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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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I think you question is already answered, but I still want to make a comment: From my point of view, the subscript 0 of p0 should be at x, x0. Then the result is correct. –  user16138 Nov 20 '12 at 13:50
    
hi, thanks for your answer. can you explain it a little further? –  Jorge Nov 20 '12 at 19:00

2 Answers 2

up vote 5 down vote accepted

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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it's because $(\hat{p})^{2i}|\textbf{p}\rangle=\textbf{p}^{2i}|\textbf{p}\rangle, i=0,1,...$ –  Jorge Nov 19 '12 at 20:53
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it's because $\exp [-i\hat{p}^2]|p\rangle=\sum_{n}(-i)^n(1/n!)(\hat{p}^2)^n|p\rangle=(1-i \hat{p} \hat{p} -\frac{1}{2}\hat{p}\hat{p}\hat{p}\hat{p}+...)|p\rangle=\exp [-ip^2]|p\rangle$ –  kηives Nov 19 '12 at 23:28

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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