I Is quantum weirdness really weird?

[forcefield]

Does anyone have a guess how 19th century physicists would have responded to:
Does "If Alice and Bob both perform experiment X they will get the same result." imply "If Alice performs X and gets value 1 and Bob performs Y and gets 2; he would have gotten 1 if he had performed X instead."?

I think they all would have said yes. Yet wouldn't those on this thread all say no?

Weirdly, no answer to the latter question yet... I'd say "not necessarily because Alice/Albert might be different from Bob/Bohr and also (because there are two experiments) those experiments might have to be done either at a different time or at a different place".

Hmm, that looks like Pauli exclusion principle.

 

[Zafa Pi]

Weirdly, no answer to the latter question yet... I'd say "not necessarily because Alice/Albert might be different from Bob/Bohr and also (because there are two experiments) those experiments might have to be done either at a different time or at a different place".

Hmm, that looks like Pauli exclusion principle.

My question was:
Does "If Alice and Bob both perform experiment X they will get the same result." imply "If Alice performs X and gets value 1 and Bob performs Y and gets 2; he would have gotten 1 if he had performed X instead."? I'm asking if the first quoted sentence implies the second quoted sentence.

I would like you to guess how 19th century physicists would have answered and how you answer.

 

[secur]

Since your question is itself counterfactual I thought perhaps it was a sophisticated joke. But apparently not ...

The problem with your question is the involvement of "19th century physicists". Men like Gauss, Riemann, (William) Thomson, Maxwell. Asking how they would answer is like asking how Mozart would develop a simple theme into a symphony. It would take a genius of equal stature to hazard a guess.

Gauss wouldn't answer directly. He'd take your question as a hint (assuming you convinced him to take it seriously, that is). He'd ask a few penetrating questions, derive QM off the top of his head, solve all the problems puzzling us today - and not bother to publish, leaving it as a trivial exercise for the reader. Well, that may be an exaggeration. But however he responded, he'd blow you away. You'd feel like you opened the door to let the cat in, but it turned out to be a hungry lion.

 

[Zafa Pi]

Since your question is itself counterfactual I thought perhaps it was a sophisticated joke. But apparently not ...

The problem with your question is the involvement of "19th century physicists". Men like Gauss, Riemann, (William) Thomson, Maxwell. Asking how they would answer is like asking how Mozart would develop a simple theme into a symphony. It would take a genius of equal stature to hazard a guess.

Gauss wouldn't answer directly. He'd take your question as a hint (assuming you convinced him to take it seriously, that is). He'd ask a few penetrating questions, derive QM off the top of his head, solve all the problems puzzling us today - and not bother to publish, leaving it as a trivial exercise for the reader. Well, that may be an exaggeration. But however he responded, he'd blow you away. You'd feel like you opened the door to let the cat in, but it turned out to be a hungry lion.

LOL. Hold your horses, oops I mean lions, it was merely a yes or no question. My guess is that all those luminaries you named would have agreed with Einstein (of 1899) and answered yes. I.e. CFD would have been accepted. When I finish my time machine we'll have the answer.
BTW, I just provided a counter example to your statement, "It would take a genius of equal stature to hazard a guess."

 

[secur]

BTW, I just provided a counter example to your statement, "It would take a genius of equal stature to hazard a guess."

Fools rush in where angels fear to tread :-)

 

[Zafa Pi]

Fools rush in where angels fear to tread :-)

If you're including Johnny Mercer with your other science luminaries then I say, "And here am I throwing caution to the wind."
Are we getting off topic yet?

 

[Simon Phoenix]

Does "If Alice and Bob both perform experiment X they will get the same result." imply "If Alice performs X and gets value 1 and Bob performs Y and gets 2; he would have gotten 1 if he had performed X instead."?

I think they all would have said yes. Yet wouldn't those on this thread all say no?

It is perfectly possible to make certain counterfactually definite statements within the context of QM - but that doesn't mean QM is a counterfactually definite theory (in the sense that QM cannot be described, or replaced perhaps, with a theory that only involves [local] counterfactual variables).

Suppose I prepare a spin-1/2 particle in the 'up' eigenstate of spin-z. I could perform a measurement of spin-x and I'd get either the result spin-x 'up' or spin-x 'down' with equal probability. But I'd be perfectly entitled to make the statement "if I'd measured spin-z instead I would have obtained the result spin-z 'up' with unit probability".

