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![Solving the Schrödinger equation
𝜓 𝑥 = 𝐴(𝑥)𝑒 𝑖𝜙(𝑥)
Using prime to denote the derivative with respect to
𝑥, we find:
𝑑𝜓
𝑑𝑥
= (𝐴′+𝑖𝐴𝜙′)𝑒 𝑖𝜙
and
𝑑2 𝜓
𝑑𝑥2
= [𝐴′′
+2𝑖𝐴′
𝜙′
+ 𝑖𝐴𝜙′′
− 𝐴 𝜙′ 2
]𝑒 𝑖𝜙
Putting all these into
𝑑2 𝜓
𝑑𝑥2 = −
𝑝2
ℏ2 𝜓 (From Schrödinger
equation ) , we get](https://image.slidesharecdn.com/thewkbapproximation-151222120014/75/The-wkb-approximation-11-2048.jpg)
![𝐴′′
+ 2𝑖𝐴′
𝜙′
+ 𝑖𝐴𝜙′′
− 𝐴 𝜙′ 2
= −
𝑝2
ℏ2
𝐴
Solving for real and imaginary parts we get,
𝐴′′ − 𝐴 𝜙′ 2 = −
𝑝2
ℏ2
𝐴
𝐴′′ = 𝐴 (∅′)2−
𝑝2
ℏ2
𝐴′′
𝐴
= (∅′)2−
𝑝2
ℏ2
The above equation cannot be solved in general, so we use WKB
approximation: we assume amplitude A varies slowly, so that the
A’’ term is negligible. We assume that 𝐴′′
𝐴 << [(∅′)2−
𝑝2
ℏ2].
Therefore, we drop that part and we get
𝜙 𝑥 = ±
1
ℏ
𝑝 𝑥 𝑑𝑥](https://image.slidesharecdn.com/thewkbapproximation-151222120014/75/The-wkb-approximation-12-2048.jpg)
![∴ 1 ⟹
𝑑𝑆0
𝑑𝑥
2
− 𝑘2
ℏ 2
+ 2
𝑑𝑆0
𝑑𝑥
𝑑𝑆1
𝑑𝑥
− 𝑖
𝑑2 𝑆0
𝑑𝑥2 ℏ = 0 ;
(neglecting higher powers of ℏ).
where 𝑘2
ℏ 2
= 2𝑚[𝐸 − 𝑉(𝑥)]
For the above equation to be valid, the coefficient
of each power of ℏ must vanish separately,
𝑑𝑆0
𝑑𝑥
2
− 𝑘2
ℏ 2
= 0 𝑜𝑟
𝑑𝑆0
𝑑𝑥
= ±𝑘ℏ
⟹ 𝑆0 x = ± 𝑘ℏ𝑑𝑥](https://image.slidesharecdn.com/thewkbapproximation-151222120014/75/The-wkb-approximation-16-2048.jpg)
![For 𝐸 < 𝑉 𝑥 : the Schrodinger equation is ,
𝑑2 𝜓
𝑑𝑥2 − 𝜅2
𝜓 = 0 ; 𝜅2
=
2𝑚[𝑉 𝑥 −𝐸]
ℏ2
Proceeding in a similar fashion , the solution can be
obtained as
𝜓 =
𝐵
𝜅
exp(± 𝜅𝑑𝑥) ; where B is a
normalization constant.](https://image.slidesharecdn.com/thewkbapproximation-151222120014/75/The-wkb-approximation-18-2048.jpg)
![Barrier to the right of turning point
1
𝑘
exp[−
𝑥
𝑎
𝑘𝑑𝑥] →
2
𝑘
sin[
𝑎
𝑥
𝑘𝑑𝑥 +
𝜋
4
]
And,
1
𝑘
exp[+ 𝑥
𝑎
𝑘𝑑𝑥] →
1
𝑘
cos[ 𝑎
𝑥
𝑘𝑑𝑥 +
𝜋
4
]](https://image.slidesharecdn.com/thewkbapproximation-151222120014/75/The-wkb-approximation-25-2048.jpg)
More Related Content
The wkb approximation..
