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Ladder operators and harmonic oscillator

  • Thread starter cleggy
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  • #1
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1. Explain why any term (such as AA†A†A†)with unequal numbers of raising and lowering operators has zero expectation value in the ground state of a harmonic oscillator.

Explain why any term (such as AA†A†A) with a lowering operator on the extreme right has zero expectation value in the ground state of a harmonic oscillator.

2. Homework Equations



3. The Attempt at a Solution

Please help as I'm totally stuck with this. Does it have to do with orthonormality of different eigenfunctions?
 

Answers and Replies

  • #2
malawi_glenn
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Yes it has, and also consider the ground state as the problem tells you to do.

what is the effect of A on |0> ? And what about it's dual? i.e. <0|A†

Now give it a try, star with your first operator AA†A†A† and tell us what you get and why
 
  • #3
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with AA†A†A† I get [tex]\psi[/tex][tex]_{}n+2[/tex]. The first to terms convert [tex]\psi[/tex][tex]_{}n[/tex] into different eigenfunctions. Because [tex]\psi[/tex][tex]_{}n[/tex] is orthonormal to these eigenfunctions, these terms can be dropped from the integral.

Correct?


Also since the lowering operator is first to act on the ground-state harmonic iscillator, this takes it below the ground-state which cannot be allowed.

On the right track?
 
  • #4
malawi_glenn
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I have a hard tome to understand your language..

with AA†A†A† acting on the state to the right (remember that this is sandwiched between [itex]\psi^* _0 [/itex] and [itex]\psi _0 [/itex], the two first will give you [itex]\psi _2 [/itex] so what is left is

[itex]\psi^* _0 AA^{\dagger} \psi_2 [/itex]

do the rest of the operation:
[itex]\psi^* _0 \psi_2 [/itex]
thus integral gives zero

And yes, you can't go further down with the ground state, thus A|0> = 0
 

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