋﻨﺩ ﺘﺤﻘﻕ ﺍﻟﺸﺭﻁ α a >> 1ﻨﺠﺩ ﺃﻥ:
. α a >> 1
2
⎛ e aα − e −aα ⎞ 1 2aα 2 ⎜ = )→ 0, sinh (aα ⎟ ≈ 4e 2 ⎝ ⎠
ﺒﺎﻟﺘﺎﻟﻲ
−aα
e
ﻓﺈﻥ: −1
⎛ ⎞ V o2 T = ⎜1 + ⎟ ) sinh 2 (α a ⎝ 4 E (Vo − E ) ⎠ ) 4 E (Vo − E 16 E (Vo − E ) −2α a −2 a e ≈ ≈ sinh ( α ) 2 2 Vo
V o
)e −2aα
ﺍﻟﺤﺎﻟﺔﺍﻟﺜﺎﻨﻴﺔ
E V o
(1 −
16 Vo
=
E > V o
ﻤﻌﺎﺩﻻﺕ ﺸﺭﻭﺩﻨﺠﺭ ﻓﻲ ﺍﻟﻤﻨﻁﻘ ﺘﻴﻥ ) ( I ﻭ ) ( III ﻟﻥ ﺘﺘﻐﻴﺭ ،ﻭﺒﺎﻟﺘﺎﻟﻲ ﺴﻭﻑ ﻨﺴﺘﺨﺩﻡ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ﻭﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ ﺍﻟﻼﺯﻤﻨﻴﺔ ﻓﻲ ﻤﻨﻁﻘﺔ ﺤﺎﺠﺯ ﺍﻟﺠﻬﺩ ) ، 0 ≤ x ≤ a ، ( II ﺴﻭﻑ ﺘﻌﺭﻑ ﺒﺎﻟﺘﺎﻟﻲ:
27
) (2ﻭ).(6
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
−
2
2
d ψ II
2m dx 2
+V oψ II = E ψ II ⇒
2
d ψ II dx
2
:( ﻫﻭ12) ﻟﻠﻤﻌﺎ ﺩﻟﺔ ψ II ( x ) = C e
i β x
(12)
= − β ψ II 2
ﻭﺍﻟﺤل ﺍﻟﻌﺎﻡ . β
+D :ﻫﻲ
2
e
=
2m 2
( E
−V o )
−i β x
ﺤﻴﺙ (13)
ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭلﺍﻟﻤﻘﺒﻭﻟﺔ ﻟﺫﻟﻙ ﻨﺠﺩ ﺃﻥ
+ B e− ikx i β x + D e − i β x ψ II ( x ) = C e (14) ikx ψ III ( x ) = G e :ﻭﻤﺸﺘﻘﺎﺘﻬﺎ ﻭﻫﻲ ﻭﺒﺘﻁﺒﻴﻕ ﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻭﺩﻴﺔ ﻟﻠﺩﻭﺍل ψ I ( x) = A eikx
ψ ( x I= 0) = ψ ( x II= 0)
= 0) = ψ ' ( x II= 0) ψ ( x II= a) = ψ ( xIII= a) ( xIII= a) ψ' (x= II a) = ψ ' ψ ' ( x
I
(15)
:ﻨﺤﺼل ﻋﻠﻰ (11) ( ﻭ10)
ﺒﺎﻟﻤﻌﺎﺩﻟﺘﻴﻥ α = i β ﻭﻭﻀﻊ
-1
⎡ 4 E ( E −V o ) ⎤ R = ⎢1 + 2 ⎥ 2 ⎣ V o sin ( β a ) ⎦ −1 ⎛ ⎞ V o2 sin 2 ( β a ) ⎟ T = ⎜1 + ⎝ 4 E ( E −V o ) ⎠
(16)
(17)
:ﻭﻫﻲ
ﻭﻟﻨﺎ ﻫﻨﺎ ﺒﻌﺽ ﺍﻟﺘﻌﻠﻴﻘﺎﺕ .ﻗﺎﻨﻭﻥ ﺤﻔﻅ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻭﻫﻭT + R = 1 ( ﻨﺠﺩ ﺃﻥ17) ( ﻭ16) ﻤﻥ ﺍﻟﻌﻼﻗﺘﻴﻥ- 1 :ﻨﺤﺼل ﻋﻠﻰ ﻭﻤﻨﻬﺎ sin( β a ) ≈ β a ﻨﺠﺩ ﺃﻥ E → V ﻟﻠﻘﻴﻤﺔ - 2 o
