Does quantum physics imply the existence of randomness?

[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, [itex]B_1, B_2, ..., B_n[/itex], 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 [itex]B_j[/itex] 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, [itex]B_1, B_2, ..., B_n[/itex], 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 [itex]B_j[/itex] 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 [itex]P(A,B,t)[/itex] means the probability of going from A to B in time t, then

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

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 [itex]t \rightarrow 0[/itex].

 

[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 [itex]P(A,B|\alpha, \beta)[/itex] for Alice and Bob to both measure spin-up, given Alice's setting [itex]\alpha[/itex] and Bob's setting [itex]\beta[/itex] cannot be factored in the form:

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

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:

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

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 [itex]P(A,B|\alpha, \beta)[/itex] for Alice and Bob to both measure spin-up, given Alice's setting [itex]\alpha[/itex] and Bob's setting [itex]\beta[/itex] cannot be factored in the form:

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

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:

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

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:

[itex]P(A,B|\alpha, \beta)[/itex]

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

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

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 [itex]\lambda[/itex]) 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 [itex]\psi(A, B|\alpha, \beta)[/itex] be the joint amplitude for the EPR experiment, where the amplitude is related to the probability via:

[itex]P(A, B|\alpha, \beta) = |\psi(A,B|\alpha, \beta)|^2[/itex]

So [itex]\psi(A,B|\alpha, \beta)[/itex] 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 [itex]\lambda[/itex] and amplitude functions [itex]\psi_{hv}(\lambda)[/itex], [itex]\psi_A(A|\alpha, \lambda)[/itex], [itex]\psi_B(B|\beta, \lambda)[/itex] such that:

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

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:

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

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

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

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

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

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

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

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.