Is the More Operator Correct for \langle x'|\hat{p} \hat{x} | x \rangle?

[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?

 

[meopemuk]

2. x is a c-number so can be pulled out

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

Eugene.

 

[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'.

 

[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.

 

[Hurkyl]

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.

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

<br /> \langle x&#039; | \hat{p} \hat{x} | x \rangle<br /> =<br /> \int_{-\infty}^{+\infty} \delta(x&#039; - \xi) \left(-i \hbar \frac{\partial}{\partial \xi} \right)<br /> \xi \delta(x - \xi) \, d\xi

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