Amplitudes, Probabilities and EPR

[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.