I Looking for straight explanation of entanglement

1. Aug 3, 2016

KFC

Hi all,
I start to read some popular science on quantum physics. The term "entanglement" appears everywhere in quantum communication and other related fields. I am trying to understand this concept based on my undergraduate knowledge on physics. From some online resource, I have a feeling that an entangled system is a system of two or more particles under some interaction. An typical form of an entangled system I seen everywhere is something like

$|0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle$

To my level, it looks like a system of a superposition of something. However, there is a weird math symbol $\otimes$ that confusing me. Could anyone explain in plain english what's above expression really stands for. And what's the difference between an entangle state and superposed state? Thanks.

2. Aug 3, 2016

Lucas SV

You are on the right track, and yes the example you gave is an entangled state.

Take two quantum systems, each has a hilbert space, and you may define the composite system as the tensor product $\mathcal{H}_1 \otimes \mathcal{H}_2$ of the two hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$. An entangled state is defined as any state of the composite system which cannot be writen as a direct product of a state from each system. If a state remains a product state at all times, the two states of the product, are essentially evolving independently, that is the particles are not interacting.

A simple example is the 2 quibit system, which you were considering. Each state can be writen as a suporposition of the basis vectors $|i\rangle \otimes |j\rangle$, with the $i,j$ running from 0 to 1. The ones which are not entangled, can be writen as $a^i b^j|i\rangle \otimes |j\rangle$. So as an exercise you may attempt to prove your example state is entangled, by first assuming it is not, writting it as a product state, and getting a contradiction.

Time evolution, as you know, is determined by the Hamiltonian. An interaction term may be added to the sum of the hamiltonians of each subsystem. Even if one prepares the system in a product state initially, the interaction hamiltonian will be responsable for coupling the two subsystems, and the state will evolve into an entangled state. In this way interactions can induce entanglement.

Last edited: Aug 3, 2016
3. Aug 3, 2016

Lucas SV

It is a superposition state, yes. One way to think about it physically is to consider two electrons. We take ket $|0\rangle_1$ to mean the state in which the first electron has spin $-\frac{1}{2}$ (i.e. if we measure the spin of the electron in state $|0\rangle_1$ we are 100% certain to get $-\frac{1}{2}$). We also take the ket $|1\rangle_1$ to mean the state in which the first electron has spin $+\frac{1}{2}$. We define $|0\rangle_2$ and $|1\rangle_2$, similarly for the second electron.

When we combine the two systems (electron 1 and electron 2), we can define the state $|0\rangle_1 \otimes |0\rangle_2$ to mean that when you measure the spin of both electrons simultaneously, you are guaranteed to find both electrons with spin $-\frac{1}{2}$ each. You can define any other $|i\rangle_1 \otimes |j\rangle_2$ in a similar way, where $i=0,1$ and $j=0,1$. Now, in order to simplify notation, we simply drop the subscripts, with the understanding that the first ket represents the first electron and the second ket represents the second electron, and we get $|i\rangle \otimes |j\rangle$. Sometimes even the tensor product is dropped, the kets are combined and we write $|i j\rangle$.

Now let us consider the superposition state $\psi=\frac{1}{\sqrt[]{2}}\left(|0\rangle_1 \otimes |0\rangle_2+|1\rangle_1 \otimes |1\rangle_2\right)$. Suppose you prepare the system in state $\psi$ and measure the spin of both electrons simultaniously. The spin of the first electron will be either $-\frac{1}{2}$ or $+\frac{1}{2}$, with a 50% chance each. Here is where you see physical effects of entanglement: if the spin of electron 1 has been found to be $-\frac{1}{2}$, it is guaranteed that the spin of electron 2 is found to be $-\frac{1}{2}$ also. If instead you find $+\frac{1}{2}$ for electron 1, you will find $+\frac{1}{2}$ for electron 2 also.

Now I cheated a little bit by considering the 2-electron system. Since the electron is a fermion, the state $\psi$ is not possible, by Pauli Exclusion Principle. But I just wanted to illustrate what the tensor product notation means. Suskind goes over entanglement of simple systems in his lecture series found on youtube

He will teach you the minimum math necessary to understand it, as you go along the lectures. Hope this helps.

