nomadreid said:
Thank you, vanhees71. Bear with me if my understanding is somewhat basic; which one of the following would be correct?
(a) the "time" with respect to which the derivative is being taken need not be the "time" of either one of the particles in an entangled pair, but rather a time which is peculiar to the state itself;
(b) the "time" is with respect to whichever particle you happen to be measuring; or
(c) neither one of the two above statements is correct.
(No, I did not life this multiple choice from an exercise; it is just my habit to make answering easier.)
It depends upon what you measure. To explain this it's better to work in the Heisenberg picture. There the statistical operator is time-independent and all time dependence is described by the operators that represent observables and the so defined time-dependence of their eigenvectors.
Then it's easy to define coincidence measurements at different times. Consider, e.g., an entangled photon pair as an initial state coming from some localized source, e.g., a BBO crystal where they are created by parametric down conversion (note that "localized" here already means "localized on a macroscopic scale", i.e., you know that the two photons come from a place localized at a precision roughly given by the spatial extent of the crystal). Then the state can be described by a pure state with a state ket given by, e.g.,
$$|\Psi \rangle=\frac{1}{2} (\hat{A}^{\dagger}(\vec{k}_1,\lambda_1=1) \hat{A}^{\dagger}(\vec{k_2},\lambda_2=-1)- \hat{A}^{\dagger}(\vec{k}_1,\lambda_1=-1) \hat{A}^{\dagger}(\vec{k_2},\lambda_2=1) |\Omega \rangle.$$
Here ##\hat{A}(\vec{k},\lambda)## are annihilation operators (##\hat{A}^{\dagger}## thus corresonding creation operators) for photons which have a quite well-defined momentum and a helicity ##\lambda##. The momenta are more or less accurately determined by the "phase-matching condition", ##\vec{k}_1+\vec{k}_2=\vec{K}##, where ##\vec{K}## is the momentum of the incoming photon (within a coherent state describing the laser "light") being down-converted to the two photons.
Now you place two detectors at some arbitrary (but not too close) position from the source. Then it makes sense to ask for the joint probability for registering a photon with a given polarization (determined by e.g., a polarization filter) at time ##t_1## and ##t_2## in either detector. As one can show this probability is determined by the two-photon correlation function
$$G_{a_1,a_2,b_1,b_2}(x_1,\ldots,x_4)=\langle \Psi |E_{a_1}^{(-)}(x_1) E_{a_2}^{(-)}(x_2) E_{a_3}^{(+)}(x_3) E_{a_4}^{(+)}(x_4)|\Psi \rangle,$$
where ##x_1,\ldots,x_4## are four-vectors ##(c t_j,\vec{x}_j)##, and ##E_{a}^{(-)}(x)## is the part of the electric-field operator with annihilation operators and ##E_{a}^{(+)}(x)## the part with creation operators in the field decomposition of free-photon fields.
Now when having detectors at places ##\vec{x}_1## and ##\vec{x}_2## measuring linear-polarized photons with polarication direction ##\vec{n}_1## and ##\vec{n}_2## respectively the probability for registering a photon at ##t_1## and ##t_2##, respectively, is proportional to
$$P(t_1,\vec{x}_1;t_2 \vec{x}_2) \propto n_{1 a_1} n_{2 a_2} n_{1 a_3} n_{2 a_4} G_{a_1,a_2,a_3,a_4}(x_1,x_2,x_1,x_2).$$
The assumption behind this is that the photo detector works via the photoelectric effect with the interaction Hamiltonian between photons and the atoms within the detector material at place ##\vec{x}## is given by the dipole approximation ##\hat{H}_{\text{int}}=-\vec{p} \cdot \vec{E}##.
For a very good introduction to quantum optics, where this is carefully explained in detail, see
J. Garrison, R. Chiao, Quantum optics, Oxford University Press, New York (2008).
https://dx.doi.org/10.1093/acprof:oso/9780198508861.001.0001
