Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
3rd Edition
ISBN: 9781107189638
Author: Griffiths, David J., Schroeter, Darrell F.
Publisher: Cambridge University Press
Question
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Chapter 9, Problem 9.13P
To determine

Obtain the allowed energies of general power -law potential using the WKB approximation.

Expert Solution & Answer
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Answer to Problem 9.13P

Allowed energies are En=(n12)ω(n=1,2,3......).

Explanation of Solution

Write the quantization condition for quantized energies for a potential well with two sides.

  x1x2p(x)dx=(n12)π20x22m(Eαxv)dx=(n12)π22mE0x2(1αExv)dx=(n12)π        (I)

Consider the integral 0x2(1αExv)dx.

  Let αExv=zx=(zEα)1vdx=(Eα)1v1vz1v1dz

Rewrite the left-hand side of equation (1) by substituting the above value of dx.

    22mE20x2(1αExv)dx=22mE(Eα)1v1v0x2z1v11zdz=22mE(Eα)1v1vΓ(1/v)Γ(3/2)Γ(1v+32)=22mE(Eα)1v1vΓ(1/v)Γ(3/2)Γ(1v+32)=22mE(Eα)1v1vΓ(1v+1)12πΓ(1v+32)=2πmE(Eα)1vΓ(1v+1)Γ(1v+32)

Equate with the right-hand side of equation (I).

    2πmE(Eα)1vΓ(1v+1)Γ(1v+32)=(n12)πE1v+12=(n12)π2πmα1vΓ(1v+32)Γ(1v+1)En=(n12)π2mα[Γ(1v+32)Γ(1v+1)](2vv+2)α

Substitute 2 for v in the above equation.

    En=(n12)π2mα[Γ(2)Γ(32)]α=(n12)2αm

Conclusion:

Substitute 12mω2 for α in the above equation.

    En=(n12)π2mα[Γ(2)Γ(32)]α=(n12)ω(n=1,2,3......)

Thus, the allowed energies are En=(n12)ω(n=1,2,3......).

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