Learn About 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 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.)
Table of Contents
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 the 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 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.
[QUOTE="Jilang, post: 5682702, member: 492883"]Hi edguy99, I would be interested in how you see anything "slightly random" in this an how a wobble would work in so much as opposite angles produce opposite results. (NB the Insight considers spins not polarisations).[/QUOTE]I will try. The use of polarization angle refers to the orientation of the spin axis as in Jones Vector. Consider the spin of a toy gyroscope or top. Once the top is spun up, the spin axis will start to precess around the vertical axis. Imagine looking at a spinning precessing top from the front as if it is coming towards you. What you see in that 2D picture is an axis of spin that is wobbling back and forth past the vertical axis. Ie, not spinning right, not spinning left, but not completely vertical either.The randomness comes in that Bob and Alice may receive entangled photons, but measure their orientation differently due to the wobble. Ie, if they both measure along the basis vectors of 0 or 90, they will always get matching results. If they measure along in between angles, the "wobble" is enough to cause their results to differ occasionally, even on entangled pairs. In the experiment in question, these pairs are being thrown out as "noise" and only the "matching" entangled pairs are being used.
[QUOTE="edguy99, post: 5682521, member: 114935"]Thanks for bumping this. I forgot how truly great this insight is where [USER=372855]@stevendaryl[/USER] asks us to consider a model of the photon, where the probability amplitude of detecting a photon at a particular angle outside of its basis vectors (vertical or horizontal) is slightly random and non-linear, specifically:ψ(A,B|α,β) ∼ 1/√2*sin((β–α)/2) if A=B, andψ(A,B|α,β) ∼ 1/√2*cos((β–α)/2) if A≠B.[/QUOTE]Hi edguy99, I would be interested in how you see anything "slightly random" in this an how a wobble would work in so much as opposite angles produce opposite results. (NB the Insight considers spins not polarisations).
[QUOTE="DrChinese, post: 5682603, member: 323"]Those cannot be thrown (for being a mismatch) out in an actual experiment. That would defeat the purpose. They can be analyzed for tuning purposes. Sometimes, they help determine the proper time window for matching. Photons that arrive too far apart (per expectation) are much less likely to be entangled.[/QUOTE]You are correct that it defeats the purpose. From the experiment "The detectors, two single-photon counting modules (SPCMs), are preceded by linear polarizers and red filters to block any scattered laser light. Even so, it is necessary to use coincidence detection to separate the downconverted photons from the background of other photons reaching the detectors."Clearly they are thowing out any "non-coincidence" detection as noise.
[QUOTE="edguy99, post: 5682521, member: 114935"]In the experiment here, the [USER=372855]@stevendaryl[/USER] model works since Dehlinger and Mitchell consider all mismatched photons (when α = β) to be noise and throw them out. As far as I can see most other experiments consider mismatched photons as noise, does anyone have a counter-example?[/QUOTE]Those cannot be thrown (for being a mismatch) out in an actual experiment. That would defeat the purpose. They can be analyzed for tuning purposes. Sometimes, they help determine the proper time window for matching. Photons that arrive too far apart (per expectation) are much less likely to be entangled.
[QUOTE="Greg Bernhardt, post: 5682032, member: 1"]Nice Insight [USER=372855]@stevendaryl[/USER]![/QUOTE]Thanks for bumping this. I forgot how truly great this insight is where [USER=372855]@stevendaryl[/USER] asks us to consider a model of the photon, where the probability amplitude of detecting a photon at a particular angle outside of its basis vectors (vertical or horizontal) is slightly random and non-linear, specifically:ψ(A,B|α,β) ∼ 1/√2*sin((β–α)/2) if A=B, andψ(A,B|α,β) ∼ 1/√2*cos((β–α)/2) if A≠B.Think of this as a photon coming straight at you that has a wobble. It's been prepared vertical (90º) except that it wobbles back and forth a bit. If you measure it vertically, it will always be vertical. If you measure it horizontal, it will never be horizontal. If you measure it at 45º, randomly it will be 50% vertical, 50% horizontal. But, if you measure it at 60º, it will have MORE then a 66% chance of being vertical.Whether using probabilities or amplitudes, as [USER=588176]@secur[/USER] has pointed out, this is a non-bell model since "Another way to put it, your scheme doesn't guarantee that if α = β then their results will definitely be opposite."In the experiment here, the [USER=372855]@stevendaryl[/USER] model works since Dehlinger and Mitchell consider all mismatched photons (when α = β) to be noise and throw them out. As far as I can see most other experiments consider mismatched photons as noise, does anyone have a counter-example?
