# Quantum Amplitudes, Probabilities and EPR

This is a little note about quantum amplitudes. Even though quantum probabilities seem very mysterious, with weird interference effects and seemingly nonlocal effects, the mathematics of quantum amplitudes are completely straight-forward. (The amplitude squared gives the probability.) As a matter of fact, the rules for computing amplitudes are almost exactly the same as the classical rules for computing probabilities for a memoryless stochastic process. (Memoryless means that future probabilities depend only on the current state, not on how it got to that state.)

**Probabilities for stochastic processes:**

If you have a stochastic process such as Brownian motion, then probabilities work this way:

Let [itex]P(i,t|j,t’)[/itex] be the probability that the system winds up in state [itex]i[/itex] at time [itex]t[/itex], given that it is in state [itex]j[/itex] at time [itex]t'[/itex].

Then these transition probabilities combine as follows: (Assume [itex]t’ < t” < t[/itex])

[itex]P(i,t|j,t’) = \sum_k P(i,t|k,t”) P(k,t”|j,t’)[/itex]

where the sum is over all possible intermediate states [itex]k[/itex].

There are two principles at work here:

- In computing the probability for going from state [itex]j[/itex] to state [itex]k[/itex] to state [itex]i[/itex], you multiply the probabilities for each “leg” of the path.
- In computing the probability for going from state [itex]j[/itex] to state [itex]i[/itex] via an intermediate state, you add the probabilities for each alternative intermediate state.

These are exactly the same two rules for computing transition amplitudes using Feynman path integrals. So there is an analogy: amplitudes are to quantum mechanics as probabilities are to classical stochastic processes.

Continuing with the analogy, we can ask the question as to whether there is a local hidden variables theory for quantum amplitudes. The answer is * YES*.

**Local “hidden-variables” model for EPR amplitudes**

Here’s a “hidden-variables” theory for the amplitudes for the EPR experiment.

First, a refresher on the probabilities for the spin-1/2 anti-correlated EPR experiment, and what a “hidden-variables” explanation for those probabilities would be:

In the EPR experiment, there is a source for anti-correlated electron-positron pairs. One particle of each pair is sent to Alice, and another is sent to Bob. They each measure the spin relative to some axis that they choose independently.

Assume Alice chooses her axis at angle [itex]\alpha[/itex] relative to the x-axis in the x-y plane, and Bob chooses his to be at angle [itex]\beta[/itex] (let’s confine the orientations of the detectors to the x-y plane, so that orientation can be given by a single real number, an angle). Then the prediction of quantum mechanics is that probability that Alice will get result [itex]A[/itex] (+1 for spin-up, relative to the detector orientation, and -1 for spin-down) and Bob will get result [itex]B[/itex] is:

[itex]P(A, B | \alpha, \beta) = \frac{1}{2} sin^2(\frac{\beta-\alpha}{2}) [/itex] if [itex]A = B[/itex]

[itex]P(A, B | \alpha, \beta) = \frac{1}{2} cos^2(\frac{\beta-\alpha}{2}) [/itex] if [itex]A \neq B[/itex]

A “local hidden variables” explanation for this result would be given by a probability distribution [itex]P(\lambda)[/itex] on values of some hidden variable [itex]\lambda[/itex], together with probability distributions

[itex]P_A(A | \alpha, \lambda)[/itex]

[itex]P_B(B | \beta, \lambda)[/itex]

such that

[itex]P(A, B | \alpha, \beta) = \sum P(\lambda) P_A(A|\alpha, \lambda) P_B(B|\beta, \lambda)[/itex]

(where the sum is over all possible values of [itex]\lambda[/itex]; if [itex]\lambda[/itex] is continuous, the sum should be replaced by [itex]\int d\lambda[/itex].)

The fact that the QM predictions violate Bell’s inequality proves that there is no such hidden-variables explanation of this sort.

But now, let’s go through the same exercise in terms of amplitudes, instead of probabilities. The amplitude for Alice and Bob to get their respective results is basically the square-root of the probability (up to a phase). So let’s consider the amplitude:

[itex]\psi(A, B|\alpha, \beta) \sim \frac{1}{\sqrt{2}} sin(\frac{\beta – \alpha}{2})[/itex] if [itex]A = B[/itex], and

[itex]\psi(A, B|\alpha, \beta) \sim \frac{1}{\sqrt{2}} cos(\frac{\beta – \alpha}{2})[/itex] if [itex]A \neq B[/itex].

(I’m using the symbol [itex]\sim[/itex] to mean “equal up to a phase”; I’ll figure out a convenient phase as I go).

