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I don't understand this equality $$\int \!d^3p~\langle\textbf{x}|e^{-i(\hat{\textbf{p}}^2/2m)t}|\textbf{p}\rangle\langle\textbf{p} | \textbf{x}_0 \rangle ~=~\int\! \frac{d^3p}{(2\pi)^3}~e^{-i(\textbf{p}^2/2m)t}e^{i\textbf{p}\cdot(\textbf{x}-\textbf{x}_0)}. $$ In particular that $$\langle\textbf{x}|e^{-i(\hat{\textbf{p}}^2/2m)t}|\textbf{p}\rangle~=~e^{-i(\textbf{p}^2/2m)t}\langle \textbf{x}|\textbf{p}\rangle.$$ It's in the second chapter of Peskin et. al. An Introduction to QFT. |
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Try expanding the exponential into a power series of the momentum operator. It should then become clear that all these powers of $\mathbf{\hat p}$ acting on $|\mathbf p\rangle$ just produce powers of the eigenvalue $\mathbf p$. The series can then be reassembled into an exponential of the eigenvalue. |
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In general, $$f(\hat{\textbf{p}})|\textbf{p}\rangle=f(\textbf{p})|\textbf{p}\rangle,$$ and the constant factor can be taken out of the innner product. |
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