# hw205a - Physics 512 Homework Set#5 Solutions Winter 2003 1...

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Physics 512 Winter 2003 Homework Set #5 – Solutions 1. This problem is essentially Merzbacher, Chapter 18, Problem 4 [or Sakurai, Chapter 5, Problem 1]. Consider a one-dimensional harmonic oscillator perturbed by a constant force H = p 2 2 m + 1 2 2 x 2 Fx a ) Show that the Frst order perturbation in the energy levels vanishes Since we will need to fnd the second order perturbation in part b ) , let us consider the general matrix element V mn = F h m (0) | x | n (0) i The operator x is given in terms oF creation and annihilation operators by x = p ¯ h/ (2 )( a + a ) . Thus V mn = F r ¯ h 2 h m (0) | ( a + a ) | n (0) i = F r ¯ h 2 [ n +1 δ m,n +1 + j,n 1 ] This indicates that the x operator changes the oscillator number by ± 1 . In par- ticular, V nn =0 , which demonstrates that the frst order energy perturbation vanishes. b ) Now calculate the eigenenergies E n up to second order in the perturbation. Using the matrix element calculated above, we fnd E (2) n = X k 6 = n | V kn | 2 E (0) n E (0) k = | V n +1 ,n | 2 E (0) n E (0) n +1 + | V n 1 ,n | 2 E (0) n E (0) n 1 = F 2 ¯ h 2 ± n ¯ + n ¯ ² = F 2 2 2 Combined with the zeroth order energy, this gives E n =( n + 1 2 F 2 2 2 c ) Show that the second-order perturbation result gives the exact eigenenergies (which may be obtained by completing the square in H ). Explain why this happens. 1

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By completing the square, we fnd H = p 2 2 m + 1 2 2 x F 2 2 F 2 2 2 = p 2 2 m + 1 2 2 x 2 F 2 2 2 where x = x F/ ( 2 ) is a shiFted coordinate. In terms oF this new coordinate, we have an ordinary harmonic oscillator, however with a constant o±set in the potential. We easily read o± the eigenenergies as E n =( n + 1 2 F 2 2 2 (1) which is identical to the perturbation result oF b ) . OF course, in general we would apply perturbation theory only when we do not know the exact answer. How- ever, in this toy example, aFter completing the square, we see that F enters the Hamiltonian only quadratically as an overall o±set. In particular, we know that the correct energies, (1) , only has F 2 , and no other powers oF F in it. Now re- call that each order in the perturbation theory corresponds to precisely that order in a series expansion oF the small perturbation parameter (which would be F in this case); frst order perturbation gives O ( F ) , second order gives O ( F 2 ) , and so on. As a result, only the second order perturbation to the energy can contribute towards the correct answer, and this is in Fact what happens. 2. Sakurai, Chapter 5, Problem 2. In nondegenerate time-independent perturbation theory, what is the probability of Fnding in a perturbed energy eigenstate | ψ n i the corresponding unperturbed eigenstate | ψ (0) n i ? Solve this up to terms of order g 2 .
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hw205a - Physics 512 Homework Set#5 Solutions Winter 2003 1...

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