# Comparison between quantum entanglement and a classical version

1. Sep 12, 2015

What I understood from Quantum Entanglement (QE), is that measuring the spin of one of two entangled particles in one location gives the spin of the other particle in other location no matter how far is the later. What I can also understand is that the same concept is applicable in the classical physics. For example, if we have a way to create two spinning tops at one point in the time and space, they will be spinning at different directions to maintain the law of conservation of angular momentum. Now measuring the direction of one of them in one location ( by just watching it), gives information about the direction of the other spinning top in the other location. So what is the difference between both QE and the classical version?

2. Sep 12, 2015

### Mentz114

In QM the two tops can be in a mixed state before the measurement is made. Thus until one is actually measured neither is known with certainty. If the states are also entangled then measuring either one projects the other into the opposite state (or the same state depending the setup).

I found this Wiki article very illuminating

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

Last edited: Sep 12, 2015
3. Sep 12, 2015

I went quickly through the article and I found that the difference between both versions ( quantum and classical) is firstly, the 2 states are in a mixed state in the QE but they are not in classical one. Secondly, the random outcome of one particle spin measured along a particular axis ( because of quantum mechanics) will maintain the anti correlation by keeping the outcome of the other particle spin in the opposite spin measured along the same axis.
While the first difference between the two versions can be cancelled if we assume a zero knowledge of information about the state of spinning tops before the measurement, the second difference is really weird.

4. Sep 12, 2015

### bhobba

Last edited: Sep 12, 2015
5. Sep 12, 2015

### Mentz114

Yes, it is very weird. As bhobba has pointed it is a profound difference.

But QT is non-local so it does not require any additional mind stretching.

6. Sep 12, 2015

### DirkMan

To quote Unruh , "exactly in what sense is it non-local?"

7. Sep 12, 2015

### Heinera

In the operational sense that "non-local" means it predicts a violation of Bell's inequality. I guess most use the phrase "non-local" in that sense, without implying anything more specific than that (which would anyway take us to interpretations).

8. Sep 12, 2015

### Strilanc

> So what is the difference between both QE and the classical version?

In quantum entanglement, you can interfere and generally "mix" the various cases to get interesting results. This prevents you from just assuming it was one of the cases beforehand, like you can with classical correlations.

- You can't do superdense coding with classical correlations, but you can with an entangled qubits.
- You can't win the Mermin-Peres game 100% of the time using classically correlated bits, but with entangled qubits you can.
- And of course the CHSH game used in Bell-inequality experiments are also an example of being able to win a game more often by using entangled qubits than you could with correlated bits.

It's difficult to give an intuitive idea of why it works. It really comes down to the way the math behaves, like the ways unitary matrices differ from stochastic matrices. I could never understand what the heck people were talking about until I started from the bottom and built upward; the high-level analogies are just too fragile.

9. Sep 12, 2015

### stevendaryl

Staff Emeritus
Bell's theorem is all about the testable difference between quantum entanglement and any analogous classical correlation.

Classically, if two measured values $A$ and $B$ are obtained far enough apart so that there is no causal influence of one measurement on the other, then any correlations between those values must be due to some unknown facts ("hidden variables") common to both measurements. A classical version of EPR might look like this:

1. You produce a pair of particles. One particle goes to Alice, another goes to Bob.
2. Alice picks a direction $\vec{a}$
3. Bob picks a direction $\vec{b}$
4. Alice measures the spin $\vec{s_A}$ of her particle relative to $\vec{a}$, and writes $A=+1$ if $\vec{s_A} \cdot \vec{a} > 0$. Otherwise, she writes down $A=-1$
5. Bob measures the spin $\vec{s_B}$ of his particle relative to $\vec{b}$, and writes $B=+1$ if $\vec{s_B} \cdot \vec{b} > 0$. Otherwise, he writes down $B=-1$
6. Then they compare the results $A$ and $B$
They find that the results are correlated, in the sense that if $\vec{a} = \vec{b}$, then $A = -B$

That's a classical correlation. It is easily explained using "hidden variables". You just assume that $\vec{s_A}$ and $\vec{s_B}$ are fixed at the time of the creation of the pair of particles in such a way that $\vec{s_A} = - \vec{s_B}$. The pair $(\vec{s_A}, -\vec{s_A})$ is the "hidden variable".

What Bell showed is that the correlations predicted by quantum mechanics cannot be explained by any such hidden-variable model (unless we allow faster-than-light influences, or back-in-time influences, or some other exotic possibility).

10. Sep 13, 2015

So, the quantum modification of the same experiment could be; instead of labeling a definite + or - for A and B, Alice and Bob define a probability of finding + or - relative to their measurement directions. In addition, corresponding readings will always yield a correlation which means each time Alice writes A=+1 Bob writes B=-1.

If this is the case, then the source of puzzle is not in the characteristic of the particles ( entanglement) but rather in the way of measuring them.

11. Sep 13, 2015

### stevendaryl

Staff Emeritus
The particular predictions of quantum mechanics for spin-1/2 twin-pair version of EPR is this: If Alice and Bob both use spin detectors oriented in x-y plane, and Alice chooses orientation $\alpha$ (relative to the x-axis) and Bob chooses orientation $\beta$, then they will get the same result (both spin-up or both spin-down) with probability $sin^2(\beta - \alpha)$, and will get opposite results with probability $cos^2(\beta - \alpha)$. The details of what is going on in the measurement process seem irrelevant; the only thing that is relevant is the angles $\alpha$ and $\beta$.

