What is the x operator in momentum space?

[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.

 

[mark726]

The solution to this problem is not unique. Here is one possible solution:

We have $$[x^i , p_j ] = i \delta^i_j$$

Let's try to find the function $f_i (k)$ that satisfies this condition. We can do this by using the Leibniz rule for derivatives:

$$[x^i , p_j ] = \frac{\partial x^i}{\partial k_j} - \frac{\partial x^j}{\partial k_i}$$

Using the given expression for $x^i$, we get:

$$i \delta^i_j = \frac{\partial}{\partial k_j} \left(i\frac{\partial}{\partial k_i} + f_i(k)\right) - \frac{\partial}{\partial k_i} \left(i\frac{\partial}{\partial k_j} + f_j(k)\right)$$

Expanding this out and simplifying, we get:

$$i \delta^i_j = \frac{\partial f_i}{\partial k_j} - \frac{\partial f_j}{\partial k_i}$$

This is a set of three equations (one for each value of $i$) that must be satisfied by the function $f_i (k)$. These can be solved by inspection. One solution is:

$$f_i (k) = \frac{1}{2} \epsilon_{ijk} k^j k^k$$

where $\epsilon_{ijk}$ is the Levi-Civita symbol. This solution is not unique, but it satisfies all the conditions that we have imposed so far.