Nice Insight @stevendaryl!
[QUOTE="Jilang, post: 5640513, member: 492883"]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.[/QUOTE]It's tempting to think that real number probabilities are the "real" (in the sense of genuine) type of probability.However, it's worthwhile remembering that the standard formulation of probability theory in terms of real numbers is also just a convenience. It is convenient because real valued probabilities resemble observed frequencies and there are analogies between computations involving probabilities and computations involving observed fequencies.Even people who have studied advanced probability theory tend to confuse observed frequencies with probabilities, However, standard probability theory gives no theorems about observed frequencies except those that talk about the probability of an observed frequency. So probability theory is exclusively about probability. It is circular in that sense.In trying to apply probability theory to observations, the various statistical methods that are used likewise are computations whose results give the probabilities of the observations or parameters that cause them. Furthermore, in mathematical probability theory ( i.e. the Kolmogorov approach) there is no formal definition of an "observation" , in the sense of an event that "actually happens". There isn't even an axiom that says it is possible to take random samples. The closest one gets to the concept of a "possibility" that "actually happens" in the definition of conditional probability and that definition merely defines a "conditional distribution" as a quotient and uses the terminology that an event is "given". The definition of conditional probability doesn't define "given" as a concept by itself. (This is analogous to the fact that the concept of "approaches" has no technical definition within the definition ##lim_{xrightarrow a} f(x)##. even though the word "approaches" appears when we verbalize the notation )The intuitive problem with using complex numbers as a basis for probability theory seems (to me) to revolve around the interpretation of conditional (complex) probabilities. They involve a concept of "given" that is different from the conventional concept of "given". This is a contrast between intuitions, not a contrast between an intutition and a precisely defined mathematical concept because conventional probability theory has no precisely defined concept of "given" – even though it's usually crystal clear how we want to define "given" when we apply that theory to a specific problem.
[QUOTE="stevendaryl, post: 5641864, member: 372855"]Well, it's more dramatic with complex amplitudes, but interference effects would show up even if all amplitudes are positive real numbers.Suppose you do a double slit experiment with positive real amplitudes. A photon can either go through the left slit, with probability [itex]p[/itex], or the other slit, with probability [itex]1-p[/itex]. If it goes through the left slit, say that it has a probability of [itex]q_L[/itex] of triggering a particular photon detector. If it goes through the right slit, say that it has a probability of [itex]q_R[/itex] of triggering that detector. Then the amplitude for triggering the detector, when you don't observe which slit it goes through, is:[itex]psi = sqrt{p} sqrt{q_L} + sqrt{1-p}sqrt{q_R}[/itex]leading to a probability[itex]P = |psi|^2 = p q_L + (1-p) q_R + 2 sqrt{p(1-p)q_L q_R}[/itex]That last term is the interference term, and it seems nonlocal, in the sense that it depends on details of both paths (and so in picturesque terms, the photon seems to have taken both paths). Without negative numbers, the interference term is always positive, so you don't have the stark pattern of zero-intensity bands that come from cancellations, but you still have a similar appearance of nonlocality.[/QUOTE]You obviously mean that a "nonlocal term" with dependence on the two paths is indeed there(that is no longer really an "interference term" since as you wrote the pattern is lost without cancellations). This observation is of course true but one should wonder where does this term come from to begin with. And the only reason is that a 2-norm is being used to compute the probabilities, if we just used the one-norm of real valued probabilities only the probabilities from each path(without cross-term) would be summed to 1, as all probabilities must sum up. It is only because the quadratic 2-norm of a complex line(Argand surface) is being used that an additional term that includes both paths appears, and their squares is what is summed to 1.So I'm afraid you can't get rid of complex numbers as they are needed to explain the appearance of a cross-term in the first place.