- 1. Introduction: • Generally, theWKB approximation or WKB method is a method for finding approximate solutions to linear differential equations with spatially varying coefficients. • In Quantum Mechanics it is used to obtain approximate solutions to the time- independent Schrodinger equation in one dimension.
- 2. Applications in QuantumMechanics In quantum mechanics it is useful in 1. Calculating bound state energies (Whenever the particle cannot move to infinity) 2. Transmission probability through potential barriers. These are given in next slides.
- 3. Main idea: • Ifpotential 𝑽(𝒙) is constant and energy 𝐸 of the particle is 𝑬 > 𝑽, then the particle wave function has the form 𝜓 𝑥 = 𝐴𝑒±𝑖𝑘𝑥 where 𝑘 = √(2𝑚(𝐸−𝑉)) ℏ (+) sign indicates : particle travelling to right (-) sign indicates : particle travelling to left • General solution : Linear superposition of the two. • The wave function is oscillatory, with fixed wavelength, 𝜆 = 2𝜋 𝑘 . • The amplitude (A) is fixed.
- 4. • If V(x) is not constant, but varies slow in comparison with the wavelength λ in a way that it is essentially constant over many λ, then the wave function is practically sinusoidal, but wavelength and amplitude slowly change with x. This is the inspiration behind WKB approximation. In effect, it identifies two different levels of x-dependence :- rapid oscillations, modulated by gradual variation in amplitude and wavelength.
- 5. • If E<Vand V is constant, then wave function is 𝜓 𝑥 = 𝐴𝑒±𝜅𝑥 where κ = √(2𝑚(𝑉−𝐸)) ℏ If 𝑽(𝒙) is not constant, but varies slowly in comparison with 1 𝜅 , the solution remain practically exponential, except that 𝐴 and 𝜅 are now slowly-varying function of 𝑥.
- 6. Failure of thisidea There is one place where this whole program is bound to fail, and that is in the immediate vicinity of a classical turning point, where 𝐸 ≈ 𝑉. For here 𝜆(𝑜𝑟 1 𝜅)) goes to infinity and 𝑉(𝑥) can hardly be said to vary “slowly” in comparison. A proper handling of the turning points is the most difficult aspect of the WKB approximation, though the final results are simple to state and easy to implement. The diagram showing turning points is given in next slide.
- 7.
- 8. The Classical Region •Let's now solve the Schrödinger equation using WKB approximation −ℏ2 2𝑚 𝑑2 𝜓 𝑑𝑥2 + 𝑉 𝑥 𝜓 = 𝐸𝜓 can be rewritten in the following way: 𝑑2 𝜓 𝑑𝑥2 = − 𝑝2 ℏ2 𝜓 ; 𝑝 𝑥 = 2𝑚(𝐸 − 𝑉(𝑥)) is the classical formula for the momentum of a particle with total energy 𝐸 and potential energy 𝑉(𝑥). Let’s assume 𝐸 > 𝑉(𝑥), so that 𝑝(𝑥) is real. This is the classical region , as classically the particle is confined to this range of 𝑥. The classical and non-classical region is shown in the diagram on the next slide.