So if we define 'counterfactual definiteness' as the ability to make meaningful statements of the form "I measured X and got the value x, but if I had measured Y then I would have got the value y" then QM allows us to make such statements, in certain circumstances, but not all.

 

[Zafa Pi]

It is perfectly possible to make certain counterfactually definite statements within the context of QM - but that doesn't mean QM is a counterfactually definite theory (in the sense that QM cannot be described, or replaced perhaps, with a theory that only involves [local] counterfactual variables).

Suppose I prepare a spin-1/2 particle in the 'up' eigenstate of spin-z. I could perform a measurement of spin-x and I'd get either the result spin-x 'up' or spin-x 'down' with equal probability. But I'd be perfectly entitled to make the statement "if I'd measured spin-z instead I would have obtained the result spin-z 'up' with unit probability".

So if we define 'counterfactual definiteness' as the ability to make meaningful statements of the form "I measured X and got the value x, but if I had measured Y then I would have got the value y" then QM allows us to make such statements, in certain circumstances, but not all.

I agree with every last word of your post. But I asked if one proposition, say p, implied another, say q; now it seems you agree that my q (post #202) doesn't always follow from my p. Thus the answer to my question: "Does p imply q?" is no. And I think most of us on this thread would agree. As opposed to 19th century physicists.

The point I am trying to make is: My question "Does p imply q?" is a short, simple, abstract English (no math or physics) question whose answer has changed among physicists (but not scientists in general). And I didn't get into trying to define CFD. If it can be done in a simpler fashion that would be cool.

 

[Simon Phoenix]

"Does p imply q?" is a short, simple, abstract English (no math or physics) question whose answer has changed among physicists

But the example you give doesn't really illustrate what you mean.

proposition 1 : If Alice and Bob both perform experiment $X$ they always obtain the same result

Assuming proposition 1 is true, can we then make the statement that : if Alice performs $X$ and Bob $Y$ getting the result $(x,y)$, Bob would have obtained the result $x$ if he had performed the experiment $X$ (given that Alice obtained the result $x$)?

Can you think of an example from QM where this statement is not true (given that proposition 1 is true)?

I suspect you're thinking of performing experiments on entangled states of the form $|00\rangle +|11\rangle$ expressed in some basis, let's say spin-z.

So let's imagine the following. Alice is given one of these entangled objects and she performs a measurement of spin-z and obtains the result 0.

Bob is then given the partner object from the entangled state and performs a measurement of spin-$\theta$, and for the purposes of the argument let's suppose he obtains the result 1. Is it legitimate to make the statement in QM that if Bob had measured spin-z instead he would have obtained the result 0?

Of course it is

 

[zonde]

So let's imagine the following. Alice is given one of these entangled objects and she performs a measurement of spin-z and obtains the result 0.

Bob is then given the partner object from the entangled state and performs a measurement of spin-$\theta$, and for the purposes of the argument let's suppose he obtains the result 1. Is it legitimate to make the statement in QM that if Bob had measured spin-z instead he would have obtained the result 0?

Of course it is

Not necessarily. For simplicity let's say Alice's measurement is after Bob's measurement (they are timelike separated). If you accept that they are not independent you can say that Bob's measurement could have changed Alice's measurement.

 

[zonde]

Not necessarily. For simplicity let's say Alice's measurement is after Bob's measurement (they are timelike separated). If you accept that they are not independent you can say that Bob's measurement could have changed Alice's measurement.

If with your phrase

Bob is then given ...

you mean that the the situation is exactly the other way around (Bob's measurement is after Alice's) then yes, the answer is "of course it is legitimate".

 

[Simon Phoenix]

For simplicity let's say Alice's measurement is after Bob's measurement

OK let's look at how this pans out.

Let's suppose Bob measures first in our example above - and let's suppose again he measures spin-$\theta$ and obtains the result 1. Then Alice makes a measurement of spin-z and obtains the value 0.

Given the condition that Alice obtains the result 0 then we can say that had Bob measured spin-z he would have obtained the result 0.

Look at the original example -

If Alice performs X and gets value 1 and Bob performs Y and gets 2;

Here we've specified Alice's result - and that fixes Bob's "would have" result - independently of the order in which Alice and Bob perform their measurements.

