We consider two spin 1/2 systems that are described by the following Hamiltonian: $$ H^{12} = jZ^{1} \otimes Z^{2} \tag{1} $$

The composite system is initialised in the state: $|\psi^{12}(0)\rangle=|x_+,x_+\rangle,$ which written in the $z$ basis ($z_+$ being spin up state along the z dimension), is: $$|\psi^{12}(0)\rangle = 1/2(|z_+,z_+\rangle+|z_+,z_-\rangle+|z_-,z_+\rangle+|z_-,z_-\rangle).\tag{2} $$

Given the Hamiltonian in $(1)$, the unitary time evolution is $U(t)=e^{-iH^{12}t/\hbar},$ and after a time $t_0=\pi \hbar/4j,$ $U(t)$ becomes

$$ U(t_0)=e^{-i\frac{\pi}{4}Z^1\otimes Z^2} \tag{3} $$

Knowing $U$ at $t_0,$ the composite state evolved by this Hamiltonian upto $t_0$ becomes:

\begin{align*} |\psi^{12}(t_0)\rangle =& U(t_0)|\psi^{12}(0)\rangle \\ =& 1/2(e^{-i\pi/4}|z_+,z_+\rangle+e^{i\pi/4}|z_+,z_-\rangle+e^{i\pi/4}|z_-,z_+e^{-i\pi/4}\rangle+e^{-i\pi/4}|z_-,z_-\rangle) \tag{4} \end{align*}

So it seems that just by evolving the separable initial state $(2)$ that was not an eigenstate, for a certain amount of time, the composite system suddenly becomes entangled $(4).$

  • Once the system has become entangled, will further evolving the system according to $U(t)$ dis-entangle the system again at some point in time? I.e. do we expect regular transitions between separable to entangled states of the composite system? Or, once the system has become entangled, it maintains its entangled state?
  • If the change between separable and entangled is expected to occur during the time evolution of this system, then it means the involved subsystems are constantly undergoing transitions from pure (when in separable state) to mixed states (when the composite state is entangled), is such behaviour physically allowed?
  • $\begingroup$ I find it rather misleading to say that the system "suddenly" becomes entangled. That is an artificial distinction and it is caused by having a measure for entanglement that is discontinuous with respect to the state, and which selects out a measure-zero set as separable states. (And, ditto for the rank of the reduced density matrices). A much better guide is plotting the purity and eigenvalues of the reduced density matrices as functions of $t$ - you'll see them oscillate smoothly. $\endgroup$ Nov 28, 2017 at 13:28
  • $\begingroup$ More generally, it always pays to keep in mind that the real world always has noise and that you're never really in a completely pure state, i.e. there is always some 'fuzz' about your state, or in other words your density matrix always has some minor contributions from residual noise and it never has rank 1. Any result or tool that strictly requires a completely pure state (which does occasionally happen) should be treated with extreme care when interfacing with the real world. $\endgroup$ Nov 28, 2017 at 13:31
  • $\begingroup$ @EmilioPisanty Thanks a lot. If I may ask: a) is purity here the off-diagonal elements of the density matrix? Additionally, how do we physically interpret the eigenvalues and eigenvectors of the density matrix in general? b) Slightly unrelated, physically, what are the eigenvectors of the unitary time evolution operator? are they the same as the energy (thus Hamiltonian) eigenstates? Thanks again $\endgroup$
    – user929304
    Nov 29, 2017 at 17:00
  • $\begingroup$ Purity refers strictly to $\mathrm{Tr}(\rho^2)$, which is $1$ for pure states and strictly smaller (though bounded below by $1/d$ for a $d$-dimensional system) for mixed states. The rest you should ask separately. $\endgroup$ Nov 29, 2017 at 18:05

3 Answers 3


Following up on ACuriousMind's answer: it's certainly possible for entangled states to evolve into product states, but it's unlikely for a generic process to do this very often, because "almost all" states in the Hilbert space are entangled, and only a vanishingly tiny fraction are product states.

You can see this just from counting degrees of freedom. For simplicity, assume the composite system state is pure (as in your example). If the two subsystem Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$ have dimensions $m$ and $n$ respectively, then a generic pure state of the total system takes the form

$$| \psi⟩ = \sum_{i=1}^m \sum_{j=1}^n \psi_{ij}\ |i⟩_A⊗|j⟩_B.$$

This state has $m \times n$ degrees of freedom (the amplitudes $\psi_{ij}$), and so lives in an $(m \times n)$-dimensional Hilbert space.