- 9. The WKB approximation V(x) E Classicalregion (E>V) Non-classical region (E<V) Fig: Classically, the particle is confined to the region where 𝑬 ≥ 𝑽(𝒙)
- 10. The function 𝜓 Ingeneral, 𝜓 is some complex function; we can express it in terms of its amplitude 𝐴(𝑥), and its phase, 𝜙(𝑥) – both of which are real : 𝜓 𝑥 = 𝐴(𝑥)𝑒 𝑖𝜙(𝑥)
- 11. Solving the Schrödingerequation 𝜓 𝑥 = 𝐴(𝑥)𝑒 𝑖𝜙(𝑥) Using prime to denote the derivative with respect to 𝑥, we find: 𝑑𝜓 𝑑𝑥 = (𝐴′+𝑖𝐴𝜙′)𝑒 𝑖𝜙 and 𝑑2 𝜓 𝑑𝑥2 = [𝐴′′ +2𝑖𝐴′ 𝜙′ + 𝑖𝐴𝜙′′ − 𝐴 𝜙′ 2 ]𝑒 𝑖𝜙 Putting all these into 𝑑2 𝜓 𝑑𝑥2 = − 𝑝2 ℏ2 𝜓 (From Schrödinger equation ) , we get
- 12. 𝐴′′ + 2𝑖𝐴′ 𝜙′ + 𝑖𝐴𝜙′′ −𝐴 𝜙′ 2 = − 𝑝2 ℏ2 𝐴 Solving for real and imaginary parts we get, 𝐴′′ − 𝐴 𝜙′ 2 = − 𝑝2 ℏ2 𝐴 𝐴′′ = 𝐴 (∅′)2− 𝑝2 ℏ2 𝐴′′ 𝐴 = (∅′)2− 𝑝2 ℏ2 The above equation cannot be solved in general, so we use WKB approximation: we assume amplitude A varies slowly, so that the A’’ term is negligible. We assume that 𝐴′′ 𝐴 << [(∅′)2− 𝑝2 ℏ2]. Therefore, we drop that part and we get 𝜙 𝑥 = ± 1 ℏ 𝑝 𝑥 𝑑𝑥
- 13. And from secondequation, we get 2𝐴′ 𝜙′ + 𝐴𝜙′′ = 0 𝐴2 𝜙′ ′ = 0 𝐴 = 𝐶 𝜙′ Where C is real constant.
- 14. Thus from theprevious slides from the equations and making ‘C’ a complex constant, we get 𝜓(𝑥) ≅ 𝐶 𝑝(𝑥) 𝑒± 𝑖 ℏ 𝑝 𝑥 𝑑𝑥 And the general solution can be written as 𝜓(𝑥) ≅ 1 𝑝(𝑥) 𝐶1 𝑒 𝑖 ℏ 𝑝 𝑥 𝑑𝑥 + 𝐶2 𝑒 −𝑖 ℏ 𝑝 𝑥 𝑑𝑥 where 𝐶1 and 𝐶2 are constants.
