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I understand that when I have two separate states that their combination state increases the Hilbert Space to $|\psi_1\rangle \otimes |\psi_2\rangle$

For example, looking at a simple example where we are considering two possible states, this can be expanded to: $(a|H_1\rangle+b|V_1\rangle)\otimes(c|H_2\rangle+d|V_2\rangle)$.

This can be then be written as $(ac|H_1\rangle |H_2\rangle + ad|H_1\rangle |V_2\rangle + bc|V_1\rangle |H_2\rangle + bd|V_1\rangle |V_2\rangle)\frac{1}{2}$

Now entanglement is defined as when we get something different than this. We have entanglement when the state can not be written as simply a Kroniker product of any superposition state of its component states ($\psi \neq |\psi_1\rangle \otimes |\psi_2\rangle$)

There are a number of different procedures for checking if a given state is entangeled, but how are entanglement states created in the first place?

I'm looking for examples of entanglement in which the mechanism that creates the entanglement is explicit.

The only example I can think of is Hong-Ou-Mendel interference creating NOON states like, $|2,0\rangle + |0, 2\rangle$. I get that generally identical possible outcomes can sometimes destructively interfere, but I'm looking for something a little bit more clear generally. In particular I'd like to build some intuition so that when I see am looking at given physical system I'll have an idea if entanglement could be generated.

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  • $\begingroup$ When a particle decays, you usually get quantum entanglement. As a super simple example when $A$ decays to two $B$'s, the momenta of the $B$'s are entangled, because they have to add up to the initial momentum of $A$. $\endgroup$ – knzhou Dec 26 '17 at 23:42
  • $\begingroup$ generally speaking, almost any global unitary evolution, acting on a separable state, will create an entangled state. This is to say that if you apply a random unitary evolution to a pair of modes, you will almost certainly get an entangled state. Practically speaking, you must be very careful to not make the two modes interact at all, otherwise you will almost certainly get some kind of entangled state. Of course, such "accidental entanglement" will also generally be unusable in practice, but that is another matter. $\endgroup$ – glS Dec 27 '17 at 0:06
  • $\begingroup$ If you connect to electrical resonators through a capacitor you get a $\sigma_x \otimes \sigma_x$ term in the Hamiltonian. That gives entanglement. $\endgroup$ – DanielSank Dec 27 '17 at 3:46
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Any procedure in a quantum system can be described by a unitary operator $U$ (quantum evolution) and/or a projection operator $P$ (measurement). If you want to bring two isolated subsystems in a state $|\psi_1\rangle\otimes|\psi_2\rangle$ into an entangled state $|\psi\rangle$ you need to ask what type of unitary operator $U$ and/or projection operator $P$ you should use such that: $$ P\left(U\left(|\psi_1\rangle\otimes|\psi_2\rangle\right)\right)=|\psi\rangle $$ As an example, imagine two $1/2-$spin systems in an initial state $|\uparrow\rangle \otimes|\uparrow\rangle$, doing the following procedures:

  1. A measurement of $\vec{S}_1\cdot \vec{S}_2=\frac{1}{2}\left[(\vec{S}_1+\vec{S}_2)^2 - \vec{S}_1^{2}-\vec{S}_2^{2}\right]=\frac{1}{2}(\vec{S}_1+\vec{S}_2)^2-\frac{3}{4}$.
  2. Or a evolution by an hamiltonian $H\propto\vec{S}_1\cdot \vec{S}_2$ by $\Delta t\neq T$, where $T$ is the period of precession.

you are going to get an entangled state.

More generally, any measurement of a global observable like $\vec{S}_1\cdot \vec{S}_2$ produces an entangled state.

For the $U$ operators, any hamiltonian that cannot be writen as $H\neq H_1\otimes1 + 1\otimes H_2$ will produce entangled states for times different than the period of oscillation, if there is any. This means that is sufficient to have an interaction between this two subsystems and avoid intervals of time $\Delta t=T$, where $T$ is some period of the system.

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  • $\begingroup$ I get the idea, and I'd like to work out the math explicitly to make sure I get it, but I'm stuck on something simple that I should know. Isn't $|1,1>$ both an eigenstate of $(S_1+S_2)^2$, $S_1^2$, and $S_2^2$? $S_1^2 |1, 1\rangle = (S_{1}+1)S_{1}|1,1\rangle = (1/2+1)1/2|1,1\rangle=3/4|1,1\rangle$? $S_2^2 |1, 1\rangle = (S_{2}+1)S_{2}|1,1\rangle = (1/2+1)1/2|1,1\rangle=3/4|1,1\rangle$? $(S_1+S_2)^2 |1, 1\rangle = (S_{tot}+1)S_{tot}|1,1\rangle = (1+1)1|1,1\rangle=2|1,1\rangle$? $\endgroup$ – Steven Sagona Dec 27 '17 at 1:42
  • $\begingroup$ @StevenSagona it is impossible to have an state that is eigenstate of both $(S_1+S_2)^2$ and $S_{z2} $ or $S_{z1}$, since they does not commute. The initial state that I gave is $|\uparrow\uparrow\rangle$, witch means $S_z=+1/2$ for both spins. This is not an eigenstate of $(S_1+S_2)^2$. To obtain the eigenstate of this operator try to learn how to sum spins in Quantum Mechanics. $\endgroup$ – Nogueira Dec 27 '17 at 1:54
  • $\begingroup$ Note that both $S_1^2$ and $S_2^2$ in these system are proportional to the identity operator, this means that there is not avaliable procedure in this system capable of change these values. $\endgroup$ – Nogueira Dec 27 '17 at 1:57
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One of the most popular methods to create entangled states in quantum optics is with the aid of spontaneous parametric down-conversion (SPDC). It is a nonlinear optical process where one photon (the pump photon) is converted into two photons (signal and idler). Conservation of momentum and energy implies that the two photons are entangled.

The nature of the entanglement is determined by the type of phase matching that is used. For type I phase matching one gets entanglement in the spatial degrees of freedom. With type II phase matching one also gets entanglement in the polarization degrees of freedom.

One way to look at the entanglement of a pure state is to express it as a Schmidt expansion $$ |\psi\rangle = \sum_n \lambda_n |\phi_n\rangle_s |\phi_n\rangle_i . $$ where $\lambda_n$ denotes the Schmidth coefficients and $|\phi_n\rangle_s$ and $|\phi_n\rangle_i$ are the Schmidt bases in the signal and idler systems, respectively. If the state is entangled, then there would be more than one nonzero $\lambda_n$. For collinear SPDC, the Schmidt bases are orbital angular momentum (OAM) eigenstates. This is a property that is often exploited in quantum optics, because it allows one to prepare entangled states in an infinite dimensional Hilbert space.

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  • $\begingroup$ I was more looking for the specifics for what exactly is the mechanism that's creating the entanglement. I've worked with these SPDC crystals before and I've always been unsatisfied with these blackbox depictions of entanglement generation. $\endgroup$ – Steven Sagona Dec 27 '17 at 23:02
  • $\begingroup$ The mechanism that produces the entanglement in this case is momentum conservation. It gives a relationship that the momenta of the two photons must obey, thus removing a degree of freedom and producing the correlation. $\endgroup$ – flippiefanus Dec 28 '17 at 3:18

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