−1
ﻴﺘﻀﺢ ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻜﻤﺎ،ﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ ﻴﺘﺫﺒﺫﺏ ﻨﺠﺩ ﺃﻥ
⎛ mV o a 2 ⎞ T = ⎜ 1 + ⎟ 2 2 ⎠ ⎝ E ( E > V o ) ﻁﺎﻗﺔ ﺍﻟﺠﺴﻴﻡ ﺍﺯﺩﻴﺎﺩ
ﻤﻊ
.(17)
28
-3
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
- 4ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (17ﻨﺠﺩ ﺃﻥ T = 1ﻋﻨﺩﻤﺎ ﻴﺘﺤﻘﻕ ، β a = nπ , n =1,2,3, ﻭﻤﻨﻪ: = nπ
a
ﺍﻟﺸﺭﻁ:
) 2m (E −V o 2
ﺃﻭ: λ 2
a=n
= nπ
⇒
a
2π λ
ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﻋﺭﺽ ﺍﻟ ﺤﺎﺌل ﻴﺘﺴﺎﻭﻯ ﻤﻊ ﺃﻋﺩﺍﺩ ﺼﺤﻴﺤﺔ ﻤﻥ ﻨﺼﻑ ﺍﻟﻁﻭل ﺍﻟﻤﻭﺠﻲ ﻟﺩﺒﺭﻭﻟﻴﻪ ﺩﺍﺨل ﺍﻟﺤﺎﺌل .ﻭﺒﺘﺭﺒﻴﻊ ﺍﻟﻤﻌﺎﺩﻻﺕﺍﻟﺴﺎﺒﻘﺔ ﻨﺤﺼل ﻋﻠﻰ: 2 2 ⎡ ⎤ 2 π E = Vo ⎢1 + n ⎥ ⎦ 8mV o a ⎣
- 5ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (17ﻨﺠﺩ ﺃﻴﻀﹰﺎ ﺃﻥ T ﺘﺼل ﺇﻟﻰ ﻨﻬﺎﻴﺔ ﺼﻐﺭﻯ ﻋﻨﺩﻤﺎ ﻴﺘﺤﻘﻕ ، β a = ( 2n + 1) π , n = 0,1, 2, 2 ﻭﻤﻨﻬﺎ ﻨﺤﺼل ﻋﻠﻰ:
ﺍﻟﺸﺭﻁ:
2 2 ⎡ ⎤ 2 π )E = Vo ⎢1 + (2 n+ 1 ⎥ 8 mV a ⎣ ⎦ o
ﻭﺃﻗل ﻗﻴﻤﺔ ﻟﻤﻌﺎﻤلﺍﻟﻨﻔﺎﺫﻴﺔ: −1
⎛ ⎞ V o2 T = T min = ⎜ 1 + ⎟ E E V − 4 ( ) ⎠ o ⎝
- 6ﺤﺘﻰ ﺒﻭﺠﻭﺩ ﺍﻟﺸﺭﻁ ﻤﻥ ﻨﺘﺎﺌﺞ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ. ﺍﻟﺸﻜل ﺍﻟﺘﺎﻟﻲ ﻴﻭﻀﺢ ﺘﻐﻴﺭ ﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ ﻤﻊ ) >V o
( E
ﻨﺠﺩ ﺃﻥ ﻫﻨﺎﻙﺍﺤﺘﻤﺎﻟﻴﺔ ﻻﻨﻌﻜﺎﺱ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻤﻥ ﺍﻟﺤﺎﺌل ،ﻭﻫﺫﻩ ﻨﺘﻴﺠﺔ E ) ( V o
ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ،ﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﻤﺘﺼل ،ﻭﻓﻲ ﺍﻟﻨﻅﺭﻴﺔ
ﺍﻟﺘﻘﻠﻴﺩﻴﺔ ،ﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﻤﺘﻘﻁﻊ .ﻨﻼﺤﻅ ﺃﻥ ﺴﻠﻭﻙ ﻤﻌﺎﻤل ﺍﻟﻨﻔﺎﺫﻴﺔ ﻓﻲ ﻤﻴﻜﺎﻨﻴﻜﺎ ﺍﻟﻜﻡ ﻴﻌﻁﻲ ﻗﻴﻤﺎﹰ ﻓﻲ ﺍﻟﻤﺩﻯ )< 1
E
(.