- 15. Alternate approach In thisapproach, the wave function is expanded in powers of ℏ . Let, the wave function be: 𝜓 = 𝐴𝑒 𝑖 ℏ 𝑆(𝑥) Using this in: 𝑑2 𝜓 𝑑𝑥2 + 𝑘2 𝜓 = 0, we get 𝑑𝑆 𝑑𝑥 2 − 𝑖ℏ 𝑑2 𝑆 𝑑𝑥2 − 𝑘2 ℏ2 = 0 --------(1) Expanding S(x) in powers of ℏ2 : 𝑆 𝑥 = 𝑆0 𝑥 + 𝑆1 𝑥 ℏ + 𝑆2 𝑥 ℏ2 2 + ⋯
- 16. ∴ 1 ⟹ 𝑑𝑆0 𝑑𝑥 2 −𝑘2 ℏ 2 + 2 𝑑𝑆0 𝑑𝑥 𝑑𝑆1 𝑑𝑥 − 𝑖 𝑑2 𝑆0 𝑑𝑥2 ℏ = 0 ; (neglecting higher powers of ℏ). where 𝑘2 ℏ 2 = 2𝑚[𝐸 − 𝑉(𝑥)] For the above equation to be valid, the coefficient of each power of ℏ must vanish separately, 𝑑𝑆0 𝑑𝑥 2 − 𝑘2 ℏ 2 = 0 𝑜𝑟 𝑑𝑆0 𝑑𝑥 = ±𝑘ℏ ⟹ 𝑆0 x = ± 𝑘ℏ𝑑𝑥
- 17. and, 2 𝑑𝑆0 𝑑𝑥 𝑑𝑆1 𝑑𝑥 − 𝑖 𝑑2𝑆0 𝑑𝑥2 = 0 𝑜𝑟, 𝑑𝑆1 𝑑𝑥 = 𝑖 2𝑘 𝑑𝑘 𝑑𝑥 ; using the value of 𝑆0(x) 𝑜𝑟, 𝑆1 = 𝑖 2 𝑙𝑛𝑘 or, 𝑒 𝑖𝑆1 = 𝑘−1 2 ∴ 𝜓 = 𝐴𝑒 𝑖 ℏ 𝑆(𝑥) = 𝐴𝑒 𝑖 ℏ 𝑆0(𝑥)+𝑆1(𝑥)ℏ = 𝐴𝑒 𝑖 ℏ 𝑆0 𝑒 𝑖𝑆1 = 𝐴 1 𝑘 𝑒±𝑖 𝑘𝑑𝑥 𝑜𝑟 𝜓 = 𝐴 𝑘 𝑒±𝑖 𝑘𝑑𝑥 where A is a normalization constant.
- 18. For 𝐸 <𝑉 𝑥 : the Schrodinger equation is , 𝑑2 𝜓 𝑑𝑥2 − 𝜅2 𝜓 = 0 ; 𝜅2 = 2𝑚[𝑉 𝑥 −𝐸] ℏ2 Proceeding in a similar fashion , the solution can be obtained as 𝜓 = 𝐵 𝜅 exp(± 𝜅𝑑𝑥) ; where B is a normalization constant.
- 19. Validity of WKBsolution The zeroeth order WKB solution is: 𝜓0 𝑥 = 𝐴𝑒±𝑖 𝑥 𝑘 𝑥 𝑑𝑥 𝑜𝑟, 𝜓0 𝑥 = 𝐴𝑒 𝑖 𝑥 𝑘 𝑥 𝑑𝑥 ; considering positive part only. 𝑑𝜓0(𝑥) 𝑑𝑥 = 𝑖𝑘(𝑥)𝜓0(𝑥) 𝑜𝑟, 𝑑2 𝜓0 𝑑𝑥2 = 𝑖 𝑑𝑘 𝑑𝑥 𝜓0 + (𝑖𝑘)2 𝜓0 𝑜𝑟, 𝑑2 𝜓0 𝑑𝑥2 + 𝑘2 + 𝑖 𝑑𝑘 𝑑𝑥 𝜓0 = 0
- 20. But we areinterested in solving the following eqn. 𝑑2 𝜓0 𝑑𝑥2 + 𝑘2 𝜓0 = 0 Hence, 𝑑𝑘 𝑑𝑥 << 𝑘2 ⟹ 1 𝑘 𝑑𝑘 𝑑𝑥 ≪ 𝑘(𝑥) i.e. k(x) should not vary so rapidly This is the Validity condition for WKB approximation.
