# Can very distant particles be entangled?

1. Feb 19, 2016

### naima

Entangled particles give no interference. What happens in the Fraunhofer approximation when the source of entangled pairs is far away?
If it depends on distance what about the apparent collapse?

2. Feb 19, 2016

### vanhees71

As long as nothing disturbes the entangled particles, entanglement stays intact. This is clear from the fact that the time evolution of a closed system is a unitary (linear) transformation of the state (in the Schrödinger picture).

3. Feb 19, 2016

### naima

far distant source of incoherent light can give interferences. The lack of coherence is the usual explanation for the lack of interferences. In the Fraunhofer approximation interferences reappear. Isn't it as if we were in the focal plane of an Heisenberg lens where which path is erased?

Last edited: Feb 19, 2016
4. Feb 19, 2016

### DrChinese

Great question. And perhaps if you did coincidence counting you would be able to see that interference pattern.

5. Feb 19, 2016

### vanhees71

I'm not sure what you are referring too. Do you mean the Hanbury-Brown-Twiss effect? Then have a first look here

https://en.wikipedia.org/wiki/Hanbury_Brown_and_Twiss_effect

Of course, the article is a bit sloppy in invoking old-fashioned ideas about "wave-particle duality" which doesn't exist, but the general ideas are right.

6. Feb 19, 2016

### naima

I did not know this effect but it looks like the Fraunhofer interference behind two slits.
My idea came from a link that DrChinese gave me.
Experiment and the foundations of quantum physics
look at fig 3. With coincidence counting an interference pattern can be observed in the focal plane of the heisenberg lens (detector D1)
Zeilinger explains that in the focal plane the which-way information is erased.
What surprises me is that the coincidence counting is done with the D2 detector behind the slits, where there is no interference pattern!
Suppose that the heisenberg lens and the slits are at the same distance of the source. the mixed state at these points is (|1><1| +|2><2|)/2
then one evolves thru the slits and give no interference pattern. The Heisenberg device transforms the other state and gives (taking account of coincidence logic) the pattern. What is the mathematics behind this transformation? And when we are not in the focal plane the fringe visibility decreases.
Did entanglement disappear?

Last edited by a moderator: May 7, 2017
7. Feb 19, 2016

### DrChinese

Did you notice that you are mixing ideas which are not really alike and attempting to synthesize something that is neither?

Light from distant stars is generally not entangled. The Hanbury-Brown-Twiss effect is fascinating, I think of it as something like entangled histories.

When you attempt to say: what if I took entangled photons, and send them across the universe, and make them coherent, and then run them through a double slit apparatus, and then use them to send signals FTL: you are oversimplifying the ideas. The devil is in the details.

Coherence (or lack thereof) of entangled pairs is a complex subject. Entanglement involves things such as fixed photon numbers (Fock states) which are not necessarily present when you also consider the Hanbury-Brown-Twiss effect across large distances (where there is bunching). The full treatment is way beyond my pay grade.

8. Feb 19, 2016

### naima

I readily admit that i mix different situations and ideas, but in any case i suppose ftl signals. In which sentence? In this post i read what Zeilinger wrote. Can you help me for the formulas giving the interference pattern in the focal plane?
I recall that this pattern only appears when the no coinciding are forgotten.
The reminding ones are locally in a mixed state. those of the slits give no interference pattern. Those with the lens give an interference pattern but Alice must wait for a classical message to wash her screen. So no FTL.
the relationship between the Zeilinger paper and my title is that the simple fact that the source of entangled particles is far away could give a natural Heisenberg lens where we are in the focal plane.

Last edited: Feb 19, 2016
9. Feb 19, 2016

### DrChinese

I guessed you were going in that direction. Like I say, there are plenty of issues to address and none can be glossed over. I generally stay away from detail discussions of the eraser setup you are referring to for the simple reason that they are almost impossible to easily explain in the context of a series of posts. So I am not going to start here.

I will remind you of this: you get interference when there is no way to obtain which path information from the entangled partner. That tells you everything you need to know. The photons registering in the Heisenberg lens will not yield which slit information, and there is no way to place them in that state at will. They will be true even in your "natural Heisenberg lens" concept.

10. Feb 19, 2016

### vanhees71

I've no clue what "entangled histories" should be. It's usually called intensity correlations or more generally in the QFT context second-order coherence:

https://en.wikipedia.org/wiki/Degree_of_coherence

For a pedagogic introduction to HBT, see

http://arxiv.org/abs/nucl-th/9804026

11. Feb 22, 2016

### sciencejournalist00

Teleportation can be applied not just to pure states, but also mixed states, that can be regarded as the state of a single subsystem of an entangled pair. The so-called entanglement swapping is a simple and illustrative example.

If Alice has a particle which is entangled with a particle owned by Bob, and Bob teleports it to Carol, then afterwards, Alice's particle is entangled with Carol's.