V o
29
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
- 5. Vﺘﻁﺒﻴﻕ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭﻓﻲ ﺜﻼﺜﺔ ﺃﺒﻌﺎﺩ ﻤﺜﺎل :ﺍﺩﺭﺱ ﺤﺭﻜﺔ ﺠﺴﻴﻡ ﺩﺍﺨل ﺼﻨﺩﻭﻕ ﻤﻘﻔل ﺫﻯ ﺜﻼﺜﺔ ﺃﺒﻌﺎﺩ ﻓﻲ ﻤﺠﺎل 0〈 x 〈 a, 0〈 y 〈 b, 0〈 z 〈 c
0,
otherwise
∞,
ﺠﻬﺩ ﻴﻭﺼﻑ ﺒﺎﻟﺘﺎﻟﻲ:
⎧ ⎪ ⎨ = ) V (x , y , z ⎪ ⎩ z
ﺃﺒﻌﺎﺩ(
ﺸﻜل 2.4ﺠﺴﻴﻡ ﺩﺍﺨل ﺼﻨﺩﻭﻕ )ﺫﻯ ﺜﻼﺜﺔ ﻁﺎﻗﺔ ﺍﻟﻭﻀﻊ ) (V ﻤﺘﻨﺎﻫﻴﺔ ﻓﻲ ﺍﻟﻜﺒﺭ ﺇﻻ ﺩﺍﺨل ﺍﻟﺼﻨﺩﻭﻕ ﻗﻴﻤﺘﻬﺎ ﺼﻔﺭ.
∞=V
x
∞=V
V=0
c b
a
ﺍﻟﺤل :ﻴﺘﻭﺍﺠﺩﺍﻟﺠﺴﻴﻡ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ ﻓﻲ ﺜﻼﺜﺔ ﺃﺒﻌﺎﺩ ﺘﺄﺨﺫ ﺍﻟﺼﻭﺭﺓ: )(1
) 〈 ( x , y , z ) 〈 (a, b , c
⎛ ∂2 ∂2 ⎞ ∂2 − + + ⎜ ) ⎟ψ ( x , y , z ) = Eψ ( x , y , z ⎠ 2m ⎝ ∂x 2 ∂y 2 ∂z 2 2
ﻭ ﺒﺫﻟﻙ ψ ﺴﻭﻑ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻹﺤﺩﺍﺜﻴﺎﺕ ﺍﻟﺜﻼﺙ ﺍﻟﻤﺘﻐﻴﺭﺍﺕ ،ﻭﻫﻲ ﻜﺎﻟﺘﺎﻟﻲ: -1ﻨﻔﺘﺭﺽ ﺃﻥ )(2
)(0,0, 0
ﻓﻘﻁ ،ﻭﻟﺫﻟﻙ ﻓﺈﻥ ﻤﻌﺎﺩﻟﺔ ﺸﺭﻭﺩﻨﺠﺭ
). ( x, y, z
) = X ( x )Y ( y ) Z (z
ﻟﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ) (1ﻨﺴﺘﺨﺩﻡ ﻁﺭﻴﻘﺔ ﻓﺼل
) ψ ( x , y , z
30
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
- 2ﺒﺎﻟﺘﻌﻭﻴﺽ )(3
ﻤﻥ ) (2ﻓﻲ ) (1ﻭﺍﺴﺘﺨﺩﺍﻡ + E y + E z for all y , and z
) ψ (0, y , z ) = ψ ( a, y , z
for all x , and z
) ψ ( x , 0, z ) = ψ ( x , b, z
for all x , and y
)ψ ( x , y , 0) = ψ ( x , y , c
ﺒﺎﻹﻤﻜﺎﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻰ
)(4
x
، E = Eﻭﺍﻟﺸﺭﻭﻁ ﺍﻟﺤﺩﻭﺩﻴﺔ
2mE x
,
2
,
2
.
ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ:
2mE y
2mE z 2
2
) d X (x
=+ k x2 X ( x ) = 0, k x2
2
=+ k y2 Y ( y ) = 0, k y2
dx 2
) d Y (y 2
dy 2
=+ k z2 Z ( z ) = 0, k z2
) d Z (z 2
dz
ﺒﺎﻟﺘﺎﻟﻲ ﻓﻨﺤﻥ ﻏﻴﺭﻨﺎ ﻤﻥ ﻤﻌﺎﺩﻟﺔ ﺘﻔﺎﻀﻠﻴﺔ ﻓﻲ ﺜﻼﺜﺔ ﻤﺘﻐﻴﺭﺍﺕ ﺇﻟﻰ ﺜﻼﺙ ﻤﻌﺎﺩﻻﺕ ﺘﻔﺎﻀﻠﻴﺔ ﻜل ﻭﺍﺤﺩﺓ ﺘﻌﺘﻤﺩ ﻋﻠﻰ ﻤﺘﻐﻴﺭ ﻭﺍﺤﺩ ﻓﻘﻁ .ﻜل ﻤﻌﺎﺩ ﻟﺔ ﻤﺸﺎﺒﻬﺔ ﻟﻤﻌﺎﺩﻟﺔ ﺠﺴﻴﻡ ﻓﻲ ﺼﻨﺩﻭﻕ ﺫﻯ ﺒﻌﺩ ﻭﺍﺤﺩ ) ،( 2. II .9ﻭﻟﺫﻟﻙ ﻨﺤﺼل ﻋﻠﻰ ﺍﻟﺤﻠﻭل ﺍﻟﺘﺎﻟﻴﺔ: = 1, 2,
x
n
,
y
n
,
= 1, 2,
)(5
= 1, 2,
z
n
n x π a n y π b
,
=
x
=y
n z π c
=z
k
sin k xx,
k
sin k yy ,
k
sin k zZ
2 a
2 b
2 c
= )X ( x = ) Y (y = )Z ( z
ﻭﺃﻴﻀﺎﹰ ⎞ ⎟ ⎠⎟ c 2
nz 2
)(6
+
n y 2 b2
+
2π 2 ⎛ n x 2
⎜⎜ 2 ⎝a
2m
=y + E z
E = E x+ E
ﻭﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻤﻴﺯﺓ ﺘﺼﺒﺢ: = 1,2, )n y = 1, 2, (7 n z = 1,2,
n x
⎞ z⎟ , ⎠
⎛ n π ⎞ ⎛ n y π ⎞ ⎛ n z π = )ψ ( x, y, z ⎜ y ⎟ sin ⎜ sin ⎜ x x ⎟ sin abc ⎝ a ⎠ ⎝ a ⎠ ⎝ a 8
ﺘﻌﻠﻴﻘ ﺎﺕ:
-1ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (6ﻨﺭﻯ ﺃﻥ ﻗﻴﻡ ﺍﻟﻁﺎﻗﺔ ﻏﻴﺭ ﻤﺘﺼﻠﺔ ) ﻤﻜﻤﺎﺓ(، -2ﺍﻷﻋﺩﺍﺩ ) (n , n , nﻫﻲ ﺃﻋﺩﺍﺩ ﺍﻟﻜﻡ ﻓﻲ ﺍﻻﺘﺠﺎﻫﺎﺕ ﺍﻟﺜﻼﺜﺔ .ﻭﻫﺫﻩ ﺍﻷﻋﺩﺍﺩ ﻤﺴﺘﻘﻠ ﺔ ،ﺃﻱ ﻻ ﻴﻌﺘﻤﺩ ﺒﻌﻀﻬﺎ ﺍﻟﺒﻌﺽ ﺍﻵﺨﺭ. - 3ﺸﺭﻁ ﺍﻟﻤﻌﺎﻴﺭﺓ ﻴﺘﻁﻠﺏ: z