- 21. The WKB approximation V(x) E Classicalregion (E>V) Non-classical region (E<V) Non-classical region (E<V) Turning points
- 22. The WKB approximation E V x ( ) ( ) i p x dx WKB C x e p x h E V x Excluding the turning points: 1 ( ) ( ) p x dx WKB C x e p x h
- 23. Patching region The WKBapproximation V(x) E Classical region (E>V) Non-classical region (E<V) ( ) ( ) i p x dxC x e p x h 1 ( ) ( ) p x dxD x e p x h
- 24. Connection Formulae 𝑘2 (𝑥) 𝑥 WKB soln notvalid a Trigonometric WKB soln Exponential WKB soln Turning point Barrier to the right of turning point
- 25. Barrier to theright of turning point 1 𝑘 exp[− 𝑥 𝑎 𝑘𝑑𝑥] → 2 𝑘 sin[ 𝑎 𝑥 𝑘𝑑𝑥 + 𝜋 4 ] And, 1 𝑘 exp[+ 𝑥 𝑎 𝑘𝑑𝑥] → 1 𝑘 cos[ 𝑎 𝑥 𝑘𝑑𝑥 + 𝜋 4 ]
- 26. Barrier to theleft of turning point WKB soln not valid 𝑘2 (𝑥) 𝑥 Trigonometric WKB soln Exponential WKB soln b Turning point
- 27. Barrier to theleft of turning point 2 𝑘 sin 𝑥 𝑏 𝑘𝑑𝑥 + 𝜋 4 → 1 𝑘 𝑒𝑥𝑝 − 𝑏 𝑥 𝑘𝑑𝑥 And, 1 𝑘 𝑐𝑜𝑠 𝑥 𝑏 𝑘𝑑𝑥 + 𝜋 4 → 1 𝑘 𝑒𝑥𝑝 + 𝑏 𝑥 𝑘𝑑𝑥
- 28.
- 29. Example 1 Potential Squarewell with a Bumpy Surface
- 30. Potential Square wellwith a Bumpy Surface Suppose we have an infinite square well with a bumpy bottom as shown in figure: and 𝑉 𝑥 = 𝑠𝑜𝑚𝑒 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛, 𝑖𝑓 0 < 𝑥 < 𝑎 ∞, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝑽 𝒙 𝒙𝟎
- 31. Inside the well,(𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝐸 > 𝑉 𝑥 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡), we have or, 𝜓 𝑥 ≅ 1 𝑝(𝑥) 𝐶1 𝑠𝑖𝑛 1 ℏ 0 𝑥 𝑝 𝑥 𝑑𝑥 + 𝐶2 𝑐𝑜𝑠 1 ℏ 0 𝑥 𝑝 𝑥 𝑑𝑥 𝜓(𝑥) must go to zero at 𝑥 = 0 and 𝑥 = 𝑎. So, putting the values we get respectively, 𝐶2 = 0 and 0 𝑎 𝑝 𝑥 𝑑𝑥 = 𝑛𝜋 ℏ ,this quantization condition determines the allowed energies. 𝜓(𝑥) ≅ 1 𝑝(𝑥) 𝐶1 𝑒 𝑖 ℏ 0 𝑥 𝑝 𝑥 𝑑𝑥 + 𝐶2 𝑒 −𝑖 ℏ 0 𝑥 𝑝 𝑥 𝑑𝑥
- 32. Special Case: If thewell has a flat bottom i.e. (𝑉 𝑥 = 0), then 𝑝 𝑥 = 2𝑚𝐸 and from quantization equation, we get 𝑝𝑎 = 𝑛𝜋ℏ. Solving these, we get value of : 𝐸 𝑛 = 𝑛2 𝜋2ℏ2 2𝑚𝑎2 which is the formula for the discrete energy levels of the infinite square well.
- 33.
- 34. In the non-classicalregion (𝐸 < 𝑉), 𝜓(𝑥) ≅ 𝐶 𝑝(𝑥) 𝑒± 1 ℏ 0 𝑥 𝑝(𝑥) 𝑑𝑥 ; 𝑝(𝑥) is complex Let us consider the following example : problem of scattering from a rectangular barrier also called tunneling.
