# Not sure I understand commutation relations

1. Oct 19, 2016

### sa1988

1. The problem statement, all variables and given/known data

Firstly, I'm looking at this:

I'm confused because my understanding is that the commutator should be treated like so:

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

but the working in the above image looks like it only goes as far as

$$aa^{\dagger}$$

This surely isn't the case, so I think I may have got my understanding wrong regarding how to take the correct steps when working out a commutation relation

Secondly, I have this question, which I think may further shed light on where I'm going wrong, if anyone may be kind enough to have a look for me:

I work through it like so:

$$[a^2,a^{\dagger}] = \Bigg[\frac{m}{2\hbar\omega}(\omega \hat{x} + \frac{i}{m}\hat{p})^2, \sqrt{\frac{m}{2\hbar\omega}}(\omega \hat{x} - \frac{i}{m}\hat{p}) \Bigg]$$
$$= \frac{m}{2\hbar\omega}(\omega \hat{x} + \frac{i}{m}\hat{p})^2 \sqrt{\frac{m}{2\hbar\omega}}(\omega \hat{x} - \frac{i}{m}\hat{p}) - \frac{m}{2\hbar\omega}(\omega \hat{x} + \frac{i}{m}\hat{p})^2 \sqrt{\frac{m}{2\hbar\omega}}(\omega \hat{x} - \frac{i}{m}\hat{p})$$
taking this part alone:
$$(\omega \hat{x} + \frac{i}{m}\hat{p})^2$$
I'm getting:
$$\omega^2[\hat{x},\hat{x}] + \frac{\omega i}{m}[\hat{x},\hat{p}] + \frac{\omega i}{m}[\hat{p},\hat{x}] - \frac{1}{m}[\hat{p},\hat{p}]$$
$$= 0 + \frac{\omega i}{m}(i \hbar) + \frac{\omega i}{m}(-i \hbar) + 0$$
$$= 0$$

Which would give a final result:

$$[a^2,a^{\dagger}] = 0$$

The only error I can think of is that I'm supposed to put a dummy function alongside the operators i.e. I'm supposed to actually show that:

$$[a^2,a^{\dagger}]\Psi(x) = 2a\Psi(x)$$

This worked in a previous exercise where I demonstrated

$$[\hat{x},\hat{p}]\Psi(x) = i\hbar \Psi(x)$$

But this isn't how it's done in the first image I've embedded in this post where there isn't any kind of "dummy function" involved.

Or perhaps I'm not expanding the brackets properly? This is probably more likely...

Any help much appreciated, thank you.

2. Oct 19, 2016

### andrewkirk

I agree with you. The working shown looks wrong, in exactly the way you point out. What is the source of that working?

For the next bit:
That looks wrong to me. I can't see any reason for putting commutators in. There should just be composition of the operators. I get:
$$\omega^2\hat{x}^2 + \frac{\omega i}{m}\hat{x}\hat{p} + \frac{\omega i}{m}\hat{p}\hat{x} - \frac{1}{m}\hat{p}^2$$
which is not necessarily zero.

3. Oct 20, 2016

### strangerep

Huh? Looks ok to me. It's just using linearity properties of the commutator, i.e., $$[A+B \,,\, C] ~=~ [A,C] + [B,C]$$
The commutator is a derivation, meaning that it satisfies the Leibniz product rule: $$[AB, C] ~=~[A,C]B + A[B,C] ~.$$ You can therefore use the result of the 1st part of your post to make the 2nd part easy. I.e., think of the $a^2$ as a product. With this hint, the solution should be a 1-liner.

4. Oct 21, 2016

### sa1988

Thanks for the replies.

So I was right in that I simply didn't know enough about commutators, i.e their properties.

Cheers :)