# So what exactly is entanglement?

1. May 23, 2013

### daveyrocket

It seems to me that when we talk about entanglement, in a first-quantized kind of language we would say that the wavefunction is not separable into products of single particle wavefunctions, ie. $\psi(x_1,x_2) \neq \psi_1(x_1) \psi_2(x_2)$.

This would mean that any state with identical particles would be entangled... say for 1/2-spin Fermions in a singlet state, $\psi_{1,2} = \tfrac{1}{\sqrt(2)} \left[ \uparrow\downarrow - \downarrow \uparrow \right]$ is an entangled state. So for any state, you write out a Slater determinant and you get an entangled state, right? Is this a kind of "trivial" entanglement which is not very interesting, compared to the stuff you read about in pop science articles about quantum entanglement? Is there some way to take an entangled state and measure how interesting it will be?

2. May 23, 2013

### VantagePoint72

"Interesting" is subjective. The state you mention is hardly trivial though: it is the entangled state used in the Bell inequality. It can also be used for quantum teleportation, non-local measurements, and pretty much all the "interesting" things we do with entanglement! The non-separability definition you mention is the only one there is: for $|\psi\rangle$ in $H_1 \times H_2$, $|\psi\rangle$ is a product state if $\exists |\psi_1\rangle \in H_1$ and $\exists |\psi_2\rangle \in H_2$ such that $|\psi\rangle = |\psi_1\rangle \otimes |\psi_2\rangle$; otherwise, it is entangled. There's no further categorization of "trivially" entangled.

I don't know if this something like what you are looking for, but there is a notion of degrees of entanglement, which corresponds to how much entropy the individual components of the state have. The state you mentioned is known as "maximally entangled", since either of the two fermions (taken individually) has a state described by a reduced density matrix that is a multiple of the unit matrix. Thus, a spin measurement along any axis of one the fermions will yield spin up or spin down with 50% probability each. Up to an overall phase, there are four unique maximally entangled states of two two-level systems (qubits): they are the Bell states, and in my opinion they are the most interesting two-particle entangled states. They show up very often in quantum information theory.

3. May 23, 2013

### daveyrocket

Well, I know interesting is subjective, and I was being a little facetious there, but that's because I'm missing a way to describe entanglement that is not very interesting.

The thing that's bugging me is that any Slater determinant is automatically going to be a non-product state. Which means any system that you invent that has at least two electrons or any other set of identical particles is going to be entangled. So I could write a Hamiltonian like $H = -\frac{e^2}{r_1} + -\frac{e^2}{r_2-R}$, ie. two hydrogen atoms separated by a vector R which could be anywhere from the Bohr radius to the size of the galaxy. The solution is something like the state I wrote above. It doesn't matter what the separation is, as soon as I write down the Slater determinant for my wavefunction I get an entangled state.

But that's weird, because if R is taken as very large, I might as well approximate these as two separate hydrogen atoms. The history of these two atoms is irrelevant. It doesn't matter if they have always been at opposite edges of the galaxy, they're automatically entangled via the symmetrization condition of the wavefunction.

Of course, realistically if two hydrogen atoms are separated by a great distance, even if we stick to the H$_2$ molecule description, any excitation from the singlet state to a triplet state would cost virtually zero energy. So any perturbation that walks by one of the atoms would easily alter the wavefunction. Although, the antisymmetrization requirement would still give an entangled state. But this case, to me, seems like it could be called "maximally boring entanglement," since it's certainly not the type of entanglement that PRL's get written about.

Last edited: May 23, 2013
4. May 23, 2013

### Charles Wilson

"Weird" doesn't equal "False", however. If you push your math here, what does it tell you - as in, "You"?

From the imagined dialogue between Bell and Jauch in _Age of Entanglement_:

"Bell had a faraway look on his face and he was nodding slightly. "Terrible things happen in the Bohm theory," he said amusingly. "The trajectories assigned to the elementary particles instantaneously change when anyone moves a magnet anywhere in the universe." "

Well, so what?!??
Keep pushing, daveyrocket. Keep pushing.

CW

5. May 23, 2013

### daveyrocket

I'm not sure what you mean with that question. I thought I had described what it was telling me.

6. May 23, 2013

### Charles Wilson

I'm very intrigued with your analysis. Just a little encouragement. I think you're onto something, something that's the opposite of boring.
[[Edit: I'm removing some side issue speculations.]]
To me, your post is very good.

CW

Last edited: May 23, 2013
7. May 24, 2013

### Jano L.

In the calculations of $\psi$ functions for atoms and molecules, this is called by the boring term correlation - the electrons do not move independently. The electrons never get far from each other, so it is hard to check out these correlations directly. Perhaps that is why they are not so "interesting". They are however considered daily in most quantum-chemical calculations.

Perhaps the correlations of two state models are more "interesting" because they are mathematically simpler and can be used to model weak radiation and do some interesting correlation experiments with it.

8. May 24, 2013

### daveyrocket

I guess it sort of bothers me that entanglement is such a buzzword, but it's a basic property of any multi-particle system.