# Spin difference between entangled and non-entangled

1. Sep 12, 2014

### Alien8

What exactly entangled-spin pairs do in Stern–Gerlach experiment that ordinary particles pairs with opposite magnetic dipole moments do not?

What equation represents the blue line and what equation if for the red line?

Last edited by a moderator: May 6, 2017
2. Sep 12, 2014

### atyy

The experiment is a test of the Bell inequalities. An explanation is given in http://arxiv.org/abs/quant-ph/0205171. That article plots a version of the different curves in Fig. 4, with the equations for the entangled and unentangled cases given by Eq. 10 and 19 respectively.

The Bell inequalities assume that the experimental results are causally explained by local hidden variables and local measurement settings. The causal relationship assumed by the Bell inequalities is diagrammed in Fig. 19 of http://arxiv.org/abs/1208.4119. However, the predictions of quantum mechanics are incompatible with the causal relationship of Fig. 19. In other words, quantum mechanics predicts that the Bell inequalities will be violated by experiments. Experimental results thus far are consistent with the predictions of quantum mechanics. Some alternative causal relationships are diagrammed in Fig. 25, 26 and 27, and are known as non-locality, superdeterminism and retrocausation. Because the non-locality alternative is usually considered the most natural for a scientific theory that can be used by human beings to make predictions, it is often said that quantum mechanics is nonlocal.

Last edited: Sep 12, 2014
3. Sep 13, 2014

### Alien8

The first question is about particle spin and magnetic dipole moment. If particle A has a magnetic dipole moment vector mA=(1,1,1) and particle B has mB(-1,-1,-1), then if particle A is measured along x,y, or z axis it will be north up, and if particle B is measured along -x,-y, or -z axis it will also be north up, or spin up, that is they will be correlated because they were created correlated. So how are two ordinary particles with opposite magnetic dipole moments different than two quantum entangled particles with opposite spins?

Eq. 10:
$P_{VV}(\alpha,\beta) = 1/2 * cos^2(\beta - \alpha)$

What quantum mechanics principle or equation is this prediction related to (derived from)?

4. Sep 13, 2014

### Staff: Mentor

There is a subtle problem with the way you've asked this question. A quantum mechanical property like spin has no definite value until is measured, so when you say "a particle pair with opposite magnetic dipole moments" you're really saying "a pair of particles that we just measured on the same axis and found one to be up and the other to be down". Unless and until we've made that measurement, there's no such thing as an "ordinary particle pair with opposite magnetic dipole moments".

The difference between an entangled pair and a non-entangled pair is that we know that if we measure them both on the same axis, we will always get opposite results with an entangled pair but we won't always get opposite results with a non-entangled pair.

It's also important to remember that after we've measured them, there is no difference whatsoever between the pairs that had been entangled and unentangled pairs that just happened to give opposite results. After the measurement, they're all just what you said: "ordinary pairs with opposite magnetic dipole moments". But that's after the measurement; before we make the measurement none of them have definite spins.

5. Sep 13, 2014

### Staff: Mentor

In quantum mechanics, the result of measuring any observable will be an eigenvalue of the operator corresponding to that observable. This formula is derived from the relationship between the operators corresponding to measurements of the spin along different axes.

I understand that this is a completely unhelpful answer... But until you have a basic familiarity with the math behind QM there is no way of answering your question, and the only way of acquiring that familiarity is to spend some serious quality time with a decent textbook.

6. Sep 13, 2014

### Alien8

I understand QM interpretation. I don't understand what is supposed to be classical interpretation and how are they different.

There's no such thing as an "ordinary particle pair with opposite magnetic dipole moments"? Is there some reason we can not make neutrons orient their magnetic moments in a specific desired direction and then send them off to be measured?

If we send today 100 neutrons to Alice each of which with definite magnetic moment orientation (Xi,Yi,Zi) that we select at random, and if tomorrow we send 100 neutrons to Bob each of which with definite magnetic moment orientation (-Xj,-Yj,-Zj), then how are we not supposed to always get opposite result for i=j pairs if we measure them along the same axis?

Last edited: Sep 13, 2014
7. Sep 13, 2014

### Alien8

I'm actually looking for a historical answer. When did "entanglement" become a part of QM, what is the actual "entanglement" equation and what experiment was it originally inferred from?

8. Sep 13, 2014

### Staff: Mentor

It's been there pretty much from the beginning, but it took a while (decades!) for the implications to be fully appreciated. If you're looking for a historical answer and don't feel like grinding through a few years of differential equations and linear algebra on the way then I recommend "The Age of Entanglement" by Louisa Gilder.

The idealized entanglement thought experiment is a single particle with zero spin that decays into two particles flying in opposite directions; conservation of angular momentum requires that if we measure the spins of the two daughter particles on the same axis the sum must also be zero. The fundamental equation of QM, the Schrodinger equation, applies to the entire system, so as far as the math of QM is concerned, we have a single system with zero net angular momentum before and after the decay. Entanglement (this detector registered spin-up; therefore the entire system is in a state such that the other detector will register spin-down if we perform that measurement) follows from there.

