# Learn About Quantum Amplitudes, Probabilities and EPR

Common Topics: amplitudes, probabilities, state, hidden, quantum

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 is completely straightforward. (The amplitude squared gives the probability.) As a matter of fact, the rules for computing amplitudes are almost 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 $P(i,t|j,t’)$ be the probability that the system winds up in state $i$ at time $t$, given that it is in state $j$ at time $t'$.

Then these transition probabilities combine as follows: (Assume $t’ < t” < t$)

$P(i,t|j,t’) = \sum_k P(i,t|k,t”) P(k,t”|j,t’)$

where the sum is over all possible intermediate states $k$.

There are two principles at work here:

1. In computing the probability for going from state $j$ to state $k$ to state $i$, you multiply the probabilities for each “leg” of the path.
2. In computing the probability for going from state $j$ to state $i$ 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 $\alpha$ relative to the x-axis in the x-y plane, and Bob chooses his to be at angle $\beta$ (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 the result $A$ (+1 for spin-up, relative to the detector orientation, and -1 for spin-down) and Bob will get result $B$ is:

$P(A, B | \alpha, \beta) = \frac{1}{2} sin^2(\frac{\beta-\alpha}{2})$ if $A = B$
$P(A, B | \alpha, \beta) = \frac{1}{2} cos^2(\frac{\beta-\alpha}{2})$ if $A \neq B$

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

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

such that

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

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

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:

$\psi(A, B|\alpha, \beta) \sim \frac{1}{\sqrt{2}} sin(\frac{\beta – \alpha}{2})$ if $A = B$, and
$\psi(A, B|\alpha, \beta) \sim \frac{1}{\sqrt{2}} cos(\frac{\beta – \alpha}{2})$ if $A \neq B$.

(I’m using the symbol $\sim$ 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 $\lambda$ with associated functions $\psi(\lambda)$, $\psi_A(A|\lambda, \alpha)$, and $\psi_B(B|\lambda, \beta)$ such that:

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

where the sum ranges over all possible values for the hidden variable $\lambda$.
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 $\lambda$ ranges over the two-element set, $\{ +1, -1 \}$
2. The amplitudes associated with these are: $\psi(\lambda) = \frac{\lambda}{\sqrt{2}} = \pm \frac{1}{\sqrt{2}}$
3. When $\lambda = +1$, $\psi_A(A | \alpha, \lambda) = A \frac{1}{\sqrt{2}} e^{i \alpha/2}$ and $\psi_B(B | \beta, \lambda) = \frac{1}{\sqrt{2}} e^{-i \beta/2}$
4. When $\lambda = -1$, $\psi_A(A | \alpha, \lambda) = \frac{1}{\sqrt{2}} e^{-i \alpha/2}$ and $\psi_B(B | \alpha, \lambda) = B \frac{1}{\sqrt{2}} e^{i \beta/2}$

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

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

$= \pm 1 \frac{i}{\sqrt{2}} sin(\frac{\alpha – \beta}{2})$

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

$= \pm 1 \frac{1}{\sqrt{2}} cos(\frac{\alpha – \beta}{2})$

So we have successfully reproduced the quantum predictions for amplitudes (up to the phase $\pm 1$).

## 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 that 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.

Tags:
83 replies
1. stevendaryl says:

[QUOTE="zonde, post: 5637534, member: 129046"]For your model to work in EPR case you would have to model post-processing using amplitudes i.e. detection results would have to remain hidden variables until coincidences are obtained and "measured".[/QUOTE]I don't understand your point about "post-processing", but I think I've said everything that there is to say: The goal was simply to write the amplitude for joint measurements as an amplitude-weighted sum of uncorrelated (product) amplitudes. It's just a mathematical exercise—I'm not making any claims about having any new interpretation of QM. It's just a way of describing the usual QM that parallels what is done with hidden-variables models (except that probabilities are replaced by amplitudes).