Last edited: Aug 3, 2016
4. Aug 4, 2016

KFC

I appreciate your explanation and the link. Very good video. But I have one quick question. Does entanglement only happen in a system of multiple particles? In some case like creating a superposition state of two momentum state of ONE particle (I remember I saw some paper on BEC to create a state of superposition of two momentum). Can I say the superposed state of two momenta is an entangled state?

5. Aug 4, 2016

Lucas SV

Well, as far as i know, to make an entangled state you need a composite system, but it should be clarified what a composite system means. You could consider a particle in one dimension (without spin). The effects of entangled are obtained when you find subsystems of your system. But if the particle is elementary, there is no subsystem. If you happen to find entanglement, what it really means is that the particle was not elementary in the first place. Or maybe it means you have found a new quantum number (analogous to spin).

There is a caveat. I said the particle is one dimension. If instead the particle is in three dimensions, it can be decomposed into 3 subsystems (roughly the directions of space). the position of the particle is uncertain with probabilities determined by the wavefunction. That is because the state of the particle will be in a superposition of position states (a position state has a spike as a wavefunction, technically the dirac delta function). If you measure the $z$ coordinate of the position to be 0, you can determine probabilities of the other coordinates by considering the wavefunction restricted to the $xy$ plane. This is similar to the discussion on spin in which a measurement in a subsystem gives information about measurements in the other subsystems. Maybe you can call this coordinate entanglement.

Not sure how this applies to BEC. It is a system of many bosons, however as in many systems in condensed matter physics, you can describe the whole system by a few quasi-particles. These behave like quantum mechanical particles but are not physical particles, but usually a collective excitation of many physical particles in the system. Maybe they are talking about entanglement of quasi-particles. But again I only know some basics of condensed matter physics, so don't quote me on this.

Last edited: Aug 4, 2016
6. Aug 4, 2016

Staff: Mentor

You might want to give Giancarlo Ghirardi's "Sneaking a look at God's cards" a try. It's as easy to read and way more complete/accurate than the popular stuff out there.

7. Aug 4, 2016

Staff: Mentor

It does not have to involve a two-particle system.

Backing up a bit....
Any time that you have a quantum system with two commuting observables, there are states in which both observables have definite values (more formally, these are states that are eigenstates of both observables). For example, if our two observables are $A$ and $B$, it may be possible to prepare the system in a state such that $A$ has the value $\alpha_1$ and $B$ has the value $\beta_1$; we'll write that state as $|\alpha_1\beta_1\rangle$. There might be another state in which the two observables have the values $\alpha_2$ and $\beta_2$, and we'd write that one as $|\alpha_2\beta_2\rangle$.

Yet another possible state would be a superposition of these first two, which we might write as $|\alpha_1\beta_1\rangle+|\alpha_2\beta_2\rangle$. This is an entangled state; if the system is in this state, any measurement of either $A$ or $B$ will collapse the wave function into one of the two possibilities and both $A$ and $B$ will take on definite values. $A$ and $B$ can be pretty much any observable properties of the quantum system; all that matters is that they commute so that they'll have common eigenstates.

Now, if observable $A$ is "The spin measured at the left-hand detector" and observable $B$ is "The spin measured at the right-hand detector", we have the ubiquitous textbook example of an entangled two-particle system. But we didn't need a two-particle system - it's just that when $A$ and $B$ are measured at spatially separated locations we get an especially striking example of how the entanglement works. If $A$ and $B$ were the energy and momentum of a single particle it wouldn't surprise anyone that a measurement of either could determine the value of both, so that example wouldn't be any fun.

8. Aug 11, 2016

KFC

This is quite interesting. I never think it that way. Base on your statement, I read back to a text and my understanding is as follows: if I have two observables $A$ and $B$, if $[A,B]=0$, if I solve the system to find the eigenstates, I could use the eigenstates to expand either $A$ or $B$, is that right? I am not quite sure I understand it correctly but from this point, it seems that $A$ and $B$ are basically the same thing or may be just different with a constant multiplier? So when you write $|\alpha_1\beta_1\rangle$, can I say that value $\alpha_1$ and $\beta_1$ are independent such that when you measure $A$, $B$ is not affected? So the notation $|\alpha_1\beta_1\rangle$ indicate some commuting relation there?