12. Sep 13, 2015

### Staff: Mentor

Try it.... You will not be able to construct such a model, and have the probabilities produce correlations that match the quantum mechanical predictions, unless you allow for the probability that Alice gets a given result on her particle to vary with the direction that Bob chooses to measure on - not just the result that Bob gets, but the direction he gets it on.

It sounds like it's possible, and people have spent enormous amounts of time on ever more complicated models that try to avoid this basic weirdness of quantum entanglement.... but it cannot be done.

13. Sep 13, 2015

So according to the example of the hidden variable you gave where both Alice and Bob orient their detectors along the same direction, the quantum interpretation of probability of opposite results =$cos^2(\beta-\alpha)$ will be 1 which will be equivalent to the hidden variable theory result. I mean there will no be any sort of weirdness if both observers align their detectors in the same direction? If the answer is no, then the source of general weirdness of QE is in the way both observers align their detectors not in the idea of superposition.

Last edited: Sep 13, 2015
14. Sep 13, 2015

### DrChinese

When the detectors are aligned the same, there is no obvious "weirdness". That much you say is correct. But that does not mean that superposition is not "weirdness", it only means that its "weirdness" does not show up in that example. At most any other relative angle setting, there will be an inequality that violates a classical explanation.

In other words: superpositions of particles that are space-like separated will demonstrate quantum non-locality at all angles. A few of those angles won't seem weird, but others will.

15. Sep 13, 2015

### Mentz114

I agree.
It seems that to calculate the quantities in the CHSH inequality one needs two settings (directions) on each detector, as in the Stern-Gerlach experiment. In that case, as you say, some settings give violations. I could never work out how, with only one setting on the detectors one can do more than measure correlations and hope to get $\pm 1$.

16. Sep 13, 2015

### stevendaryl

Staff Emeritus
You have to actually go through the mathematics to see why that number, $cos^2(\beta - \alpha)$, cannot be explained through normal local mechanisms. It can be explained through a nonlocal mechanism easily enough:
1. Initially, the spin directions for Alice's and Bob's particles are completely undetermined.
2. When Alice measures the spin of her particle, she randomly gets $\pm 1$, with 50/50 probability of each outcome.
3. If Alice measures +1 at direction $\alpha$, then Bob's particle "collapses" to the state with spin direction $\phi = \alpha - \pi$.
4. If Alice measures -1 at direction $\alpha$, then Bob's particle "collapses" to the state with spin direction $\phi = \alpha$.
5. Later, when Bob measures the spin of his particle at direction $\beta$, he gets +1 with probability $cos^2(\frac{\beta - \phi}{2})$ and -1 with probability $sin^2(\frac{\beta - \phi}{2})$.
This nonlocal model (the "collapse" interpretation) explains the probabilities, but at the cost of nonlocality; Bob's particle's state changes instantaneously when Alice finishes her measurement.

[edit: probabilities in 5 corrected; originally there was a missing factor of 2]
[edit 2: sign of $\phi$ in 3 was changed]

Last edited: Sep 15, 2015
17. Sep 13, 2015

### bohm2

You may find these papers comparing classical versus quantum entanglement worth reading as they point out the limits of classical analogues of entanglement:

Brownian Entanglement
http://arxiv.org/pdf/quant-ph/0412132v1.pdf
https://staff.fnwi.uva.nl/t.m.../BrownianEntanglement-1-10.ppt [Broken]
A Classical Analogy of Entanglement
http://www.science.uva.nl/research/aplp/eprints/Spr98.pdf

Entanglement in classical Brownian motion
https://esc.fnwi.uva.nl/thesis/centraal/files/f292681290.pdf

Last edited by a moderator: May 7, 2017
18. Sep 14, 2015

I am confused after reviewing this post compared with yours #11.
Suppose Bob chose his detector to be aligned at angle α which is the same as Alice (α=β). Now, the probability of Bob measuring +1 is cos2(β-Φ)=cos2(α-Φ)=cos2(α-π+α)=cos2(2α-π). So the probability of getting the same result (+1) ≠ 0 unless α=π/2. In fact if α is chosen to be 0, then that probability is 1.
In post #11, the probability of getting the same result is sin2(β-α)=0?

Last edited: Sep 14, 2015
19. Sep 14, 2015

### Staff: Mentor

There's a missing division by two in steps 4 and 5 - it's a typo. Try this instead:
1) Initially, the spin directions for Alice's and Bob's particles are completely undetermined
2) When Alice measures the spin of her particle, she randomly gets ±1, with 50/50 probability of each outcome.
3) If Alice measures +1 at direction $\alpha$ then Bob's particle "collapses" to the state with spin direction $\phi=\pi-\alpha$
4) If Alice measures -1 at direction α, then Bob's particle "collapses" to the state with spin direction $\phi=\alpha$
5 )Later, when Bob measures the spin of his particle at direction $\beta$, he gets +1 with probability $\cos^2\frac{\beta−\phi}{2}$ and -1 with probability $\sin^2\frac{\beta−\phi}{2}$.

These are the rules for spin-entangled electrons. You'll also see people explaining Bell's theorem using experiments with polarization-entangled photons instead; the argument is the same but that division by two is not there (maximum anti-correlation effect is found at 90 degrees instead of 180 degrees). However, the all-important dependence on $\beta-\alpha$ is still there - that's what makes the result at one detector change when we change the setting of the other detector.

20. Sep 14, 2015

### Staff: Mentor

You might want to give this paper a try: http://www.quantum3000.narod.ru/papers/edu/cakes.pdf

Although it argues by analogy, it does a pretty good job of identifying the essential difference between quantum and classical models, and doesn't require understanding the QM treatment of the singlet state as the price of admission.