A more symmetric way to describe the situation is the following: Alice has one particle, Bob two, and Carol one. Alice's particle and Bob's first particle are entangled, and so are Bob's second and Carol's particle.

Now, if Bob does a projective measurement on his two particles in the Bell state basis and communicates the results to Carol, as per the teleportation scheme described above, the state of Bob's first particle can be teleported to Carol's. Although Alice and Carol never interacted with each other, their particles are now entangled.

https://en.wikipedia.org/wiki/Quantum_teleportation#Entanglement_swapping

12. Feb 22, 2016

### DrChinese

Could you give us an example of that? Your example of teleportation via entanglement swapping would not apply, since the states being teleported are pure states (being entangled).

13. Feb 22, 2016

### Strilanc

The standard teleportation procedure works on mixed states.

(A mixed state came be eigen-decomposed into a weighted sum of pure states. When you apply the teleportation operations, they will distribute over that sum and into each the pure state cases. So the final state will be a weighted sum of teleported pure states, which is just a teleported mixed state.)

14. Feb 22, 2016

### DrChinese

Are you sure? I admit that I open to enlightenment on the point, but don't believe I have seen an example previously.

15. Feb 22, 2016

### StevieTNZ

I asked William Wootters whether you can teleport a horizontally polarized photon (photon A) to another photon (photon C; entangled with photon B) - he replied yes

16. Feb 22, 2016

### DrChinese

I just can't picture C ending up horizontally polarized after such operation, except perhaps half the time at most. Unless there is some kind of classical operation occurring.

17. Feb 22, 2016

### Strilanc

Yes, I'm sure.

Given a teleportation operator $T(\left| \psi \right\rangle \left\langle \psi \right| \otimes R_\text{epr}) = R_\text{junk} \otimes \left| \psi \right\rangle \left\langle \psi \right|$ and a mixed state $M = \sum_k \lambda_k \left| \psi_k \right\rangle \left\langle \psi_k \right|$ we find:

$T(M \otimes R_\text{epr})$
$= T((\sum_k \lambda_k \left| \psi_k \right\rangle \left\langle \psi_k \right|) \otimes R_\text{epr})$
$= \sum_k \lambda_k T(\left| \psi_k \right\rangle \left\langle \psi_k \right| \otimes R_\text{epr})$
$= \sum_k \lambda_k R_\text{junk} \otimes \left| \psi_k \right\rangle \left\langle \psi_k \right|$
$= R_\text{junk} \otimes \sum_k \lambda_k \left| \psi_k \right\rangle \left\langle \psi_k \right|$
$= R_\text{junk} \otimes M$

We can also confirm by doing a full calculation of the teleportation process given an unknown mixed state.

Suppose Alice has a qubit in the state represented by the density matrix $M = \begin{bmatrix} a & b \\ \overline{b} & c \end{bmatrix}$. It might be mixed. She also shares an EPR pair $P = \frac{1}{\sqrt{2}} \left| 00 \right\rangle + \frac{1}{\sqrt{2}} \left| 11 \right\rangle$ with Bob. The density matrix for state of the system as a whole is:

$\psi_1 = M \otimes PP^\dagger = \frac{1}{2} \begin{bmatrix} M & 0 & 0 & M \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ M & 0 & 0 & M \end{bmatrix} = \frac{1}{2} \begin{bmatrix} a & b & 0 & 0 & 0 & 0 & a & b \\ \overline{b} & c & 0 & 0 & 0 & 0 & \overline{b} & c \\ 0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\ a & b & 0 & 0 & 0 & 0 & a & b \\ \overline{b} & c & 0 & 0 & 0 & 0 & \overline{b} & c \end{bmatrix}$

Now we apply the teleportation operations.

First, a controlled-not of Alice's qubit containing $M$ onto the first qubit of $P$. The odd-index rows and columns (at indices 1, 3, 5, 7; you may be more used to calling them 2nd, 4th, 6th, and 8th) correspond to states where $M$ is ON. Toggling the first qubit of $P$ corresponds to pairing columns and rows whose index in binary have matching bits everywhere except in the second position (0-2, 1-3, 4-6, and 5-7) and swapping them. So a CNOT of $M$ onto the first qubit of $P$ swaps the 2nd and 4th columns, the 6th and 8th columns, the 2nd and 4th rows, and finally the 6th and 8th rows:

$\psi_{2} = \frac{1}{2} \begin{bmatrix} a & 0 & 0 & b & 0 & b & a & 0 \\ 0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\ \overline{b} & 0 & 0 & c & 0 & c & \overline{b} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0& 0 & 0\\ \overline{b} & 0 & 0 & c & 0 & c & \overline{b} & 0 \\ a & 0 & 0 & b & 0 & b & a & 0 \\ 0 & 0 & 0 & 0 & 0 & 0& 0 & 0 \end{bmatrix}$

Second, a Hadamard operation is applied to the qubit that started off storing $M$. Group the rows by everything except $M$ (0 with 1, 2 with 3, 4 with 5, 6 with 7), then give the even-index row of each pair the pair's sum while the odd index gets the difference. Repeat for the columns. Gain a factor of 1/2.

