quantumapplitudes

Quantum Amplitudes, Probabilities and EPR

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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:

  1. 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.
  2. 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:

  1. The parameter [itex]\lambda[/itex] ranges over the two-element set, [itex]\{ +1, -1 \}[/itex]
  2. The amplitudes associated with these are: [itex]\psi(\lambda) = \frac{\lambda}{\sqrt{2}} = \pm \frac{1}{\sqrt{2}}[/itex]
  3. 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]
  4. 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.

 

 

81 replies
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  1. Stephen Tashi
    Stephen Tashi says:

    In a non-QM context, complex numbers have been used in the place of real number transition probabilites in Markov chains.  The simplest example, I know of this is the paper :  http://bidabad.com/doc/complex-prob.pdf.   The general idea is that the we have data for a Markov process who steps occur increments of time T and we wish to have a model that proceeds in smaller steps of time or a model that is a continuous time Markov process.

    I don't know what approach this paper takes, but it's an oft-cited work:

    D.R. Cox, A use of Complex Probabilities in the Theory of Stochastic Processess

    https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/div-classtitlea-use-of-complex-probabilities-in-the-theory-of-stochastic-processesdiv/3DE2C9013903EDD218F5B85129F65B2C

    I don't subscribe to that site and I haven't been able to find a free article that gives that citation and also explains the concept of the paper.

  2. RockyMarciano
    RockyMarciano says:

    One can derive the Shroedinger equation from the assumption that the process of state evolution is like a Markov process except with conditional probabilities replaced by conditional amplitudes.

    Sure, but that was a more general case, the nice thing here is that it is simplified to the two-state system in Bell's inequalities, and it is much easier to see how the amplitudes being complex can convey just the right amount of info about phase when squared so that the nonlocal statistical correlations of the physics can be correctly predicted. 

    While it doesn't explain why nature is like this it shows how the math gets it right and how this cannot be done with classical probabilities.

  3. lavinia
    lavinia says:

    Sure, but that was a more general case, the nice thing here is that it is simplified to the two-state system in Bell's inequalities, and it is much easier to see how the amplitudes being complex can convey just the right amount of info about phase when squared so that the nonlocal statistical correlations of the physics can be correctly predicted.

    While it doesn't explain why nature is like this it shows how the math gets it right and how this cannot be done with classical probabilities.

    There is a general principle being described here.

    A good exercise by the way is to see how discrete Brownian motion leads to the Heat Equation.

  4. lavinia
    lavinia says:

    I'm not sure what general principle you are referring to, can you state it explicitly?

    Just that the time evolutions of QM systems are like stochastic processes with amplitudes and conditional amplitudes instead of probabilities and conditional probabilities.  The reason for this is that the passage of time itself is a linear operator for any time increment. The Shroedinger equation for a free particle is only one example.

  5. RockyMarciano
    RockyMarciano says:

    Just that the time evolutions of QM systems are like stochastic processes with amplitudes and conditional amplitudes instead of probabilities and conditional probabilities.  The reason for this is that the passage of time itself is a linear operator for any time increment. The Shroedinger equation for a free particle is only one example.

    Absolutely, the importance of the amplitudes being complex in the shift from classical to quantum theories has been known from the beginning and was underlined by Feynman more than half a century ago.

    What I was trying to convey is that by using a pure state bipartite system something more about how what you call general principle actually works can be deduced. But maybe it would lead to depart slightly from the conclusions in the (now) insights article of the OP.

    I mean let's pretend that, as secur pointed out earlier, the conclusion that all the difference between the classical and the quantum correlations lies in the process of squaring the amplitudes is not completely correct and the difference lies actually in the amplitudes. Since stevendaryl showed that the only difference between the square root amplitudes and the usual quantum amplitudes is a +/- sign, i.e. a global phase, let's pretend(please bear with me) that this difference normally considered irrelevant is somehow not irrelevant in this case. Can anybody think of a mathematical reason global phase might be relevant for the argument of complex numbers?

    It is relevant at least formally in the analysis of EPR correlations in the form of one of the three angles needed for the analysis, that many peolple finds odd as they think it would be enough with two angles for the difference between the polarizers. This might give a clue for the question above.

  6. secur
    secur says:

    Rocky, my understanding is that the maths is just a convenience. Complex numbers have the ability to reduce two real solutions to one complex one.

    I think that's not the right way to look at QM. It's true that anywhere in math you can always reformulate complex numbers in terms of reals. But that doesn't mean the complex numbers have no physical meaning. Contrast QM to EM. You can do EM calculations using complex numbers but when finally getting the solution you take the real part. In that case the complex numbers are indeed "just a convenience" and EM (Maxwell's eqns) are naturally expressed with reals. But that's not the case with QM, where "i" has a deep physical meaning.

    Consider two and three dimensions. We could say they're "just a convenience": they can be represented as tensor products of two and three 1-dimensional real number lines. But not only is that very awkward, also it doesn't negate the fact that the two and three dimensions have very important physical relevance. Or, consider transcendental numbers like pi and e. For any given problem we can get an arbitrarily accurate answer by representing pi and e as finite rational numbers, with enough decimal places. But still, the exact transcendental numbers have very important physical meaning (circumference of circle, Euler's number). The point: the fact we can get rid of "i" in QM by awkwardly using coupled real number equations (actually it's even more trouble than that), doesn't mean it has no physical significance.

    My favorite way of seeing that significance comes from Paul Dirac. The Hilbert space vector which represents a pure state always has norm 1, of course (that's why we're dealing with projective Hilbert Space). Thus the wavefunction represents a point on the unit sphere (in infinite dimensions). Now, what is the time derivative of that vector? It can only move on the unit circle: that means it can only move orthogonally to its direction. So the time derivative must be 90 degrees from the state vector's direction. 90 degrees rotation is represented by multiplying by "i". So the time derivative is i times the direction of the vector, and that's what the "i" on the left side of Schroedinger's eqn is for. Over-simplifying a bit.

    BTW David Hestenes with his Geometric Algebra or Space-Time Algebra strongly makes the point that "i" is unnecessary in QM. But that doesn't contradict what I said above. He just substitutes a different square root of -1. (There are infinitely many square roots of -1 in geometric algebra.) So he still agrees that a quantity similar to i, playing the same role, has deep physical significance.

  7. edguy99
    edguy99 says:

    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:

    Can you clarify? I may be misunderstanding this.

    Do you mean "when their detector angles are equal, they will always detect the opposite" as a statement of how nature works or as a way of classifying whether an experiment is "bell-type" or not?

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