B Does quantum physics imply the existence of randomness?

[LaserMind]

What law of physics prevents a measurable / observable property from actually having a specific infinitely precise value? I read your statements as though this is axiomatic or obvious from inspection, it is not obvious to me, at least.

I thought a property could not have an infinitely precise value because infinity is never ending. Kinda obvious!

 

[DrChinese]

He means that an exact value is a theoretical phenomenon, not a practical one. Theoretically, you can balance a pin on its tip; but it is practically infeasable.

He might mean this (that's how I interpreted too). But the only practical limitation is in us making a measurement, which certainly does not constrain an observable itself in any way. It could be completely random, and in fact behaves as such. There is no particular reason to believe collapse is not completely random - other than by pure assumption.

 

[DrChinese]

I thought a property could not have an infinitely precise value because infinity is never ending. Kinda obvious!

So obvious that... it is not obvious.

Suppose I have an observable that can take on 1 of 2 values. Are you saying that observable cannot be completely random because the value does not have an infinite number of decimal places? Because I wouldn't agree with that.

And I wouldn't agree for an observable that is continuous either.

 

[Isaac0427]

Suppose I have an observable that can take on 1 of 2 values. Are you saying that observable cannot be completely random because the value does not have an infinite number of decimal places? Because I wouldn't agree with that.

I could be wrong, but I think he's saying that if you had an observable, such as position, that could take on one of an infinite number of values (I mean, there are an infinite amount of values between 0 and 1), the observable can't be completely random. What I got from his message, in a more mathematical form, the wavefunction $\psi=Ne^{ix}$ where N is a constant satisfying normalization is "completely random," but there is no possible value of N, and thus the wavefunction is not practical.

 

[DrChinese]

I could be wrong, but I think he's saying that if you had an observable, such as position, that could take on one of an infinite number of values (I mean, there are an infinite amount of values between 0 and 1), the observable can't be completely random. What I got from his message, in a more mathematical form, the wavefunction $\psi=Ne^{ix}$ where N is a constant satisfying normalization is "completely random," but there is no possible value of N, and thus the wavefunction is not practical.

If someone asserted that, I would challenge it. The number of possible outcomes does not change whether something is or is not random.

 

[Isaac0427]

If someone asserted that, I would challenge it. The number of possible outcomes does not change whether something is or is not random.

I think complete randomness would be defined as there are no values that the particle is more or less likely to be in. Again, not too relevant to this discussion, but it is true that by that definition, complete randomness is impossible in quantum mechanics for the mathematical reason in my previous post.

 

[entropy1]

What law of physics prevents a measurable / observable property from actually having a specific infinitely precise value? I read your statements as though this is axiomatic or obvious from inspection, it is not obvious to me, at least.

I was thinking about virtual particles for instance; they would bump into the balanced pin and push it over. Space is not exactly empty. Another example would be that a 'trapped particle' would gain infinite momentum and its position wouldn't be exactly measured. That sort of thing.

 

[Grinkle]

I think complete randomness would be defined as there are no values that the particle is more or less likely to be in.

I am not able to be precise with my language, so maybe I just can't make more progress here. I will try, and I appreciate any help.

Complete randomness might mean that given a set of possible values, there are no values in that set which the particle is more or less likely to be in. There may be other values not in the set of possibilities that the particle has zero chance of being in.

Is that a bad / unuseful definition of complete randomness? I would use the term "even probability distribution" instead of complete randomness to describe what I am saying.

I think you are saying that complete randomness means there is no value whatsoever excluded from the possible value set, and in addition there is a perfectly even probability distribution.

I never studied statistics. There must be some math to show whether or not a set containing infinitely many members can have an even probability distribution that sums to 1? Is that the mathematical issue, or am I way off base?

 

[Nugatory]

I think he's saying that if you had an observable, such as position, that could take on one of an infinite number of values (I mean, there are an infinite amount of values between 0 and 1), the observable can't be completely random.

If that's what we're talking about, it's not right. There is a perfectly satisfactory theory of continuous probability distributions and what "random" means in that context. You may not encounter it until a few years into college because, unlike the simpler discrete cases, you need a moderate amount of calculus just to get started, but it's there.

 

[Grinkle]

I was thinking about virtual particles for instance; they would bump into the balanced pin and push it over. Space is not exactly empty. Another example would be that a 'trapped particle' would gain infinite momentum and its position wouldn't be exactly measured. That sort of thing.

