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GR0177 #94
Problem
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Quantum Mechanics$\Rightarrow$}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 $a^2 = a \sqrt{n}|n-1\rangle = \sqrt{n}\sqrt{n-1}|n-2\rangle$.)

Alternate Solutions
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sjayne
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)=5V
risyou
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.
Donofnothing
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?
 keradeek2011-08-26 01:03:37 yeah, you just got super lucky.
 FutureDrSteve2011-11-04 17:34:55 Awesome! My plan is to get lucky on 100 problems in a row.... :S
 yummyhat2017-10-27 05:19:08 same\r\n

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