This is a sequel of this question, and assumes similar notation.

In the previous question I essentially asked if the process of entanglement is just a formal idealization, in which there's some sort of sudden change

$(\sum_i a_i \psi_i,\varphi_0) \mapsto \sum_i a_i \psi_i\otimes\varphi_i$

or if it's realized by unitary evolution. The answer is that it is realized gradually by unitary evolution $U(t)((\sum_ia_i\psi_i)\otimes\varphi_0)$, occurring at all times in the Hilbert space $\mathcal H =\mathcal{H}_A\otimes\mathcal{H}_B$ of the joint system, and at a given time $t=T$ we finally have:

$U(T):\;\sum_ia_i\psi_i\otimes\varphi_0 \mapsto \sum_ia_i\psi_i\otimes\varphi_i$

where $U(T)$ is the time evolution at time $t=T$ corresponding to a Hamiltonian $\hat H\in\mathcal L(\mathcal H)$. Also, the resulting time-evolved states $U(t)((\sum_ia_i\psi_i)\otimes\varphi_0)$ will be non-entangled if and only if the Hamiltonian is of the form $\hat H = \hat H_A\otimes 1 + 1\otimes \hat H_B$.

Now I would like to have a concrete example of this unitary entanglement process.

More specifically, I would like to see a toy model comprising of the following:

  • Hilbert spaces $\mathcal H_A$, $\mathcal H_B$, and $\mathcal H =\mathcal H_A\otimes\mathcal H_B$,
  • distinct states $\psi_i\in \mathcal H_A$, $\varphi_0$ and $\varphi_i$ in $\mathcal H_B$, and complex numbers $a_i$,
  • an explicit Hamiltonian $\hat H\in \mathcal L(\mathcal H)$, of the form $\hat H = \hat H_A\otimes 1 + 1\otimes \hat H_B + H_{\mathrm{int}}\;,$

such that, for a time $T>0$, $U(T)((\sum_ia_i\psi_i)\otimes\varphi_0)=\sum_ia_i\psi_i\otimes\varphi_i$ where $U(t)$ is as above.

If possible, I would also like to have $\langle\psi_i\mid\psi_j\rangle=0$ for $i\neq j$, and $\langle\varphi_i\mid\varphi_j\rangle=0$ for $i\neq j$.

Also: is it possible to do the above literally, or some sort of approximation is needed? (for example the initial state not being literally separable but only approximately so?). Is it possible with the Hamiltonian being a bounded operator (so not having to use rigged Hilbert spaces, generalized eigenvectors and the like)?


1 Answer 1


Consider a two-qubit Hilbert space. Let the initial state be given as $|\psi(t=0)\rangle = |0\rangle|0\rangle $. We can evolve the system using the unitary operator

$U(t) = \exp(i\sigma_{x}\otimes\sigma_{x}\,t)\quad$ with $\quad H=\sigma_{x}\otimes\sigma_{x}$.

Now, it is easy to show: $U(t) = \cos(t) + i\sin(t)\sigma_{x}\otimes\sigma_{x}$. Then under time-evolution the final states is

$|\psi(t)\rangle = \cos(t)|0\rangle|0\rangle + i\sin(t)|1\rangle|1\rangle$

Therefore, we see that the system under time evolution keeps switiching from a separable state to entangled state and back. For example:

$|\psi(t=\pi/4)\rangle = \frac{1}{\sqrt{2}}(|0\rangle|0\rangle + i|1\rangle|1\rangle)$

Which is an entangled state.

  • 3
    $\begingroup$ Great answer. For perspective, this might be worth mentioning: For any given factorization of the Hilbert space, almost every state is entangled, and almost every Hamiltonian will evolve every initially unentangled state into an entangled state. $\endgroup$ Commented Nov 11, 2020 at 15:09
  • $\begingroup$ Yes, that seems to be true. $\endgroup$ Commented Nov 11, 2020 at 17:19
  • $\begingroup$ Thanks a lot. Would you also know how to write down an example in which two initially non interacting particles get close and become entangled? $\endgroup$
    – Qfwfq
    Commented Nov 11, 2020 at 20:57
  • $\begingroup$ @Chiral Anomaly: "almost every state is entangled" - probably the locus of decomposable tensors is a closed quadratic cone as it happens in finite dimensions (e.g. in this example). --- " almost every Hamiltonian will evolve every initially unentangled state into an entangled state" - this also sounds true cause the locus of hamiltonians of the form $A\otimes 1 + 1\otimes B$, I think, looks like a closed cone too. $\endgroup$
    – Qfwfq
    Commented Nov 11, 2020 at 22:33
  • $\begingroup$ @Qfwfq: You can work with the same example as above after slight modification. Two particles initially free and approaching each other can be described using time dependent Hamiltonian. For example, $H_{int}(t)=\theta(t)\theta(t_{0}-t) \sigma_{x}\otimes \sigma_{x} $ would mimic an interaction among two particles for a time interval $0$ to $t_{0}$. Now, if $t_{0}$ is chosen suitably, then the final state will be entangled. $\endgroup$ Commented Nov 12, 2020 at 4:13

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