Quantum state variable for entangled system

[nomadreid]

TL;DR Summary

Should a state be dependent on a single time variable or possibly more time variables?

(This question is on the elementary side...) In the Schrödinger picture, the state is dependent on time. If you have a state composed of several particles, generally you can break them up, with each one depending on local time. But in an entangled system, say of two particles, you can no longer break them up; since the particles can be widely separated in spacetime, the time variable would be different for each part of the system, no? If so, would one say that the state depends on (t1, t2) in some way? Or am I looking at this completely incorrectly?

 

[vanhees71]

In the Schrödinger picture all time-dependence is on the state, i.e., you have a statistical operator $\hat{\rho}(t)$ obeying the von Neumann equation of motion
$$\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}]+\partial_t \hat{\rho}=0,$$
where $\hat{H}$ is the Hamilton operator of the system; $\partial_t$ denotes the time derivative with respect to the explicit time dependence of the statistical operator.

 

[Nugatory]

since the particles can be widely separated in spacetime, the time variable would be different for each part of the system, no?

If you’re working with non-relativistic QM, the time is the same everywhere so there’s only one time no matter how widely separated the particles are.

 

[nomadreid]

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.)

 

[nomadreid]

Thanks, Nugatory. Ah, good point. I had automatically assumed the problem with comparing times in a relativistic system. As a side question: since entanglement can occur not only across space but across time, then to have the same time for any collection of entangled particles would require a static universe, no?

 

[PeroK]

Thanks, Nugatory. Ah, good point. I had automatically assumed the problem with comparing times in a relativistic system. As a side question: since entanglement can occur not only across space but across time, then to have the same time for any collection of entangled particles would require a static universe, no?

No. It just means that QM, as a theory, is non-relativistic. If you are studying aeronautics, for example, you are making no assumptions or have any requirement for a static universe.

You only have to consider an expanding universe when you come to deal with cosmological phenomena.

QFT is the extension to QM that incorporates SR and flat spacetime. But, still not GR and curved spacetime.

You could turn this whole question round, in fact, and say that GR requires a non-quantum universe!

Each theory Classical Mechanics, QM, QFT, SR, GR has its realm of applicability. It doesn't make a lot of sense to demand that CM, say, requires a non-quantum, non-relativistic universe. It simply requires the effect of those aspects of the universe to be negligible within the phenomena you are studying using classical mechanics.

 

[DrChinese]

If so, would one say that the state depends on (t1, t2) in some way?

My comment is not a direct reply to the above. However, you can see from this example how difficult it is to factor time into any entangled system. Specifically: entangled systems can consist of components that have never coexisted. This sounds impossible, and yet this has already been demostrated:

https://arxiv.org/abs/1209.4191

 

[vanhees71]

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

 

[Demystifier]

Summary: Should a state be dependent on a single time variable or possibly more time variables?

(This question is on the elementary side...) In the Schrödinger picture, the state is dependent on time. If you have a state composed of several particles, generally you can break them up, with each one depending on local time. But in an entangled system, say of two particles, you can no longer break them up; since the particles can be widely separated in spacetime, the time variable would be different for each part of the system, no? If so, would one say that the state depends on (t1, t2) in some way? Or am I looking at this completely incorrectly?

Yes, quantum mechanics can be formulated in that way, it's called many-time formalism and has been developed by Dirac, Tomonaga and others. See e.g. http://de.arxiv.org/pdf/1407.8058v1 and Refs. [12], [22-26] therein.