Useful Raising/Lowering Operator Equation, Quick derivation help

[logic smogic]

Problem

Show that,

$$a (a^{ \dagger})^{n} = n (a^{\dagger})^{n-1}+(a^{\dagger})^{n} a$$

Formulae

$$a = \sqrt{\frac{m \omega}{2 \hbar}}(x+\frac{\imath p}{m \omega})$$

$$a^{\dagger} = \sqrt{\frac{m \omega}{2 \hbar}}(x-\frac{\imath p}{m \omega})$$

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

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 $$a^{\dagger}$$’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?

 

[Dick]

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.

 

[logic smogic]

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.

 

[logic smogic]

So then,

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

Consider,

$$|n>=\frac{1}{\sqrt{n!}}(a^{\dagger})^{n}|0>$$

where

$$a|0>=0$$

then,

$$a^{2}|n>=\frac{1}{\sqrt{n!}}a^{2}( a^{\dagger})^{n}|0>$$
$$= \frac{1}{\sqrt{n!}}( a n (a^{\dagger})^{n-1} + n (a^{\dagger})^{n-1} a<br /> )| 0>$$
$$= \frac{1}{\sqrt{n!}} a n (a^{\dagger})^{n-1}| 0>$$
$$=\frac{1}{\sqrt{n!}} n (n-1) (a^{\dagger})^{n-2}|0>$$
$$= \sqrt {n (n-1)} |n-2>$$

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!