y
x
31
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM c
)(8
b
a
∫ |ψ | dxdydz = ∫ | X (x ) | dx ∫ |Y ( y ) | dy ∫| Z ( z ) | dz = 1 2
2
2
0
- 4ﻓﻲ ﺤﺎﻟﺔ )(9
∞
2
0
∞-
0
ﺍﻟﻤﻜﻌﺏ ) ( a = b = c = Lﻓﺈﻥ: 2π 2
2mL2
E1 =
2π 2
( n + n + n ) = n E1 , 2mL2 2
2 z
2 x
2 y
=
E
n2
- 5ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ) (9ﺴ ﻨﻌﺭﻑ ﺩﺭﺠﺔ ﺍﻻﻨﺘﻤﺎﺀ )ﺃﻭ ﺍﻟﺘﺤﻠل( ﺒﺄﻨﻬﺎ ﻋﺩﺩ ﺍﻟﺩﻭﺍل ﺍﻟﺘﻲ ﻟﻬﺎ ) ﺘﻨﺘﻤﻲ ﺇﻟﻰ( ﻨﻔﺱ ﺍﻟﻁﺎﻗﺔ .ﻫﺫﺍ ﺍﻻﻨﺘﻤﺎﺀ ﻨﺎﺘﺞ ﻤﻥ ﺘﻤﺎﺜل ﺍﻟﻤﻜﻌﺏ ﻭﻴﺯﻭل ﺘﻤﺎﻤﺎﹰ ﺒﺘﻐﻴﺭ ﺃﻁﻭﺍل ﺍﻟﺼﻨﺩﻭﻕ .ﻭ ﺍﻟﺠﺩﻭل ﺍﻟﺘﺎﻟﻲ ﻴﺼﻑ ﺩﺭﺠﺎﺕ ﺍﻻﻨﺘﻤﺎﺀ ﺍﻟﺨﺎﺼﺔ ﺒﺎﻟﻤﻜﻌﺏ. ψ n x ,n y ,n z
n z
n y
n x
n2
ﺩﺭﺠﺔ ﺍﻻﻨﺘﻤﺎﺀ
ψ 1,1,1
1
1
1
3
1
ψ 1,1,2
2 1 1
1 2 1
1 1 2
6 6 6
3
ψ 1,2,2
2 2 1
2 1 2
1 2 2
9 9 9
ψ 1,1,3
3 1 1
1 3 1
1 1 3
11 11 11
ψ 2,2,2
2
2
2
12
ψ 1,2,1 ψ 2,1,1 ψ 2,1,2 ψ 2,2,1 ψ 1,3,1 ψ 3,1,1
32
3
3
1
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
ﻤﻨﻭﻋﺔ ﻤﺴﺎﺌل 7 :ﺍﻟﻜﻡ
ﻤﻴﻜﺎﻨﻴﻜﺎ ﺸﺭﻭﻁ ﺘﺤﻘﻕ ﺃﻨﻬﺎ ﻟﺘﺭﻯ ﺍﺭﺴﻡ ﺍﻟﺩﻭﺍلﺍﻟﻤﻭﺠﻴﺔ ﺍ ﻵﺘﻴﺔ،ﻗﻭﺴﻴﻥ ﻓﻲ ﺍﻟﻤﺩﻯ ﺍﻟﻤﺤﺩﺩ ﺒﻴﻥ-1 sin( x ) (a) (c) e− x (0, ∞) ِ (0, ∞) (b) a x (−5,5) x :ﻟﻤﺅﺜﺭﺍﺕ ﺍ ﻵﺘﻴ ﺔ ﻤﻌﺎﺩﻻﺕ ﺍ ﺘﺤﻘﻕ ﻤﻥ ﺼﺤﺔ- 2 3