- 35. To the leftof the barrier (𝑥 < 0), 𝜓 𝑥 = 𝐴𝑒 𝑖𝑘𝑥 + 𝐵𝑒−𝑖𝑘𝑥 where 𝐴 is the incident amplitude and 𝐵 is the reflected amplitude, and 𝑘 = 2𝑚𝐸 ℏ . To the right of the barrier (𝑥 > 𝑎), 𝜓 𝑥 = 𝐹𝑒 𝑖𝑘𝑥 where 𝐹 is the transmitted amplitude. In the tunneling region (0 ≤ 𝑥 ≤ 𝑎) WKB approximation gives, 𝜓 𝑥 ≅ 𝐶 𝑝 𝑥 𝑒 1 ℏ 0 𝑥 𝑝 𝑥 𝑑𝑥 + 𝐷 𝑝(𝑥) 𝑒 −1 ℏ 0 𝑥 𝑝(𝑥) 𝑑𝑥
- 36. The transmission probabilityis 𝑇 = 𝐹 2 𝐴 2 ∵ 𝐹 𝐴 ~ 𝑒− 1 ℎ 0 𝑎 𝑝 𝑥 𝑑𝑥 ∴ 𝑇 ≅ 𝑒− 2 ℎ 0 𝑎 𝑝 𝑥 𝑑𝑥 where T is transmission probability.
- 37. Example 3 Eigen valueequation for Bound State
- 38. Eigen value equationfor Bound State Here, a and b are the classical turning points. 𝜓𝐼 = 𝐴 𝜅 𝑒− 𝑥 𝑏 𝜅 𝑥 𝑑𝑥 ; 𝜅2 = −𝑘2 = 2𝑚 ℏ2 𝑉 𝑥 − 𝐸 Using connection formula at 𝑥 = 𝑎 𝜓𝐼𝐼 = 2𝐴 𝑘 𝑠𝑖𝑛 𝑎 𝑥 𝑘(𝑥)𝑑𝑥 + 𝜋 4
- 39. For 𝑥 =𝑏: 𝑎 𝑥 𝑘(𝑥)𝑑𝑥 + 𝜋 4 = 𝑎 𝑏 𝑘𝑑𝑥 − 𝑥 𝑏 𝑘𝑑𝑥 + 𝜋 4 + 𝜋 2 = 𝜋 2 + 𝜃 − 𝑥 𝑏 𝑘𝑑𝑥 + 𝜋 4 ∴ 𝑠𝑖𝑛 𝑎 𝑥 𝑘(𝑥)𝑑𝑥 + 𝜋 4 = 𝑐𝑜𝑠 𝜃 − 𝑥 𝑏 𝑘𝑑𝑥 + 𝜋 4 ∴ 𝜓𝐼𝐼 = 2𝐴 𝑘 𝑐𝑜𝑠 𝜃 − 𝑥 𝑏 𝑘𝑑𝑥 + 𝜋 4
- 40.
- 41. 𝜃 = 𝑛+ 1 2 𝜋 ; 𝑛 = 0,1,2,3 … 𝑎 𝑏 𝑘𝑑𝑥 = 𝑛 + 1 2 𝜋 This is the Eigen value equation for bound state using WKB approximation. Now, using this equation the energy Eigen values for Linear Harmonic Oscillator (LHO) can be calculated as shown in next slide :
- 42. LHO energy Eigenvalues For LHO potential is 𝑉 𝑥 = 1 2 𝑚𝜔2 𝑥2 ∴ 𝑘2 𝑥 = 2𝑚 ℏ2 𝐸 − 1 2 𝑚𝜔2 𝑥2 = 2𝑚 ℏ2 1 2 𝑚𝜔2 𝑥0 2 − 𝑥2 where, 𝑥0 2 = 2𝐸 𝑚𝜔2 is the classical amplitude or turning points.
- 43. 𝑜𝑟, 𝑘2 𝑥 = 𝑚2 𝜔2 ℏ2 𝑥0 2 −𝑥2 Using this value of 𝑘(𝑥) in Eigen value equation for bound state, −𝑥0 +𝑥0 𝑚𝜔 ℏ 𝑥0 2 − 𝑥2 𝑑𝑥 = (𝑛 + 1 2 )𝜋 On solving the above equation, 𝐸 ℏ𝜔 = (𝑛 + 1 2 ) or, 𝐸 𝑛 = (𝑛 + 1 2 )ℏ𝜔 ; 𝑛 = 0,1,2, … - which are the energy Eigen values for a Linear Harmonic Oscillator.