Those are measurement issues, not existence issues, aren't they?

I thought you might talking about something equivalent to asking if space-time is discrete or continuous (the answer is not obvious to me, I don't have any leaning one way or the other on which is more likely true) but maybe that is not what you are getting at.

 

[entropy1]

Those are measurement issues, not existence issues, aren't they?

Indeed, things won't let themselves measure exactly, in practice. Everything at least has a little vibration and/or uncertainty. (Recall I was responding to Lasermind with respect to this)

 

[DrChinese]

Complete randomness might mean that given a set of possible values, there are no values in that set which the particle is more or less likely to be in. There may be other values not in the set of possibilities that the particle has zero chance of being in.

Coming close to yours is the following, from Wikipedia: "In situations where a population consists of items that are distinguishable, a random selection mechanism requires equal probabilities for any item to be chosen."

However, that is not the case for QM, where there is randomness but some values may be more likely than others (same source): "According to several standard interpretations of quantum mechanics, microscopic phenomena are objectively random.[6] That is, in an experiment that controls all causally relevant parameters, some aspects of the outcome still vary randomly. For example, if you place a single unstable atom in a controlled environment, you cannot predict how long it will take for the atom to decay—only the probability of decay in a given time.[7] Thus, quantum mechanics does not specify the outcome of individual experiments but only the probabilities."

 

[Mark Harder]

The collapse of the wavefunction could not be *completely random* in practice because that would mean an exact point-position to an infinite number of decimal places.

Certainly an exact location of a quantum particle is impossible. The UP guarantees that. But I don't understand why a "completely random" collapse would imply exactness. In fact, it sounds to me that just the opposite is true: if the collapse is random (completely or not), then any future evolution of the system could not be determined. With random initial conditions, how could a system trajectory be determined exactly? Since the wavefunction is linear, we are guaranteed that any 2 solutions cannot diverge without limit. That's the best we can do when we attempt to specify future states from imprecise initial states.

 

[write4u]

Certainly an exact location of a quantum particle is impossible. The UP guarantees that. But I don't understand why a "completely random" collapse would imply exactness. In fact, it sounds to me that just the opposite is true: if the collapse is random (completely or not), then any future evolution of the system could not be determined. With random initial conditions, how could a system trajectory be determined exactly? Since the wavefunction is linear, we are guaranteed that any 2 solutions cannot diverge without limit. That's the best we can do when we attempt to specify future states from imprecise initial states.

Question: If the observer has a fixed position, would that not play a part in any equation of the quantum function?

IOW, the observer *determines* the final position of the particle even as the observer cannot predict the exact position of the particle while it is in motion.

Question: How could we construct precisely focused lasers if total randomness existed.

As I understood one of the questions, once the wave function has collapsed and the particle is manifest it has a precise location, relative to the observer. The problem lies in the *uncertainty* of the particle's position while in motion, but apparently we can control this randomness (to an extend).

 

[TheMeInTeam]

You bring up what seems to me to be the deepest question in this topic. It seems that when physical processes are deterministically chaotic, probably theory is only a model of something that, though deterministic, is so complex that in practice is practically impossible to calculate. In such cases - like tossing a coin or casting dice, - choosing probability as a model for the process is a kind of "fudging". Here, probability describes non-stochastic systems for which we can have incomplete knowledge only. On the other hand, assigning probabilities to a quantum event is an appropriate description of a truly random process that exists in nature. The physical meaning of stochastic variables is not the same for all cases, in other words.

Is it really certain that the dice example isn't comparable to the quantum event one? From what I can tell reading up on this there's a good chance we're "fudging" in both cases.

 

[Grinkle]

Is it really certain that the dice example isn't comparable to the quantum event one?

That is re-stating the title of the thread, imo.

 

[TheMeInTeam]

That is re-stating the title of the thread, imo.

True, but as far as I can tell the answer is "no", because we don't know, don't even have evidence to prefer a conclusion. Pick your interpretation and the answer changes, right? We don't know which interpretation is correct, if any, but that's good enough reject a conclusion of "implied existence of randomness".

 

[Grinkle]

don't even have evidence to prefer a conclusion

See post 31. To me, non-locality is also weird. If quantum determinism exists, then non-locality must also exist. I think.