2 3 ⎞ d 3 d 2 d ⎛ d 1⎞ ⎛d ⎞x ⎛ d 2 2 + ⎟ = ⎜ 2 + 2 x + x + 1⎟ b) ⎜ + ⎟ = 3 + (a) ⎜ 2 x dx dx dx ⎝ dx x ⎠ dx ⎝ ⎠ ⎝ dx ⎠
(c) (e)
⎛ xd ⎞ ⎜ dx ⎟ ⎝ ⎠ ⎛d ± ⎜ dx ⎝
2
d2
2
2 d d ⎛ ⎞ 2 d = x 2+ x x⎟ = x + 3 x +1 (d) ⎜ 2 dx dx dx dx ⎝ dx ⎠ ⎞⎛ d ± ⎞ = ⎛ d 2 − 2 ∓ 1⎞ x⎟⎜ x⎟ ⎜ 2 x ⎟ ⎠⎝ dx ⎠ ⎝ dx ⎠
2
d
ﻤﻌﺎﺩﻟﺔﺍﻟﻘﻴﻡ ﺍﻟﻤﻤﻴﺯﺓ ﺘﺤﻘﻕ ﺤﺩﺩ ﺃﻴﺎ ﻤﻥ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍ ﻵﺘﻴﺔ-3 (a)
⎛ π x ⎞ sin ⎜ ⎟ dx ⎝ 2 ⎠ d
:ﺍ ﻵﺘﻴ ﺔ
d n
⎛ π x ⎞ sin (b) ⎜ 2 ⎟ (c) 2 dx ⎝ ⎠ d2
dx
n
e α x (d) -i
ﺘﺤﻘﻕ ﻤﻥ ﺍﻟﻤﻌﺎﺩﻻﺕ ψ = 45 xy ﻭﺍﻟﺩﺍﻟﺔ 2
ˆ = 1 C
( )
ﻭ
⎛ 1 ⎞ ⎜ ⎟=− 1 2 2 ⎜ ( 45 xy 2 ) ⎟ 45x y ⎝ ⎠ ⎞ d ⎛ 1 1 ˆ ˆ ˆ ˆ − ⎜ ⎟ = (b) C(A(ψ ))=C( C( ) = −45 x 2 y 2 2 2 2 dx ⎜ ( 45 xy ) ⎟ 45x y ⎝ ⎠ ˆ ˆ ψ )= d (a) A(C dx
33
∂ ikx e ∂ x ˆ = d A dx
( ) ﺒﻤﻌﺭﻓﺔ ﺍﻟﻤﺅﺜﺭﻴﻥ-4
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM
:ﺒﺎﻟﺘﺎﻟﻰ ﻭﻴﻭﺼﻑ ﺒﺎﻟﺸﻜل ﻤﺘﻤﺎﺜل ﻜﻤﺎ ﺠﻬﺩ
V=∞
V=0
II
I
-a
ﻤﺠﺎل ﺠﺴﻴﻡ ﻓﻲﺔ ﺤﺭﻜ ﺍﺩﺭﺱ ⎧0,
V=∞
V ( x) = ⎨
-5
−a ≤ x ≤ a
⎩ ∞, otherwise
III
ﻭﺃﺜﺒﺕ ﺃﻥ x
a
0
⎧ A sin( nπ x ) n is even 2a ⎪⎪ ψ I ( x ) = ⎨ ⎪ n π ⎪⎩ Bcos( 2a x) nis odd E=
π 2 2 2ma 2
n2 ,
A= B =
1 a
ﻟﻠﺩﺍﻟﺔ -6 ⎧ A,
f ( x ) = ⎨
⎩0,
a≤ x
≤b
otherwise
:ﺍﻵﺘﻲ
= A
1 b
−a
,<
x> =
b
+a 2
,<
b
2
