4
$\begingroup$

[Warning: I'm not a physicist]

Let $A$ and $B$ be microscopic systems, with corresponding Hilbert spaces of state vectors given by $H_A$ and $H_B$ respectively.

Let's say $A$ is in a state $\psi\in H_A$ and $B$ is in a state $\phi\in H_B$, possibly both evolving unitarily in time. When the two systems "interact", a new Hilbert space is formed, for the combined system, which is the (completed) tensor product $H=H_A\otimes H_B$. Likewise, the new combined state becomes $\Psi=\sum_i\psi_i\otimes \phi_i \in H_A\otimes H_B$. Well, or maybe just $\psi\otimes \phi\in H_A\otimes H_B$, but this is a separable state, right? so I don't know why one would say the two systems "get entangled"! In case it's not just $\psi\otimes\phi$, I'll ask:

Q1. Is $\Psi$ determined by the pair $(\psi,\phi)$ or one would need to also know the details of how the two systems interact?

Anyway, it seems to me that the above is just a mathematical idealization: before the "interaction", we have a pair of Hilbert spaces with given state vectors, and just after the "interaction" the Hilbert spaces have magically changed into a tensor product, and likewise the state vectors. This reminds me of the magical collapse of the wave function that is supposed to happen upon measurement according to the Copenhagen interpretation.

But, if quantum mechanics has to hold globally (say for the whole universe, or at least the combined system) and at all times, then the whole "interaction" giving rise to entanglement (or, at least, tensor products) has to take place within a larger Hilbert space $H$ (maybe $H_A\otimes H_B$ is sufficient for this purpose) and according to unitary evolution $\dot{\Psi}(t)=-\frac{\mathrm{i}}{\hbar} \hat{H}\Psi(t)$ where $\Psi(t)\in H$ is a state vector (varying with time). Independently from any interpretation of QM, I would assume the evolution, in $H$, is unitary because no measurement is performed on the combined system "$A+B$" from the outside (or is this wrong cause the two systems somehow "measure each other"?). I imagine the Hamiltonian $\hat{H}$ should depend on the Hamiltonians $\hat{H}_A\in \mathcal{L}(H_A)$ and $\hat{H}_B\in \mathcal{L}(H_B)$ of the two systems, and on how the two systems are supposed to interact.

Q.2 Is there a theory that describes the evolution of $\Psi(t)\in H$ such that $\Psi(0)$ somehow "corresponds" to the datum of noninteracting $\psi$ and $\phi$, and $\Psi(\infty)$ corresponds to $\sum_i\psi_i\otimes \phi_i$ (or $\Psi(T)$ corresponds approximately to $\sum_i\psi_i\otimes\phi_i$ for $T$ some sufficiently large time)? How does the encoding $(\psi,\phi)\mapsto\Psi$ work?

Also, in the case $B$ is a macroscopic measuring device, does the above question have something to do with the Measurement Problem? (If yes, since the Problem is allegedly unsolved, then I guess I'm not expecting a definitive answer but just a justification for the link)

Maybe, the encoding is just $(\psi(t),\phi(t))\mapsto\psi(t)\otimes\phi(t)$ at all times. But then one would need a definition of what it means that somehow, at $t=0$, $\psi(0)\otimes\phi(0)$ describes two separate systems while, at $t=T$, $\psi(T)\otimes\phi(T)$ describes a state of the combined system. Maybe, a "measure of combinedness" $\mu$ such that $\mu(\psi(0)\otimes\phi(0),H_A\otimes H_B)=0$ and it is $>0$ at $t=T$?

Another guess: maybe, the "combinedness" of the systems depends on the choice of an observable $X$, and we have $X(t)=a(t)\otimes 1+1\otimes b(t)+K(t)$ where $X(t)=e^{-\mathrm{i}\hat{H}t/\hbar} X(0)e^{\mathrm{i}\hat{H}t/\hbar }$ etc., and $a$ and $b$ are observables on $A$ and $B$ respectively while $K$ is an observable of the combined system, and $K(0)=0$ (or $||K(0)||\ll 1$).

$\endgroup$
3
$\begingroup$

Imagine a system which can be thought of has having two parts $A$ and $B$, corresponding to Hilbert spaces $\mathcal H_A$ and $\mathcal H_B$. The Hilbert space of the whole system is $\mathcal H_A \otimes \mathcal H_B$, as you say.

If the Hamiltonian for the entire system can be written

$$\hat H = \hat H_A \otimes \mathbf 1 + \mathbf 1 \otimes \hat H_B$$

then the time-evolution operator for the system is (setting $\hbar =1$)

$$\hat U(t) = \exp(-it\hat H) = \exp(-it\hat H_A)\otimes \exp(-it\hat H_B)\equiv \hat U_A(t)\otimes \hat U_B(t)$$ and so the systems evolve unitarily and independently from one another. Separable states evolve to separable states, and the two parts of the system never magically become entangled with one another.

The same is not true if the Hamiltonian has an interaction term. In this case, the (unitary) time evolution of the full system cannot simply be viewed as unitary time evolution of each part separately. A separable state will generally evolve into a non-separable state, and it is in this way that entanglement occurs.