Its not evidence to prefer one or the other, but it is certainly not so easy to dismiss quantum randomness if that implies a rational person must then be accepting non-locality as the explanation for the Bell theorem experiments.

 

[TheMeInTeam]

See post 31. To me, non-locality is also weird. The existence of both quantum determinism and non-locality is contradicted by experiment. I think.

Its not evidence to prefer one or the other, but it is certainly not so easy to dismiss quantum randomness if that implies a rational person must then be accepting non-locality as the explanation for the Bell theorem experiments.

Any of it can "seem weird" when thinking in terms of every day experiences of what we're used to thinking.

Why do we necessarily need hidden variables to get to determinism? Going back to the dice example, the eye is not enough instrumentation to detect that the outcome a roll is, depending exactly on how the dice is thrown and how it interacts with the landing surface, deterministic.

Just as non-locality seems weird to you (and I'm not sure it's required in principle, but maybe I'm missing something), I find it strange that this point is the only thing in physics I've heard of that lacks clear causality and is somehow accepted by many regardless...even as we don't completely understand it yet. This one topic is exceptional with how we know things work otherwise, posited because we don't understand it! Quite a bit of the discussion here just assumes a wave function collapse outright...which might or very well might not be an accurate framework to use.

To me the best answer is "we don't know yet."

 

[carllooper]

Why is red light red? It's silly to say "we don't know yet" because the very thing we do know, if only for ourselves, is that red light gives us a particular sensation we can call red. What we don't know is how to formulate that in mathematical terms. I might see red light as cyan, and you otherwise see it as red, but we will both still call it "red". And in the same way, the mathematics of colour, based on the frequency of light, and otherwise used in technologies of colour (such as colour photography) is not in any way altered by this ambiguity in how we otherwise perceive colour. The same model works for both of us, whether you or I perceive red light as red or as cyan. A mathematical model, as much as words such as "red", are unable to make any distinction between sensations as otherwise personally experienced, even if we can otherwise entertain just such distinctions.

It becomes a useful thing that a model can work regardless of personal perception, but it also demonstrates how a physical/mathematical model may not ever be able to encode the full perceptual situation. There may very well be a limit to the scope of what physics and mathematics alone, can describe.

Randomness is particularly resistant to mathematical description. Indeed one might say, in the context of mathematics, there is no such thing as randomness. It may very well be that mathematics is inherently faulty with respect to randomness. In statistics, on the other hand, randomness is at least an operating assumption, regardless of whether there is a mathematical solution (a pseudo-random solution) or not. What matters in statistics is the aggregate effects regardless of the the precise nature of the causes (or lack thereof).

An absence of a causal model for wave function collapse doesn't mean we fail to see, in a given experiment, what is otherwise meant by wave function collapse. The collapse still happens, so to speak. It doesn't need us and our models, or lack thereof, in order to take place. We still see individual particle detections, and we still see the pattern of those detections, and we still see the mathematical correlation between the pattern and a wave function as derived from the geometry of the setup.

 

[TheMeInTeam]

The same model works for both of us, whether you or I perceive red light as red or as cyan.

That is true, but we don't have alternative models calling red different colors, or a sound. It's a sensory experience that, while it may appear different to all of us, elicits an experience we can consistently react to in the same fashion. If someone paints the roses red, others can identify them as red. As such I don't think it's a fair analogy to the thread topic, where people are doing something more similar to interpreting red as cyan, a sound, and the experience you get when jumping about 7 feet onto foam padding.

There may very well be a limit to the scope of what physics and mathematics alone, can describe.

That might be the case, but for the time being we don't have a good reason to prefer that conclusion.

Randomness is particularly resistant to mathematical description. Indeed one might say, in the context of mathematics, there is no such thing as randomness. It may very well be that mathematics is inherently faulty with respect to randomness.

This all assumes that true randomness (not just apparent randomness because we can't perceive causes at the micro level in real time, or anything close to that) exists at all. It might. Statistics makes models based on incomplete information/noisy measures and makes practical estimations given the constraints of our knowledge, but it's hardly something that will settle whether there is randomness.

An absence of a causal model for wave function collapse doesn't mean we fail to see, in a given experiment, what is otherwise meant by wave function collapse. The collapse still happens, so to speak.

