Axiomatization of quantum mechanics and physics in general ?

[billschnieder]

Here we are considering the quantum formalism in which the addition of expectation values is true.

Okay so we can linearly combine them. But it is nontrivial for non-commuting observables. The issue is not that we can not combine them but that we don't expect a trivial substitution to work. Like I said, there is no problem with the substitution if we interpret is as corresponding to 4 isolated systems, but as you can see above, the inequality is different, S <= 4. The problem only arises if you interpret the substitution as pertaining to the same system for which S <=2. So when it is said that QM violates the S <= 2 inequality, it is suspect because if we carry that argument, we would have to treat the substitution as pertaining to the same system and we end up with an expression that has no solution because it is impossible to find an eigenvector for that specific combination of observables.

You can factorize that linear combination and end up with an expression of the form
$\langle A_a(B_b - B_{b'}) + A_a'(B_b + B_{b'})\rangle$ which is an expression of the form
$\hat{A}\otimes\hat{B} + \hat{C}\otimes\hat{D}$
If it has a solution, should satisfy

$\hat{A}|\phi\rangle\otimes\hat{B}|\chi\rangle + \hat{C}|\phi\rangle\otimes\hat{D}|\chi\rangle = \alpha(|\phi\rangle\otimes|\chi\rangle)$
But the LHS cannot be factored since the $[\hat{A},\hat{C}] ≠0$ and $[\hat{B},\hat{D}] ≠0$ so there is no solution. It is a meaningless expression for a single system.

 

[atyy]

commuting/non-commuting is irrelevant if you have ψ1 and ψ2 (isolated systems). But it is surely relevant if you have just ψ (one system). I can prepare one system, measure x precisely, then prepare a different one very similarly and measure p precisely. The two observables would commute, that won't be the case if you have just ψ. It is a distinction between "similar" and "the same".

No, that is simply not true in the quantum formalism. There is no ψ1 and ψ2. The quantum calculation simply assumes ψ. Within the quantum formalism <ψ|x+p|ψ> has the same value as <ψ|x|ψ>+<ψ|p|ψ>.

This is not true. You can easily verify that the expression:

ab - a'b + a'b + a'b' <= 2

is only valid "the same" system because you can factorize a(b-b') + a'(b+b') and show that whenever (b-b') = 0, (b+b) = -2 or 2 and vice-versa for a,b,a'b' = {+1, -1}. This is not true for separate systems because you cannot factorize the expression (a1b2 - a2'b2 + a3'b3 + a4'b4') and the RHS is necessarily 4.

I believe we agree, except we are using different terms. Here by "same system" you mean that the distribution over the hidden variable is the same for different measurement settings. However, by "different systems", I mean that the distribution over the hidden variable is the same for different measurement settings, but that the different measurement settings are performed on different trials. And yes, it is a known loophole that if the measurement settings depend on the hidden variable, or if the hidden variable depends on the measurement setting, then one can have a local variable explanation of a Bell inequality violation.

 

[billschnieder]

No, that is simply not true in the quantum formalism.

I think there is a problem somewhere. If I have two entangled spin-1/2 particles send one to Alice and the other to Bob and I measure their spins, I always get opposite results according to QM, yes?
Now if I take a different pair of entangled spin-1/2 particles similar to the first pair, and measure them both at exactly the same angles as the first two, according to QM, I still get the exact same relationship between them, opposite results again.

But does QM say that The first particle from the first pair will have exactly opposite results from the second particle of the second pair?

I believe we agree, except we are using different terms. Here by "same system" you mean that the distribution over the hidden variable is the same for different measurement settings.

No, I mean the exact same set of particles, not some different set with similar distribution of hidden variables.

However, by "different systems", I mean that the distribution over the hidden variable is the same for different measurement settings, but that the different measurement settings are performed on different trials.

This is what I mean by different systems. The same properties but not the same individual. The same type of particle pair but not the same particle pair. The same "type of" wavefunction but not the same wavefunction. That is what I mean by ψ1 and ψ2 vs ψ.

And yes, it is a known loophole that if the measurement settings depend on the hidden variable, or if the hidden variable depends on the measurement setting, then one can have a local variable explanation of a Bell inequality violation.

I'm not talking about loopholes at all. I'm not even talking about distributions of hidden variables and what may depend on settings or not depend on settings.