[QUOTE="RockyMarciano, post: 5641842, member: 585697"]True, but it is in the context of complex numbers that you can integrate those nonpositive amplitudes in a coherent mathematical way.I think we basically agree that all the weirdness is due to using complex numbers instead of reals as inputs(as commented by Lavinia in previous post this is nothing new), so maybe my point is just a nitpicking that might seem pedantic, but mathematically I think it is important to remark that the difference between classical and EPR correlations is not just the squaring, but as you say the squaring combined with more things that conform the complex structure of QM.[/QUOTE]Well, it's more dramatic with complex amplitudes, but interference effects would show up even if all amplitudes are positive real numbers.Suppose you do a double slit experiment with positive real amplitudes. A photon can either go through the left slit, with probability [itex]p[/itex], or the other slit, with probability [itex]1-p[/itex]. If it goes through the left slit, say that it has a probability of [itex]q_L[/itex] of triggering a particular photon detector. If it goes through the right slit, say that it has a probability of [itex]q_R[/itex] of triggering that detector. Then the amplitude for triggering the detector, when you don't observe which slit it goes through, is:[itex]psi = sqrt{p} sqrt{q_L} + sqrt{1-p}sqrt{q_R}[/itex]leading to a probability[itex]P = |psi|^2 = p q_L + (1-p) q_R + 2 sqrt{p(1-p)q_L q_R}[/itex]That last term is the interference term, and it seems nonlocal, in the sense that it depends on details of both paths (and so in picturesque terms, the photon seems to have taken both paths). Without negative numbers, the interference term is always positive, so you don't have the stark pattern of zero-intensity bands that come from cancellations, but you still have a similar appearance of nonlocality.
[QUOTE="stevendaryl, post: 5641822, member: 372855"]Well, it's the combination of nonpositive amplitudes and squaring that leads to interference effects. (You don't need complex amplitudes for that, just negative ones).[/QUOTE]True, but it is in the context of complex numbers that you can integrate those nonpositive amplitudes in a coherent mathematical way. I think we basically agree that all the weirdness is due to using complex numbers instead of reals as inputs(as commented by Lavinia in previous post this is nothing new), so maybe my point is just a nitpicking that might seem pedantic, but mathematically I think it is important to remark that the difference between classical and EPR correlations is not just the squaring, but as you say the squaring combined with more things that conform the complex structure of QM.
[QUOTE="RockyMarciano, post: 5641815, member: 585697"]To be specific, the probability amplitudes used to obtain probability densities are different from the amplitudes up to sign obtained from the square root of the probabilities. Namely, only the former have a complex phase, so it seems it is this complex phase rather than their squaring that is responsible for the differences between classical and quantum correlations.[/QUOTE]Well, it's the combination of nonpositive amplitudes and squaring that leads to interference effects. (You don't need complex amplitudes for that, just negative ones).
[QUOTE="Jilang, post: 5640513, member: 492883"]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.[/QUOTE]In general you are right that math is just a convenient tool to describe the physics, but I'm not questioning this when I try to anlayze the role of the complex structure of amplitudes in the context of classical probabilties versus EPR.I think if we are invited to think about and draw conclusions from the clear set up in the OP we have to consider the role of the complex structure in this particular case, not necessarily to derive anything about nature but about the mathematical meaning of the variables involved here and therefore wich are the valid conclusions to draw if any..To be specific, the probability amplitudes used to obtain probability densities are different from the amplitudes up to sign obtained from the square root of the probabilities. Namely, only the former have a complex phase, so it seems it is this complex phase rather than their squaring that is responsible for the differences between classical and quantum correlations. It would be interesting to know if somebody disagrees with this or thinks it is irrelevant and if so why.
It's how nature works, assuming we have "this typical Bell-type experiment" – i.e., the particular experiment that [USER=372855]@stevendaryl[/USER] proposed. The two particles are entangled with opposite spins – sometimes called a "Bell state". Therefore "when their detector angles are equal, they will always detect the opposite".The term "Bell-type" is vague. I don't think there's any official definition. To me it does not necessarily mean entanglement with opposite spins, although that's most common, and that's how Bell originally did it. They could instead be in the "twin state" so that they must have the same spin for the same angles. I even use that term sometimes when I'm not talking about Bell's inequality at all, but something similar like CHSH inequality. Almost any experiment that demonstrates the conclusion Bell came up with (ruling out realist, local hidden-variables model) might be referred to, loosely, as "Bell-type". The meaning should be clear from context.If others disagree with my use of this term "Bell-type", I won't argue, maybe they're right.
[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]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?
[QUOTE="Jilang, post: 5640513, member: 492883"]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.[/QUOTE]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.
