Amplitudes, Probabilities and EPR

[lavinia]

Thanks but I was meaning the derivation itself.

Feynmann has the derivation. I will write it out when I have time.

 

[RockyMarciano]

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

Sure, but that was a more general case, the nice thing here is that it is simplified to the two-state system in Bell's inequalities, and it is much easier to see how the amplitudes being complex can convey just the right amount of info about phase when squared so that the nonlocal statistical correlations of the physics can be correctly predicted.
While it doesn't explain why nature is like this it shows how the math gets it right and how this cannot be done with classical probabilities.

 

[lavinia]

Sure, but that was a more general case, the nice thing here is that it is simplified to the two-state system in Bell's inequalities, and it is much easier to see how the amplitudes being complex can convey just the right amount of info about phase when squared so that the nonlocal statistical correlations of the physics can be correctly predicted.
While it doesn't explain why nature is like this it shows how the math gets it right and how this cannot be done with classical probabilities.

There is a general principle being described here.

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

 

[RockyMarciano]

There is a general principle being described here.

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

 

[lavinia]

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

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

 

[RockyMarciano]

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

Absolutely, the importance of the amplitudes being complex in the shift from classical to quantum theories has been known from the beginning and was underlined by Feynman more than half a century ago.
What I was trying to convey is that by using a pure state bipartite system something more about how what you call general principle actually works can be deduced. But maybe it would lead to depart slightly from the conclusions in the (now) insights article of the OP.

I mean let's pretend that, as secur pointed out earlier, the conclusion that all the difference between the classical and the quantum correlations lies in the process of squaring the amplitudes is not completely correct and the difference lies actually in the amplitudes. Since stevendaryl showed that the only difference between the square root amplitudes and the usual quantum amplitudes is a +/- sign, i.e. a global phase, let's pretend(please bear with me) that this difference normally considered irrelevant is somehow not irrelevant in this case. Can anybody think of a mathematical reason global phase might be relevant for the argument of complex numbers?
It is relevant at least formally in the analysis of EPR correlations in the form of one of the three angles needed for the analysis, that many peolple finds odd as they think it would be enough with two angles for the difference between the polarizers. This might give a clue for the question above.

 

[Jilang]

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.

 

[secur]

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

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

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

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

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

 

[edguy99]

Seems there's some confusion. I think the best way to straighten it out is, please address the other point I made, which is very simple.

In this typical Bell-type experiment, QM says A and B must always be opposite (product is -1) when their detector angles are equal. A valid hidden-variable model must reproduce that behavior. But that's not the case with your model:

Can you clarify? I may be misunderstanding this.

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

 

[secur]

It's how nature works, assuming we have "this typical Bell-type experiment" - i.e., the particular experiment that @stevendaryl 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.

 

[RockyMarciano]

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.

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 which 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.

 

[stevendaryl]

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.

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).

 

[RockyMarciano]

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).

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.

 

[stevendaryl]

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.

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 $p$, or the other slit, with probability $1-p$. If it goes through the left slit, say that it has a probability of $q_L$ of triggering a particular photon detector. If it goes through the right slit, say that it has a probability of $q_R$ of triggering that detector. Then the amplitude for triggering the detector, when you don't observe which slit it goes through, is:

$\psi = \sqrt{p} \sqrt{q_L} + \sqrt{1-p}\sqrt{q_R}$

leading to a probability

$P = |\psi|^2 = p q_L + (1-p) q_R + 2 \sqrt{p(1-p)q_L q_R}$

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.

 

[RockyMarciano]

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 $p$, or the other slit, with probability $1-p$. If it goes through the left slit, say that it has a probability of $q_L$ of triggering a particular photon detector. If it goes through the right slit, say that it has a probability of $q_R$ of triggering that detector. Then the amplitude for triggering the detector, when you don't observe which slit it goes through, is:

$\psi = \sqrt{p} \sqrt{q_L} + \sqrt{1-p}\sqrt{q_R}$

leading to a probability

$P = |\psi|^2 = p q_L + (1-p) q_R + 2 \sqrt{p(1-p)q_L q_R}$

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.

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.

 

[Stephen Tashi]

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.

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 $lim_{x\rightarrow a} f(x)$. 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.

 

[Greg Bernhardt]

Nice Insight @stevendaryl!

 

[edguy99]

Nice Insight @stevendaryl!

Thanks for bumping this. I forgot how truly great this insight is where @stevendaryl 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, and
ψ(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 @secur 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 @stevendaryl 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?

 

[DrChinese]

In the experiment here, the @stevendaryl 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?

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.

 

[edguy99]

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.

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.

 

[Jilang]

Thanks for bumping this. I forgot how truly great this insight is where @stevendaryl 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, and
ψ(A,B|α,β) ∼ 1/√2*cos((β–α)/2) if A≠B.

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).

 

[edguy99]

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).

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.