 

[bhobba]

But it is nontrivial for non-commuting observables.

Sorry - but its utterly trivial.

Its a simple consequence of Born's rule - commuting, non commuting - no difference.

Thanks
Bill

 

[bhobba]

But does QM say that

Of course it does - its a basic property of Bell States:
http://en.wikipedia.org/wiki/Bell_state

Pick any Bell state - say the first one.

The result of the observation will be |1A>|1B> or |0A>|0B>. If A gets state 1 B must be state 1 and conversely.

Its a simple result of what entanglement means.

Thanks
Bill

 

[billschnieder]

Of course it does - its a basic property of Bell States:
http://en.wikipedia.org/wiki/Bell_state

Pick any Bell state - say the first one.

The result of the observation will be |1A>|1B> or |0A>|0B>. If A gets state 1 B must be state 1 and conversely.

Its a simple result of what entanglement means.

Thanks
Bill

Maybe you misunderstood, each member of a pair is entangled with the other the second pair is not entangled with the first pair.

 

[billschnieder]

Sorry - but its utterly trivial.

Its a simple consequence of Born's rule - commuting, non commuting - no difference.

Thanks
Bill

So you think that expression has a solution?

 

[bhobba]

So you think that expression has a solution?

I have zero idea what you are getting at.

I was addressing the linearity of expectations.

Thanks
Bill

 

[bhobba]

Maybe you misunderstood, each member of a pair is entangled with the other the second pair is not entangled with the first pair.

So?

The second pair when observed will do the same thing.

I think there is a problem somewhere. If I have two entangled spin-1/2 particles send one to Alice and the other to Bob and I measure their spins, I always get opposite results according to QM, yes? Now if I take a different pair of entangled spin-1/2 particles similar to the first pair, and measure them both at exactly the same angles as the first two, according to QM, I still get the exact same relationship between them, opposite results again.But does QM say that The first particle from the first pair will have exactly opposite results from the second particle of the second pair?

Yes it does say that. The same experiments give the same results - that applies to any area of science - its so trivial its rarely if ever stated.

Thanks
Bill

 

[kith]

But in this case we are dealing with non-commuting observables. If you measure A on one system ψ1 and B on a different system ψ2, you will not necessarily get the same result as what you should expect if you had measured A and B on the same system. Because even though <ψ1|A|ψ1> always commutes with <ψ2|B|ψ2> <ψ|A|ψ> does not commute with <ψ|B|ψ>. If I derive an inequality for a single system which contains both A and B, it will not be correct to trivially substitute in values from different systems, would it?

Expectation values always commute because they are numbers, so your math is messed up. Do you think what you say fits within standard QM? If yes, you should be able to use standard QM notation. If no, you should clarify where your symbols deviate from the standard notation.

In QM, the state is associated with a preparation procedure. If we apply the same preparation procedure to two systems, they are in the same state. Is your question motivated by the idea what happens if the preparation procedure is unsharp, so that it prepares slightly different states in different trials?

Would it be appropriate to call your objection the induction-is-impossible loophole?

 

[bhobba]

Because even though <ψ1|A|ψ1> always commutes with <ψ2|B|ψ2> <ψ|A|ψ> does not commute with <ψ|B|ψ>.

Expectation values always commute because they are numbers, so your math is messed up.

Well spotted - I didn't even notice it.

I looked at it, looked at it again, scratched my head, scratched it some more - still no closer to understanding his issue.

Thanks
Bill

 

[kith]

Should there be a natural distinction between classical and quantum mechanics for continuous variables?

There are many ways to state fundamental differences between classical mechanics and QM, so I would say that these are certainly fundamentally different theories. Sure, we can supplement QM with Bohmian hidden variables to make at least some of these differences go away, but then we don't compare classical mechanics and QM but classical mechanics and Bohmian mechanics.

I don't really like the statement that QM is an effective theory because it suggests that there's something fundamentally wrong with QM while there's no experimental hint of this. Even in areas where QM gets in trouble conceptually -like dealing with gravity- the alleged more fundamental Bohmian mechanics doesn't seem to lead to different solutions than QM but again provides "only" an unobservable supplement. (To be fair I have to say that I may simply not know enough about possibly observable consequences from Bohmian mechanics which are incompatible with QM. Also I put "only" in " because I do value Bohmian mechanics as a different way of looking at QM. I just don't think it is superior.)