2. zonde says:

[QUOTE="stevendaryl, post: 5636842, member: 372855"]The EPR case is only more complicated, in that the observable being measured is a pair of values, one measured by Alice and one measured by Bob, and the initial state is a product state.[/QUOTE]For your model to work in EPR case you would have to model post-processing using amplitudes i.e. detection results would have to remain hidden variables until coincidences are obtained and "measured".

3. stevendaryl says:

[QUOTE="zonde, post: 5636652, member: 129046"]Let me ask a bit more about role of ##lambda## in your model. Do your model assumes that each detection of pair has only one of the possible values of ##lambda## attached to it? Or does every observation of pair have both values of ##lambda## attached to it?In other words do we get averaged amplitude $psi(A, B|alpha,beta)$ for every coincidence or do we get it only when we average over ensemble of pairs?[/QUOTE]My description in terms of $lambda$ is just an alternative way to describe the standard quantum mechanical computation of transition amplitudes, to make the description more parallel to a classical hidden-variables probabilistic model.The traditional approach is this:

• Let $|phi_lambdarangle$ be a complete basis for the system of interest.
• Let $H$ be its Hamiltonian.
• Let $O_alpha$ be a set of possible observables.
• Let $|chi_{alpha, j}rangle$ be a different complete basis in which the operator $O_alpha$ is diagonal. Let $a_{alpha, j}$ be the eigenvalue of $O_alpha$ for state $|chi_{alpha, j}rangle$. That is, $O_alpha |chi_{alpha, j}rangle = a_{alpha, j} |chi_{alpha, j}rangle$. For simplicity, let's assume that there is no degeneracy; there is only state $|chi_{alpha, j}rangle$ with the eigenvalue $a_{alpha,j}$.
• Let $gamma_{alpha, a, lambda}(t)$ be the transition amplitude: $langle chi_{alpha,j} | e^{-iHt} |phi_lambda rangle$, where $j$ is the unique index such that $a_{alpha, j} = a$

With all that, we can compute probability amplitudes for getting certain measurement results from certain initial states (Fix the time $t$ that the measurement will be performed, for simplicity):

• Write the initial state $|psirangle$ (at time $t=0$) as a superposition of basis states: $|psirangle = sum_lambda c_lambda |phi_lambdarangle$
• Then if the observer measures observable $O_alpha$ at time $t$, the probability amplitude of measuring value $a$ is given by: $sum_lambda c_lambda gamma_{alpha,a,lambda}(t)$

The translation to the "hidden variables" approach is trivial:

• Let the hidden variable, $lambda$, be the index of the complete basis used to describe the initial state.
• Let the initial amplitude, $psi(lambda)$ be given by $psi(lambda) = c_lambda$, the coefficient of the basis element $|phi_lambda$ of the initial state.
• Let the conditional amplitude be given by $psi(a | alpha, lambda) = gamma_{alpha,a,lambda}(t)$.
• Then the amplitude for getting result $a$ at time $t$ is given by: $psi_a(a | alpha) = sum_lambda psi(lambda) psi(a | alpha, lambda)$

The EPR case is only more complicated, in that the observable being measured is a pair of values, one measured by Alice and one measured by Bob, and the initial state is a product state.

4. zonde says:

[QUOTE="stevendaryl, post: 5636175, member: 372855"]The summation $sum_lambda psi(lambda) psi_A(A|alpha, lambda) psi_B(B|beta, lambda)$ does not represent any kind of post-processing, because $lambda$ is hidden, so nobody ever knows what its value was. The meaning of that sum is that this is a model explaining how the amplitude $psi(A, B|alpha,beta)$ might arise. So I'm not sure I understand why you say it's nonlocal.I guess in general, it's sort of nonlocal to add probabilities, or probability amplitudes, but that seems like a necessary step in order to talk about nonlocal correlations, which is what hidden variables are supposed to explain. I don't understand why you think it's more nonlocal to talk about amplitudes than to talk about probabilities.[/QUOTE]Let me ask a bit more about role of ##lambda## in your model. Do your model assumes that each detection of pair has only one of the possible values of ##lambda## attached to it? Or does every observation of pair have both values of ##lambda## attached to it?In other words do we get averaged amplitude $psi(A, B|alpha,beta)$ for every coincidence or do we get it only when we average over ensemble of pairs?

