# I Amplitudes, Probabilities and EPR

Tags:
1. Dec 1, 2016

### stevendaryl

Staff Emeritus
Last edited by a moderator: Dec 9, 2016
2. Dec 1, 2016

Nice job.

3. Dec 1, 2016

### Mentz114

@stevendaryl

I like that a lot. Gourmet food for thought. There is something in the two-valuedness of amplitude that is elusive.
[later]
I deleted this bit. Still thinking about it.

Last edited: Dec 1, 2016
4. Dec 1, 2016

### mikeyork

I think yes. There are two diverse concepts of probability. The primary one that we think of all the time is that which corresponds to frequency counting. It can be determined only asymptotically as the number of possible events approaches infinity. But there is also a theoretical concept of probability which we use all the time before even a single event is detected. Historically we try to predict probability as that asymptotic limit itself. But, any quantity from which the asymptotic limit is uniquely computable is an encoding of theoretical probability. In QM the amplitude is such an encoding. As such a probability encoding it enables the probabilistic prediction of quantum states without detecting any events. It has all the requirements of a theoretical probability encoding while at the same time describing the physical reality of a single system. Frequency counting, on the other hand, has meaning only for large numbers of systems/events. We can think of QM and the probability amplitude, therefore, as the primary encoding of theoretical probability and frequency counting as secondary.

5. Dec 2, 2016

### zonde

We can talk about probabilities when assumption that ensemble is i.i.d. holds. When this assumption holds probabilities of individual events independently contribute to total probability. Amplitudes obviously are not independent as two opposite amplitudes can cancel out.

It is interesting that in your model the final amplitude is calculated by adding/subtracting amplitudes from two different subsets of pairs (with $\lambda$ +1 and -1). So each separate pair does not produce correct amplitude. And they add up not just statistically but in a way that suggest some interdependence on the level of ensemble of pairs.

6. Dec 2, 2016

### stevendaryl

Staff Emeritus
I'm not sure I understand your point. Amplitudes add in the same way that probabilities do. The reason that some amplitudes cancel others is because they aren't guaranteed to be positive.

As I said, I think it's the same issue with classical probabilities (except for the big difference that classical probabilities are positive reals numbers, while quantum amplitudes are complex numbers).

Let me give a made-up classical probability problem to illustrate. Suppose I'm trying to figure out the probability that a certain newborn baby will grow up to be left-handed. This is not true, but I'm going to pretend that there is a "left-handed gene" such that if you have this gene, there is a 90% chance that you will be left-handed, and if you lack this gene, then there is a 90% chance that you will be right-handed. Let's furthermore assume that the mother lacks this gene, but the father received the gene from one parent but not the other. Then genetics would say that the father has a 50% chance of passing on the gene to the baby. So letting $\lambda = +1$ to indicate having the gene, and $\lambda = -1$ to indicate not having the gene, we can compute:

$P(LH) = \sum_\lambda P(\lambda) P(LH | \lambda) = P(+1) P(LH | +1) + P(-1) P(LH | -1) = .50 \cdot .90 + .50 \cdot .10 = .50$

So the baby has a 50% chance of being left-handed.

That involves summing over different values of $\lambda$ in exactly the same way as the amplitudes were computed in my EPR example.

7. Dec 2, 2016

### stevendaryl

Staff Emeritus
I skipped over this first line without asking: What does "i.i.d." stands for?

8. Dec 2, 2016

### Demystifier

Yes, that's true! A similar conclusion is also drawn in
https://arxiv.org/abs/0707.2319

9. Dec 2, 2016

### zonde

10. Dec 2, 2016

### DrChinese

Great analysis for the original post!

11. Dec 2, 2016

### secur

Sorry, this is not a hidden-variable model as understood in Bell experiment. The problem is that A and B results must be +-1. When you calculate

$\psi(A, B|\alpha, \beta) = \sum \psi(\lambda) \psi_A(A | \alpha, \lambda) \psi_B(B | \beta, \lambda)$

the numbers that appear for $\psi_A(A | \alpha, \lambda)$ and $\psi_B(B | \beta, \lambda)$ can't be complex; can't even be any real numbers, except +-1. That's because this calculation must be performed on their actual results, after the experiment is concluded.

Another way to put it, your scheme doesn't guarantee that if $\alpha$ = $\beta$ then their results will definitely be opposite. A and B must "flip a coin" based on their amplitudes, and record a definite +1 or -1. In general with these probability amplitudes they will often get the same result, even though the angles are equal.

