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

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