I tend to think more along Copenhagen-inspired lines that every physical theory is somehow effective. So if QM is a very good effective theory, Bohmian mechanics is a slightly worse effective theory because it still has the outstanding experimental support of QM but also introduces additional unobservable elements.

 

[billschnieder]

So?

The second pair when observed will do the same thing.

A bell state has two particles not 4. Now you
Are saying a measurement of one particle of a Bell state affects a particle not part of *the* Bell state. I'm surprised you don't see a problem.

Yes it does say that. The same experiments give the same results - that applies to any area of science - its so trivial its rarely if ever stated.

But that is the problem what you are suggesting disagrees with experiment.

 

[bhobba]

A bell state has two particles not 4. Now you Are saying a measurement of one particle of a Bell state affects a particle not part of *the* Bell state

Where you get that from has me beat - I specifically stated otherwise.

You measure one entangled pair and get one result.

Measure another entangled pair and get another result.

QM describes the outcome of both independent of each.

Like I said this is basic science not just QM - its the idea of independent experiments giving reproducible results.

I think there is a problem somewhere. If I have two entangled spin-1/2 particles send one to Alice and the other to Bob and I measure their spins, I always get opposite results according to QM, yes? Now if I take a different pair of entangled spin-1/2 particles similar to the first pair, and measure them both at exactly the same angles as the first two, according to QM, I still get the exact same relationship between them, opposite results again.But does QM say that The first particle from the first pair will have exactly opposite results from the second particle of the second pair?

Lets be clear. What you wrote and I have put in bold is saying if I repeat the same experiment you get the same result - of course you do. Here same result is you will get the same correlations.

Thanks
Bill

 

[billschnieder]

But does QM say that The first particle from the first pair will have exactly opposite results from the second particle of the second pair?

But does QM say that

Of course it does - its a basic property of Bell States:
http://en.wikipedia.org/wiki/Bell_state

That is why I suggested that you misunderstood me, because you were (are) implying that the result of one pair is correlated with the result from a different pair. This is not what is observed experimentally. The reproducibility which I'm not questioning is the fact that the correlation between the members of the first pair is exactly the same correlation between the members of the second pair. Each pair is a Bell state after all. But you surely don't mean that you can mix and match one member of each pair and still have a bell state. Then you end up with a situation in which once the first pair is measured, the outcomes of the rest will be well defined and then the randomness from pair to pair usually obtained in experiments will not happen. That is why I say you are contradicting experiment.

 

[billschnieder]

Expectation values always commute because they are numbers, so your math is messed up.

Well spotted - I didn't even notice it.

So then what was Bell complaining about the nontriviality of additivity of expectation values?

The essential assumption can be criticized as follows. At first sight, the required additivity of expectation values seems very reasonable, and is rather the non-additivity of allowed values (eigenvalues) which requires explanation. Of course the explanation is well known: A measurement of a sum of noncommuting observables cannot be made by combining trivially the results of separate observations on the two[individual] terms -- it requires a quite distict experiment. ... But this explanation of the non-additivity of allowed values also establishes the nontriviality of the additivity of expectation values

That is why I said earlier:
Starting from
$E_\psi(x,z) = \langle\psi|(\sigma_L\cdot x)(\sigma_R\cdot z)|\psi\rangle = -x\cdot{z}$

$S^{\psi} = |\langle\psi|(\sigma_L\cdot{a})(\sigma_R\cdot{b}) - (\sigma_L\cdot{a})(\sigma_R\cdot{b'}) + (\sigma_L\cdot{a'})(\sigma_R\cdot{b}) + (\sigma_L\cdot{a'})(\sigma_R\cdot{b'})|\psi\rangle|$
Do all those spin observables commute?

 

[wle]

I'm not talking about the derivation. I'm talking about the demonstration of QM violation of the inequality. The part where expectations are linearly combined.

For example, how do you show that QM violates Bell's inequality. It is this calculation I'm talking about.

I think you're misunderstanding what the CHSH correlator is and what the local bound on it actually means.

Formally, the CHSH correlator is by definition the linear combination
$$S = \sum_{abxy} (-1)^{a + b + xy} P(ab \mid xy) \,,$$
where the variables $x, y \in \{0, 1\}$ are Alice's and Bob's choice of measurement settings, $a, b \in \{0, 1\}$ are the measurement results, and $P(ab \mid xy)$ is the prior probability that Alice and Bob get the result $(a, b)$ given that they chose the measurement pair $(x, y)$ in a CHSH-type experimental setting.