Another way to express this might be to consider the parity of the 2 measurements. In our example above with Bob obtaining 1 from a spin-$\theta$ measurement and Alice obtaining 0 from a spin-z measurement (so an odd parity) we can make the counterfactual statement that had Bob measured spin-z instead the parity would be even.

 

[zonde]

Does anyone have a guess how 19th century physicists would have responded to:
Does "If Alice and Bob both perform experiment X they will get the same result." imply "If Alice performs X and gets value 1 and Bob performs Y and gets 2; he would have gotten 1 if he had performed X instead."?

I think they all would have said yes. Yet wouldn't those on this thread all say no?

Do you imply that Alice's and Bob's experiments are to be considered independent?

 

[Zafa Pi]

But the example you give doesn't really illustrate what you mean.

proposition 1 : If Alice and Bob both perform experiment $X$ they always obtain the same result

Assuming proposition 1 is true, can we then make the statement that : if Alice performs $X$ and Bob $Y$ getting the result $(x,y)$, Bob would have obtained the result $x$ if he had performed the experiment $X$ (given that Alice obtained the result $x$)?

Can you think of an example from QM where this statement is not true (given that proposition 1 is true)?

I suspect you're thinking of performing experiments on entangled states of the form $|00\rangle +|11\rangle$ expressed in some basis, let's say spin-z.

So let's imagine the following. Alice is given one of these entangled objects and she performs a measurement of spin-z and obtains the result 0.

Bob is then given the partner object from the entangled state and performs a measurement of spin-$\theta$, and for the purposes of the argument let's suppose he obtains the result 1. Is it legitimate to make the statement in QM that if Bob had measured spin-z instead he would have obtained the result 0?

Of course it is

Yay, we are back at. No it ain't. It's true that they would agree, but why at (0,0)? Alice got 0 when Bob measured spin-θ, how would you go about testing she would have got 0 if he measured spin-z instead. You are merely stating a consequence of realism. I'm not saying he couldn't get 0, it's just that there is no way you can claim that he must. You can't have her measure again because now her object is an eigenstate of spin-z, not an object from an entangled pair. If she starts over she'll get 0 with probability ½. You can't just have Bob measure again since his object is now an eigenstate of spin-θ.
This has been covered (perhaps not to your satisfaction) see posts #161 & #165.

 

[Simon Phoenix]

Alice got 0 when Bob measured spin-θ, how would you go about testing she would have got 0 if he measured spin-z instead

I don't - I'm going from the specified conditions. You wrote :

"If Alice and Bob both perform experiment X they will get the same result." imply "If Alice performs X and gets value 1 and Bob performs Y and gets 2; he would have gotten 1 if he had performed X instead."? I'm asking if the first quoted sentence implies the second quoted sentence.

In the second part here you specify that Alice obtains the value 1 and consider Bob's "would have" result.

For the entangled example, specifying Alice's result as 0 fixes Bob's "would have" result to be 0.

If you now want to change the question to consideration of Alice's "would have" result at the same time we consider Bob's "would have" result - then you've just moved the goalposts from their original location :-)

 

[Zafa Pi]

Do you imply that Alice's and Bob's experiments are to be considered independent?

Read the bottom #161. I'm not sure what your answer is to my question of #202, but I say no and Simon says yes, so who ever is wrong must admit QM is weird.

 

[zonde]

Read the bottom #161.

zonde your definition of independent is untestable. Would you have Bob measure again to see if his value didn't change with a new photon? You know he will get a different answer with P = ½.

Definitions are not tested. It either can be meaningfully applied or not.
My definition obviously does not apply to reality as you have pointed out. But it can apply to some models of reality. QM is not such a model as it predicts only statistics but does not talk about particular outcomes.
But we can ask if there can be a model to which we can apply my definition of "independent" and which reproduces QM predictions.

 

[Zafa Pi]

For the entangled example, specifying Alice's result as 0 fixes Bob's "would have" result to be 0.

One last time for me, then we'll have leave it alone for a while in spite of it being fascinating.
If A & B both measure X and A gets 1 then B begets 1. If A measures X and gets 1 and B measures Y and gets 2, then in this case A got 1 when B measured with Y. Why would A get 1 if B measured with X instead? It's a different experiment. Show me how you would test your claim.

 

[Zafa Pi]

QM is not such a model as it predicts only statistics but does not talk about particular outcomes

False. Consider GMZ. Over and out for the evening.

