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conantee
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« on: January 08, 2010, 02:11:46 AM »
96. A particle of mass M is an infinitely deep square well potential V where
V = 0 for -a \leqslant x \leqslant a, and
V = \infty for x<-a, a<x

A very small perturbing potential V^\prime is superimposed on V such that
V^\prime = \epsilon (\frac{a}{2} - |x|) for  \frac{-a}{2} \leqslant x \leqslant \frac{a}{2}, and
V^\prime = 0 forx<\frac{-a}{2}, \frac{a}{2}<x

If \psi_0, \psi_1, \psi_2, \psi_3, ... are the energy eigenfunctions for a particle in the infinitely deep square well potential, with \psi_0 being the ground state, which of the following statements is correct about the eigenfunction \psi_0^\prime of a particle in the perturbed potential V +V^\prime?
(A) \psi_0^\prime = a_{00} \psi_0, a_{00} \neq 0
(B) \psi_0^\prime =\displaystyle \sum^{\infty }_{n=0} a_{0n} \psi_n with a_{0n} = 0 for all odd values of n
(C) \psi_0^\prime =\displaystyle \sum^{\infty }_{n=0} a_{0n} \psi_n with a_{0n} = 0 for all even values of n
(D) \psi_0^\prime =\displaystyle \sum^{\infty }_{n=0} a_{0n} \psi_n with a_{0n} \neq 0 for all values of n
(E) None of the above
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conantee
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« Reply #1 on: October 16, 2010, 11:49:14 AM »
This question is difficult for who do not know perturbation theory. However, you can eliminate (D) and (E) since they are not famous for correct answers (no thinking, no relation with symmetry of perturbed potential). (A) is also wrong since scaling old ground state wave function yields same ground state energy - not the new perturbed ground state energy. Now, you can guess between (B) and (C).

Actually, the coefficient a_{0n} is equal to \frac{<\psi_n | V^\prime |\psi_0>}{E_0 - E_n}. Since V^\prime is even function and \psi_0, the ground state of a particle in a infinite square well, is even, \psi_n needs to be odd so that <\psi_n | V^\prime |\psi_0> becomes zero. As a result, we needs n to be odd to make a_{0n} = 0.

The correct answer is therefore (B). 23 of 100 people answered this question correctly.
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