5. stevendaryl says:

[QUOTE="zonde, post: 5636062, member: 129046"]Your Story 2 is quite counterintuitive but it's not the problem I see there. It is a step further where I see the problem.In your description of entangled pair, amplitudes are attached to different coincidences not simply local measurements (as in your example with babies). And then amplitudes of coincidences can cancel out. The problem is that coincidences are not basic measurements but rather derived by postselection. So it would seem that this cancelation should happen somewhere in the process of postselection.Imagine experimentalist who compares two sheets of paper where there are "clicks" of detectors with timestamps from two locations. He then on the third paper counts coincidences in two columns as being the same polarization or opposite polarization. But then he sometimes subtracts coincidence from one or the other column (amplitudes canceling out). That is the strange part in it.[/QUOTE]The summation $sum_lambda psi(lambda) psi_A(A|alpha, lambda) psi_B(B|beta, lambda)$ does not represent any kind of post-processing, because $lambda$ is hidden, so nobody ever knows what its value was. The meaning of that sum is that this is a model explaining how the amplitude $psi(A, B|alpha,beta)$ might arise. So I'm not sure I understand why you say it's nonlocal.I guess in general, it's sort of nonlocal to add probabilities, or probability amplitudes, but that seems like a necessary step in order to talk about nonlocal correlations, which is what hidden variables are supposed to explain. I don't understand why you think it's more nonlocal to talk about amplitudes than to talk about probabilities.

6. zonde says:

[QUOTE="stevendaryl, post: 5636010, member: 372855"]I don't understand why you say the sign of the summands implies something about locality, although I do agree that there is something fishy about canceling amplitudes.Let me go through a pair of "stories", one about probabilities, and one about probability amplitudes, and maybe I can get at the reason that you think there is something weird about amplitudes.Story 1: Suppose that there is a "left-handed" gene, such that if you have it from either of your parents, you're 81% likely to be left-handed, and if you lack it, you are 81% likely to be right-handed. So a couple has a baby, and by Mendelian genetics, we figure that the baby has a 50% chance of getting the gene from the father. So we compute that he has a 50% chance of being-lefthanded: $.5 cdot .81 + .5 cdot .19 = .5$. Presumably, we could test this empirically by checking many babies in the same genetic situation.Story 2: Suppose that it works by amplitudes, rather than probabilities. If the baby has the gene, he has an amplitude of 0.9 of being left-handed, and 0.44 of being right-handed. If he lacks the gene, the amplitudes are switched. Now, suppose that we compute that he has an amplitude of +0.44 of having the gene, and -0.9 of lacking the gene. Then the amplitude that he is left-handed is $.44 cdot .9 + (-.9) cdot .44 = 0$. So he has ZERO chance of being left-handed.The mathematical analysis is very similar in both cases. However, in the first case, each set of parents can reason that really the baby either has the gene, or doesn't, and that the probability reflects their lack of knowledge about the true state of their baby. In the second case, it's hard to see how certainty (that the baby will not be left-handed) can arise from lack of knowledge.[/QUOTE]Your Story 2 is quite counterintuitive but it's not the problem I see there. It is a step further where I see the problem.In your description of entangled pair, amplitudes are attached to different coincidences not simply local measurements (as in your example with babies). And then amplitudes of coincidences can cancel out. The problem is that coincidences are not basic measurements but rather derived by postselection. So it would seem that this cancelation should happen somewhere in the process of postselection.Imagine experimentalist who compares two sheets of paper where there are "clicks" of detectors with timestamps from two locations. He then on the third paper counts coincidences in two columns as being the same polarization or opposite polarization. But then he sometimes subtracts coincidence from one or the other column (amplitudes canceling out). That is the strange part in it.