On the other hand, a generic product state takes the form $$|\psi⟩ = |\psi^A\rangle \otimes |\psi^B\rangle = \left( \sum_{i=1}^m \psi^A_i |i\rangle_A \right) \otimes \left( \sum_{i=1}^m \psi^B_j |j\rangle_B \right) = \sum_{i=1}^m \sum_{j=1}^n \psi^A_i \psi^B_j\ |i⟩_A⊗|j⟩_B,$$ where in the last step I simply distributed over the tensor product. The difference is that in the special case of a product state, the $m \times n$ generic components $\psi_{ij}$ must "factorize" into a product of a set of $m$ components $\psi^A_i$ and a set of $n$ components $\psi^B_j$. (If you think of the amplitudes $\psi_{ij}$ as forming an $m \times n$ matrix, then for a product state the matrix is rank-1 - i.e., the outer product of a vector $\psi^A_i$ and a vector $\psi^B_j$.) So you only need to specify the $m$ numbers $\psi^A_i$ and the $n$ numbers $\psi^B_j$ in order to completely specify the entire state, so there are $n+m$ degrees of freedom.

The set of product states forms an $(m+n)$-dimensional subset of the full $mn$ dimensional Hilbert space (subset but not subspace, because it isn't closed under addition). So it is measure zero if $m + n < mn$, which is always the case unless $m=n=2$ (because the dimensionality of a nontrivial Hilbert space must be $\geq 2$, because a 1D Hilbert space is trivial and only contains one physical state with no degrees of freedom).

There are several intuitive ways to see why a product state only has $m+n$ degrees of freedom. The basically idea is that the two subsystems of a product state are completely independent and uncorrelated, so you need only describe "what subsystem A is doing by itself" ($m$ pieces of information/degrees of freedom) and separately "what subsystem B is doing by itself" ($n$ pieces of information), without needing to specify "what they're doing together". It's like you took the two vector vectors $\psi^A$ and $\psi^B$ and combined them together in the simplest possible way, by just stacking them on top of each other to get a longer vector (although this is not actually what's happening mathematically - that would be the direct sum $\oplus$, not the direct product $\otimes$).


In general, there's nothing we "expect" about how often transitions between separable and entangled states happen, except that if the Hamiltonian is non-interacting, i.e. of the form $H = H_1\otimes\mathbf{1}_2 + \mathbf{1}_1\otimes H_2$, then separable states will never become entangled. However, the set of separable states has zero measure in the total space of states regardless of what measure you pick, so we probably shouldn't expect entangled states to become separable again unless the time evolution is cyclic (which it is in your case).

As for whether transitions between pure and mixed states are "allowed", this is asking the wrong question: Since your systems are interacting, you cannot meaningfully talk about the evolution of one of the systems without talking about the other, that is, it doesn't really make sense to claim that the subsystems are undergoing transitions. Sure, with ordinary unitary time evolution such transitions are forbidden, but when you try to restrict the evolution operator to one of the subsystems, you don't get something unitary - your subsystem is open, and none of the usual assumptions hold. Just keep looking at the full system, it's much easier.

  • 1
    $\begingroup$ Thank you very much for the prompt answer. This is exactly along the lines I was looking for, your arguments make perfect sense. I just feel that your point at the end of the first paragraph, about the zero measure of separable states and cyclic evolution, are very crucial in fully understanding the situation, but a priori I don't fully understand them. It would be extremely neat if you could expand on these two points a bit further, such that a QM amateur can see a bit more clearly either mathematically or physically, why the separable states have zero measure in Hilbert space. $\endgroup$
    – user929304
    Nov 27, 2017 at 14:29
  • $\begingroup$ @tparker thank you very much! I see the reasoning now. If I may, from a technical point of view: a) I don't understand the difference between the two expressions of $|\psi\rangle$ that you've written. How s the second expression capturing only the product states? b) additionally, is there an intuitive way of seeing why the product state subspace only has dimensions n+m? $\endgroup$
    – user929304
    Nov 27, 2017 at 23:56
  • $\begingroup$ @user929304 I expanded my comments into an answer that addresses your follow-up questions. $\endgroup$
    – tparker
    Nov 28, 2017 at 1:28

A) Yes, it is possible to evolve the entangled state back into the unentangled state, this is referred to as a quantum revival and happens to any quantum system. This is also the reason that quantum chaos is significantly different the classical chaos. We can see this by looking at the eigenvalues of $Z^{1}\otimes Z^{2}$ which are $\pm 1$. Therefore if we set the time $t$ to $\frac{jt}{\hbar} = 2\pi$ then the evolution operator in question is the identity.

B) This behaviour is not a mathematical artifact and is physically allowed, having been observed experimentally. However in most physical systems this won't be perfect because of coupling to the environment causing the revival strength to decay over time.


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