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Quantum Mechanics}Perturbation Theory

The energy for first-order perturbation theory of H=H_0+\Delta H (H_0 is the known Hamiltonian and \Delta H is the perturbed Hamiltonian) is given by E_1=\langle \psi_0 | \Delta H \psi \rangle_0, where the wave-functions are the unperturbed ones.

Thus, the problem amounts to calculating E_1=\langle n | V(a+a^\dag)^2 | n \rangle. This is just raising and lowering operator mechanics.

(a+a^\dag)^2=a^2+a^{\dag^2}+aa^\dag + a^\dag a. But, after bra-ketting, one finds that the expectation value of a^2 and a^{\dag^2} are 0, since n and n+2, n-2 are orthogonal. Thus, the problem becomes,

E_1=\langle n | (a a^\dag + a^\dag a) | n \rangle. Applying the given eigen-equations, one finds that E_1=\langle n | (a a^\dag + a^\dag a) | n \rangle = (2n+1)V. For n=2, one finds E_1=5V, as in choice (E).

(Note that: a^\dag a |n\rangle = a^\dag \sqrt{n} | n-1 \rangle = n | n \rangle and a a^\dag |n\rangle = a \sqrt{n+1} | n+1 \rangle = (n+1)|n\rangle and a^2 = a \sqrt{n}|n-1\rangle = \sqrt{n}\sqrt{n-1}|n-2\rangle.)

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Alternate Solutions
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2015-06-23 11:22:35
I did it the same way. a ended up being \sqrt{2}|1> and adagger was \sqrt{3}|3> . Foiling it out any of the terms where |1> and |3> multiply go to 0, and in any of the terms where the kets are the same the kets go to 1. So you are left with V(2+3)=5VNEC
2012-11-07 22:49:18
It just looks like the harmonic oscillator if you have tried to express them with the up down operator.
So I just put n=2 to it.. no answer so I give it up.
I feel so tried after doing 70+ question.
2010-10-08 12:04:06
this doesn't make much sense. I got the same answer from facotring out the (a+adagger) term, ignoring the (a*adagger), and using only a^2 and a^dagger squared. did i just get lucky, or is this alternate?
2011-08-26 01:03:37
yeah, you just got super lucky.
2011-11-04 17:34:55
Awesome! My plan is to get lucky on 100 problems in a row.... :S
2017-10-27 05:19:08

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