7. Mentz114 says:

[QUOTE="stevendaryl, post: 5636010, member: 372855"]I don't understand why you say the sign of the summands implies something about locality, although I do agree that there is something fishy about canceling amplitudes.Let me go through a pair of "stories", one about probabilities, and one about probability amplitudes, and maybe I can get at the reason that you think there is something weird about amplitudes…..The mathematical analysis is very similar in both cases. However, in the first case, each set of parents can reason that really the baby either has the gene, or doesn't, and that the probability reflects their lack of knowledge about the true state of their baby. In the second case, it's hard to see how certainty (that the baby will not be left-handed) can arise from lack of knowledge.[/QUOTE]It looks like you've introduced interference which rules out one possibility. Lack of information is easy to define in quantum systems but (apparently) has a different interpretation between CM and QM. I found this paper very interesting on this subject arXiv:quant-ph/9601025v1 25 Jan 1996

8. stevendaryl says:

[QUOTE="zonde, post: 5635934, member: 129046"]Compare these two expressions:##P(A,B|alpha, beta) = sum_lambda P(lambda) P_A(A|alpha, lambda) P_B(B|beta, lambda)## (1)##psi(A,B|alpha, beta) = sum_lambda psi(lambda) psi_A(A|alpha, lambda) psi_B(B|beta, lambda)## (2)In these two sums we combine non-local summands. There is no problem as these summands are acquired from local values by postprocessing.The question however is if this summation can be viewed as part of postprocessing. In case of (1) summands are nonnegative and can be viewed as independent elements in postprocessing.But in (2) summands do not combine statistically as they can cancel each other out so it can't be just a step in postprocessing. So this summation is not consistent with locality.[/QUOTE]I don't understand why you say the sign of the summands implies something about locality, although I do agree that there is something fishy about canceling amplitudes.Let me go through a pair of "stories", one about probabilities, and one about probability amplitudes, and maybe I can get at the reason that you think there is something weird about amplitudes.Story 1: Suppose that there is a "left-handed" gene, such that if you have it from either of your parents, you're 81% likely to be left-handed, and if you lack it, you are 81% likely to be right-handed. So a couple has a baby, and by Mendelian genetics, we figure that the baby has a 50% chance of getting the gene from the father. So we compute that he has a 50% chance of being-lefthanded: $.5 cdot .81 + .5 cdot .19 = .5$. Presumably, we could test this empirically by checking many babies in the same genetic situation.Story 2: Suppose that it works by amplitudes, rather than probabilities. If the baby has the gene, he has an amplitude of 0.9 of being left-handed, and 0.44 of being right-handed. If he lacks the gene, the amplitudes are switched. Now, suppose that we compute that he has an amplitude of +0.44 of having the gene, and -0.9 of lacking the gene. Then the amplitude that he is left-handed is $.44 cdot .9 + (-.9) cdot .44 = 0$. So he has ZERO chance of being left-handed.The mathematical analysis is very similar in both cases. However, in the first case, each set of parents can reason that really the baby either has the gene, or doesn't, and that the probability reflects their lack of knowledge about the true state of their baby. In the second case, it's hard to see how certainty (that the baby will not be left-handed) can arise from lack of knowledge.

9. stevendaryl says:

[QUOTE="mikeyork, post: 5635899, member: 22888"]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).[/QUOTE]Sorry, I misunderstood what you were saying. You're exactly right–the $psi(lambda)$ are Clebsch-Gordan coefficients, and the $psi_A(A|alpha, beta)$ are the coefficients relating spin-up in one direction to spin-up in another direction.