Last edited: Sep 13, 2014
9. Sep 13, 2014

### stevendaryl

Staff Emeritus
From Wikipedia on the subject:

10. Sep 13, 2014

### Staff: Mentor

We can. We take a beam of neutrons and pass them through a S-G apparatus oriented in a partcular direction, and then if we only want the spin-up direction we only use the neutrons in the part of the beam that was deflected upwards. But that's a measurement; the things did not have any orientation before then (and that is actually an experimentally verifiable fact - google for "Bell's Theorem" and "Alain Aspect").

There is no way of producing a beam of neutrons that is known to be in a particular spin orientation without somehow "processing" them to get them into the desired state. The neat thing about entangled particles is that no matter what orientation we process one member of the pair into, we know that a measurement of the other member will produce the opposite result; in effect we're always "processing" the entire pair.

Let's say I want to produce two beams of neutrons, both spin-up, but going in opposite directions. So I set up two S-G devices, one on each side of my neutron source, orient them both vertically, and then block off the downwards-deflected beam from both. If my neutron source is generating entangled pairs, then I will never (except by random chance when multiple pairs arrive at the SG devices at the same time) find a neutron passing through the left-hand device at the same moment that a neutron passes through the right-hand device. Many neutrons will make it through the left-hand SG device, and many neutrons will make it through the right-hand device, but for each pair, only one member will make it through.

Last edited: Sep 13, 2014
11. Sep 13, 2014

### Alien8

If the concept of "entanglement" didn't exist before Einstein and co. invented it for their EPR paper, then what were they referring to, what is it they based their premise on? In other words, how could there be QM prediction if there was no prior QM theory which to base that prediction on?

Now, looking at Wikipedia I think this might be the answer:
A beam of light incident on a half-silvered mirror has 50% probability to pass through, so we may describe those photons with the same wave function. That seem to be the original empirical root I'm asking about, which gave birth to the whole "entanglement" concept. It seems like "entanglement" is the same thing as "equal probability". But this doesn't quite explain if these photons are supposed to be in entangled state before or after they interact with the mirror.

Last edited: Sep 13, 2014
12. Sep 13, 2014

### DrChinese

You have already received some answers which may or may not directly address you questions. First, there has never really been a classical theory which matches the red line. The red line represents the CLOSEST any classical theory could every come to the QM prediction for entangled particle pairs, shown in blue. The red represents a boundary condition, in other words.

Second, you asked about the difference between entangled particles and particle pairs which are have opposite spins. The difference can easily be seen in their spin statistics. The first has "entangled state" statistics in which measurements will be completely anti-correlated at ANY angle chosen. The second has "product state" statistics in which measurements will be variably anti-correlated at ANY angle chosen. The predictions are different, and experiment matches predictions on both.

Historically, EPR (1935) did not know much about entanglement other than a few key basics. I don't believe experimental versions appeared until much later. Even theoretical treatments didn't go far until the 1950's (Bohm comes to mind). There was so much going on in QM during this period that it took a while to put all the pieces together on entanglement. Bell's Theorem probably another big turning point, although that too took years to fully sink in.

But the essential point is the there is a conservation factor and there is an indistinguishability factor. Classical particles are distinguishable in all respects even when they obey conservation as a pair. This "detail" makes all the difference.

I would skip the reference to half-silvered mirrors when considering entanglement, as that is more of an example of single particle superposition. Entangled particles are in a superposition, true enough.

Last edited by a moderator: May 6, 2017
13. Sep 13, 2014

### atyy

I'm not sure whether this is the historical route, but a simple way to get entanglement from elementary quantum mechanics is to consider a system of two partcles, eg. the electrons in a helium atom.

For two distinguishable particles, entanglement means that the wave function cannot be written as a product state.

For one particle, an arbitrary wave function $\psi(x)$ can be written as a superposition of basis states:

$\psi(x) = \sum\limits_{n} \phi_{n}(x)$.

For two particles, an arbitrary wave function $\psi(x,y)$ can be written as a superposition of basis states:

$\psi(x,y) = \sum\limits_{n,m} \phi_{n}(x)\phi_{m}(y)$.

So entanglement arises from the fact that for the two particles basis states are built of products of the one particle basis, and that the general state is formed by a superposition of basis states.

Last edited: Sep 13, 2014
14. Sep 13, 2014

### atyy

In Pauli's Nobel lecture http://www.nobelprize.org/nobel_prizes/physics/laureates/1945/pauli-lecture.html he mentions that already in 1926, Heisenberg wrote papers about wave function symmetrization for identical particles. That requires the idea that the two-particle Hilbert space is the tensor product of the one particle Hilbert spaces. So the placement of entanglement within the quantum formalism goes back to within a year or two of quantum mechanics proper. Of course, many consequences were only worked out later. I should also say that in the previous post, I only talked about entanglement for distinguishable particles. For identical particles, one needs a different definition. There seem nowdays to be many refinements in the classification of entanglement, depending on what operations one is interested in.

15. Sep 13, 2014

### Alien8

That's how it works, but I'm asking how classical physics fails to explain it. What is spin for QM in classical mechanics is fully 3-dimensional magnetic vector field which can have arbitrary orientation and simultaneous definite magnitude component along x,y, and z axis.