I am open to the possibility that I'm behind the times, but I have yet to see any evidence that a collapse necessarily happens, and it's the only thing in physics like that. It doesn't just lack a clear causal backing, it's a theoretical point that adds a (unique!) extra detail while the point it occurs has been a moving target historically. Absent that, I'm not willing to buy a conclusion of randomness that makes assumptions on details we don't have, especially when wave function collapse is unique even within quantum physics and every prior instance of "randomness" we see in the classical sense is instead determinism we can't keep up with.

It might actually happen, we might even manage to find a direct cause or at least strong evidence to prefer that interpretation. If we have it and I've missed it, please enlighten me. I have no stake here other than interest and a better understanding of reality.

 

[carllooper]

This all assumes that true randomness (not just apparent randomness because we can't perceive causes at the micro level in real time, or anything close to that) exists at all. It might. Statistics makes models based on incomplete information/noisy measures and makes practical estimations given the constraints of our knowledge, but it's hardly something that will settle whether there is randomness.

Yes, that's right. We can only assume there is true randomness. And we can only assume there isn't. Mathematics, as it is currently constituted, enforces one of these assumptions. Statistics, on the other hand doesn't care, which allows either cases (information vs noise) to remain in play.

I am open to the possibility that I'm behind the times, but I have yet to see any evidence that a collapse necessarily happens, and it's the only thing in physics like that. It doesn't just lack a clear causal backing, it's a theoretical point that adds a (unique!) extra detail while the point it occurs has been a moving target historically. Absent that, I'm not willing to buy a conclusion of randomness that makes assumptions on details we don't have, especially when wave function collapse is unique even within quantum physics and every prior instance of "randomness" we see in the classical sense is instead determinism we can't keep up with.

Wave function collapse is an heuristic. A way of speaking. A poetic turn of phrase. It refers to the disjuncture we can entertain between a wave function as a description of a particle detection, prior to a detection, and the same wave function as a description, following detection. The heuristic can be regarded as referencing this disjuncture. By saying wave function collapse "still happens, so to speak" it is only to suggest that despite the absence of a causal model for such a heuristic (despite the heuristic being a heuristic) it doesn't in anyway prevent the particle detections from occurring. Which should be an obvious point of course, but it's one that can be inadvertently forgotten.

It might actually happen, we might even manage to find a direct cause or at least strong evidence to prefer that interpretation. If we have it and I've missed it, please enlighten me. I have no stake here other than interest and a better understanding of reality.

Indeed we might. Assuming there is a direct cause of course. Until then there is only the assumption of a direct cause, and no theoretical model with which to experiment.

 

[rkastner]

According to the Transactional Interpretation (TI), the 'collapse' is real and it is genuinely indeterministic. I've extended TI into the relativistic domain and I explicitly describe the quantum state as a possibility wave, so I call this extended version "possibilist TI" or PTI.
For the basics of this model, which technically is a slightly different theory from standard QM at the relativistic level, see this blog post:
https://transactionalinterpretation...tivistic-and-non-relativistic-quantum-theory/
For an application of this to Feynman's sum-over-paths picture, see my latest blog post:
https://transactionalinterpretation...ts-possible-paths-from-source-to-destination/
In the latter, see especially the introductory quote from Feynman which observes that 'the real glory of science is that we can find a way of thinking such that the law is evident." This is what TI does for us concerning the Born Rule: it just drops out of the physics rather than being just an empirically observed law.
(I also have a 2015 book for the general reader which presents PTI in math-free form.) Comments/questions welcome.

 

[carllooper]

According to the Transactional Interpretation (TI), the 'collapse' is real and it is genuinely indeterministic. I've extended TI into the relativistic domain and I explicitly describe the quantum state as a possibility wave, so I call this extended version "possibilist TI" or PTI.
For the basics of this model, which technically is a slightly different theory from standard QM at the relativistic level, see this blog post:
https://transactionalinterpretation...tivistic-and-non-relativistic-quantum-theory/
For an application of this to Feynman's sum-over-paths picture, see my latest blog post:
https://transactionalinterpretation...ts-possible-paths-from-source-to-destination/
In the latter, see especially the introductory quote from Feynman which observes that 'the real glory of science is that we can find a way of thinking such that the law is evident." This is what TI does for us concerning the Born Rule: it just drops out of the physics rather than being just an empirically observed law.
(I also have a 2015 book for the general reader which presents PTI in math-free form.) Comments/questions welcome.

