Momentum operator on position eigenket

Question

What is the result of:
$$mathbf{langle x'|f(hat{P})|xrangle}$$
I obtained this result:
$$mathbf{langle x'|f(hat{P})|xrangle=tilde{f}(x'-x)}$$
where $tilde{f}$ is the Fourier transform of $f$.

Michael Riberdy · Answer

Working in momentum space we find:
$$langle x'|f(P)|xrangle=int e^{-ipx}f(p)e^{ipx}dp=int f(p)e^{-ip(x'-x)}dp=tilde{f}(x'-x)$$
So you are indeed correct.

Momentum operator on position eigenket

One Answer

Add your own answers!

Ask a Question