+ ab + a 2 3
,
−a
b
∆ x= < x2 > − < x> 2 =
x> =
2
ﺘﺤﻘﻕ ﻤﻥ .ﺜﺎﺒﺕ A ﺤﻴﺙ
12
:ﺍﻵﺘﻲ
− x Ne
ﺘﺤﻘﻕ ﻤﻥ ψ ( )x=
2
/ 2a
− ∞ ≤ x≤ ∞ ﺠﺎﻭﺱ ﻟﺩﺍﻟ ﺔ -7
,
1/ 4
2 a ⎛ 1 ⎞ , 2 2 N= ⎜ x 0, x , p 0, p , < > = < > = < >= < >= ⎟ π a 2 2 a ⎝ ⎠ a
∆x= < x >−< x> = 2
< E=
p
2
2m
2
>=
2
2
1
2m
2a
,
∆ p= < p > − < p > = 2
∞
: ﺍﻟﺘﻜﺎﻤلﺍﻟﻘﻴﺎﺴﻲ
∫ ﺍﺴﺘﺨﺩﻡe −∞
−ax 2
dx
=
2
2a
, and
∆ ∆x p=
,
2
2 ∞
π a
,
∫
2
x e
−∞
−ax 2
dx
=
1
π
2
a
3
,
ﻟﻠﺩﺍﻟﺔ -8 ⎧cx (a − x ), 0 ≤ x ≤ a otherwise ⎩0,
f (x ) = ⎨
:ﺍﻵﺘﻲ
34
ﺘﺤﻘﻕ ﻤﻥ
© Drs. I. Nasser and Afaf Abdel-Hady
Chapter 2
Schrödinger equation and its applications
10/21/2008 8:09 AM c
=
30 a5
a
2
2
2
6
a2
< x >= , < x 2 >= a2 , < p >= 0, < p 2 >= 10
,
7
∆ x = < x 2 > − < x >2 =
14
∆ p = < p2 > − < p >2 =
a,
, 10 a
, and
2 ∂2 < E = − >= 5 2 2m ∂x 2 ma
2
∆ x ∆p = 0.6 >
2
ﺘﺤﻘﻕ ﻤﻥ 0 ≤ x ≤ L ﻭﻓﻲ ﺍﻟﻤﺩﻯψ 1 ( x) =
:ﺍﻵﺘﻲ x=
L ⎛ 2π
2
L ;
p= 0;
x
=
2
n− 3 ⎞
2
2
= ⎜ π 2 ⎟ ; ⎝ ⎠ 2 2 2 2 L ⎛ 2 nπ − 3 ⎞ L 2 2 ∆ x= < x > − < x> = ⎜ ⎟− 6n 2 ⎝ π 2 ⎠ 4 2
2
6n 2
2
∆ p= < p > − < p> = 2
∆ x ∆p =
2
π 2 n 2 − 2
2
3
2
2
n π L
= .57 〉
2
2
35
ˆ 2p
2
2
n π L2
2 L
⎛ π x ⎞ ⎟ ﻟﻠﺩﺍﻟﺔ -9 ⎝L ⎠
sin ⎜
