Useful Raising/Lowering Operator Equation, Quick derivation help

  • #1
Problem

Show that,

[tex]a (a^{ \dagger})^{n} = n (a^{\dagger})^{n-1}+(a^{\dagger})^{n} a[/tex]

Formulae

[tex] a = \sqrt{\frac{m \omega}{2 \hbar}}(x+\frac{\imath p}{m \omega})[/tex]

[tex] a^{\dagger} = \sqrt{\frac{m \omega}{2 \hbar}}(x-\frac{\imath p}{m \omega})[/tex]

[tex] [a,a^{\dagger}]= a a^{\dagger}-a^{\dagger}a=1[/tex]

Attempt

This is just one step in a long derivation from another problem. The author (Goswami, pg 147) uses it without proof, but I would like to modify it – and hence I need to understand where he got it from.

I can see how you might pull [tex] a^{\dagger}[/tex]’s out from the first and third term, and try to use the commutation relation, but the n’s don’t seem to work out right.

Any thoughts?
 
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Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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The formula you are trying to prove is the same as [a,(a^+)^n]=n*(a^+)^(n-1), right? For any three operators A,B and C we know [A,BC]=[A,B]C+B[A,C]. This looks like the product rule for differentiation with [A, acting like a derivative. You can extend this to any number of operators, [A,BCDEF...]. So what does this say about [A,B^n] when [A,B]=1? BTW I think [a,a^+]=1, not the order you wrote.
 
  • #3
Ah, I didn't think to look up some commutator identities. And, yes, we proved the differentiation/commutation relationship last homework assignment. Right, I'll try it again. Thanks.
 
  • #4
So then,

[tex][a^{2}, (a^{\dagger})^{n}] = a [a, (a^{\dagger})^{n}]+[a, (a^{\dagger})^{n}]a= a n (a^{\dagger})^{n-1} + n (a^{\dagger})^{n-1} a[/tex]

Consider,

[tex]|n>=\frac{1}{\sqrt{n!}}(a^{\dagger})^{n}|0>[/tex]

where

[tex]a|0>=0[/tex]

then,

[tex]a^{2}|n>=\frac{1}{\sqrt{n!}}a^{2}( a^{\dagger})^{n}|0> [/tex]
[tex]= \frac{1}{\sqrt{n!}}( a n (a^{\dagger})^{n-1} + n (a^{\dagger})^{n-1} a
)| 0>[/tex]
[tex]= \frac{1}{\sqrt{n!}} a n (a^{\dagger})^{n-1}| 0>[/tex]
[tex]=\frac{1}{\sqrt{n!}} n (n-1) (a^{\dagger})^{n-2}|0>[/tex]
[tex]= \sqrt {n (n-1)} |n-2>[/tex]

I know that I may have not provided enough of a primer in the problem to have anyone check this, but just puttin’ it out there!
 
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