 

[Simon Phoenix]

Why would A get 1 if B measured with X instead?

No reason at all - which is why I was careful to explicitly state "given Alice's result".

So given that Alice and Bob get the same results if they perform experiment X, and given that Alice's result is 1 for experiment X and Bob's is whatever for experiment Y, then we can make the counterfactual statement "given Alice's result of 1, Bob would have obtained this result also if he had measured X".

So I don't think I'm 'claiming' what you appear to think I am

What we can say is that Bob "would have" obtained the same result as Alice - whatever that result is.

 

[zonde]

False. Consider GMZ.

In GHZ experiment for particular three particle measurements QM predicts that one of the four combinations will be measured and other four combinations won't. But QM does not say which one of the four combinations will be measured, instead it gives equal probabilities for these four combinations. It's still statistics.
It's basically the same type of certainty that you can get with measurement of entangled particles with the same measurement settings.

 

[secur]

General Rule: If a scientist (let's call him Bob) is able, theoretically, to know what result an experiment would or will have, then it would or will have that result. Otherwise it's unknown.

"Theoretically" means: it doesn't matter whether Bob actually does know. If the laws of physics - and math, logic, all of science - make it possible to know the result, then that result will happen. It also doesn't matter if there's an actual person (Bob) involved or just impersonal nature.

The "laws of physics" include classical, quantum, whatever's relevant.

Note, of course our current knowledge of physics is not complete and is wrong in some cases (although none of us knows what those cases are). So I'm talking about the ideal, "true" laws. Alternatively, we can say the Rule applies to the extent that our current knowledge is correct.

It doesn't matter whether it's a classical, quantum, or whatever type of experiment.

It doesn't matter when the experiment is conducted relative to us. We could be talking about an experiment to be done tomorrow, or this instant; or an experiment that wasn't done yesterday but could have been. The above General Rule always works.

Ok, let's apply this Rule to some specific situations.

So given that Alice and Bob get the same results if they perform experiment X, and given that Alice's result is 1 for experiment X and Bob's is whatever for experiment Y, then we can make the counterfactual statement "given Alice's result of 1, Bob would have obtained this result also if he had measured X".

If Bob's measurement was made after Alice then you're right. "After" means they're timelike separated, and a photon sent from Alice when she did the experiment has time to reach Bob. OTOH if Bob does his measurement at the same time, or before, Alice, then you're wrong.

Let's be more explicit. Suppose Alice and Bob measure their entangled particles at the same time, A at 0 degrees and B at 30. A gets "1". Now, suppose Bob had instead measured at 0, would he have gotten 1? Nobody knows, or can know. 50/50 to get 1 or -1. This means, obviously, that if he'd used 0 degrees we also can't say Alice would have gotten 1.

BTW I realize this statement is controversial. With some very far-fetched assumptions, like MWI, or "consistent histories", we could possibly deny the statement. It's a bit difficult, if anyone wants to discuss it further. If you don't then there's no difficulty :-)

What we can say is that Bob "would have" obtained the same result as Alice - whatever that result is.

Yes, laws of physics assure us their results would (or will) match, when they measure twin-state entangled particles at same angle, regardless who went first (or, if spacelike separated, neither went "first"). The laws also assure us we can't know whether result will be 1 or -1. Except if Bob goes after Alice, and she got 1, then he must also.

BTW it may be impossible to definitely say they are measuring "at the same time". But that's not important at the moment.

Suppose I prepare a spin-1/2 particle in the 'up' eigenstate of spin-z. I could perform a measurement of spin-x and I'd get either the result spin-x 'up' or spin-x 'down' with equal probability. But I'd be perfectly entitled to make the statement "if I'd measured spin-z instead I would have obtained the result spin-z 'up' with unit probability".

That's right

So if we define 'counterfactual definiteness' as the ability to make meaningful statements of the form "I measured X and got the value x, but if I had measured Y then I would have got the value y" then QM allows us to make such statements, in certain circumstances, but not all.

Classical and quantum physics are identical re. CFD, they both follow the same "General Rule". You can say classical/quantum physics "is" or "is not" counterfactual. If you guys can agree which, I'll be happy to go along. If you all agree to call physics a petunia, fine, I have no problem with that. All that matters is the results of experiments. Not terminology. Not theory. Not Einstein's opinion, nor mine, nor yours. None of that means anything concerning physics. All that matters is the results of experiments.