Thanks Ruth. Had a quick read. That's really quite interesting. I must follow it up in more detail. I spent some time with the original TI theory many years ago and found it quite interesting at the time. A nice feature of such is the time symmetric structure of such.

 

[lavinia]

Here is an illustration of why the time evolution of quantum states reminds one of a Markov process.

For simplicity take the case of finitely many states. For a continuous time Markov process let $C_{i}(t)$ denote the probability of finding the random variable in state $i$ at time $t$. For a quantum mechanical system let $C_{i}(t)$ denote the amplitude that the system will be found in state $i$ at time $t$. In bra-ket notation this is $<i|ψ(t)> = C_{i}(t)$. Then one has the equation,

Then $C_{i}(t+Δt) = Σ_{j}T_{ij}(t,t+Δt)C_{j}(t)$

where for the Markov process $T_{ij}(t,t+Δt)$ is the conditional probability that the random variable will be in state $i$ at time $t+Δt$ given that it is in state $j$ at time $t$ and for the quantum mechanical system $T_{ij}(t,t+Δt)$ is the conditional amplitude that the system will be in state $i$ at time $t+Δt$ given that it is in state $j$ at time $t$. This amplitude is $<i|A(t,t+Δt)|j>$ where $A$ is the passage of time operator.

For both the Markov process and the QM system one has for small $Δt$

$T_{ij}(t,t+Δt) = δ_{ij} + K_{ij}(t)Δt +o(Δt)$ where $δ_{ij}$ is the Kronecker $δ$.

For the QM system the matrix $K$ is -i$/h$ times the Hamiltonian $H$.

In the limit one gets the usual equation i$hdC_{i}(t)/dt = Σ_{j}H_{ij}(t)C_{j}(t)$

References: Feynman's Lectures on Physics Book 3 section 8-4
https://en.wikipedia.org/wiki/Continuous-time_Markov_chain

 

[stevendaryl]

Here is an illustration of why the time evolution of quantum states reminds one of a Markov process.

Yes! All the strangeness of quantum probability (entanglement, particularly) seems to vanish if instead of focusing on probabilities, we focus on probability amplitudes. The rules for computing quantum amplitudes are almost exactly the same as the rules for computing probabilities for a random process such as Brownian motion:

• The probability/amplitude for going from A to B and then to C is just the product of the probability/amplitude for going from A to B and the probability/amplitude for going from B to C.
• If there are a number of mutually exclusive for an intermediate state, B_1, B_2, ..., B_n, then the probability/amplitude for going from A to C via one of those intermediate states is the sum over j of the probability/amplitude for going from A to B_j and then to C.

The mysterious part is that amplitudes can be complex, and that you have to square them to get a probability.

 

[stevendaryl]

Yes! All the strangeness of quantum probability (entanglement, particularly) seems to vanish if instead of focusing on probabilities, we focus on probability amplitudes. The rules for computing quantum amplitudes are almost exactly the same as the rules for computing probabilities for a random process such as Brownian motion:

• The probability/amplitude for going from A to B and then to C is just the product of the probability/amplitude for going from A to B and the probability/amplitude for going from B to C.
• If there are a number of mutually exclusive for an intermediate state, B_1, B_2, ..., B_n, then the probability/amplitude for going from A to C via one of those intermediate states is the sum over j of the probability/amplitude for going from A to B_j and then to C.

The mysterious part is that amplitudes can be complex, and that you have to square them to get a probability.

I guess another difference between quantum amplitudes and probabilities for a random process is that there can be amplitudes associated with same-time transitions. When it comes to a random process such as Brownian motion, we have a limiting case: If P(A,B,t) means the probability of going from A to B in time t, then

lim_{t \rightarrow 0} P(A,B,t) = \delta_{AB}

The corresponding limit isn't true for quantum amplitudes: Two states can be "overlapping", and so the transition amplitude can be nonzero even in the limit as t \rightarrow 0.

 

[Zafa Pi]

Yes! All the strangeness of quantum probability (entanglement, particularly) seems to vanish if instead of focusing on probabilities, we focus on probability amplitudes. :

I'll make this as simple as possible.
1) Alice and Bob meet for a strategy session and are then sent to their respective labs and not allowed to communicate.
2) A is then given a uniform random bit (0 or 1) x. B is given uniform random bit y that is independent of x. (e.g. flip a fair coin to get x, flip again to get y)
3) A selects bit a. B selects bit b.
4) A and B win the game if a = b if and only if x and y are not both 1.