10. zonde says:

[QUOTE="Nugatory, post: 5635397, member: 382138"]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….[/QUOTE]Compare these two expressions:##P(A,B|alpha, beta) = sum_lambda P(lambda) P_A(A|alpha, lambda) P_B(B|beta, lambda)## (1)##psi(A,B|alpha, beta) = sum_lambda psi(lambda) psi_A(A|alpha, lambda) psi_B(B|beta, lambda)## (2)In these two sums we combine non-local summands. There is no problem as these summands are acquired from local values by postprocessing.The question however is if this summation can be viewed as part of postprocessing. In case of (1) summands are nonnegative and can be viewed as independent elements in postprocessing.But in (2) summands do not combine statistically as they can cancel each other out so it can't be just a step in postprocessing. So this summation is not consistent with locality.

11. 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.

12. 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 $|Phirangle$ be the composite two-particle spin state. Then

• $lambda = +1 Rightarrow |Phirangle = |u_z d_zrangle$. 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.
• $lambda = -1 Rightarrow |Phirangle = |d_z u_zrangle$. 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.
• $lambda = +1 Rightarrow psi(lambda) = frac{1}{sqrt{2}}$
• $lambda = -1 Rightarrow psi(lambda) = frac{-1}{sqrt{2}}$

So this is just a decomposition of the usual spin-zero state: $frac{1}{sqrt{2}} (|u_z d_zrangle – |d_z u_zrangle)$. The amplitudes for $lambda = +1$ and $lambda = -1$ can be read off immediately.Then the other amplitudes:$psi_A(A|alpha, lambda) =$ the probability amplitude for measuring spin $A/2$ in the direction $hat{x} cos(alpha) + hat{y} sin(alpha)$, given that it was prepared to have spin $lambda/2$ in the direction $hat{z}$.$psi_B(B|alpha, lambda) =$ the probability amplitude for measuring spin $B/2$ in the direction $hat{x} cos(alpha) + hat{y} sin(alpha)$, given that it was prepared to have spin $-lambda/2$ in the direction $hat{z}$.

13. 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

14. 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.

15. 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 ?

16. 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.

17. 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….

18. 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.

19. 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]

20. 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/55/1/10.1119/1.14965Space-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?

21. rubi says:

[QUOTE="Jilang, post: 5635356, member: 492883"]Without hidden variables or interactions over space time distances what is another explanation?[/QUOTE]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[Mentor's note:  This post has been edited to remove a reply to a deleted post]

22. mikeyork says:

[QUOTE="Jilang, post: 5635356, member: 492883"]Without hidden variables or interactions over space time distances what is another explanation?[/QUOTE]Here's one: space-time distances are a creation of the observer, not fundamental to reality. We already know from relativity that space-time is observer-dependent. So what is it in the absence of an observer?

23. Jilang says:

[QUOTE="rubi, post: 5635291, member: 395236"]I don't think that this is how people argue for locality of QM. The argument for locality is that a hidden parameter is not the only possible explanation for the correlations, because mathematically, the assumption of a hidden parameter is a non-trivial restriction on the set of models (i.e. hidden variable models aren't the most general models). In order for a particular model to be local, that model just needs to offer an explanation for how the correlations can come about without invoking interactions over space-like distances.[/QUOTE]Without hidden variables or interactions over space time distances what is another explanation?

24. zonde says:

[QUOTE="stevendaryl, post: 5633728, member: 372855"]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.[/QUOTE]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?

25. rubi says:

[QUOTE="stevendaryl, post: 5635253, member: 372855"]And as a matter of fact, when people give rigorous mathematical proofs of the locality of quantum mechanics or quantum field theory, they are really showing that amplitudes behave locally, even if probabilities do not.[/QUOTE]I don't think that this is how people argue for locality of QM. The argument for locality is that a hidden parameter is not the only possible explanation for the correlations, because mathematically, the assumption of a hidden parameter is a non-trivial restriction on the set of models (i.e. hidden variable models aren't the most general models). In order for a particular model to be local, that model just needs to offer an explanation for how the correlations can come about without invoking interactions over space-like distances.