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.
[QUOTE="lavinia, post: 5639692, member: 243745"]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.[/QUOTE]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.
[QUOTE="RockyMarciano, post: 5639564, member: 585697"]I'm not sure what general principle you are referring to, can you state it explicitly?[/QUOTE]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.
[QUOTE="lavinia, post: 5639085, member: 243745"]There is a general principle being described here.[/QUOTE]I'm not sure what general principle you are referring to, can you state it explicitly?
[QUOTE="RockyMarciano, post: 5639042, member: 585697"]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.[/QUOTE]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.
[QUOTE="lavinia, post: 5638852, member: 243745"]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.[/QUOTE]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.
[QUOTE="Jilang, post: 5638916, member: 492883"]Thanks but I was meaning the derivation itself.[/QUOTE]Feynmann has the derivation. I will write it out when I have time.
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 Processesshttps://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/3DE2C9013903EDD218F5B85129F65B2CI 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.
[QUOTE="lavinia, post: 5638906, member: 243745"]Feynmann's Lecture Volume 3. There is another post where I wrote out what Feynmann said. Forget what it was called.[/QUOTE]Thanks but I was meaning the derivation itself.
[QUOTE="Jilang, post: 5638904, member: 492883"]That's interesting. Do you have link for that?[/QUOTE]Feynmann's Lecture Volume 3. There is another post where I wrote out what Feynmann said. Forget what it was called.
[QUOTE="lavinia, post: 5638852, member: 243745"]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.[/QUOTE]That's interesting. Do you have link for that?
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.
[QUOTE="lavinia, post: 5638241, member: 243745"]The Shroedinger Equation for a free particle is just the Heat Equation with an ##i## thrown in. It is no surprise that it should behave like a complex diffusion – meaning that conditional amplitudes replace conditional probabilities.[/QUOTE]This is indeed no surprise, but the OP shows a little more than this. At least I thought that what it was highlighting by using the amplitudes obtained from the squared root of the probabilities(i.e. going backwards with respect to the usual process from complex amplitudes to probabilities) and therefore up to phase, which led to a local hidden variables model of those amplitudes, was the contrast with the usual forward procedure from complex amplitudes being multiplied to their complex conjugates and where some info about phase makes it to the probabilities. This can't happen with classical probabilities and real valued inputs for obvious mathematical reasons.
The Shroedinger Equation for a free particle is just the Heat Equation with an ##i## thrown in. It is no surprise that it should behave like a complex diffusion – meaning that conditional amplitudes replace conditional probabilities.
[QUOTE="secur, post: 5638106, member: 588176"]The only problem (as I indicated before) is that using this approach, you can make it result in the correct probabilities, namely, 1/2 cos^2 and sin^2 of ([itex]alpha[/itex] – [itex]beta[/itex])/2. That seems to undermine the conclusion that "The squaring process is in some sense responsible for the weirdness of QM correlations".[/QUOTE]That sentence may be ambiguous in its meaning, I think that rather than making certain operation responsible for any perceived weirdness, what is shown is how the formalism (Born rule)attains nonlocal correlations from local amplitudes not connected to measurements directly. It is of course nature that is responsible for the lack of predetermined values of measurements, not any mathematical operation.
The only problem (as I indicated before) is that using this approach, you can make it result in the correct probabilities, namely, 1/2 cos^2 and sin^2 of ([itex]alpha[/itex] – [itex]beta[/itex])/2. That seems to undermine the conclusion that "The squaring process is in some sense responsible for the weirdness of QM correlations".
[QUOTE="stevendaryl, post: 5633307, member: 372855"]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.[/QUOTE][QUOTE="stevendaryl, post: 5634793, member: 372855"]. 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][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]I do think these observations might contribute to a better understanding in that they analyze a specific quantum situation in wich it is very clear that the amplitudes cannot correspond to measurements, given the bipartite system prepared as pure states used, so only the probabilities are relevant, so while it is clean enough that can never raise any doubt about Bell's theorem it gives us hints about how complex amplitudes being squared erase any trace of hidden variables in QM and therefore helps clarify the mathematical device that the formulism uses to achieve this.
[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 think you are under a similar confusion that secur initially had concerning what the OP model intends. If it "worked" in the way that you seem to think it would it would be a counterexample to Bell's theorem and that is not at all its aim.