According to that this is how I suppose classical theory should work, please point out the step where it goes astray:

1. We can send neutrons A to Alice with precisely defined, randomly chosen, magnetic moment orientation in 3D space, which can be represented with 3D vectors (x,y,z)

2. The next day for each neutron A we can send neutron B to Bob with the opposite magnetic moment orientation, which can be represented with 3D vector (-x,-y,-z)

3. Therefore if we measure both corresponding neutrons along the same axis we will always measure the opposite spin simply because they were sent with that particular orientation to begin with

16. Sep 13, 2014

### atyy

A very good introduction to the difference between classical and quantum mechanics regarding entanglement is given by http://arxiv.org/abs/1303.3081. The key is the derivation of the Bell inequalities. Under a number of assumptions, one can show that no classical strategy of pre-established agreement (such as a process that always prepares both spins in the same direction) can violate the Bell inequalities. Yet quantum mechanics predicts the violation of the Bell inequalities.

Two features of quantum mechanics that are not present in classical mechanics that enable it to violate the Bell inequalities are non-commuting observables, and entanglement.

17. Sep 13, 2014

### stevendaryl

Staff Emeritus
I don't know whether anyone conclusively answered the question as to what the red and blue curves represent in the figure.

The blue curve is the quantum prediction for correlation between measurements of the two particles in a spin-1/2 twin pair EPR experiment. If Alice measures the spin of one particle along axis $a$ and Bob measures the spin of the other particle along axis $b$, then the correlation $E(a,b)$ is the average value, over many trials, of $AB$ where $A$ is $\pm 1$, depending on whether Alice measures spin-up or spin-down, and $B$ is $\pm 1$, depending on whether Bob measures spin-up or spin-down.

The quantum prediction is: $E(a,b) = -cos(\theta)$ where $\theta$ is the angle between $a$ and $b$. So it's -1 at $\theta = 0$ and 0 at $\theta = 90^o$

So what's the red line? It's a little bit misleading to call that the classical prediction. It's the prediction of a particular classical model. There are lots of different possible classical models, and they make different predictions. But the specific model that that's a graph for is one describe by Bell in his discussion of EPR. It has the nice feature that $E(a,b) = -1$ when $\theta = 0$ and $E(a,b) = 0$ at $\theta = 90^o$ and $E(a,b) = +1$ at $\theta = 180^o$, just like the quantum predictions. This classical model is the following:

Assume that when a twin pair is produced, there is, attached to the particles, an associated spin vector $\vec{S}$ pointing in a random direction.

Alice then chooses an axis $\vec{a}$ to measure the spin relative to. She gets $A= +1$ if the angle between $\vec{S}$ and $\vec{a}$ is less than 90°. She gets $A = -1$ if the angle is more than 90°.

Bob chooses an axis $\vec{b}$, and gets $B = \mp 1$ depending on whether the angle between $\vec{b}$ and $\vec{S}$ is less than or more than 90°. (The opposite of the rule for Alice.)

With this classical rule, if $\vec{a} = \vec{b}$, (relative angle $\theta = 0$), they will always get opposite results, for a correlation of $-1$. If $\vec{a} = -\vec{b}$ (relative angle $\theta = 180^o$), they will always get the same result, for a correlation of $+1$. If $\vec{a}$ and $\vec{b}$ are at right angles (relative angle $\theta = 90^o$), they will get the same result 50% of the time, and opposite results 50% of the time, for a correlation of 0.

So this classical model makes the same predictions as QM for the cases of relative angles of $0^o, 90^o, 180^o$, but makes different predictions for other angles.

18. Sep 13, 2014

### stevendaryl

Staff Emeritus
The way that spin measurements work is that you are only able to measure one component of the spin. Spin is a vector $\vec{S}$ (or can be thought of as a vector) with 3 components, but you can only measure the spin relative to a chosen axis $\vec{a}$. And that answer always gives the answer $+1/2$ or the answer $-1/2$. That fact by itself shows that quantum spin is very different from a classical vector. If a classical vector $\vec{S}$ is pointing in the x-direction, and I measure its component in the y-direction, I'll get zero. Quantum-mechanically, you never get zero for a spin-1/2 particle, you always get $\pm 1/2$ (in units of $\hbar$).

19. Sep 13, 2014

### DrChinese

And as I mentioned in my post, the statistical predictions are quite different between the classical situation you describe and the entangled stats. The classical is a product state. Let's take your example:

I have 2 electrons oriented at x=+ and x=- respectively. If I measure them later at a 90 degree angle to x, I get NO correlation.

On the other hand, entangled electrons would show perfect anti-correlation at that measurement angle.

20. Sep 13, 2014

### Alien8

That's it. So let's put it in perspective. For example take ordinary unpolarized sunlight incident on some polarizer. Half of the photons will pass through, so each of them may be represented with the same wave (probability) function. Does that mean they were all entangled before, or only those that went through are entangled now, or neither?

The thing is that's all still only about one light beam interacting with only one polarizer, where does it say this can be applied to two separate polarizers?