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.

 

 

83 replies
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  1. J
    Jilang says:

    [QUOTE="rubi, post: 5635382, member: 395236"]That depends on the model. There are several manifestly local quantum mechanical models. One example would be consistent histories. A careful analysis of the EPR paradox is done in the following paper:

    http://scitation.aip.org/content/aapt/journal/ajp/55/1/10.1119/1.14965

    Space-time is not observer dependent. Relativity doesn't claim that.[/QUOTE]

    Sorry, I don't have a registration with that provider. Can the third alternative (fourth-sorry Mike) be summarised here?

  2. J
    Jilang says:

    [QUOTE="zonde, post: 5635313, member: 129046"]Let's say I am giving you apples. Every time I give you apples we describe this event with positive (or at least non negative) integer. Every such event can be viewed as independent because it's different apples every time. But now let's say that event of me giving you apples can be described by any integer (positive, negative or zero). If I give you negative number of apples it actually means I am taking apples from you. Obviously event of taking away apples is not independent from event of giving you apples as the same apples participate in both events.

    But how would you model "negative" click in detector?[/QUOTE]

  3. J
    Jilang says:

    [QUOTE="zonde, post: 5635313, member: 129046"]Let's say I am giving you apples. Every time I give you apples we describe this event with positive (or at least non negative) integer. Every such event can be viewed as independent because it's different apples every time. But now let's say that event of me giving you apples can be described by any integer (positive, negative or zero). If I give you negative number of apples it actually means I am taking apples from you. Obviously event of taking away apples is not independent from event of giving you apples as the same apples participate in both events.

    But how would you model "negative" click in detector?[/QUOTE]

    Zonde, SD already stressed that it is not the amplitude that gets measured, You don't need to worry about negative clicks.

  4. Nugatory
    Nugatory says:

    [QUOTE="zonde, post: 5635313, member: 129046"]But how would you model "negative" click in detector?[/QUOTE]

    We don't.  We have negative (or even complex) amplitudes for positive or zero numbers of clicks.

    Stevendaryl's point about us not having an intuition for what it means to select a result according to an amplitude, as opposed to a probability, is looking pretty good right now….

  5. S
    secur says:

    [USER=372855]@stevendaryl[/USER]'s approach seems to be unaffected by this issue. It still works if we treat the amplitude the normal way, when it comes to selecting an actual result, since that's not crucial in his scheme. I.e., square the amplitude (complex norm) and use that as the probability. Perhaps I'm missing something.

  6. S
    Stephen Tashi says:

    [QUOTE="stevendaryl, post: 5635253, member: 372855"]

    The screwy thing about the amplitude story is that we have an intuitive idea about what it means to choose a value according to a certain probability distribution (rolling dice, for instance), but we don't have an intuitive idea about what it means to choose a value according to a certain amplitude.[/QUOTE]

    Can we make the description of the intuitive difficulty more precise?

    Mathematically,  it is easy to imagine choosing a value according to any sort of input variable.    You just need an algorithm that maps values of the input to a value that defines a probability.  You can use that probability to make you final choice.

    So doesn't the intuitive problem begin in step 4 or 5 instead of in step 1 and 2 ?

  7. M
    mikeyork says:

    stevendaryl: forgive me if I have misunderstood. But don't we already know what ##lambda## is? Isn't it the eigenvalue of the composite state? So if ##A,B## are individual spins, then ##lambda## is the composite spin. And your ##psi(A,B;alpha,beta,lambda)## are essentially Clebsch-Gordon coefficients  — apart from the rotation which takes the orientation of one detector into the other.

  8. R
    rubi says:

    [QUOTE="Jilang, post: 5635385, member: 492883"]Sorry, I don't have a registration with that provider. Can the third alternative (fourth-sorry Mike) be summarised here?[/QUOTE]

    Unfortunately, I don't think it can be understood easily without understanding consistent histories first. The CH answer is that the EPR argument is invalid, because it mixes incompatible frameworks. If you are interested in CH, you should check out Griffiths book "Consistent Quantum Theory". He also has some slides on his homepage: http://quantum.phys.cmu.edu/CHS/histories.html

  9. stevendaryl
    stevendaryl says:

    [QUOTE="mikeyork, post: 5635532, member: 22888"]stevendaryl: forgive me if I have misunderstood. But don't we already know what ##lambda## is? Isn't it the eigenvalue of the composite state? So if ##A,B## are individual spins, then ##lambda## is the composite spin. And your ##psi(A,B;alpha,beta,lambda)## are essentially Clebsch-Gordon coefficients  — apart from the rotation which takes the orientation of one detector into the other.[/QUOTE]

    That's on the right track, but not exactly right. In the EPR experiment, the composite spin is zero, so there is only one possible value for that.

    No, the meanings of the various amplitudes is this: Let [itex]|Phirangle[/itex] be the composite two-particle spin state. Then

    • [itex]lambda = +1 Rightarrow |Phirangle = |u_z d_zrangle[/itex]. The spin state of the first particle (the positron, say) is spin-up in the z-direction, and the spin state of the other particle (the electron) is spin-down in the z-direction.
    • [itex]lambda = -1 Rightarrow |Phirangle = |d_z u_zrangle[/itex]. The spin state of the first particle is spin-down in the z-direction, and the spin state of the other particle is spin-up in the z-direction.
    • [itex]lambda = +1 Rightarrow psi(lambda) = frac{1}{sqrt{2}}[/itex]
    • [itex]lambda = -1 Rightarrow psi(lambda) = frac{-1}{sqrt{2}}[/itex]

    So this is just a decomposition of the usual spin-zero state: [itex]frac{1}{sqrt{2}} (|u_z d_zrangle – |d_z u_zrangle)[/itex]. The amplitudes for [itex]lambda = +1[/itex] and [itex]lambda = -1[/itex] can be read off immediately.

    Then the other amplitudes:

    [itex]psi_A(A|alpha, lambda) = [/itex] the probability amplitude for measuring spin [itex]A/2[/itex] in the direction [itex]hat{x} cos(alpha) + hat{y} sin(alpha)[/itex], given that it was prepared to have spin [itex]lambda/2[/itex] in the direction [itex]hat{z}[/itex].

    [itex]psi_B(B|alpha, lambda) = [/itex] the probability amplitude for measuring spin [itex]B/2[/itex] in the direction [itex]hat{x} cos(alpha) + hat{y} sin(alpha)[/itex], given that it was prepared to have spin [itex]-lambda/2[/itex] in the direction [itex]hat{z}[/itex].

  10. M
    mikeyork says:

    [QUOTE="stevendaryl, post: 5635795, member: 372855"]In the EPR experiment, the composite spin is zero, so there is only one possible value for that.[/QUOTE]

    That's just a special case. Your ##lambda## can be in any other basis, but I think your  ##psi(lambda)## functions will be that of a superposition in that new basis equivalent to the composite spin state.

    As regards the differing orientations ##alpha## and ##beta##, that is simply a matter of a frame transformation (a rotation) of the spin projection direction for each detector and that is handled by D-functions (the simplest example being ##d(alpha)## and ##d(beta)## — rotations about the y-axis chosen to be perpendicular to the plane of the z-axis and your direction of projection).

    So although you are correct that your ##lambda## quantum numbers are not simply the composite spin, they are mathematically derived from it.

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