$\psi_{3} = \frac{1}{4} \begin{bmatrix} a & a & b & -b & b & -b & a & a \\ a & a & b & -b & b & -b & a & a \\ \overline{b} & \overline{b} & c & -c & c & -c & \overline{b} & \overline{b} \\ -\overline{b} & -\overline{b} & -c & c & -c & c & -\overline{b} & -\overline{b}\\ \overline{b} & \overline{b} & c & -c & c & -c & \overline{b} & \overline{b} \\ -\overline{b} & -\overline{b} & -c & c & -c & c & -\overline{b} & -\overline{b}\\ a & a & b & -b & b & -b & a & a \\ a & a & b & -b & b & -b & a & a \end{bmatrix}$

Third, a CNOT of $P$'s first qubit onto its second qubit is performed. Swap the 3rd and 7th columns. And rows. Also 4th and 8th. Technically a measurement is supposed to happen beforehand, since $P$'s first qubit was originally with Alice and now we're conditioning on its value in Bob-land. But you can defer measurement when performing calculations; you'll get the same result at the end as long as the measured qubits are only used as controls in the interim.

$\psi_{4} = \frac{1}{4} \begin{bmatrix} a & a & a & a & b & -b& b & -b \\ a & a & a & a & b & -b& b & -b \\ a & a & a & a & b & -b& b & -b \\ a & a & a & a & b & -b& b & -b \\ \overline{b} & \overline{b}& \overline{b} & \overline{b} & c & -c & c & -c \\ -\overline{b} & -\overline{b}& -\overline{b} & -\overline{b} & -c & c & -c & c\\ \overline{b} & \overline{b}& \overline{b} & \overline{b} & c & -c & c & -c \\ -\overline{b} & -\overline{b}& -\overline{b} & -\overline{b} & -c & c & -c & c \end{bmatrix}$

The final non-measurement operation is a controlled-Z of the qubit that started off storing $M$ onto $P$'s second qubit. This negates the columns and rows whose indices are of the form 1X1. So negate the 6th and 8th columns, and rows:

$\psi_{4} = \frac{1}{4} \begin{bmatrix} a & a & a & a & b & b& b & b \\ a & a & a & a & b & b& b & b \\ a & a & a & a & b & b& b & b \\ a & a & a & a & b & b& b & b \\ \overline{b} & \overline{b}& \overline{b} & \overline{b} & c & c & c & c \\ \overline{b} & \overline{b}& \overline{b} & \overline{b} & c & c & c & c\\ \overline{b} & \overline{b}& \overline{b} & \overline{b} & c & c & c & c \\ \overline{b} & \overline{b}& \overline{b} & \overline{b} & c & c & c & c \end{bmatrix}$

Which factors:

$\psi_{4} = \frac{1}{4} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \otimes \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \otimes \begin{bmatrix} a & b \\ \overline{b} & c \end{bmatrix}$

Measuring the first two qubits will give a uniformly random result (since they are each in the state $\frac{1}{\sqrt{2}} \left| 0 \right\rangle + \frac{1}{\sqrt{2}} \left| 1 \right\rangle$), and drop them into the maximal mixed state. The actual protocol performs measurement earlier, but we used the deferred measurement principle to get the same final state despite delaying the measurement calculations until now:

$\psi_{5} = \frac{1}{4} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} a & b \\ \overline{b} & c \end{bmatrix} = \frac{1}{4} I_2 \otimes I_2 \otimes M$

As you can see, the third qubit (which corresponds to the second qubit of $P$; i.e. Bob's qubit) has ended up in the state $M$.

Also you can see that it's much easier to use the intuition that "if it works on pure states then it must work on mixed states since mixed states are like not knowing which pure state you're in" than it is to run out the full calculation.

Last edited: Feb 22, 2016
18. Feb 22, 2016

### StevieTNZ

I guess it depends on what kind of bell-state you get for photons A and B (I think if it is |H>|V> - |V>|H>, then no operation needs to be performed on photon C to make it horizontally polarized). But yes, it only happens a certain percentage of the time.

19. Feb 22, 2016

### DrChinese

Thanks. Usually, I would expect one Bell state to produce H, another to produce V. Thus no FTL signalling is possible.

I am still looking Strilanc's derivation to understand how it is both correct and not in violation of signalling.

20. Feb 22, 2016

### Strilanc

The teleportation I described requires sending two classical bits (the results of measuring Alice's qubits) from Alice to Bob, so it can't be used to signal faster than light.

If you're not familiar with quantum teleportation from the perspective of quantum information, you might find this video by Michael Nielson useful. Also, if you're thinking in terms of photons, then the wikipedia article on linear optical quantum computing is probably a good resource.