For the purpose of what Bell actually proved originally -- that no locally causal theory can make the same predictions as quantum physics -- it's not an error that the $P(ab \mid xy)$s are all defined on the same particle pair because the CHSH correlator doesn't need to be a physically meaningful or measurable quantity. Quantum mechanics is a theory that is defined mathematically and Bell defined the class of locally causal theories mathematically. As such, they can be compared mathematically without ever needing to do an actual experiment. In this context, the CHSH correlator is just an intermediate variable in the proof of Bell's theorem and doesn't need to "mean" anything beyond that.

 

[billschnieder]

I think you're misunderstanding what the CHSH correlator is and what the local bound on it actually means. ...
For the purpose of what Bell actually proved originally -- that no locally causal theory can make the same predictions as quantum physics -- it's not an error that the P(ab∣xy)s are all defined on the same particle pair because the CHSH correlator doesn't need to be a physically meaningful or measurable quantity.

I don't think I am. I think you are misunderstanding what the issue is. The CHSH *is* indeed derived for a single system. The problem I see is that the demonstration of QM violation of the CHSH uses 4 expectation values with two possible interpretations of the QM substitution, either one not good for Bell. The two options are:
1) The first option is that the QM expression, like the CHSH correlator applies to a single system. This is problematic because then it would be making the same mistake as von Neumann which Bell criticized.

Yet the Von Neumann proof, if you actually come to grips with
it, falls apart in your hands. There’s nothing to it. It’s not
just flawed. . .it’s silly. When you translate his assumptions into
physical significance, they’re nonsense. You may quote me on
this. The proof of Von Neumann is not just false, it’s foolish

2) The second option is that QM expression, unlike the CHSH correlator applies to 4 different (but similar) systems. This is problematic because then, he won't be able to demonstrate violation, since the 4-system upper bound is 4.

 

[wle]

I don't think I am. I think you are misunderstanding what the issue is. The CHSH *is* indeed derived for a single system. The problem I see is that the demonstration of QM violation of the CHSH uses 4 expectation values with two possible interpretations of the QM substitution, either one not good for Bell. The two options are:
1) The first option is that the QM expression, like the CHSH correlator applies to a single system. This is problematic because then it would be making the same mistake as von Neumann which Bell criticized.

I'm not familiar with von Neumann's argument so I don't know offhand what the issue with it was. von Neumann may well have added terms from quantum mechanics together in a way that didn't make sense in the context of his argument. But there's nothing problematic with the CHSH correlator. Like I said in my previous post, it's just defined as the linear combination of probabilities

$$S = \sum_{abxy} (-1)^{a + b + xy} P(ab \mid xy) \,, \qquad a, b, x, y \in \{0, 1\} \,.$$
Alternatively, it can be written more compactly as the scalar product

$$S = \bar{I} \cdot \bar{P}$$
with the vector $I$ defined by the components $I_{abxy} = (-1)^{a + b + xy}$. There's no problem evaluating this quantity (which is just a variable defined for convenience) for a given set of measurement operators on a given quantum state in quantum physics. The proof of Bell's theorem just uses that if $\bar{I} \cdot \bar{P}_{1} \neq \bar{I} \cdot \bar{P}_{2}$ then necessarily $\bar{P}_{1} \neq \bar{P}_{2}$.

 

[bhobba]

I'm not familiar with von Neumann's argument so I don't know offhand what the issue with it was.

Its simple. If you assume expectations are additive then Born's Rule follows. Its not hard.

First its easy to check <bi|O|bj> = Trace (O |bj><bi|).

O = ∑ <bi|O|bj> |bi><bj| = ∑ Trace (O |bj><bi|) |bi><bj|

Now we use the linearity assumption ie expectations are additive and if f is that expectation

f(O) = ∑ Trace (O |bj><bi|) f(|bi><bj|) = Trace (O ∑ f(|bi><bj|)|bj><bi|)

Define P as ∑ f(|bi><bj|)|bj><bi| and we have f(O) = Trace (OP).

P, by definition, is called the state of the quantum system. The following are easily seen. Since f(I) = 1, Trace (P) = 1. Thus P has unit trace. f(|u><u|) is a positive number >= 0 since |u><u| is an effect. Thus Trace (|u><u| P) = <u|P|u> >= 0 so P is positive.

