# More operator questions

1. Oct 31, 2007

### BeauGeste

Consider
$$\langle x'|\hat{p} \hat{x} | x \rangle$$.
Are these steps correct?
1. $$\hat{x}$$ operates on the x eigenstate to get
$$\langle x'|\hat{p} x | x \rangle$$.
2. x is a c-number so can be pulled out to get
$$x \langle x'|\hat{p} | x \rangle$$.
3. $$x \langle x'|\hat{p} | x \rangle = x \frac{\partial}{\partial x'}\delta (x'-x)$$

The part I'm most wary on is 2 where I pulled x out. It seems like the p operator is acting on it so I shouln't be able to do that.

Alternatively, could I do the following:
1. have p operate on $$\hat{x}$$ first to get
$$\frac{\partial}{\partial x'}x' \delta (x'-x)$$
2. Which becomes
$$\delta (x'-x) + x' \delta'(x'-x)$$

Is one of these two routes incorrect?

2. Oct 31, 2007

### meopemuk

Yes, you can do that. Operators do not act on c-numbers.

Eugene.

3. Oct 31, 2007

### Avodyne

Both are correct (except you dropped a factor of -i*hbar in line 3, but you also dropped it in the 2nd part). You can check that your two final expressions are equal by integrating them against a test funtion of either x or x'.

4. Nov 1, 2007

### BeauGeste

Won't you get the same thing for
$$\langle x'|\hat{x} \hat{p} | x \rangle$$
though? That would imply the two operators commute.

Last edited: Nov 1, 2007
5. Nov 1, 2007

### Hurkyl

Staff Emeritus
It's not a problem because $\hat{p}$ is not a differential operator: $\hat{p}$ is a linear operator acting on states. Because it's linear, it commutes with all complex numbers.

$\hat{p}$ only turns into a differential operator when everything's written out in the position representation -- in that case, you would have

$$\langle x' | \hat{p} \hat{x} | x \rangle = \int_{-\infty}^{+\infty} \delta(x' - \xi) \left(-i \hbar \frac{\partial}{\partial \xi} \right) \xi \delta(x - \xi) \, d\xi$$

and now it's differently obvious, since x clearly commutes with $\partial / \partial \xi$.

Last edited: Nov 1, 2007