26. stevendaryl says:

Just to expand a little bit about the analogy between classical probabilities and quantum amplitudes:Here is the (false, as shown by tests of Bell's inequality) classical nondeterministic local hidden-variables story for correlated measurements:

1. There is a source of twin particles. Each twin particle is associated with a hidden parameter, $lambda$. The source randomly chooses a value of $lambda$ according to some probability distribution $P(lambda)$.
2. Alice chooses a detector setting $alpha$ for her measurement. Her detector randomly chooses a value for the output, $A$, according to a probability distribution $P_A(A|alpha, lambda)$, which depends on both $lambda$ and $alpha$.
3. Bob chooses a detector setting $beta$ for his measurement. His detector randomly chooses a value for the output, $B$, according to a probability distribution $P_B(B|beta, lambda)$, which depends on both $lambda$ and $beta$.
4. Steps 2&3 are independent, so the joint probability is just a product of the individual probabilities: $P(A,B|alpha, beta, lambda) = P_A(A|alpha, lambda) P_B(B|beta, lambda)$.
5. Since Alice and Bob don't know $lambda$, we average over all possible values, weighted by the probability $P(lambda)$, to get a correlated joint probability distribution: $P(A,B|alpha, beta) = sum_lambda P(lambda) P(A,B|alpha, beta, lambda)$

So this story would explain the correlation in Alice's and Bob's measurements as being due to a common (though unknown) hidden variable, $lambda$. That's what a local hidden variable theory would do, if there were one. Note, that even though this model is nondeterministic, all choices being made—which value of $lambda$, which value of $A$, which value of $B$—are made using only local information.Here's the analogous story for amplitudes:

1. There is a source of twin particles. Each twin particle is associated with a hidden parameter, $lambda$. The source randomly chooses a value of $lambda$ according to some amplitude $psi(lambda)$.
2. Alice chooses a detector setting $alpha$ for her measurement. Her detector randomly chooses a value for the output, $A$, according to an amplitude $psi_A(A|alpha, lambda)$, which depends on both $lambda$ and $alpha$.
3. Bob chooses a detector setting $beta$ for his measurement. His detector randomly chooses a value for the output, $B$, according to an amplitude $psi_B(B|beta, lambda)$, which depends on both $lambda$ and $beta$.
4. Steps 2&3 are independent, so the joint amplitude is just a product of the individual probabilities: $psi(A,B|alpha, beta, lambda) = psi_A(A|alpha, lambda) psi_B(B|beta, lambda)$.
5. Since Alice and Bob don't know $lambda$, we average over all possible values, weighted by the amplitude $psi(lambda)$ to get a correlated joint probability distribution: $psi(A,B|alpha, beta) = sum_lambda psi(lambda) psi(A,B|alpha, beta, lambda)$

The amplitude story has one final step:6. We compute a probability from the amplitude, according to the rule $P(A,B|alpha, beta) = |psi(A,B|alpha,beta)|^2$The hidden-variables amplitude story sounds as local as the hidden-variables probability story. And as a matter of fact, when people give rigorous mathematical proofs of the locality of quantum mechanics or quantum field theory, they are really showing that amplitudes behave locally, even if probabilities do not.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.

27. mikeyork says:

[QUOTE="mfb, post: 5634988, member: 405866"]It looks like you talk about phases, not amplitudes.[/QUOTE]Yes, of course; the magnitude is already physically significant in giving the probability.

28. secur says:

[QUOTE="stevendaryl, post: 5634793, member: 372855"]The point, which I made in the very first post, is that1.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.[/QUOTE]Yes it's interesting, and now it's a lot clearer to me, as delineated in my …[QUOTE="secur, post: 5634486, member: 588176"]previous post.[/QUOTE]Initially I thought you were talking about the standard Bell experiment hidden-variables situation. You're actually doing something a bit different, and unique, AFAIK. That's a problem with an original idea: many people will mistake it for the "same old thing" they've heard before. Reviewing OP the misunderstanding seems pretty natural. On the plus side we've been able to bring out some of the subtleties of the standard Bell HV model, by contrast. Also, no doubt others thought the same as I did, so it was worthwhile to straighten that out. Thanks for the stimulating thread!