Does anyone have a guess how 19th century physicists would have responded to: Does "If Alice and Bob both perform experiment X they will get the same result." imply "If Alice performs X and gets value 1 and Bob performs Y and gets 2; he would have gotten 1 if he had performed X instead."? I think they all would have said yes.

We can guess. But it's impossible, by laws of physics, logic etc, to know what they would have said.

A couple more examples to beat the horse to death.

Suppose Alice went first, and measured "1" (as did Bob). Then - if Bob had gone first by a few seconds, instead - would he have gotten "1"? No one knows. 50/50 it's +-1.

Suppose Alice measures "1". Suppose that, instead, she'd done the experiment a second before or a second after, or an inch to the left or right. Would she have gotten 1? It's 50/50.

Suppose I asked Alice - yesterday, today, tomorrow - what her name is. Suppose she's been asked that question thousands of times and always answered "Alice". Can I assume she will, or would have, said "Alice"? No. She might say "my name is Alice", or "Jeez we've worked together for years, don't you know my name yet already?" She might sing the Star-Spangled Banner, mangling that high F. It's impossible to know, because the laws of physics (and science, etc) can't predict with certainty what a human being will do or would have done, and probably never will.

 

[Simon Phoenix]

OTOH if Bob does his measurement at the same time, or before, Alice, then you're wrong.

Read what I wrote more carefully. The critical part is the statement "given Alice's result is 1". With this result fixed (i.e a given) then what I'm saying is not wrong.
So I'm saying that the answer to the question "what result would Bob have obtained if he'd measured X instead AND Alice had also obtained the result 1?" is 1. This is independent of the time ordering of Alice's and Bob's measurements.

There is a reason why I stressed Alice's result as a 'given', you know, and you've outlined it very nicely because if we don't fix things for Alice then we have to consider "would have" for both parties.

Without this 'given' all we can say is that Alice's and Bob's "would have" results match (so we can make a counterfactual statement about the 'parity' observable).

 

[forcefield]

My question was:
Does "If Alice and Bob both perform experiment X they will get the same result." imply "If Alice performs X and gets value 1 and Bob performs Y and gets 2; he would have gotten 1 if he had performed X instead."? I'm asking if the first quoted sentence implies the second quoted sentence.

I would like you to guess how 19th century physicists would have answered and how you answer.

At second chance I say "no" because same result does not imply independence. I guess 19th century physicists might have known about common cause.

 

[Zafa Pi]

Yes, laws of physics assure us their results would (or will) match, when they measure twin-state entangled particles at same angle, regardless who went first (or, if spacelike separated, neither went "first"). The laws also assure us we can't know whether result will be 1 or -1. Except if Bob goes after Alice, and she got 1, then he must also.

It doesn't matter if Bob goes after Alice. If she gets 1 so will he. I assume we are referring to the physical set up of post #161, if not I'm lost.

 

[secur]

@Simon Phoenix,

Did you delete a previous post? Perhaps you realized that we don't get "weird" behavior with the correlated density matrix? Apart from that I agreed with your other points. Yes, the weirdness is partly due to essential randomness of measurements. That, plus non-commutativity. Entanglement, I would say, doesn't really add to that. Rather, it's the best place to come to grips with the "weirdness" because we can make two separate measurements on the same wave function.

To review a bit. Clearly intuitive "reality" assumptions must be modified somehow. We're concentrating on the assumption of counterfactual definiteness. The first impulse is to disallow it entirely. Then you certainly can't write a Bell-type inequality, which depends on believing CFD. But it seems too restricting: there are natural, non-paradoxical CFD statements we would like to be able to make. So we're searching for the minimal necessary rejection of CFD.

This is a well-trodden path, people have been puzzling over this for 100 years. We're not going to come up with a brilliant new way to treat it. But the attempt makes it easier to understand what the fuss is all about. I believe "consistent histories" can be viewed in this light. They define "legal" branches by rules which ensure that, as long as you stick to defined branches, you won't run into the Bell-inequality "paradox". Don't know if they really came up with the minimal necessary rules but probably they did better than we're going to.

Possibly we could characterize all the interpretations by the way they deal with this question.