The trouble is while experiment shows expectations of quantum observables are additive, and its very intuitive anyway, it does not necessarily apply to hidden variables. That's the key point Bell showed. He was not the only one - but due to Von Neumann's reputation (of course he is correctly regarded as a mathematician/mathematical physicist of the highest calibre - many have him in the top 10 greatest of all time - as do I) they were ignored:
http://mpseevinck.ruhosting.nl/seevinck/Aberdeen_Grete_Hermann2.pdf

Stronger proofs came along later - Gleason's theorem probably being the deepest. The real key is non-contextuality - its also tied up with locality and Bells Theorem in a subtle way:
http://people.maths.ox.ac.uk/tillmann/CATlect2013SA4.pdf

Thanks
Bill

 

[wle]

I had a quick skim through Bell's article on the subject [Rev. Mod. Phys. 38, 447 (1996)]. If I've understood the issue I'd explain it as follows:

Suppose $E(A \mid \Psi; \lambda)$ is the expectation value associated with an observable $A$ given a state vector $\lvert \Psi \rangle$ and some additional variables $\lambda$ according to some hidden variable theory, presumably satisfying some condition like $\int \mathrm{d} \lambda \rho(\lambda) E(A \mid \Psi; \lambda) = \langle \Psi \rvert A \lvert \Psi \rangle$ in order to recover quantum physics. The issue seems to be that von Neumann assumed a linearity condition along the lines of

$$E(a A + b B \mid \Psi; \lambda) = a E(A \mid \Psi; \lambda) + b E(A \mid \Psi; \lambda) \,,$$
while consistency with quantum physics would only require the weaker condition that linearity holds after averaging, i.e.,

$$\int \mathrm{d} \lambda \rho(\lambda) E(a A + b B \mid \Psi; \lambda) = a \int \mathrm{d} \lambda \rho(\lambda) E(A \mid \Psi; \lambda) + b \int \mathrm{d} \lambda \rho(\lambda) E(A \mid \Psi; \lambda) \,.$$

If that's the case, I'd agree that isn't necessarily justified, and it's equally clear to me that there is no such problem with Bell's theorem.

 

[wle]

Erratum:

$$E(a A + b B \mid \Psi; \lambda) = a E(A \mid \Psi; \lambda) + b E(A \mid \Psi; \lambda) \,,$$

$$\int \mathrm{d} \lambda \rho(\lambda) E(a A + b B \mid \Psi; \lambda) = a \int \mathrm{d} \lambda \rho(\lambda) E(A \mid \Psi; \lambda) + b \int \mathrm{d} \lambda \rho(\lambda) E(A \mid \Psi; \lambda) \,.$$

Those last terms should, of couse, be $b E(B \mid \Psi; \lambda)$ rather than $b E(A \mid \Psi; \lambda)$. (Why is there a 3 minute time limit on editing posts?!)

 

[bhobba]

If that's the case, I'd agree that isn't necessarily justified, and it's equally clear to me that there is no such problem with Bell's theorem.

.
Home run hit.

Spot on.

And indeed Bells Theorem has no such issue.

Thanks
Bill

 

[wle]

[Rev. Mod. Phys. 38, 447 (1996)].

... and that should be 1966.

 

[billschnieder]

bhobba & wle,
I think you are still missing the point. Perhaps if I ask you both a simple question: From the wikipedia page you cited earlier:

$S = \langle A(a) B(b) \rangle + \langle A(a') B(b') \rangle + \langle A(a') B(b) \rangle - \langle A(a) B(b') \rangle = \tfrac{4}{\sqrt{2}} = 2 \sqrt{2} > 2$

My question is very simple: Do each of the expectation values in that expression apply to the exact same system, or does it apply to different but similar systems?

Bhobba, I know we've discussed this previously so let me clarify what I mean by "exact same system" as opposed to "different but similar": If we are talking about two particles in a Bell state, then "exact same system" means there are only two particles in the discussion, and we are simply adding up what the exact same particle pair would do at different settings for the exact same two particles (one pair) . "different but similar" means there are 8 particle pairs in the discussion and we are adding results from one particle pair at one pair of settings, to results of a different but similar particle pair at another setting pair etc.

So which interpretation is it? Or do you not think there is a difference between both.