Ok ... so it means when you measure one of the observable, only $|\alpha_1\beta_1\rangle$ OR $|\alpha_2\beta_2\rangle$ will be returned but NOT BOTH! However, once $A$ or $B$ was measured, we will either get $\alpha$ or $\beta$ (depending on which observable we measure), and once we figure our the value of $A$ (or $B$) when measuring $A$ (or $B$), the value of $B$ (or $A$) will be known immediately because they are part of the state $|\alpha_{1/2}\beta_{1/2}\rangle$. Am I understood it correctly?

So can I conclude that a entangled state must be a composed system but either the states from two (or more) different systems or two (or more) states from the same system. The key point is in either case, the variables involved must be mutual commutable?

9. Aug 12, 2016

vanhees71

It's better to talk about entanglement between observables. You can also have entanglement of two observables of just one particle. An example is the Stern-Gerlach experiment. You run an electrically neutral particle with a magnetic moment (e.g., a neutron or an atom; in the original experiment it was a beam of silver atoms) through an inhomogeneous magnetic field. The particles are deflected according to its spin-magnetic quantum number, $m$, i.e., now the position of the particle is entangled with the value of the corresponding spin component (usually one chooses the $z$ direction).

Here (in the nonrelativistic treatment) the spin component $\hat{s}_z$ and position operators $\hat{\vec{x}}$ commute, and thus there are common eigenvectors of these observables, $|\vec{x},s_z \rangle$. To describe the state after the particles run through the magnetic field, you have to introduce wave packete $|\phi_{s_z} \rangle$, which are quite sharply peaked around the positions of the partial beams of particles depending on the value of $s_z$. Then the state is given by
$$|\psi \rangle=\sum_{s_z=-s}^{s} |\phi_{s_z},s_z \rangle.$$

You can also have entanglement for composite systems. Often one uses two-photon states, whose polarizations are entangled. The two photons may be detected at very large distances of the detectors, but as long as the photons are not disturbed by anything and just propagating freely the entanglement stays intact. Often you have a singlet state
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|H V\rangle-|VH \rangle).$$
Here I've left out the momentum part of the photon states, which have to be also antisymmetric under exchange of the photons, so that the overall state gets symmetric, as it must be for photons.

The funny thing with this type of entanglement is that, although the two photons may be detected at far distant places, their polarization states are strictly correlated, i.e., if observer A measures $H$ for her photons, observer B must find $V$ for his and vice versa. Nevertheless, the single-photon polarization states, given by the reduction of the state
$$\hat{\rho}_1=\mathrm{Tr}_2 (|\Psi \rangle \langle \Psi |)=\frac{1}{2} \hat{1}=\frac{1}{2} (|H \rangle \langle H|+|V \rangle \langle V |),$$
i.e., the single photon are completely unpolarized, i.e., you find it to be H-polarized or V-polarized with probability 50%, respectively.

10. Aug 12, 2016

Staff: Mentor

If A and B commute I can find a set of states that form a complete set of eigenstates (an "eigenbasis") of both. Note that this is not the same as saying that every eigenstate of A will be an eigenstate of B, or vice versa.

A and B are not "the same thing". For an example, we could take A to be the energy of a bound electron (more often written as H) and B to be its spin (more often written as $S_z$). The energy eigenstates are solutions of $H|\psi\rangle=E|\psi\rangle$; this equation has an infinite number of solutions, one for each of the quantized energy levels of the system, which we can call $E_n$. The spin eigenstates are solutions of $S_z|\psi\rangle=s|\psi\rangle$; $s$ can take on only the values $\pm\frac{1}{2}$. Because $H$ and $S_z$ commute, they have a set of common eigenfunctions: for every $n$ there are two states that are eigenfunctions of $H$ with eigenvalue $E_n$, one of them being also an eigenfunction of $S_z$ with eigenvalue $\frac{1}{2}$ and the other being an eigenfunction of $S_z$ with eigenvalue $-\frac{1}{2}$. Just as a matter of notation, it is convenient to write these as $|n+\rangle$ and $|n-\rangle$.

(Note that $|3+\rangle+|3-\rangle$ is an eigenvector of $H$ with eigenvalue $E_3$, but it is not an eigenvector of $S_z$).

Last edited: Aug 12, 2016
11. Aug 15, 2016

woody stanford

Thank you for this explaination. It was in plain english and helped a lot.