I am trying to evaluate $< \mathbf{p} | L_z| \phi >$: the representation of the third component of angular momentum in Cartesian coordinates in the momentum basis, where $|\phi >$ is some generic state in our Hilbert space, and $|\mathbf{x}>$ and $|\mathbf{p}>$ are eigenstates of the relevant 3D operators.
The internet seems to suggest the answer should be $-i\hbar(p_x\partial_{p_y}-p_y\partial_{p_x}) <\mathbf{p}|\phi>$.
Here's how far I've got:
Note that $L_z = XP_y – YP_x$ and $<\mathbf{x}|\mathbf{p}>$ = $e^{i \mathbf{x \cdot p}/ \hbar}/(2 \pi \hbar)^{3/2}$.
So $< \mathbf{p} | L_z| \phi >$ = $< \mathbf{p} |XP_y| \phi > – < \mathbf{p} |YP_x| \phi >$.
Consider the first term, since the second term will then follow immediately:
$< \mathbf{p} |XP_y| \phi >$ = $<\mathbf{p}| \int_{\mathbb{R}^3} |\mathbf{x}><\mathbf{x}|d^3x \ XP_y \ \int_{\mathbb{R}^3} |\mathbf{p}><\mathbf{p}|d^3p \ |\phi>$.
since $\int_{\mathbb{R}^3} |\mathbf{x}><\mathbf{x}|d^3x$ is the identity in the Hilbert space, and likewise for the momentum integral.
I think we can take $<\mathbf{p}|$ and $X$ into the first integral and similarly $P_y$ and $|\phi>$ for the second integral to give
$< \mathbf{p} |XP_y| \phi >$ = $\int_{\mathbb{R}^3} <\mathbf{p} |\mathbf{x}><\mathbf{x}| Xd^3x \ \int_{\mathbb{R}^3} P_y |\mathbf{p}><\mathbf{p}|\phi>d^3p \ $
$< \mathbf{p} |XP_y| \phi >$ = $\int_{\mathbb{R}^3} <\mathbf{p} |\mathbf{x}><\mathbf{x}| x \ d^3x \ \int_{\mathbb{R}^3} p_y |\mathbf{p}><\mathbf{p}|\phi>d^3p \ $, where $x$ and $p_y$ are eigenvalues.
$< \mathbf{p} |XP_y| \phi >$ = $i \hbar p_y \frac{\partial}{\partial{p_x}} [\int_{\mathbb{R}^3} <\mathbf{p} |\mathbf{x}><\mathbf{x}| \ d^3x] \ \int_{\mathbb{R}^3} |\mathbf{p}><\mathbf{p}|\phi>d^3p \ $.
Now I am not sure how to proceed. Please could I have some help?
Best Answer
First note $|\phi\rangle$ on the right is a canard: you might as well skip it. You are merely understanding the ineffable power of Dirac notation in momentum space.
Secondly, and crucially, your evaluations are nonsense, to the extent you are using $ \mathbf{p}$ both as a free state label, and as a dummy integration variable, when you insert a complete set of states. The correct evaluation should have been, instead, $$\langle \mathbf{p} |XP_y = \langle\mathbf{p}| \int_{\mathbb{R}^3}\!\! d^3x |\mathbf{x}\rangle \langle\mathbf{x}|\ XP_y \ \int_{\mathbb{R}^3} \!\!d^3p' |\mathbf{p'}\rangle \langle \mathbf{p'}| \\ = \int_{\mathbb{R}^3}\!\! d^3x \int_{\mathbb{R}^3}\!\! d^3p' ~x \langle\mathbf{p}|\mathbf{x}\rangle p'_y \langle\mathbf{x}|\mathbf{p'}\rangle \langle\mathbf{p'}| \\ = \frac{1}{(2\pi\hbar)^3}\int_{\mathbb{R}^3}\!\! d^3x \int_{\mathbb{R}^3}\!\! d^3p' ~ x p'_y ~ e^{i\mathbf{x} \cdot (\mathbf{p'}-\mathbf{p})/\hbar} \langle\mathbf{p'}| \\ = \int_{\mathbb{R}^3}\!\! d^3p' ~ p'_y ~ i\hbar \frac{\partial}{\partial p_x }~ \delta( \mathbf{p'}-\mathbf{p}) \langle\mathbf{p'}| = i\hbar p_y \frac{\partial}{\partial p_x }~ \langle\mathbf{p}| ~,$$ where the last line is gotten by supplanting the conjugate gradient for x before performing the integral in x to obtain the delta function in p.
You may now evaluate your second term, mutatis mutandis.