GREPhysics.NET: GR0177 #94

GR0177 #94

Problem

GREPhysics.NET Official Solution

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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 $a^2 = a \sqrt{n}|n-1\rangle = \sqrt{n}\sqrt{n-1}|n-2\rangle$ .)

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Comments

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

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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.

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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?

keradeek
2011-08-26 01:03:37

yeah, you just got super lucky.

FutureDrSteve
2011-11-04 17:34:55

Awesome! My plan is to get lucky on 100 problems in a row.... :S

yummyhat
2017-10-27 05:19:08

same\r\n

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