The situation is reminiscent of formal logic, when the field was confronted with Godel's incompleteness theorem, and Russell's paradox, etc. They saw that simple logic / set theory just didn't work. Bertrand Russell and others considered rejecting all self-referential statements but that was too restrictive (analogous to rejecting all CFD in QM). So they investigated minimal mods to existing set theory. Zermelo-Fraenkel is the best-known attempt, or perhaps ZFC, with axiom of choice included. A similar line of thought gives rise to "Quantum Logic", minimally-modified logic systems to avoid the Bell "paradox".

But, I digress. My main point is that we (or perhaps, just me) are trying to re-invent the wheel here. It's a great exercise: if you really want to understand how a wheel works, try to make one yourself. But ultimately you're better off just buying one from Firestone.

My attempt above was too optimistic, it's not that simple. At the moment the best I can come up with is: you can never assume the results of two non-commuting measurements.

Suppose I prepare a spin-1/2 particle in the 'up' eigenstate of spin-z. I could perform a measurement of spin-x and I'd get either the result spin-x 'up' or spin-x 'down' with equal probability. But I'd be perfectly entitled to make the statement "if I'd measured spin-z instead I would have obtained the result spin-z 'up' with unit probability".

That's right. You're entitled to know the result you would have seen if you'd measured z instead. But here's the rub. The spin-x measurement doesn't commute with spin-z. So if you assume the z you're no longer entitled to assume the result you got for x! It's counter-intuitive. You actually did measure x, so surely the statement "if I'd measured x I would have gotten the result I actually did get" is tautological. But you can't use both the x and z at once: one or the other only.

Bohr and his camp might express it like this. Prior to the measurement, we know spin-z, because we prepared that definite state. After the measurement we know spin-x because we measured it. But there was never a time when both were valid at once.

I still say it's not all that weird! Odd, maybe :-)

 

[Simon Phoenix]

Did you delete a previous post? Perhaps you realized that we don't get "weird" behavior with the correlated density matrix?

Yes, I asked for the post to be deleted (thanks Admin ). I fired off the post in haste - at least that's the excuse I'm sticking to - and by the time I had chance to come back to the forum I realized I'd written a load of twaddle. Well more twaddlish than my usual anyway :-)

I'm still pondering that example of the mixed density operator - which in some sense can be described as the most strongly correlated 'classical' state of 2 qubits. If we view it as a proper mixture then there's no problem with Alice's result - it wouldn't have changed if Bob had measured X instead of Y (in the terminology of the above discussion). But if we view the same state as having derived from the partial trace of a GHZ state, say, so an improper mixture - then we can't make the same claim. Of course there are absolutely no observable consequences arising from this difference in what we can claim in the 2 situations.

Yes, the weirdness is partly due to essential randomness of measurements. That, plus non-commutativity

In a way I'd say all the (alleged) weirdness of QM is essentially down to non-commutativity, liberally sprinkled with a dash of purity.

 

[secur]

You know, we're really making an effort in this thread to avoid vague language and express our ideas mathematically. That's the only way to make progress and avoid endless, meaningless debates about mere terminology. Not to give a lecture, but I really feel this is the right approach. So, I have to ask you to please provide precise mathematical definitions for "twaddle" and "twaddlish".

 

[zonde]

I'm still pondering that example of the mixed density operator - which in some sense can be described as the most strongly correlated 'classical' state of 2 qubits. If we view it as a proper mixture then there's no problem with Alice's result - it wouldn't have changed if Bob had measured X instead of Y (in the terminology of the above discussion). But if we view the same state as having derived from the partial trace of a GHZ state, say, so an improper mixture - then we can't make the same claim. Of course there are absolutely no observable consequences arising from this difference in what we can claim in the 2 situations.

I have pondered over that difference between proper and improper mixed states and my conclusion is that so called single particle pure state and multi-particle pure (entangled) state are physically distinct. Say if you take single particle pure state as fundamental you can't describe entangled state but if you take entangled state as fundamental you can't describe single particle pure state.

 

[Simon Phoenix]

So, I have to ask you to please provide precise mathematical definitions for "twaddle" and "twaddlish".

Well In SI Units (Systeme Idiotique) 1 twaddle is approximately equal to a mega drivel. Before the irrationalization of these units it would have been equal to roughly 3 and half brain farts.

