# Having trouble with commutators

1. Mar 8, 2008

### fifthrapiers

I'm having trouble with commutators. I have to solve them 2 ways. First, using $$[x,p]=i\hbar$$ and other identities/formulas, and the the second method the "direct way".

1.) $$x,\hat{H}$$

My work:

$$[x,\hat{H}]\psi &= x\hat{H}\psi - \hat{H}x\psi$$
$$= x \left ( \frac{p^2}{2m} + V(x) \right )\psi - \left ( \frac{p^2}{2m} + V(x) \right )x\psi$$
$$= \frac{xp^2\psi}{2m} + x V(x)\psi - \frac{p^2x\psi}{2m} - V(x) x\psi$$

2.) $$[\hat{p}, \hat{H} + x]$$

$$[\hat{p}, \hat{H} + x]\psi &= \hat{p}(\hat{H}+x)\psi + (\hat{H}+x)\hat{p}\psi$$
$$= i\hbar\frac{\partial}{\partial p} \left(\left( \frac{p^2}{2m} + V(x) \right ) + x\right)\psi + \left(\left( \frac{p^2}{2m} + V(x) \right ) + x\right)i\hbar\frac{\partial}{\partial p}\psi$$

Yikes.

2. Mar 8, 2008

### malawi_glenn

What is the original question? i.e the problem stated.

Use the [x,p] and other commutator idendites, you have not done that,
for example:

[x,H] = [x,pp + V] = {i skip the constant factors} = [x,pp] + [x,V] = ? continue

3. Mar 8, 2008

### fifthrapiers

The orig question was [x,h] for # 1, I just forgot to put in the brackets...

Yes, that's the whole point, I'm not sure how. I showed my work.

4. Mar 8, 2008

### malawi_glenn

do you know the commutator relation for

[A,B*C] = ??

find that, and use it.

5. Mar 8, 2008

### fifthrapiers

Ah yes! That was quite helpful. How does this look:

$$[x,\hat{H}]\psi = [x,\frac{p^2}{2m}]\psi$$
$$= [x,p\cdot\frac{p}{2m}]\psi$$
$$= [x,p]\frac{p}{2m}\psi + p[x,\frac{p}{2m}]\psi$$
$$= i\hbar\frac{p}{2m}\psi + p[x,p\cdot\frac{1}{2m}]\psi$$
$$= i\hbar\frac{p}{2m}\psi + p[x,p]\frac{1}{2m}\psi + p[x,\frac{1}{2m}]\psi$$
$$= i\hbar\frac{p}{2m}\psi + \frac{p}{2m}i\hbar + p\frac{x}{2m}\psi + \frac{x}{2m}\psi$$
$$= \boxed{2i\hbar\frac{p}{2m}\psi + 2\frac{px}{2m}\psi}$$

Not sure if it's right. The direct way, which I tried earlier, is still causing probs. I'm guessing for the second one, I should use [A,B+C] = [A,B] + [A,C]?

6. Mar 8, 2008

### kdv

So far so good but you don't need to stick in a wavefunction when you do it this way. One has to put a test function only when one uses explicit representations in terms of differential operators.

No!
$$[x,\frac{p}{2m}] = \frac{1}{2m} [x,p]$$
simply (the 1/2m is an overall constant which youcan pull out of the commutator.

7. Mar 8, 2008

### Hurkyl

Staff Emeritus
For the opening poster... technically, you weren't wrong; you were merely encumbered. ($[x, a] = 0$ for any constant a)

8. Mar 8, 2008

### kdv

Yes, that is true. My apologies if I gave the wrong impression.

Technically, it is correct to write

$$[x,P/2m] = \frac{1}{2m} [x,P] + [x, \frac{1}{2m}] P$$

Nobody woul dnormally write it this way but it is indeed correct.
The next step is to use the fact that [x,1/2m] = 0.

However, the next line in the demonstration of the OP is incorrect.