12. Dec 2, 2016

### Jilang

Secur, why can't the wavefunctions be complex?

13. Dec 2, 2016

### secur

@Jilang. the wavefunctions can, in fact, be complex. I said that in the calculation of $\psi(A, B|\alpha, \beta)$ "the numbers that appear for $\psi_A(A | \alpha, \lambda)$ and $\psi_B(B | \beta, \lambda)$ can't be complex". Realizing the point might not be too clear I went on to "Another way to put it" which is, I think, straightforward.

Consider what happens in the Bell-type experiments. Alice and Bob independently measure +1 or -1 for the spin states of their respective particles. Those measurements are based, probabilistically, on complex wavefunctions. After their experimental data is complete we bring the results together in a calculation where A and B (+-1) are multiplied together, similar to @stevendaryl's $\psi(A, B|\alpha, \beta)$. The key point: when those results are multiplied together they must have been reduced to +-1, can't still be the underlying wavefunction numbers.

Note if A and B could record the underlying spinors (as quaternions), we could easily get the right correlations for actual QM probabilities, never mind amplitudes.

I'm not very happy with this explanation, hopefully with further discussion it will get better. But consider my simpler second point, "Another way to put it". That's symptomatic of the same problem and clearly shows there's something wrong here.

14. Dec 2, 2016

### stevendaryl

Staff Emeritus
I'm afraid you're completely misunderstanding what is being claimed, and in particular, what the meanings of the various $\psi$s are. I tried to explain it in the first post, but if it was unclear, let me try again.

We have experimenters Alice and Bob that are performing measurements at a spacelike separation. Assume that their results are described by a joint probability distribution $P(A, B|\alpha, \beta)$, which gives the probability that Alice gets result $A$ (assumed to be $\pm 1$) and Bob gets $B$ (also $\pm 1$), given that Alice chooses detector setting $\alpha$ and Bob chooses detector setting $\beta$

A nondeterministic local hidden-variables theory explanation for $P(A, B|\alpha, \beta)$ would consist of:
1. A set of values for a parameter, $\lambda$
2. A probability distribution $P(\lambda)$ on the values of $\lambda$
3. Probability distributions $P_A(A|\alpha, \lambda)$ and $P_B(B|\beta, \lambda)$
such that

$P(A, B| \alpha, \beta) = \sum_\lambda P(\lambda) P_A(A|\alpha, \lambda) P_B(B|\beta, \lambda)$

A deterministic local hidden-variables theory explanation makes a stronger assumption, that $A$ and $B$ are determined by the parameters $\alpha, \beta, \lambda$. That is, it assumes that there are deterministic functions $F_A(\alpha, \lambda)$ and $F_B(\beta, \lambda)$ such that whenever Alice chooses setting $\alpha$ and the hidden variable has value $\lambda$, then Alice will deterministically get the result $F_A(\alpha, \lambda)$. Similarly, whenever Bob chooses setting $\beta$ and the hidden variable has value $\lambda$, then Bob will deterministically get the result $F_B(\beta, \lambda)$. Such a deterministic model would reproduce the joint probability distribution, provided that:

$P(A, B | \alpha, \beta) = \sum'_\lambda P(\lambda)$

where $\sum'$ means that the sum is only over those values of $\lambda$ such that
$F_A(\alpha, \lambda) = A$ and $F_B(\beta, \lambda) = B$.

Do you understand the distinction?

The distinction was not important for Bell, because it's easy to show that if there is a nondeterministic local hidden variables theory, then there is also a deterministic local hidden variables theory. So if he disproved the existence of a deterministic local hidden variables theory, that also proved that there was no nondeterministic local hidden variables theory.

But when I made my amplitude analogy, I was making an analogy to the nondeterministic theory, not the deterministic theory.

The functions $\psi(\lambda)$, $\psi_A(A|\alpha, \lambda)$, $\psi_B(B|\beta, \lambda)$ are the amplitude analogues of the probability distributions $P(\lambda)$, $P_A(A|\alpha, \lambda)$, $P_B(B|\beta, \lambda)$. They are not analogues to the deterministic functions $F_A(\alpha, \lambda)$ and $F_B(\beta, \lambda)$.

15. Dec 2, 2016

### stevendaryl

Staff Emeritus
You are misunderstanding what the functions $\psi_A(A|\alpha, \lambda)$ and $\psi_B(B|\beta, \lambda)$ are.

