I Is quantum weirdness really weird?

[Zafa Pi]

Quantum mechanics the way I see it.
(In the quantum mechanical forest few paths do not lead into nowhere land.)The above link came ... from here .
Due to discussions ... occurring here.
For what it's worth... ? ... lol ... ( a joke... kinda )

Great references. I'm sure somewhere in the wiki article there must be a dead horse.

 

[Zafa Pi]

What is special about the GHZ state is that one doesn't need statistics to disprove the existence of elements of reality.

This was in response to my question that regarding the discussion of this thread there wasn't anything special about GHZ as opposed to any other Bell doings.
I agree that it's mighty cute that you only need one trial to contradict the equality in the GHZ theorem, as opposed to the needed statistics in secur's example at post #147, or CHSH et. al.

However, I wouldn't be to fast in asserting that the GHZ result doesn't need statistics. Here is GHZ:
The set up and notation are like mine at post #161 except we introduce a third player Carol that also receives a photon from Eve who uses the state |GHZ⟩ = √½(|000⟩ + |111⟩) to send three entangled photons.
GHZ Theorem: Assume that if only one of Alice, Bob, or Carol flips heads then the product of their values is -1. Then Ah•Bh•Ch = -1.
Proof: -1 = Ah•Bt•Ct = At•Bh•Ct = At•Bt•Ch, then the product of all three products = -1 = (At•Bt•Ct)²Ah•Bh•Ch ⇒ Ah•Bh•Ch = -1 (GHZ Equality). QED

Now if Ah, Bh, and Ch are the result of measuring with (Pauli)X, while the others employ (Pauli)Y the the hypothesis of the theorem is satisfied, but Ah•Bh•Ch = 1. The GHZ Equality is violated!

Just one trial of Ah•Bh•Ch does the trick. But how do we know the hypothesis is satisfied, e.g. that Ah•Bt•Ct = -1 for sure? It's akin to my saying I have a two headed coin and all you get to see is the results of flips. I would have to make a bunch before you were really convinced. Statistics.

 

[OCR]

... there must be a dead horse.

Lol, I get it... and she's still getting beat on ...

 

[zonde]

@zonde, please look at it this way. We've got entangled particles with state |11> + |00>. (Ignoring normalization). Suppose Alice uses 0 degrees setting on her detector and Bob uses 30. Suppose Alice gets 1, spin up. We can safely make the following "CFD 1" assertion:

CFD 1: "If Bob had used 0 setting also, A & B would have read the same result."

That's a counterfactual QM statement, nevertheless it's as reliable as any in classical mechanics. However we can't say this:

CFD 2: "If Bob had used 0 setting also, he would have measured 1."

Instead, if "we could run it again", (which of course we can't), it's a new QM calculation. Yes, Alice and Bob would definitely detect the same spin. But it might be up or down, 50 / 50 chance. The fact that A got 1 before is irrelevant.

Now, putting is simply but more or less accurately: Bell-type experiments and inequalities (including CHSH, GHZ, etc) prove you can't assert the CFD 2 assertion. If you do, you can't get the right experimental results. With or without non-locality.

Personally I don't find any of this excessively weird. It's good physics.

So, what part of it don't you accept?

Everything is fine. Well, CFD1 is of course non-local statement, while CFD2 is local statement. So if a theory can't make symmetrical CFD2 type claims about A & B we can't say it's local (meaning, A and B measurements are independent).

 

[secur]

I would call it non-local also. But others say it's not, because Alice and Bob commute (they live in the suburbs). It doesn't matter, as long as we agree what happens in the actual experiments.

 

[ljagerman]

Please explain or give simple example.
My position is that we use the model exactly the same way whether we ask "what will happen?" or we ask "what would have happened?". It's exactly the same input for the model and therefore it has to produce exactly the same output.