But you're right, and I apologize : "twaddlish" is a completely made up word and has no place on these forums

 

[vanhees71]

I have pondered over that difference between proper and improper mixed states and my conclusion is that so called single particle pure state and multi-particle pure (entangled) state are physically distinct. Say if you take single particle pure state as fundamental you can't describe entangled state but if you take entangled state as fundamental you can't describe single particle pure state.

You can ponder as much as you like, but there's no difference between proper and improper mixed states, but there are just mixed states. I don't know, how you want to distinguish between different kinds of mixed states (in the lab!).

Both, pure and mixed states are described by a Statistical operator which by definition is a self-adjoint positive semidefinite operator of trace 1. That's it; nothing else. It describes the probabilistic knowledge about a system and is operationally defined as an equivalence class of preparation procedures. A statistical operator describes a pure state if and only if it is a projection operator with $\hat{\rho}=\hat{\rho}^2$. Otherwise it represents a mixed state.

 

[secur]

But you're right, and I apologize : "twaddlish" is a completely made up word and has no place on these forums

I was just kidding! Shakespeare made up words constantly, you can too - as far as I'm concerned. In fact I'm looking forward to the opportunity to use "twaddlish" myself.

You can ponder as much as you like, but there's no difference between proper and improper mixed states, but there are just mixed states. I don't know, how you want to distinguish between different kinds of mixed states (in the lab!). ... ... A statistical operator describes a pure state if and only if it is a projection operator with $\hat{\rho}=\hat{\rho}^2$. Otherwise it represents a mixed state.

AFAIK there's no observational difference between "proper" and "improper" mixed states, both described by a convex sum of pure states ∑Pi |Ai><Ai|, with ∑Pi =1. However an ensemble in a single pure state is mathematically and operationally different. The fact that it's a projection operator is non-trivial, AFAIK.

 

[vanhees71]

Indeed, I find "twaddlish" an extremely nice word (may it be made up or not). I think another very good word is "gibberish". It perfectly sounds like what it means, and it always comes into my mind when I read "philosophical texts"

The fact that a pure state is represented by a projection operator as its statistical operator is easy to understand since it just means that the state is completely determined in the sense that a complete set of compatible observables has determined values, because as statstical operator that is a projector determines the corresponding eigenvector uniquely.

That's even easily proven: The projection property means that $\hat{\rho}^2=\hat{\rho}$, which implies that $\hat{\rho}$ has only 0 and 1 as eigenvalues, and since it's positive semidefinite and $\mathrm{Tr} \hat{\rho}=1$, there is exactly one eigenvector for the eigenvalue 1 and thus
$$\hat{\rho}=|\psi \rangle \langle \psi|$$
with a normalized $|\psi \rangle$.

 

[zonde]

The fact that a pure state is represented by a projection operator as its statistical operator is easy to understand since it just means that the state is completely determined in the sense that a complete set of compatible observables has determined values, because as statstical operator that is a projector determines the corresponding eigenvector uniquely.

That's even easily proven: The projection property means that $\hat{\rho}^2=\hat{\rho}$, which implies that $\hat{\rho}$ has only 0 and 1 as eigenvalues, and since it's positive semidefinite and $\mathrm{Tr} \hat{\rho}=1$, there is exactly one eigenvector for the eigenvalue 1 and thus
$$\hat{\rho}=|\psi \rangle \langle \psi|$$
with a normalized $|\psi \rangle$.

Projection operator of pure state does not determine it's state vector uniquely. Phase factor remains uncertain. But without phase factor you can't describe interference.

 

[vanhees71]

The phase factor is irrelevant for the determination of the state. The one-to-one correspondence between the pure state and its description is the statistical operator (which is a projection operator in this case) or, equivalently, a ray in Hilbert space (i.e., "a Hilbert space vector modulo a non-zero factor").

 

[zonde]

The phase factor is irrelevant for the determination of the state. The one-to-one correspondence between the pure state and its description is the statistical operator (which is a projection operator in this case) or, equivalently, a ray in Hilbert space (i.e., "a Hilbert space vector modulo a non-zero factor").

Phase factor is relevant for observable phenomena. A model without some element (or property) that represents phase factor can not model interference.

 

[vanhees71]

That's a new theory than, not standard QT and thus outside of what should be discussed in this scientific (!) forum.

 

[zonde]

Interesting. So in standard QT phase factor has no place? That is news for me.

 

[stevendaryl]

Phase factor is relevant for observable phenomena. A model without some element (or property) that represents phase factor can not model interference.

