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How do I calculate path integral representation of $\langle q_f t_f|p(t_1)|q_i t_i\rangle $ where $t_i<t_1<t_f$? I am doing this by discretizing, the time intervals and adding a complete set of $|q_j t_j\rangle$ states at each point, then at the time $t_1$, one can add $p(t_1)|p^{\prime} t_1\rangle \langle p^{\prime} t_1|$, which is the momentum of the particle at time $t_1$. After taking the inner product with the position state at time $t_1$, and integrating I get $$\frac{1}{h} \int \mathcal{D(q)} \mathcal{D(p)} \int dp^{\prime} \frac{p^{\prime^2}}{2m} exp(-\frac{i}{\hbar}q_{t_1-1}p^{\prime}-(t_f-t_i)H(p^{\prime}, (\frac{q_{t_1-1}+q_{t_1}}{2})))exp(\frac{i}{\hbar}S)$$

This looks terribly ugly and wrong. How do I do it?

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Comment to the question (v2): Please re-check your formula. –  Qmechanic May 27 '13 at 12:39
I suggest that you rewrite your initial expression as : $\langle q_f | e^{-i H(t_f - t_1)} \, \hat p(t_1) \, e^{-i H(t_1 -t_i)}|q_i\rangle$ –  Trimok May 28 '13 at 14:29

1 Answer 1

Do the path integral with those boundary conditions and an extra source term $S = \int_{t_0}^{t_1} q(t) j(t)dt$. This is a Gaussian integral, and completely do-able (although you have to be a bit careful with the boundary conditions). Then use $Z(j)$ as a generating functional: Differentiate $Z(j)$ with respect to the 'coordinate direction' $j \mapsto j(t_1)$, and then set $j=0$. This brings down the $p(t_1)$ you wanted.

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Do you mean to say add the term $\int q(t)j(t)$?, I don't see how differentiation with $j$, would bring down $p(t_1)$? –  ramanujan_dirac May 28 '13 at 4:17
Sorry above comment should read $p(t)j(t)$ –  ramanujan_dirac May 28 '13 at 11:01
Yes. Sorry! You can also just do the difference quotient on the lattice. –  user1504 May 28 '13 at 12:24

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