Does quantum physics imply the existence of randomness?

  • #76
stevendaryl
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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.
 
  • #77
stevendaryl
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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].
 
  • #78
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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.
 
  • #79
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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.
 
  • #80
Grinkle
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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.
 
  • #81
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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
 
  • #82
stevendaryl
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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).
 
  • #83
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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.
 
  • #84
stevendaryl
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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.
 
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  • #85
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?").
 
  • #86
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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!
 
  • #87
Grinkle
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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.
 

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