Question: Is there a strategy allowing A and B to win with probability > 3/4?
I contend that prior to 1900 no body in the world could answer yes.
That is why quantum entanglement is weird, and nothing you've said has changed that.

 

[Zafa Pi]

Since in everyday life, locality is not violated, we can use the violation of a Bell inequality to guarantee randomness.

That is randomness in the model/theory called quantum mechanics. Whether randomness occurs in reality is unknown will likely remain so.

 

[Grinkle]

That is randomness in the model/theory called quantum mechanics. Whether randomness occurs in reality is unknown will likely remain so.

I hope we have not reached the limits of what can be tested regarding quantum randomness, but if we have, then true randomness exists, as far as experimental science is concerned at least. I mean that in that case we can make predictions based on that premise, and those predictions will be confirmed by any experiment we can ever do.

The Bell inequality is a profound experimental observation imo.

 

[Zafa Pi]

I hope we have not reached the limits of what can be tested regarding quantum randomness, but if we have, then true randomness exists, as far as experimental science is concerned at least. I mean that in that case we can make predictions based on that premise, and those predictions will be confirmed by any experiment we can ever do.

How could one distinguish between "true randomness" and a very good unknown algorithm. You could gather data forever and not know. You could make predictions based on a well balanced coin flipped into a wind tunnel.

"The Bell inequality is a profound experimental observation imo."

I agree and here is imo the ultimate form of Bell's Inequality:
1) A and B meet for a strategy session and are then sent to their respective labs and not allowed to communicate.
2) Then we flip a fair coin giving the result to A, then flip again and give the result to B.
3) A selects bit a (= 0 or 1). B selects bit b.
4) A and B win the game if a ≠ b when they both received heads, and a = b otherwise.

Question: Is there a strategy allowing A and B to win with probability > 3/4?

I contend that prior to 1900 no body in the world could logically answer yes.
QM can achieve 85%. That is why quantum entanglement is weird

 

[stevendaryl]

I'll make this as simple as possible.
1) Alice and Bob meet for a strategy session and are then sent to their respective labs and not allowed to communicate.
2) A is then given a uniform random bit (0 or 1) x. B is given uniform random bit y that is independent of x. (e.g. flip a fair coin to get x, flip again to get y)
3) A selects bit a. B selects bit b.
4) A and B win the game if a = b if and only if x and y are not both 1.

Question: Is there a strategy allowing A and B to win with probability > 3/4?
I contend that prior to 1900 no body in the world could answer yes.
That is why quantum entanglement is weird, and nothing you've said has changed that.

I wasn't claiming that quantum mechanics isn't weird (I have always been on the side of "quantum mechanics is weird"), but just remarking that the rules for combining amplitudes are sensible. Bell's theorem shows that in EPR the joint probability P(A,B|\alpha, \beta) for Alice and Bob to both measure spin-up, given Alice's setting \alpha and Bob's setting \beta cannot be factored in the form:

P(A,B|\alpha, \beta) = \sum_\lambda P_{hv}(\lambda) P_A(A|\alpha, \lambda) P_B(B|\beta, \lambda)

So there is no hidden-variables explanation for the joint probability. However, it's interesting (to me, anyway) that probability amplitudes don't have this problem. The joint probability amplitude does factor in exactly that way:

\psi(A, B|\alpha, \beta) = \sum_\lambda \psi_{hv}(\lambda) \psi_A(A|\alpha, \lambda) \psi_B(B|\beta, \lambda)

But when you square the amplitude to get the probability, you get cross-terms which spoil the factorization.

I don't know what, if anything, this implies about quantum mechanics, but it is interesting (again, to me).

 

[Zafa Pi]

I wasn't claiming that quantum mechanics isn't weird (I have always been on the side of "quantum mechanics is weird"), but just remarking that the rules for combining amplitudes are sensible. Bell's theorem shows that in EPR the joint probability P(A,B|\alpha, \beta) for Alice and Bob to both measure spin-up, given Alice's setting \alpha and Bob's setting \beta cannot be factored in the form:

P(A,B|\alpha, \beta) = \sum_\lambda P_{hv}(\lambda) P_A(A|\alpha, \lambda) P_B(B|\beta, \lambda)

So there is no hidden-variables explanation for the joint probability. However, it's interesting (to me, anyway) that probability amplitudes don't have this problem. The joint probability amplitude does factor in exactly that way:

\psi(A, B|\alpha, \beta) = \sum_\lambda \psi_{hv}(\lambda) \psi_A(A|\alpha, \lambda) \psi_B(B|\beta, \lambda)

But when you square the amplitude to get the probability, you get cross-terms which spoil the factorization.

