A Copenhagen: Restriction on knowledge or restriction on ontology?

[bhobba]

Quantum mechanics does not unsettle me.

Me either. Nor does classical mechanics which we now know depends crucially on QM. Things progress and people always want to lift the veil as see what future research will reveal or participate in that research. The thing is this one has proven HARD and progress is slow.

Thanks
Bill

 

[DarMM]

That's true.

There is no joint probability distribution describing $S_x$ and $S_y$. However, if you lift it to the macroscopic level, there can be a joint probability distribution describing the two situations: "I measured $S_x$ and found it to be $+\frac{1}{2}$" and "I measured $S_y$ and found it to be $+\frac{1}{2}$". Those two situations are exclusive: if one is true, the other is false. This is in contrast to the microscopic situation, where it is not possible to give simultaneous truth values to $S_x = + \frac{1}{2}$ and $S_y = + \frac{1}{2}$.

There isn't a joint probability distribution for them as the two events belong to two incompatible histories. They are exclusive correct, but the rest of the properties of Boolean logic are not well defined for them and thus without a common Boolean lattice you cannot build a probability distribution for both.

 

[stevendaryl]

There isn't a joint probability distribution for them as the two events belong to two incompatible histories. They are exclusive correct, but the rest of the properties of Boolean logic are not well defined for them and thus without a common Boolean lattice you cannot build a probability distribution for both.

They belong to different histories, but not incompatible histories.

Look, if I flip a coin and if it's heads, I go left and if it's tails I go right, then there are two possible histories: one in which I flip the coin and go left and one in which I flip the coin and go right. Those are alternative, but they are not incompatible in the sense of probability distributions. I can give a probability of 50% to each.

This is in contrast to the two histories (1) an electron has spin-up in the x-direction, and (2) an electron has spin-up in the y-direction. Those two possibilities cannot be given consistent probabilities.

 

[DarMM]

To construct a probability distribution for both you need a Boolean lattice. Where are $A \land B$ and $A \lor B$ for those two events?

 

[Josh0768]

To my mind, the spirit of the Copenhagen interpretation refers to the ontology. As Heisenberg puts it:

"However, all the opponents of the Copenhagen interpretation do agree on one point. It would, in their view, be desirable to return to the reality concept of classical physics or, to use a more general philosophic term, to the ontology of materialism. They would prefer to come back to the idea of an objective real world whose smallest parts exist objectively in the same sense as stones or trees exist, independently of whether or not we observe them."

In case one gives up the concept of 'physical realism' - a viewpoint according to which an external reality exists independent of observation – and doesn’t insist on thinking about quantum phenomena with classical ideas, quantum mechanics doesn’t unsettle anymore.

Hey, could you tell me what book/paper that quote is from?

 

[stevendaryl]

To construct a probability distribution for both you need a Boolean lattice. Where are $A \land B$ and $A \lor B$ for those two events?

$A \wedge B$ is the impossible event. You can't measure both spin in the x-direction and spin in the y-direction.

Let's make a complete list of possible histories for such an experiment:

1. We measure the spin in the x-direction, and find it is $+ \frac{1}{2}$
2. We measure the spin in the x-direction, and find it is $- \frac{1}{2}$
3. We measure the spin in the y-direction, and find it is $+ \frac{1}{2}$
4. We measure the spin in the y-direction, and find it is $- \frac{1}{2}$

I don't understand what problem you see in giving a probability distribution to those 4 possibilities. For example, give all 4 the probability of 25%. They are exclusive events, in that the intersection of any two of them is the impossible event.

 

[DarMM]

No problem thus far. Just a quick question first. The probabilities you have there are you imagining something like having a system in a $|z = +\frac{1}{2}\rangle$ state and then flipping a coin to see if you measure $S_x$ and $S_y$?

 

[stevendaryl]

No problem thus far. Just a quick question first. The probabilities you have there are you imagining something like having a system in a $|z = +\frac{1}{2}\rangle$ state and then flipping a coin to see if you measure $S_x$ and $S_y$?

Yes.

I took a peak at the Wikipedia article on consistent histories (https://en.wikipedia.org/wiki/Consistent_histories), but I'm not sure that I understood it. The criterion given there was that

$Tr(C_{H_i} \rho C_{H_j}^\dagger) = 0$ whenever $i \neq j$, where $C_{H_i}$ is the product of the Heisenberg operators corresponding to the projection operators in history $i$, and $\rho$ is the initial density matrix.