This is a source of some confusion. On the one hand, the wave function \psi(x) and the wave function e^{i \alpha} \psi(x) represent the same state. On the other hand, phases are important in calculating interference effects. I think it would be helpful if someone could give a short Physics Insights presentation reconciling these two facts. They seem contradictory, but obviously, they are not, because physics have been applying both of them, and getting consistent results, since the probabilistic interpretation of quantum mechanics was developed.

Let me just illustrate using a conceptually simple quantum-mechanical problem. Suppose we prepare an electron in state |A\rangle at time t_I and we want to know the probability P_{AD} that it will be found in state |D\rangle at time t_F. Then the quantum-mechanical rules tell us to compute this as follows:

P_{AD} = |\psi_{AD}|^2

where \psi_{AD} is the transition amplitude, which can be calculated from |A\rangle and |B\rangle using the time evolution operator e^{\frac{-i H (t_F - t_I)}{\hbar} }.

Since the physically important quantity, P_{AD}, is real, the phase of the amplitude \psi_{AD} is irrelevant. But to compute \psi_{AD} in the first place, it seems that we need to pick a phase for states |A\rangle and |D\rangle. These two points are not contradictory. The phases are important for the mathematics of working with a complete basis of the Hilbert space (a complete basis includes a choice of phases), but the phases of initial and final states drop out from the final answer. It's sort of like with electromagnetism: Often you need to deal with the electromagnetic vector potential, A^\mu, even though it is only defined up to a gauge change. So mathematically, you often have to pick a gauge, and do your calculations, even though the end result will be independent of which gauge was chosen.

 

[vanhees71]

Of course relative phases in superpositions play an observable role, but these also refer to different projection operators. Again, the statistical operators are precisely what describes the pure or mixed state of a quantum system!

 

[PeterDonis]

That's a new theory than, not standard QT

It doesn't seem like that was the intent. Stevendaryl in post #239 gives a more detailed statement of what I think zonde's underlying question is (and a sketch of an appropriate response).

 

[Zafa Pi]

Interestingly, Bell's Theorem (post #161),
zonde's independence criterion (top of #161) applied to the local set up of #161,
my implication question (#194) when answered yes and applied to the set up of #161
are all logically equivalent. That is each one can prove the other.

Any one of them can be proved via the assumption of realism. With regard to my implication question, if Bob had selected experiment X instead, the reality facing Alice at that time would be the same (Bell separation set up) and thus would have determined that she get the same value, 1. The assumption of that reality is realism (or determinism or hidden variables or CFD). The violation of Bell's Inequality forces the answer to my question to be no. The virtue of my question is the lack of any technical language, and can be asked to any Joe walking down the street.

secur thinks that attempting to guess what 19th century physicists would say is a fool's errand. I recall being asked what Einstein would have said after being made aware of Bell's Theorem and the lab results verifying the violation of the Inequality. Of course, it goes without saying that I could never know for sure, and I would love to bring him back from the dead more than any other, nevertheless I did hazard a guess: "Lord, why have you forsaken me?"
I personally find QM weird, or unintuitive is perhaps a better term, but maybe not as bad as "ZFC, with axiom of choice included" [secur], since that gives us the Tarski-Banach Theorem. I have found this thread edifying and fun.

 

[Zafa Pi]

In GHZ experiment for particular three particle measurements QM predicts that one of the four combinations will be measured and other four combinations won't. But QM does not say which one of the four combinations will be measured, instead it gives equal probabilities for these four combinations. It's still statistics.
It's basically the same type of certainty that you can get with measurement of entangled particles with the same measurement settings.

After gathering sufficient data to justify the individual parts of the hypothesis, it takes only one measurement to refute the conclusion (the GHZ Equality).

 

[Zafa Pi]

So I'm saying that the answer to the question "what result would Bob have obtained if he'd measured X instead AND Alice had also obtained the result 1?" is 1. This is independent of the time ordering of Alice's and Bob's measurements.

This correct , but not my original question.

Without this 'given' all we can say is that Alice's and Bob's "would have" results match (so we can make a counterfactual statement about the 'parity' observable).

I think this parity business is a red herring (at least as viable as a dead horse), because if in my original question I used -1 instead of 2 there would be no parity changes, but it would be the same question.

 

[PeterDonis]

Thread closed for moderation.

[Edit: The thread has run its course and will remain closed.]