In analogy with the case for probabilities, let’s say a “hidden variables” explanation for these amplitudes will be a parameter [itex]\lambda[/itex] with associated functions [itex]\psi(\lambda)[/itex], [itex]\psi_A(A|\lambda, \alpha)[/itex], and [itex]\psi_B(B|\lambda, \beta)[/itex] such that:

[itex]\psi(A, B|\alpha, \beta) = \sum \psi(\lambda) \psi_A(A | \alpha, \lambda) \psi_B(B | \beta, \lambda)[/itex]

where the sum ranges over all possible values for the hidden variable [itex]\lambda[/itex].

I’m not going to bore you (any more than you are already) by deriving such a model, but I will just present it:

- The parameter [itex]\lambda[/itex] ranges over the two-element set, [itex]\{ +1, -1 \}[/itex]
- The amplitudes associated with these are: [itex]\psi(\lambda) = \frac{\lambda}{\sqrt{2}} = \pm \frac{1}{\sqrt{2}}[/itex]
- When [itex]\lambda = +1[/itex], [itex]\psi_A(A | \alpha, \lambda) = A \frac{1}{\sqrt{2}} e^{i \alpha/2}[/itex] and [itex]\psi_B(B | \beta, \lambda) = \frac{1}{\sqrt{2}} e^{-i \beta/2}[/itex]
- When [itex]\lambda = -1[/itex], [itex]\psi_A(A | \alpha, \lambda) = \frac{1}{\sqrt{2}} e^{-i \alpha/2}[/itex] and [itex]\psi_B(B | \alpha, \lambda) = B \frac{1}{\sqrt{2}} e^{i \beta/2}[/itex]

**Check:**

[itex]\sum \psi(\lambda) \psi_A(A|\alpha, \lambda) \psi_B(B|\beta, \lambda) = \frac{1}{\sqrt{2}} (A \frac{1}{\sqrt{2}} e^{i \alpha/2}\frac{1}{\sqrt{2}} e^{-i \beta/2} – \frac{1}{\sqrt{2}} e^{-i \alpha/2} B \frac{1}{\sqrt{2}} e^{+i \beta/2})[/itex]

If [itex]A = B = \pm 1[/itex], then this becomes (using [itex]sin(\theta) = \frac{e^{i \theta} – e^{-i \theta}}{2i}[/itex]):

[itex] = \pm 1 \frac{i}{\sqrt{2}} sin(\frac{\alpha – \beta}{2})[/itex]

If [itex]A = -B = \pm 1[/itex], then this becomes (using [itex]cos(\theta) = \frac{e^{i \theta} + e^{-i \theta}}{2}[/itex]):

[itex] = \pm 1 \frac{1}{\sqrt{2}} cos(\frac{\alpha – \beta}{2})[/itex]

So we have successfully reproduced the quantum predictions for amplitudes (up to the phase [itex]\pm 1[/itex]).

**What does it mean?**

** **

In a certain sense, what this suggests is that quantum mechanics is a sort of “stochastic process”, but where the “measure” of possible outcomes of a transition is not real-valued probabilities but complex-valued probability amplitudes. When we just look in terms of amplitudes, everything seems to work out the same as it does classically, and the weird correlations that we see in experiments such as EPR are easily explained by local hidden variables, just as Einstein, Podolsky and Rosen hoped. But in actually testing the predictions of quantum mechanics, we can’t directly measure amplitudes, but instead compile statistics which give us probabilities, which are the squares of the amplitudes. The squaring process is in some sense responsible for the weirdness of QM correlations.

Do these observations contribute anything to our understanding of QM? Beats me. But they are interesting.

Nice job.

@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.

Do these observations contribute anything to our understanding of QM?

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

primaryencoding of theoretical probability and frequency counting as secondary.We can talk about probabilities when assumption that ensemble is i.i.d. holds. When this assumption holds probabilities of individual events

independentlycontribute 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.

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 [itex]lambda = +1[/itex] to indicate having the gene, and [itex]lambda = -1[/itex] to indicate not having the gene, we can compute:

[itex]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[/itex]

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

That involves summing over different values of [itex]lambda[/itex] in exactly the same way as the amplitudes were computed in my EPR example.

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

Yes, that's true! A similar conclusion is also drawn in

https://arxiv.org/abs/0707.2319

Independent and identically distributed

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

Great analysis for the original post!

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

[itex]psi(A, B|alpha, beta) = sum psi(lambda) psi_A(A | alpha, lambda) psi_B(B | beta, lambda)[/itex]

the numbers that appear for [itex] psi_A(A | alpha, lambda)[/itex] and [itex]psi_B(B | beta, lambda)[/itex] 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 [itex]alpha[/itex] = [itex]beta[/itex] 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.