In the simplest case, we have one-event histories, in which case $C_{H_i}$ is just the projection operator corresponding to the 4 possibilities I gave above.

So let $|\psi_j\rangle \langle \psi_j|$ be the projection operator corresponding to possibility $j$. Let $|\psi_0\rangle \langle \psi_0|$ be the initial density matrix (assumed to be a pure state for simplicity). Then

$C_{H_i} \rho C_{H_j}^\dagger = |\psi_i\rangle \langle \psi_i|U|\psi_0\rangle \langle \psi_0|U^{-1}|\psi_j\rangle \langle \psi_j|$

where $U$ is the evolution operator. Taking the trace gives:

$\langle \psi_j | \psi_i \rangle \langle \psi_i | U |\psi_0\rangle \langle \psi_0|U|\psi_j \rangle$

So if the matrix element $\langle \psi_j | \psi_i \rangle$ is zero, then the two histories are consistent, according to Wikipedia's criterion.

 

[DarMM]

Okay so then if you are measuring a spin-1 particle you will have nine possible macrohistories or if you are measuring two spin-$\frac{1}{2}$ particles (and keep to only measuring $S_x$ and $S_y$) you will have sixteen possible macrohistories.

Then basically there will be some states where the probabilities they give for those histories cannot result from a common measure you place over the whole set of nine/sixteen.

 

[A. Neumaier]

if you are measuring two spin-2 particles (and keep to only measuring $S_x$ and $S_y$) you will have sixteen possible macrohistories.

25, not 16?

 

[DarMM]

25, not 16?

Wrong spin! Don't want to force poor @stevendaryl to detect gravitons!

 

[stevendaryl]

Okay so then if you are measuring a spin-1 particle you will have nine possible macrohistories or if you are measuring two spin-$\frac{1}{2}$ particles (and keep to only measuring $S_x$ and $S_y$) you will have sixteen possible macrohistories.

I don't think it needs to be that complicated. I was thinking of a single particle initially spin-up in the z-direction. I flip a coin and either measure its spin in the x-direction, or in the y-direction. So there are 4 possible final states: $|X,up\rangle, |X,down\rangle, |Y,up\rangle, |Y, down\rangle$.

Then basically there will be some states where the probabilities they give for those histories cannot result from a common measure you place over the whole set of nine/sixteen.

I don't think that's true. If two possibilities are exclusive (they can't both happen), then they are consistent. It's only when two different possibilities have a nonzero overlap that there is an incompatibility. There is no overlap between "I measured spin in the x-direction and got spin-up" and "I measured spin in the y-direction and got spin-up".

 

[DarMM]

I don't think it needs to be that complicated.

It does. I've no issue with your original example for in that case one can construct an overall probability measure. However for measurements on any quantum system with Hilbert space dimension $d \geq 3$ what you are describing will not work. I simply listed $d = 3,4$ examples.

 

[DarMM]

Is this related?
Eric G. Cavalcanti
https://arxiv.org/abs/1602.07404

Just need time to absorb this, perhaps a day or two.

 

[stevendaryl]

It does. I've no issue with your original example for in that case one can construct an overall probability measure. However for measurements on any quantum system with Hilbert space dimension $d \geq 3$ what you are describing will not work. I simply listed $d = 3,4$ examples.

Give a concrete example of what you're talking about. As I understand it (which I admit I only do superficially), histories that are mutually exclusive are always compatible. Histories with different macroscopic details are mutually exclusive, so they are always compatible.

 

[DarMM]

Give a concrete example of what you're talking about. As I understand it (which I admit I only do superficially), histories that are mutually exclusive are always compatible. Histories with different macroscopic details are mutually exclusive, so they are always compatible.

For the $d = 4$ case we can take a pair of spin-$\frac{1}{2}$ particles and relate this back to entanglement.
The observables on one particle are $S_0, S_{\frac{\pi}{4}}$ and the observables on the other $S_{\frac{\pi}{2}}, S_{\frac{3\pi}{4}}$.

There are then four possible measurements and each measurement can have four possible outcomes.