Anyway, it seems to me that the above is just a mathematical idealization: before the "interaction", we have a pair of Hilbert spaces with given state vectors, and just after the "interaction" the Hilbert spaces have magically changed into a tensor product, and likewise the state vectors.

No. We always have the tensor product space as the Hilbert space for the whole system; it's just that if the Hamiltonian can be neatly decomposed as above, separable states evolve into separable states, so you can treat the evolution as occurring independently in the constituent Hilbert spaces.

In practice, what happens is that the interaction part of the full Hamiltonian is initially irrelevant - perhaps the systems are well-separated in space, so the interaction energy is negligible. As the system evolves (e.g. the particles get closer together), the interaction term becomes relevant, which allows evolution from a separable state to a non-separable (entangled) state. Even if the interaction term becomes irrelevant again, the final state remains entangled.

Also, in the case B is a macroscopic measuring device, does the above question have something to do with the Measurement Problem?

Yes. The evolution of $\mathcal H$ is unitary, but if we insist only on keeping track of $A$ or $B$ (by taking a partial trace over the undesirable degrees of freedom), we will find that the evolution is not unitary. A suggested resolution to the measurement problem is that apparently projective evolution only occurs because we are "tracing over" the state of the measurement apparatus. See e.g. here or here.


By the way, how do you get the identity $\exp(a\otimes 1 + 1 \otimes b) = \exp(a)\otimes \exp(b)$? Is it by BCH?

If $\Psi = \psi\otimes \phi$ and $U(t)\Psi = (\hat U_A(t)\psi)\otimes(\hat U_B(t)\phi)$, then to first order in $\epsilon$ we would have that $$U(\epsilon)\Psi \approx \Psi - i\epsilon \hat H\Psi = \psi\otimes \phi - i\epsilon\left((\hat H_A\psi)\otimes \phi + \psi\otimes(\hat H_B\phi)\right)$$

which implies that

$$\hat H = \hat H_A \otimes \mathbf 1 + \mathbf 1 \otimes \hat H_B$$

The reverse implication follows (as you suggest) immediately from BCH, noting that $\hat H_A \otimes \mathbf 1$ and $\mathbf 1 \otimes \hat H_B$ commute, so

$$U(t)=\exp\big(-it(\hat H_A\otimes \mathbf 1+\mathbf 1 \otimes \hat H_B)\big)$$ $$=\exp\big(-it\hat H_A\otimes \mathbf 1\big)\exp\big(-it\mathbf 1 \otimes \hat H_B\big) $$ $$= \big[\exp\big(-it\hat H_A\big)\otimes \mathbf 1\big]\big[\mathbf 1 \otimes \exp\big(-it\hat H_B\big)\big]$$ $$ = \exp\big(-it\hat H_A\big)\otimes \exp\big(-it\hat H_B\big)$$

The conclusion is that time evolution will occur for the two systems independently iff the Hamiltonian of the full system decomposes into the nice sum found above, which makes good physical sense. If you want systems to interact (which includes the possibility of becoming entangled), then the full Hamiltonian needs to include an interaction term.

$\endgroup$
5
  • $\begingroup$ Very clear, thank you. So my last guess was the correct one with $X=$ the joint Hamiltonian. $\endgroup$ – Qfwfq Oct 4 '20 at 13:45
  • $\begingroup$ When you say by tracing out $B$ the evolution is not unitary, do you mean that the time-dependent density matrix $\mathrm{tr}_B(U(t)^{-1}\rho_\Psi U(t))$ (where $\rho_\Psi = \mid \Psi\rangle\langle\Psi\mid$) is not of the form $U_A(t)\rho_A U_A(t)^{-1}$ for $U_A(t)$ the unitary time evolution in $A$ and $\rho_A$ some density matrix in $A$? $\endgroup$ – Qfwfq Oct 4 '20 at 13:45
  • 1
    $\begingroup$ @Qfwfq If you start with a density matrix $\rho$ for the full system and trace out system $B$, you're left with a density matrix $\rho_A$ on $\mathcal H_A$ which we associate with the state of system $A$. The point I was making is that $\rho(t)$ and $\rho(t')$ are related by a unitary transformation, but $\rho_A(t)$ and $\rho_A(t')$ generally are not. If a state is not separable, then a partial trace over $B$ yields a so-called mixed state of $A$, which cannot be written as $\rho_A = |\psi\rangle\langle\psi|$ for any $\psi\in \mathcal H_A$. $\endgroup$ – J. Murray Oct 4 '20 at 23:57
  • $\begingroup$ By the way, how do you get the identity $\exp(a\otimes 1 + 1\otimes b)=\exp(a)\otimes \exp(b)$? Is it by BCH? $\endgroup$ – Qfwfq Oct 5 '20 at 0:05
  • 1
    $\begingroup$ @Qfwfq I've added a section to the end of my answer to address that point. $\endgroup$ – J. Murray Oct 5 '20 at 1:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.