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Most of the time entanglement is described using some instantaneous classical measurement from an apparatus, but it is possible to describe it between two quantum systems, with two systems $O$ (for the observer) and $S$ undergoing the following process

$$\vert \psi^O \rangle \otimes\vert \psi^S \rangle \to \sum_\alpha a_\alpha \vert \psi^O_\alpha \rangle \otimes \vert \phi_\alpha \rangle$$

for a bunch of measurement with value $\alpha$ and probability $a_\alpha^2$.

But is there a toy model to show this process actually happening, where two particles, originally just a tensor product of the free states, evolve into entangled states, ideally without using some instantaneous interaction like a quantum gate? Also does this conflict with the fact that for long range interactions, particles are always interacting with each other?

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  • $\begingroup$ What do you mean by "described"? I can parse your first sentence only if I replace "describe" by something like "create". Am I missing something? $\endgroup$
    – Stéphane Rollandin
    Commented Dec 22, 2017 at 9:53
  • $\begingroup$ That this is what the wavefunction of an observer and measured system are pre and post measurement from the point of view of a second observer. $\endgroup$
    – Slereah
    Commented Dec 22, 2017 at 9:54
  • $\begingroup$ a "toy model" of QM entanglement is the pilot wave hydrodynamics model, or modelling particles as simple harmonic oscillators of space waves. it is not entirely incorrect, serves as a rough classical approximation of much qm phenomena, has some remarkable emergent properties/ consequences and how far it can be pushed is open to question/ active research area. more discussion chat.stackexchange.com/rooms/9446/theory-salon $\endgroup$
    – vzn
    Commented Dec 22, 2017 at 17:10
  • $\begingroup$ What do you mean by "for long range interactions, particles are always interacting with each other"? Are you referring to the entangled particles in a Bell or quantum teleportation experiment? $\endgroup$
    – plan
    Commented Dec 22, 2017 at 18:36

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Assume that $\left| \psi^O \right. \rangle$, $\left| \psi^S \right. \rangle$ are just two spin 1/2 particles. Then a Heisenberg interaction:

$$\vec{S^o} \cdot \vec{S^s} = \hat{S_x^o} \otimes \hat{S_x^s} + \hat{S_y^o} \otimes \hat{S_y^s} + \hat{S_z^o} \otimes \hat{S_z^s}$$

will evolve a product state into an entangled singlet state:

$$\frac{1}{\sqrt{2}} \left( \left| \uparrow^O \right. \rangle \left| \downarrow^S \right. \rangle - \left| \downarrow^O \right. \rangle \left| \uparrow^S \right. \rangle \right)$$

It is possible to generalize this to a more realistic "observer" and "system". A simple way of playing around with this is to numerically generate a random hermitian $(NM \times NM$) matrix, where $N = dim(system)$, $M=dim(observer)$. Diagonalize the matrix and find the lowest energy eigenvector. Reshape this $NM$-vector into a $(N,M)$ matrix and do a singular value decomposition of it. The entanglement entropy can then be found from the singular values $\Lambda_i$ as

$$S = \sum_i -\Lambda^2_i \log(\Lambda^2_i)$$

You will find that most Hamiltonians have an entangled ground state. This would mean that your initial product state (with sufficient overlap) would also evolve to such an entangled state.

More realistic would be to include the effect of an environment by using the theory of open quantum systems. In "most cases" the environment destroys the entanglement, which is an explanation why quantum entanglement is so fragile. See https://arxiv.org/pdf/quant-ph/0306072.pdf for more on the subject.

Maybe this at least answers the first part of your question.

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