So we would have histories like:
"I measured $S_0 = \frac{1}{2}$ on the first particle and $S_{\frac{\pi}{2}} = \frac{1}{2}$ on the second"

If you look at the probabilities the Bell state gives to these histories and you look at all 16 histories you'll see it isn't consistent with them being drawn from a single probability distribution.

This applies also outside of entanglement in the $d = 3$ case.

 

[DarMM]

Chapter 19 of Griffiths book demonstrates the difference between the macroscopic histories of a stochastic (in the Kolmogorov sense with a single sample space) world and a quantum mechanical world.

Particularly see 19.3 and 19.4 which directly discusses the counterfactual differences and the remarks on p.271

 

[stevendaryl]

For the $d = 4$ case we can take a pair of spin-$\frac{1}{2}$ particles and relate this back to entanglement.
The observables on one particle are $S_0, S_{\frac{\pi}{4}}$ and the observables on the other $S_{\frac{\pi}{2}}, S_{\frac{3\pi}{4}}$.

There are then four possible measurements and each measurement can have four possible outcomes.

So we would have histories like:
"I measured $S_0 = \frac{1}{2}$ on the first particle and $S_{\frac{\pi}{2}} = \frac{1}{2}$ on the second"

If you look at the probabilities the Bell state gives to these histories and you look at all 16 histories you'll see it isn't consistent with them being drawn from a single probability distribution.

This applies also outside of entanglement in the $d = 3$ case.

I would have to see this worked out. What you're saying doesn't seem correct to me.

We have 16 possibilities, which we can characterize by 4 numbers: $A_j, a_j, B_j, b_j$ where $A_j$ is the $j^{th}$ measurement on the first particle, $a_j$ is the $j^{th}$ result of that measurement (either "up" or "down"), $B_j$ is the $j^{th}$ measurement on the second particle, and $b_j$ is the $j^{th}$ result of that measurement. For maximally entangled anti-correlated spin-1/2 particles, if you measure one particle along axis $\overrightarrow{A}$ and the other along axis $\overrightarrow{B}$, then the probability that the results will be the same is $sin^2{\frac{\theta}{2}}$, where $\theta$ is the angle between the two axes. I'm not sure why you chose those particular orientations, but the probabilities work out to:

$$ \left[ \begin{array} \\ A & a & B & b & P(a,b|A,B)
\\ 0 & up & \frac{\pi}{2} & up & 0.25
\\ 0 & up & \frac{\pi}{2} & down & 0.25
\\ 0 & down & \frac{\pi}{2} & up & 0.25
\\ 0 & down & \frac{\pi}{2} & down & 0.25
\\ 0 & up & \frac{3\pi}{4} & up & 0.43
\\ 0 & up & \frac{3\pi}{4} & down & 0.07
\\ 0 & down & \frac{3\pi}{4} & up & 0.07
\\ 0 & down & \frac{3\pi}{4} & down & 0.43
\\ \frac{\pi}{4} & up & \frac{\pi}{2} & up & 0.07
\\ \frac{\pi}{4} & up & \frac{\pi}{2} & down & 0.43
\\ \frac{\pi}{4} & down & \frac{\pi}{2} & up & 0.43
\\ \frac{\pi}{4} & down & \frac{\pi}{2} & down & 0.07
\\ \frac{\pi}{4} & up & \frac{3\pi}{4} & up & 0.25
\\ \frac{\pi}{4} & up & \frac{3\pi}{4} & down & 0.25
\\ \frac{\pi}{4} & down & \frac{3\pi}{4} & up & 0.25
\\ \frac{\pi}{4} & down & \frac{3\pi}{4} & down & 0.25
\end{array} \right] $$

If we give a 25% probability for all 4 choices of which two measurements to perform, then you just multiply every conditional probability above by 0.25. What's wrong with considering that to be a probability space with 16 possibilities?

 

[DarMM]

Okay now take the observable $B_2$. Since this is a single sample space attempting to replicate the statistics, what value does $B_2$ take in each outcome such that its moments give you results replicating QM?

 

[stevendaryl]

I don't know what you mean. What are you saying is impossible?

The usual argument about Bell's inequality is not that it is impossible for a stochastic model to reproduce the statistics, but that it is impossible for a local model to reproduce the statistics. There is no claim that the stochastic model is local in the sense of Bell.