$\psi_A(A|\alpha, \lambda) =$ the amplitude for Alice getting result $A$, given that she chooses setting $\alpha$ and that the initial twin-particle state was described by variable $\lambda$.

$\psi_B(B|\beta, \lambda) =$ the amplitude for Bob getting result $B$, given that he chooses setting $\beta$ and that the initial twin-particle state was described by variable $\lambda$.

These are amplitudes. They are not measurement results. Amplitudes are complex numbers whose squares give probabilities for transitions.

16. Dec 2, 2016

### secur

Seems there's some confusion. I think the best way to straighten it out is, please address the other point I made, which is very simple.

In this typical Bell-type experiment, QM says A and B must always be opposite (product is -1) when their detector angles are equal. A valid hidden-variable model must reproduce that behavior. But that's not the case with your model:

Last edited: Dec 2, 2016
17. Dec 2, 2016

### secur

Another way to look at it: you don't specify the procedure whereby Alice and Bob generate their two results, A and B, which = +1 or -1. Presumably they're generated randomly based on their (known, given) amplitudes. I referred to it above hand-wavingly as "flipping a coin". Can you specify that procedure, showing how it guarantees the correct result for equal angles, namely, A * B = -1 ?

18. Dec 3, 2016

### stevendaryl

Staff Emeritus
It certainly is. The resulting probability amplitude is (the very first post):
• If $A=B=\pm 1$, then $\psi(A, B|\alpha, \beta) = \pm \frac{i}{\sqrt{2}} sin(\frac{\beta-\alpha}{2})$. This means that the probability amplitude that Alice and Bob both get the same result is proportional to $sin(\frac{\beta-\alpha}{2})$, which means it is zero when $\alpha = \beta$.
• If $A=-B=\pm 1$, then $\psi(A, B|\alpha, \beta) = \pm \frac{1}{\sqrt{2}} cos(\frac{\beta-\alpha}{2})$. This means that the probability amplitude that Alice and Bob get opposite results is proportional to $cos(\frac{\beta - \alpha}{2})$, which means that it's 0 when $\alpha - \beta = \pi$.
Look, the whole point of the first post was to reproduce the EPR spin-1/2 joint probability function.

19. Dec 3, 2016

### secur

Ok, I thought that answer would remove my confusion. This is NOT a hidden-variables model of the type addressed by Bell's theorem.

These are the amplitudes which, in usual Bell HV attempt, need to be achieved. One comes up with a scheme to generate A's and B's - two sequences of +-1's - which will result, ultimately, in these amplitudes (actually, the correlation function is usually aimed at). Alice's results can depend on her detector setting $\alpha$, the hidden variable(s), and almost anything else, except Bob's detector setting $\beta$. And similar for Bob.

But that's not the type of model you're doing. You present amplitudes - NOT sequences of detector results - which multiply (and, sum over the $\lambda$'s) to give the correct joint probability (or, amplitude). The two amplitudes, for Alice and Bob, don't depend on the other's detector setting - that's good. But you don't present correct results (indeed, any results) for each individual detection, nor (a fortiori) do you use them to get to the final joint distribution (or, correlation).

Exactly. NOT the individual results for each pair of entangled photons, as in all other attempted HV models.

That's fine. On its own terms, your model works. The only problem, however, is your assumption that a similar model can't reproduce the true QM probabilities, with squares of sin and cos.

No. It's been well-proven there's no (local, realistic) HV model of the usual sort giving the right QM predictions. But none of that work has addressed the different type of HV model you're presenting. Your model doesn't have to worry about individual photon-by-photon results, and the two factors from Alice and Bob are represented by complex numbers, not just +-1's. It may well be that a model like that can reproduce the right QM predictions. At least no one's shown otherwise. Without such demonstration this conclusion is not justified:

20. Dec 3, 2016

### stevendaryl

Staff Emeritus
Of course not. Bell proved that there was no such thing.

The point, which I made in the very first post, is that
1. We can formulate certain mathematical rules for how we think that probability ought to work, in a local realistic model.
2. We can prove that QM probabilities don't work that way.
3. However, the analogous rules for QM amplitudes do work that way.
Amplitudes work for QM in the way that we would expect probabilities to work in a local hidden variables model of the sort Bell investigated. As you say, and as I said in the very first post, amplitudes don't correspond directly to anything can measure, unlike probabilities, so it's unclear what relevance this observation is. I just thought it was interesting.