 

[atyy]

My question is very simple: Do each of the expectation values in that expression apply to the exact same system, or does it apply to different but similar systems?

They do not apply to what you are calling "the exact same system". They apply to different pairs of particles, each drawn from the same ensemble.

 

[billschnieder]

They do not apply to what you are calling "the exact same system". They apply to different pairs of particles, each drawn from the same ensemble.

Not sure I follow what you mean by "the same ensemble". Do you simply mean that they are similar? Or do you mean that the 4 expectation values apply to the "exact same ensemble". The issue does not disappear because we start talking about "ensemble" as opposed to individual particle pairs.

Perhaps it is still not clear what the difference is between "the same" and "different but similar". If i = 1,..,N represents identity for N members of a set, where the members could be particle pairs, ensembles, or whatever. "the same" means the same i, "different but similar" means different i, but still a member of the set. In computer-science-speak, it would be the difference between "equality" and "identity" or the difference between "value" and "reference".

So my question in this context will be:
Are those expectation values applicable to the same system, or to different but similar systems (whatever your definition of system is, particle pairs or ensembles).

You could then also ask the same question, of von Neumann's essential assumption:
von Neumann: Any real linear combination of any two Hermitian operators represents an observable, and the same linear combination of expectation values is the expectation of the combination.
Was he talking about the same system or different but similar system.

 

[atyy]

Not sure I follow what you mean by "the same ensemble". Do you simply mean that they are similar? Or do you mean that the 4 expectation values apply to the "exact same ensemble". The issue does not disappear because we start talking about "ensemble" as opposed to individual particle pairs.

Perhaps it is still not clear what the difference is between "the same" and "different but similar". If i = 1,..,N represents identity for N members of a set, where the members could be particle pairs, ensembles, or whatever. "the same" means the same i, "different but similar" means different i, but still a member of the set. In computer-science-speak, it would be the difference between "equality" and "identity" or the difference between "value" and "reference".

So my question in this context will be:
Are those expectation values applicable to the same system, or to different but similar systems (whatever your definition of system is, particle pairs or ensembles).

The ensemble is an infinite number of pairs of particles, prepared in such a way that if I make a measurement on a large enough subset of pairs independently drawn from the infinite number of pairs, and then do this again on a different but large enough subset, the histogram of results will be essentially identical.

You could then also ask the same question, of von Neumann's essential assumption:
von Neumann: Any real linear combination of any two Hermitian operators represents an observable, and the same linear combination of expectation values is the expectation of the combination.
Was he talking about the same system or different but similar system.

The von Neuman proof and Bell's criticism of the implication von Neumann drew is irrelevant. Here we are talking about quantum mechanics. von Neumann and Bell were talking about hidden variables for quantum mechanics.

 

[billschnieder]

The ensemble is an infinite number of pairs of particles, prepared in such a way that if I make a measurement on a large enough subset of pairs independently drawn from the infinite number of pairs, and then do this again on a different but large enough subset, the histogram of results will be essentially identical.

But you are not answering my simple question. Do the expectation values each apply to the same large enough subset of pairs, or do they each apply to a different large-enough subsets of pairs from your "ensemble"? Or do you think there is no difference. That is my question.

 

[billschnieder]

The von Neuman proof and Bell's criticism of the implication von Neumann drew is irrelevant. Here we are talking about quantum mechanics. von Neumann and Bell were talking about hidden variables for quantum mechanics.

When we write
S = ⟨A(a)B(b)⟩+⟨A(a′)B(b′)⟩+⟨A(a′)B(b)⟩−⟨A(a)B(b′)⟩
And then substitute in expectation values from QM, we are talking about the linear combination of expectation values being the expectation of the linear combination and we are using it to imply that local realistic hidden variables do not agree with QM, are we not?

 

[atyy]

But you are not answering my simple question. Do the expectation values each apply to the same large enough subset of pairs, or do they each apply to a different large-enough subset of pairs from your "ensemble"? Or do you think there is no difference. That is my question.

