# Quantum Measurement of Entangled Time-Varying Systems

Suppose System A has a dynamic, continuous variable $O_A(t)$, which we wish to measure via a quantum probe. Assume system A has a self-Hmiltonian, $H_{SA}$ so that the evolving wave function of system A, in the O basis, is $|\Psi_{O_A} (t)\rangle = exp (-i\int_0^t H_{SA}dt )|\Psi_{O_Ai}\rangle$, where $|\Psi_{O_Ai} \rangle$ is the initial wave function for System A, at $t=0$.

Suppose System B is a quantum probe with initial wave function $|\Psi_{P_Bi}\rangle$ in the momentum basis, and has no self-Hamiltonian. Also, suppose its momentum becomes entangled with $O_A$ at $t_1 > 0$, via impact interaction-Hamiltonian $H_{int}$. This creates a composite wave function of both systems for which not all states of system B's momentum are possible for all states of $O_A$ (due to the entanglement). Further, suppose $Y_B$ is the conjugate variable to $P_B$.

The composite wave function (for $t > t_1 + \tau$) then
\begin{align}|\Psi_C(Y_B, O_A, t)\rangle = exp(-i\int_{t_1}^{t_1 +\tau} H_{int} dt)|\Psi_{Y_Bi}\rangle exp(-i\int_0^t H_{SA}dt)|\Psi_{0_Ai}\rangle \\= exp(-i\int_{0}^{t} (H_{int} + H_{SA}) dt)|\Psi_{Y_Bi}\rangle |\Psi_{0_Ai}\rangle \end{align} where $H_{int} = 0$ for $t_1 > t > t + \tau$, and $\tau$ is the short duration of impact interaction between A and B, and Y is the conjugate variable to P.

$$|\Psi_C(P_B, O_A, t)\rangle = F.T. (exp(-i\int_{0}^{t} (H_{int} + H_{SA}) dt)|\Psi_{Y_Bi}\rangle |\Psi_{0_Ai}\rangle)$$ where F.T. is the Fourier Transform, over $Y_B$.
Does the above sound correct, especially the inclusion of $H_{SA}$, so that $|\Psi_C\rangle$ will evolve properly over time?

The probability that a measurement of the probe's momentum, at $t_2 > t_1 + \tau$, will yield momentum eigenstate $|p_j\rangle$, if System A is in eigenstate $|o_k\rangle$, is $$Pr(|p_j o_k\rangle) = \langle p_j o_k|\Psi_C(P_B, O_A, t_2)\rangle$$

Therefore, the total probability of the probe's momentum measurement yielding $|p_j\rangle$ is $$Pr(|p_j\rangle) = \int_{-\infty} ^\infty \langle p_j o|\Psi_C(P_B, O_A,t_2)\rangle do$$ We need the integral because we cannot generally assume that each $|p\rangle$ state will only have a single $|o\rangle$ state entangled with it. In fact, we may have a continuous range of $|o\rangle$ states that are entangled with $|p_j\rangle$

Of course, a measurement result of $|p_j\rangle$ would also pick out the corresponding set of $|o\rangle$ states that are entangled with it. Hence, a measurement of the probe's momentum yielding $|p_j\rangle$ results in an (indirect) measuremeant of $O_A$.

I find that interesting because then, at $t_2$, we are measuring $O_A$ based on how it evolved up to $t_2$, even though the probe, by which we are measuring it, had no interaction with system A after $t_1 + \tau$, when the entanglement completed. Of course, the reason is that we included $exp(-i\int_0^t H_{SA}dt)$ in the formula for $|\Psi_C(P_B, O_A,t)\rangle$, above.
Does that sound right?

In fact, if the entanglement at $t_1 + \tau$ also caused a back action to $O_A$'s conjugate variable and that, in turn, caused a back action to $O_A$ (as would be the case if $O_A$ were, for instance, position and it's conjugate were momentum), then that back-action would also be modeled in $|\Psi_C(P_B, O_A,t)\rangle$. Therefore, the measurement of the probe at $t_2$ would be a measurement of $O_A$'s complete evolution, up to $t_2$, including evolution caused by its self-hamiltonian and any back action upon it, up until $t_2$.

• What is "The probabilities of each state of $O_A$ form a Gaussian distribution" supposed to mean? $O_A$ is an observable, it doesn't have states. What exactly are you saying when you say the momentum $P_B$ "becomes entangled with $O_A$"? States can be entangled, not observables or values. When you are "measuring the probe system", what observable are you measuring, and why do you think this has anything to do with measuring $O_A$? – ACuriousMind Feb 15 '16 at 0:33
• Maybe $O_A$ can be the charge of a two parallel plate capacitor (in an LC circuit) that has no side walls and the probe is an electron that passes between those plates. The states of the charge are just the possible charge values (1 coulomb, 2 coulomb, etc.) . By $O_A$ I was not referring to an operator. Also, I meant measuring the momentum of the probe, whose possible momentum values (states) are entangled with the charge states. – David Feb 15 '16 at 0:37
• The Gaussian distribution is part the specifics of the problem I am trying to solve. Maybe it is not critical to state. It is explaining how the expectation value is changing over time. In any case, if it is not important, then please ignore it. – David Feb 15 '16 at 0:44
• Or better yet, just refrain from commenting on this question. – David Feb 15 '16 at 0:49
• I don't know what the "spirit" of this question is supposed to be because you seem to misuse a lot of technical terminology. In particular, I don't know what you mean by "getting an indirect measurement" of $O_A(t_0)$ or $O_A(t_1)$. Let's say you have some entangled state $\chi = \sum_{i,j} c_{ij}\phi_i\otimes \psi_j$ for $\phi_i$ an eigenbasis for $O_A$ and $\psi_j$ some basis of $B$, and it evolves to $\chi' = \sum_{i,j} c'_{ij}\phi_i\otimes\psi_j$. You measure $O_A$, i.e. project this state onto one particular $\phi_k$. What's the question about the resultant state? – ACuriousMind Feb 15 '16 at 0:55

Two systems are entangled at time $t_0$, with state $\xi_0$. The state evolves according to the Schrodinger equation to state $\xi_1$ at time $t_1$. You perform a measurement that projects $\xi_1$ onto some subspace. You are asking, I think, whether the resulting projection somehow depends on $\xi_0$.
The answer is yes in the sense that you can always run the Schrodinger equation backward to recover $\xi_0$ from $\xi_1$, so that "dependence on $\xi_0$'' and "dependence on $\xi_1$'' are equivalent conditions.
On the other hand, if you're asking whether knowledge of $\xi_0$ tells you anything about the new state conditional on your knowledge of $\xi_1$, then the answer of course is no --- for two reasons. First, the state is defined to embody all properties of the system, so that knowledge of anything beyond the current state can't add anything. Second, even if that weren't true, the fact that $\xi_0$ and $\xi_1$ uniquely determine each other tells you that if you know one, there's no further information in the other.