Question about proof of Weinberg-Witten theorem

Question

In proving the Weinberg-Witten theorem, there is a step where one needs to show

\begin{align*} \lim_{k' \to k}\langle k, \sigma | J^{\mu} |k', \sigma \rangle &= \frac{q k^{\mu}}{k^0}\frac{1}{(2\pi)^3} \\ \lim_{k' \to k}\langle k, \sigma | T^{\mu \nu} |k', \sigma \rangle &= \frac{ k^{\mu} k^{\nu}}{k^0}\frac{1}{(2\pi)^3}. \end{align*}

Take, say the first one. Though I think my questions will also apply to the second one.

I think it's clear enough that \begin{align*} \lim_{k' \to k}\langle k, \sigma | J^{\mu} |k', \sigma \rangle &= A k^{\mu}\\ \end{align*}

for some number $A$, since there needs to be a $\mu$ index on the RHS, and there is nothing other than $k^{\mu}$ to provide this.

So the problem is reduced to showing that

\begin{align*} \lim_{k' \to k}\langle k, \sigma | J^{0} |k', \sigma \rangle &= \frac{q}{(2 \pi)^3}. \end{align*}

So we have that

\begin{align*} q \delta^3(\vec{k} - \vec{k'})&= \langle k ,\sigma| Q |k', \sigma \rangle \\ &= \langle k , \sigma| \int d^3 x J^0(t, \vec{x}) |k', \sigma \rangle \\ &= \langle k , \sigma| \int d^3 x e^{i \vec{P} \cdot \vec{x}}J^0(t, \vec{0}) e^{-i \vec{P} \cdot \vec{x}}|k', \sigma \rangle \\ &= \int d^3 e^{i (\vec{k} - \vec{k'})\cdot \vec{x})} \langle k , \sigma|J^0(t, \vec{0})|k', \sigma \rangle \\ &= (2 \pi)^3\delta^3(\vec{k} - \vec{k'}) \langle k , \sigma|J^0(t, \vec{0})|k', \sigma \rangle. \end{align*}

What I am unsure about is the argument of the $J^0$ (or $J^{\mu}$ more generally). When we have the expression \begin{align*} \lim_{k' \to k}\langle k, \sigma | J^{\mu} |k', \sigma \rangle, \end{align*}

it is usually suppressed, but $J^{\mu}$ is a fucntion of time a space, that is $J^{\mu} = J^{\mu}(t, \vec{x})$. I know that we are working in the Schrodinger picture in canonical quantisation, so we are picking a fixed time. So I guess the $t$ argument can just be ignored. But the argument I have above only seems to hold at $\vec{x} = \vec{0}$. Is this true? Why is this enough?

The sources I'm looking at are the solutions to assignment 1, problem 1 here. Along with this essay and wikipedia.

Any clarifications would be appreciated.

I hope the notation above is clear enough from context to anyone who know about the theorem and its proof. I can clarify if anything is unclear.

Answer 1

So I guess the $t$ argument can just be ignored.

Per the link provided, $Q$ is a conserved charge. "Conserved" means that it doesn't change with time, so it is the same for all $t$: $$ \langle k ,\sigma| Q(t) |k', \sigma \rangle = \langle k ,\sigma| Q(0) |k', \sigma \rangle = \langle k , \sigma| \int d^3 x J^0(0, \vec{x}) |k', \sigma \rangle $$

But the argument I have above only seems to hold at $\vec{x} = \vec{0}$. Is this true? Why is this enough?

I don't see why you think this is the case. In fact, you have already stated that this is "clear":

I think it's clear enough that \begin{align*} \lim_{k' \to k}\langle k, \sigma | J^{\mu} |k', \sigma \rangle &= A k^{\mu}\\ \end{align*}

for some number $A$...

The right hand side above has no $\vec x$ dependence, so the left-hand side can have no $\vec x$ dependence.

This can be seen from: $$ \lim_{k' \to k}\langle k, \sigma | J^{\mu}(t,\vec x) |k', \sigma \rangle = \lim_{k' \to k}\langle k, \sigma | e^{i\vec P \cdot \vec x}J^{\mu}(t,0)e^{-i\vec P \cdot \vec x} |k', \sigma \rangle $$ $$ = \lim_{k' \to k}e^{i\vec x \cdot (\vec k' - \vec k)}\langle k, \sigma | J^{\mu}(t,0)|k', \sigma \rangle $$ $$ =\lim_{k' \to k}\langle k, \sigma | J^{\mu}(t,0)|k', \sigma \rangle\;, $$ where I set $k'=k$ in the exponent, since you are taking that limit.