# What is the x operator in momentum space?

1. Mar 21, 2017

### David Silva

The Hilbert space for a free relativistic particle has inner product (in the momentum representation) $$\langle \chi | \phi \rangle = \int \frac {d \vec k^3} {(2 \pi)^3 2 \sqrt{\vec k^2 + m^2}} \chi (\vec k) * \phi (\vec k)$$ States undergo time evolution $$i∂t|ψ \rangle = H_0 | ψ \rangle$$ with the Hamiltonian is $$H_0 = \sqrt {\vec p^2 + m^2}$$. Consider the self-adjoint operator, $\vec x$, which satisfies $$[x^i , p_j ] = i δ^{i}_{j}$$ It can be written in the form $$x^i = i (\frac {\partial} {\partial k_i} + f_i( \vec k))$$ for some real function $f_i$ . What is $fi(\vec k)$?

So what I have done so far:

I was using regular commutator methods to try and work this out using the dummy function method described in Zettili (eq. 2.310). $$[x^i, p_j] = i(\frac \partial {\partial k_i} + f_i (\vec k)) * k_j - k_j * i(\frac \partial {\partial k_i} + f_i (\vec k)) \\ = i (\frac \partial {\partial k_i} * k_j + i * f_i (\vec k) * k_j - i * k_j * \frac \partial {\partial k_i} + i * k_j * f_i (\vec k) \\ = i * \delta^i_j + i * f_i (\vec k) * k_j - i * k_j * \frac \partial {\partial k_i} - i * k _j* f_i (\vec k) \\ => f_i (\vec k) * k_j - k _j* f_i (\vec k) = k_j * \frac \partial {\partial k_i}$$ I'm stuck at this last part, I don't know how to proceed. Any thoughts? I haven't inserted the dummy function in yet, I wanted to reduce it but my prof said it should come out before I even have to do that but I just don't see what he's talking about.

2. Mar 26, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.