#### TeethWhitener

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$$H(P,Q)=\int{\frac{dx}{2\pi}\frac{dk}{2\pi} e^{ixP+ikQ}}\int{dp \text{ }dq\text{ }e^{-ixp-ikq}H(p,q)}$$

where ##P,Q## are momentum and position operators and ##H(p,q)## is the classical Hamiltonian. He then mentions that the matrix element for a single time slice of the path integral ##\langle q_2 | e^{-iH\delta t}|q_1\rangle## is given by:

$$\langle q_2 | e^{-iH\delta t}|q_1\rangle =\int{\frac{dp_1}{2\pi}e^{-iH(p_1,\overline{q}_1)\delta t}e^{ip_1(q_2-q_1)}}$$

where ##\overline{q}_1=\frac{1}{2}(q_1+q_2)##. I’m having trouble showing how the midpoint ##\overline{q}_1## enters into the equation. I expanded the propagator to first order and inserted a complete set of momentum states to get:

$$\langle q_2 | e^{-iH\delta t}|q_1\rangle =\int{dp_1\langle q_2|p_1\rangle\langle p_1|q_1\rangle}-\int{dp_1}\int{\frac{dx}{2\pi}\frac{dk}{2\pi}\langle q_2|e^{ixP}|p_1\rangle\langle p_1|e^{ikQ}|q_1\rangle} \times \int{dp\text{ }dq \text{ }e^{-ixp-ikq}H(p,q)}$$

Letting ##\langle q|p\rangle =(2\pi)^{-1/2}\exp{(ipq)}##, this simplifies to:

$$\langle q_2 | e^{-iH\delta t}|q_1\rangle =\int{\frac{dp_1}{2\pi}e^{ip_1(q_2-q_1)}}\left(1-i\delta t\int{\frac{dx}{2\pi}\frac{dk}{2\pi}e^{ixp_1+ikq_1}}\int{dp\text{ }dq\text{ }e^{-ixp-ikq}H(p,q)}\right)$$

It looks like if you just let the ##\exp{(ixp_1)}## integral become two delta functions ##\delta(p_1)## and ##\delta(x_1)##, you’ll get a matrix element dependent on the Hamiltonian ##H(p_1,q_1)## instead of the Hamiltonian at the position midpoint ##H(p_1,\overline{q}_1)##. What am I missing?