I don't know what, if anything, this implies about quantum mechanics, but it is interesting (again, to me).

OK, I retract my criticism. Now I'm confused at a different level. A pair of photons have joint state in the tensor product space. If the individual photons each had a state then their joint state is also a tensor product and the joint probabilities factor as if they were independent random variables. However if the joint state is not a tensor product (i.e. the photons/state is entangled, or EPR). In that case neither individual photon has a state, so I don't see how you can talk about their amplitudes. Of course there are clear rules for calculating joint probabilities and indeed they don't factor over the individual measurement probabilities.

 

[stevendaryl]

However if the joint state is not a tensor product (i.e. the photons/state is entangled, or EPR). In that case neither individual photon has a state, so I don't see how you can talk about their amplitudes.

Alice and Bob don't have individual amplitudes, but the joint amplitude can be written as an amplitude-weighted sum of products of individual amplitudes. Let me explain the analogy with hidden variables for probabilities.

In terms of probabilities, we have a joint probability for Alice and Bob:

P(A,B|\alpha, \beta)

where A is Alice's measurement result and B is Bob's measurement result, and \alpha is Alice's detector setting, and \beta is Bob's detector setting. A "hidden-variables" model for this joint probability would be a hidden variable \lambda and probabilities P_{hv}(\lambda), P_A(A|\alpha, \lambda), P_B(B|\beta, \lambda) such that:

P(A,B|\alpha, \beta) = \sum_\lambda P_{hv}(\lambda) P_A(A|\alpha, \lambda) P_B(B|\beta, \lambda)

If there were such a hidden-variables model, then we could explain the joint probability distribution in terms of a weighted average (averaged over possible values of \lambda) of products of single-particle probability functions. But alas, Bell proved that there was no such hidden-variables model for the joint probability distribution.

Now, let's shift the focus from probabilities to amplitudes. We let \psi(A, B|\alpha, \beta) be the joint amplitude for the EPR experiment, where the amplitude is related to the probability via:

P(A, B|\alpha, \beta) = |\psi(A,B|\alpha, \beta)|^2

So \psi(A,B|\alpha, \beta) is a joint amplitude, but Alice and Bob do not have individual amplitudes. But is there a "hidden-variables" model for this joint amplitude? By analogy with the hidden-variables model for probabilities, we say that a hidden-variables model for the joint amplitude would be a hidden variable \lambda and amplitude functions \psi_{hv}(\lambda), \psi_A(A|\alpha, \lambda), \psi_B(B|\beta, \lambda) such that:

\psi(A, B|\alpha, \beta) = \sum_\lambda \psi_{hv}(\lambda) \psi_A(A|\alpha, \lambda) \psi_B(B|\beta, \lambda)

If there were such a "hidden-variables" model for the probability amplitudes, we could interpret the joint amplitude as an amplitude-weighted sum of products of single-particle amplitudes.

It's not too hard to show that there is a hidden-variables model for amplitudes in EPR, even though there is no hidden-variable model for probabilities.

In the correlated two-photon EPR experiment, we have a joint probability distribution given by:

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

where A and B are Alice's and Bob's measurement results, each of which have possible values from the set \{ H, V \} (horizontal or vertically polarized, relative to the polarizing filter), and \alpha and \beta represent Alice's and Bob's filter orientations. In terms of amplitudes, we have:

\psi(A, B|\alpha, \beta) = \frac{1}{\sqrt{2}} cos(\beta - \alpha) (if A=B)
= \frac{1}{\sqrt{2}} sin(\beta - \alpha) (if A\neqB)

We can easily write this in the "hidden-variables" form \psi(A, B|\alpha, \beta) = \sum_\lambda \psi_{hv}(\lambda) \psi_A(A|\alpha, \lambda) \psi_B(B|\beta, \lambda) by the following model:

• \lambda has two possible values, 0 or \frac{\pi}{2}.
• \psi_{hv}(0) = \psi_{hv}(\frac{\pi}{2}) = \frac{1}{\sqrt{2}}
• \psi_A(A | \alpha, \lambda) = cos(\alpha - \lambda) (if A=H)
• \psi_A(A | \alpha, \lambda) = sin(\alpha - \lambda) (if A=V)
  
• \psi_B(B |\beta, \lambda) = cos(\beta - \lambda) (if B=H)
• \psi_B(B |\beta, \lambda) = sin(\beta - \lambda) (if B=V)

Using trigonometry, we can easily show that this satisfies the equation:
(In the case A=B=H; the other cases are equally straight-forward)

\sum_\lambda \frac{1}{\sqrt{2}} cos(\alpha - \lambda) cos(\beta - \lambda)
= \frac{1}{\sqrt{2}} cos(\alpha - 0) cos(\beta - 0) + \frac{1}{\sqrt{2}} cos(\alpha - \frac{\pi}{2}) cos(\beta - \frac{\pi}{2})
= \frac{1}{\sqrt{2}} cos(\alpha) cos(\beta) + \frac{1}{\sqrt{2}} sin(\alpha ) sin(\beta)
= \frac{1}{\sqrt{2}} cos(\beta - \alpha)

So there is a strong sense in which amplitudes for quantum mechanics work the way we expect probabilities to work in classical probability.

 

[TheMeInTeam]

How could one distinguish between "true randomness" and a very good unknown algorithm. You could gather data forever and not know. You could make predictions based on a well balanced coin flipped into a wind tunnel.

"The Bell inequality is a profound experimental observation imo."

I agree and here is imo the ultimate form of Bell's Inequality:
1) A and B meet for a strategy session and are then sent to their respective labs and not allowed to communicate.
2) Then we flip a fair coin giving the result to A, then flip again and give the result to B.
3) A selects bit a (= 0 or 1). B selects bit b.
4) A and B win the game if a ≠ b when they both received heads, and a = b otherwise.

Question: Is there a strategy allowing A and B to win with probability > 3/4?

I contend that prior to 1900 no body in the world could logically answer yes.
QM can achieve 85%. That is why quantum entanglement is weird

I don't like the idea that QM is "weird". It's inconsistent with our intuitive grasp of reality in some cases, so in that sense it's weird from our perspective...but it's still reality, happening routinely (on an enormous scale) and consistently. The ideal is that perception of it moves to normal, because reality probably isn't going to change for us. Otherwise it's too easy wind up with a similar kind of mental barrier to students who like to claim "I'm bad at math and it scares me", an emotional rejection that undercuts their ability to understand something that, in the strict sense, is simpler than many things they're already learned. To them math still seems weird ("X is a number...but it changes between problems?").

 

[Zafa Pi]

I don't like the idea that QM is "weird". It's inconsistent with our intuitive grasp of reality in some cases, so in that sense it's weird from our perspective...but it's still reality, happening routinely (on an enormous scale) and consistently. The ideal is that perception of it moves to normal, because reality probably isn't going to change for us. Otherwise it's too easy wind up with a similar kind of mental barrier to students who like to claim "I'm bad at math and it scares me", an emotional rejection that undercuts their ability to understand something that, in the strict sense, is simpler than many things they're already learned. To them math still seems weird ("X is a number...but it changes between problems?").

I am sorry you don't like weird, I love it. I'm not a physicist, but was drawn to QM because it was weird. My favorite math theorem is the Tarski-Banach Theorem in spite of being quite familiar with the proof. Minimal information problems: Terrific. Kids I know love weird as well. Being scared of math is 90% due to a crappy start (and continuation) at age 4. Same reason some people don't know how to throw a ball.
If it's inconsistent with my intuitive grasp of reality or reason, then I say BRING IT!

 

[Grinkle]

I don't like the idea that QM is "weird".

I think I get you. I think QM is weird in the same way I think a 4-d cube is weird. It is a mathematical construct that I cannot create a mental image of or intuition for. I think I really do get you, because I think a 4-d cube is much much lower on the weird scale than QM, and that is due to my much much better understanding of how a 3-d surface model can be extended to 4-d geometrically. Weirdness should not be a barrier to comprehension, even if true intuition may never be possible eg in the way our brains simply lack needed circuitry to genuinely visualize a 4-d cube.