 

[DarMM]

I don't know what you mean. What are you saying is impossible?

The usual argument about Bell's inequality is not that it is impossible for a stochastic model to reproduce the statistics, but that it is impossible for a local model to reproduce the statistics. There is no claim that the stochastic model is local in the sense of Bell.

Much simpler than that. How do you define $B_2$ on each outcome so that it is a random variable on this sample space. Ignore Bell's theorem as such for now.

 

[stevendaryl]

Much simpler than that. How do you define $B_2$ on each outcome so that it is a random variable on this sample space. Ignore Bell's theorem as such for now.

I don't understand your question. There are 16 possible events. Each event has 4 components: $A, a, B, b$. You give a probability to each of the 16 combinations so that the probabilities add up to 1.

 

[DarMM]

I get that. How do you define $B_2$ as a random variable, i.e. a function on this space. So that you can compute things like $\langle B_2 \rangle$?

 

[stevendaryl]

I get that. How do you define $B_2$ as a random variable, i.e. a function on this space. So that you can compute things like $\langle B_2 \rangle$?

I didn't make any claims about $B_2$ as a random variable. I don't know what that means. There is one random variable, which is which event out of 16 possibilities. It can take values 1 through 16. If the value is 1, then in that history, the setting of $A$ is $0$, the setting of $B$ is $\frac{\pi}{2}$, the outcome $a$ is $up$, the outcome $b$ is $up$.

 

[stevendaryl]

I didn't make any claims about $B_2$ as a random variable. I don't know what that means. There is one random variable, which is which event out of 16 possibilities. It can take values 1 through 16. If the value is 1, then in that history, the setting of $A$ is $0$, the setting of $B$ is $\frac{\pi}{2}$, the outcome $a$ is $up$, the outcome $b$ is $up$.

You can certainly look at the subset of histories in which $B = \frac{3\pi}{4}$. For those histories, you can compute the expectation value of the result $b$.

 

[DarMM]

I didn't make any claims about $B_2$ as a random variable. I don't know what that means.

Really? $B_2$ is an observable quantitiy and will be encoded in a probability model as a random variable. Thus if you are using Kolmogorov probability $B_2$, $B_{\frac{3\pi}{4}}$ as we are calling it, has to be a random variable on that space. That's just how probability theory works.

You can certainly look at the subset of histories in which $B = \frac{3\pi}{4}$. For those histories, you can compute the expectation value of the result $b$.

Exactly. There is only a subset on which $B_{\frac{3\pi}{4}}$ is defined as a random variable. Thus in truth these are four separate Kolmogorov models with their own individual sample spaces.

 

[stevendaryl]

Really? $B_2$ is an observable quantitiy and will be encoded in a probability model as a random variable.

Okay, suppose that I take 16 cards. On each index card, I write down 4 quantities: $(A, a, B, b)$ where $A$ is either 0 or $\frac{\pi}{4}$ and $a$ is either $up$ or $down$ and $B$ is either $\frac{\pi}{2}$ or $\frac{3\pi}{4}$ and $b$ is again either $up$ or $down$. So I play lots and lots of rounds where I pick a card according to some probability distribution. Tell me in that situation what it means to say that $B_2$ is a random variable.

The random variable is which card I choose. There are 16 cards. The values of $a$ and $b$ and $A$ and $B$ are functions of this one random variable.

 

[DarMM]

Tell me in that situation what it means to say that $B_2$ is a random variable.

What you're missing here is that $B_2$ is a physical quantity. It's something you can observe. In a probability model it should therefore be a random variable. Note though what you are saying isn't entirely wrong. See the next the question below, this might serve the discussion better.

The usual argument about Bell's inequality is not that it is impossible for a stochastic model to reproduce the statistics, but that it is impossible for a local model to reproduce the statistics. There is no claim that the stochastic model is local in the sense of Bell.

Where is the nonlocality in your table?

 

[stevendaryl]

What you're missing here is that $B_2$ is a physical quantity.

In my card game, what corresponds to $B_2$?

 

[DarMM]

In my card game, what corresponds to $B_2$?

A value on the card I assume. Let's just focus on the nonlocality. In your table how does it manifest?