29. mfb says:

[QUOTE="mikeyork, post: 5634970, member: 22888"]Because of an arbitrary global phase. But relative amplitudes are physically significant. Spin-statistics is an obvious example.[/QUOTE]It looks like you talk about phases, not amplitudes.

30. mikeyork says:

[QUOTE="stevendaryl, post: 5634793, member: 372855"]amplitudes don't correspond directly to anything can measure[/QUOTE]Because of an arbitrary global phase. But relative amplitudes are physically significant. Spin-statistics is an obvious example.

31. Jilang says:

That the situation is local at some level could remove an element of the spookiness (of which far too much is made).

32. stevendaryl says:

[QUOTE="secur, post: 5634486, member: 588176"]Ok, I thought that answer would remove my confusion. This is NOT a hidden-variables model of the type addressed by Bell's theorem.[/QUOTE]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.

33. secur says:

Ok, I thought that answer would remove my confusion. This is NOT a hidden-variables model of the type addressed by Bell's theorem.[QUOTE="stevendaryl, post: 5634440, member: 372855"]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 $alpha – beta = pi$.[/QUOTE]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).[QUOTE="stevendaryl"]Look, the whole point of the first post was to reproduce the EPR spin-1/2 joint probability function.[/QUOTE]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.[QUOTE="stevendaryl, post: 5633307, member: 372855"]The fact that the QM predictions violate Bell's inequality proves that there is no such hidden-variables explanation of this sort.[/QUOTE]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:[QUOTE="stevendaryl"]The squaring process is in some sense responsible for the weirdness of QM correlations.[/QUOTE]

34. stevendaryl says:

[QUOTE="secur, post: 5634338, member: 588176"]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:[/QUOTE]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.

35. secur says:

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 ?

36. secur 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:[QUOTE="secur, post: 5633911, member: 588176"]Sorry, this is not a hidden-variable model as understood in Bell experiment. The problem is that A and B results must be +-1 …… 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.[/QUOTE]

37. stevendaryl says:

[QUOTE="secur, post: 5634091, member: 588176"][USER=492883]@Jilang[/USER]. 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".[/QUOTE]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.

38. stevendaryl says:

[QUOTE="secur, post: 5633911, member: 588176"]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.[/QUOTE]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 $F_B(beta, lambda)$.

39. secur says:

[USER=492883]@Jilang[/USER]. 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 [USER=372855]@stevendaryl[/USER]'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.

40. Jilang says:

Secur, why can't the wavefunctions be complex?

41. secur says:

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.

42. DrChinese says:

[QUOTE="stevendaryl, post: 5633728, member: 372855"]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.[/QUOTE]Great analysis for the original post!

43. Demystifier says:

[QUOTE="stevendaryl, post: 5633307, member: 372855"]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.[/QUOTE]Yes, that's true! A similar conclusion is also drawn in https://arxiv.org/abs/0707.2319

44. stevendaryl says:

[QUOTE="zonde, post: 5633718, member: 129046"]We can talk about probabilities when assumption that ensemble is i.i.d. holds. [/QUOTE]I skipped over this first line without asking: What does "i.i.d." stands for?

45. stevendaryl says:

[QUOTE="zonde, post: 5633718, member: 129046"]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.[/QUOTE]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.[QUOTE]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.[/QUOTE]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.

46. zonde says:

[QUOTE="stevendaryl, post: 5633307, member: 372855"]Do these observations contribute anything to our understanding of QM? Beats me. But they are interesting.[/QUOTE]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.

47. mikeyork says:

[QUOTE="stevendaryl, post: 5633307, member: 372855"]Do these observations contribute anything to our understanding of QM?[/QUOTE]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.

48. Mentz114 says:

[USER=372855]@stevendaryl[/USER]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.