First you have to let me know what you mean by these terms. Let's take the simpler case of one particle in a harmonic potential. The energy is $H = x^{2} + p^{2}$. Let's concentrate on pure states for simplicity. Let's consider an ensemble represented by $\psi$. On one realization of the ensemble I measure $x^{2}$ and find the expectation $X^{2} = \langle\psi|x^{2}|\psi\rangle$. On a second separate realization of the same ensemble represented by $\psi$, I measure $p^{2}$ and find the expectation $P^{2} = \langle\psi|p^{2}|\psi\rangle$. On a third separate realization of the same ensemble represented by $\psi$, I measure $H$ and find the expectation $E = \langle\psi|H|\psi\rangle$

1. Will I find that $E = X^{2} + P^{2}$?

2. In you terminology, did I measure $x^{2}$ and $p^{2}$ on the same or different "large-enough subset of pairs"?

 

[billschnieder]

First you have to let me know what you mean by these terms. Let's take the simpler case of one particle in a harmonic potential. The energy is $H = x^{2} + p^{2}$. Let's concentrate on pure states for simplicity. Let's consider an ensemble represented by $\psi$. On one realization of the ensemble I measure $x^{2}$ and find the expectation $X^{2} = \langle\psi|x^{2}|\psi\rangle$. On second separate realization of the same ensemble represented by $\psi$, I measure $p^{2}$ and find the expectation $P^{2} = \langle\psi|p^{2}|\psi\rangle$. On a third separate realization of the same ensemble represented by $\psi$, I measure $H$ and find the expectation $E = \langle\psi|H|\psi\rangle$

1. Will I find that $E = X^{2} + P^{2}$?
2. In you terminology, did I measure $x^{2}$ and $p^{2}$ on the same or different "large-enough subset of pairs"?

See the three underlined statements above? I am asking you whether when you write $S = ⟨A(a)B(b)⟩+⟨A(a′)B(b′)⟩+⟨A(a′)B(b)⟩−⟨A(a)B(b′)⟩$? the terms represent "the same realization" of the ensemble, or ⟨A(a)B(b)⟩ represents "one realization", while ⟨A(a′)B(b′)⟩ represents "a second separate realization.", etc in your terminology. Or do you believe it does not matter (ie, it can be both). The issue can not be directly translated to your example with position and momentum but to answer your (2) you measured the terms on different subsets.

 

[atyy]

See the three underlined statements above? I am asking you whether when you write $S = ⟨A(a)B(b)⟩+⟨A(a′)B(b′)⟩+⟨A(a′)B(b)⟩−⟨A(a)B(b′)⟩$? the terms represent "the same realization" of the ensemble, or ⟨A(a)B(b)⟩ represents "one realization", while ⟨A(a′)B(b′)⟩ represents "a second separate realization.", etc in your terminology. Or do you believe it does not matter (ie, it can be both). The issue can not be directly translated to your example with position and momentum but to answer your (2) you measured the terms on different subsets.

Still sticking with my example, how about the answer to (1) if the answer to (2) is that I measured the terms on different subsets?

 

[billschnieder]

Still sticking with my example, how about the answer to (1) if the answer to (2) is that I measured the terms on different subsets?

Okay, you start by saying:

Let's take the simpler case of one particle in a harmonic potential. The energy is $H = x^{2} + p^{2}$.

Implying it is a relationship which applies to one particle. Then you suddenly switch to ensembles while using the same symbols

Let's consider an ensemble represented by $\psi$. On one realization of the ensemble I measure $x^{2}$ and find the expectation $X^{2} = \langle\psi|x^{2}|\psi\rangle$. On a second separate realization of the same ensemble represented by $\psi$, I measure $p^{2}$ and find the expectation $P^{2} = \langle\psi|p^{2}|\psi\rangle$. On a third separate realization of the same ensemble represented by $\psi$, I measure $H$ and find the expectation $E = \langle\psi|H|\psi\rangle$

1. Will I find that $E = X^{2} + P^{2}$?

Your question is not clear because you are using the same symbols but they mean different things. H no longer applies to one particle but to an ensemble. Maybe if you answered the question and said why the answer was relevant to my question, I would appreciate the point you are making.

 

[atyy]

Okay, you start by saying:

Implying it is a relationship which applies to one particle. Then you suddenly switch to ensembles while using the same symbols

Your question is not clear because you are using the same symbols but they mean different things. H no longer applies to one particle but to an ensemble. Maybe if you answered the question and said why the answer was relevant to my question, I would appreciate the point you are making.

I am using standard quantum mechanical language. One particle means an ensemble, each member of which is one particle. What is your answer to (1) if your answer to (2) was that I measured on different subsets?