Trying to prove that a propagator only depends on the difference of coordinates if our system has translational symmetry
As the title says, I'm trying to prove that if a system is translation invariant, then its propagator depends only on the difference of coordinates. That is to say $$K(\boldsymbol{x},\boldsymbol{x}';t - t_0) = K(\boldsymbol{x} - \boldsymbol{x}';t-t_0)$$ where $$K(\boldsymbol{x},\boldsymbol{x}';t - t_0) = \langle \boldsymbol{x}| e^{-i H (t - t_0) / \hbar} |\boldsymbol{x}' \rangle$$ I've tried to prove this in two ways, but I'm stuck in some things.
First attempt: If we have translational symmetry, then $|\boldsymbol{x} \rangle = |\boldsymbol{x} + \boldsymbol{a} \rangle = e^{-i \boldsymbol{p} \cdot \boldsymbol{a} / \hbar} |\boldsymbol{x} \rangle$ \begin{align} \Rightarrow K(\boldsymbol{x} + \boldsymbol{a},\boldsymbol{x}' + \boldsymbol{a};t - t_0) &= \langle \boldsymbol{x} + \boldsymbol{a}| e^{-i H (t - t_0) / \hbar} |\boldsymbol{x}' + \boldsymbol{a} \rangle\ &= \langle \boldsymbol{x}| e^{i \boldsymbol{p} \cdot \boldsymbol{a} / \hbar} e^{-i H (t - t_0) / \hbar} e^{-i \boldsymbol{p} \cdot \boldsymbol{a} / \hbar} |\boldsymbol{x}' \rangle \end{align} I don't know what else to do here. 669
Couldn't I just sum the exponents? I wouldn't arrive anywhere. I can't see where the dependence from the difference of coordinates could be deduced from here.
Second attempt: If there's translational symmetry, then $[H,\boldsymbol{p}] = \boldsymbol{0}$. Then \begin{align} \langle \boldsymbol{x} | e^{-i H (t - t_0) / \hbar} | \boldsymbol{x}' \rangle &= \int\limits_{\mathcal{V}} \langle \boldsymbol{x} | \boldsymbol{p} \rangle \langle \boldsymbol{p} | e^{-i H (t - t_0) / \hbar} | \boldsymbol{x}' \rangle \text{d} \boldsymbol{p}\ &= \frac{1}{(2 \pi \hbar)^{n/2}} \int\limits_{\mathcal{V}} \langle \boldsymbol{p} | e^{-i H (t - t_0) / \hbar} | \boldsymbol{x}' \rangle e^{i \boldsymbol{p} \cdot \boldsymbol{x} / \hbar} \text{d} \boldsymbol{p}\ &\stackrel{?}{=} \frac{1}{(2 \pi \hbar)^{n/2}} \int\limits_{\mathcal{V}} e^{-i H (t - t_0) / \hbar} \langle \boldsymbol{p} | \boldsymbol{x}' \rangle e^{i \boldsymbol{p} \cdot \boldsymbol{x} / \hbar} \text{d} \boldsymbol{p}\ &= \frac{1}{(2 \pi \hbar)^{n}} \int\limits_{\mathcal{V}} e^{-i H (t - t_0) / \hbar} \ e^{\frac{i}{\hbar} \boldsymbol{p} \cdot (\boldsymbol{x} - \boldsymbol{x}')} \text{d} \boldsymbol{p}\ \end{align} The step over which there's a "? 955