I also find that this way of thinking about QM makes sense: "Anything that can happen does happen," but each possible happening has a probability associated with it. The best that QM math can do is provide that probability. I can then say that a manifestation of quantum "weirdness" is something very unlikely indeed happening. Thus a plane bound to fly from New York to London can get there via Jupiter - a very weird but possible happening. A more familiar weird theoretically possible event is that for a few minutes all molecules of oxygen will be located in one remote corner of your room, and you will die from anoxia.

To make this more concrete I invoke Feynman's path integral approach to QM: Each molecule can traverse any possible path, each path has a probability, these probabilities can interfere with each other, and in our ordinary environment one path is usually by far the most likely; it enjoys the path of highest constructive interference and least destructive interference. Thus fortunately the probability that all molecules of oxygen moving to one corner of the room is extremely small in our lifetimes, but such weirdness is possible.

BTW, Feynman's system meshes neatly with what we see in two-slit experiments, the kind that are said to reveal wave-particle duality: The bands of an interference pattern that we see on a screen or photo arise from path interference.

 

[A. Neumaier]

this way of thinking about QM makes sense: "Anything that can happen does happen," but each possible happening has a probability associated with it.

This way of thinking also makes sense classically, but it explains nothing at all. Moreover, both classically and quantum mechanically, it is a gross violation of the basic philosophic principle called Ockham's razor.

 

[secur]

A more familiar weird theoretically possible event is that for a few minutes all molecules of oxygen will be located in one remote corner of your room, and you will die from anoxia.

Has nothing to do with the topic, but: no, you won't die from anoxia, in a minute or two. Instead you'll die in milliseconds from major trauma as all those air molecules immediately rush back into the rest of the room in a huge explosion. As pointed out by some PF poster a while ago; iirc, @Demystifier.

"Anything that can happen does happen", but each possible happening has a probability associated with it.

This way of thinking also makes sense classically, but it explains nothing at all. Moreover, both classically and quantum mechanically, it is a gross violation of the basic philosophic principle called Ockham's razor.

Is that true? "Anything that can happen does happen" is, in a sense, the simplest scenario, because if rejected you have to explain why particular things happen and not others. This is exactly parallel to the philosophically uncomfortable "collapse of the wave function". Why does it collapse to this branch of the wave function, and not all those others, some of which might indeed have higher probabilities? Applying Ockham to this question results directly in MWI.

 

[Zafa Pi]

This way of thinking also makes sense classically, but it explains nothing at all. Moreover, both classically and quantum mechanically, it is a gross violation of the basic philosophic principle called Ockham's razor.

William of Ockham was also known as William of Occam. Adeptly applying his razor you should use Occam.

 

[OCR]

When it's said...

Anything that can happen does happen...

And it's said...

...each possible happening has a probability associated with it.

Don't those statements become some what contradictory ?

Wouldn't it make more sense to say... "Anything that can happen... could, might, may, or even, can, etc. ... happen ?
Such ... "Anything that can happen, does can happen."

Don't you actual say that when you say... ?

The best that QM math can ...
can get there via Jupiter...
Each molecule can traverse any possible path...
these probabilities can interfere with each other...

I don't see the word does, or, do ... anywhere.

Instead you'll die in milliseconds from major trauma as all those air molecules immediately rush back into the rest of the room in a huge explosion.

I don't remember reading that... was it posted in this thread ?

 

[secur]

@OCR, yes it's contradictory to say "everything happens, but some things have less probability of happening" when talking about a single event. Indeed this is a big difficulty facing MWI proponents. But in normal QM situations (like calculating particle decays) we're talking about expectations, over many instances. Still I'm not sure that completely resolves the philosophical difficulty.

Someone - it may have been @Demystifier - corrected me in some other thread months ago, when I made the same comment about anoxia. Actually, depending how the air got into that room corner, you might have died then, if it all rushed over there very quickly.

 

[OCR]

Thank you for replying, secur ...

 

[OCR]

Adeptly applying his razor you should use Occam.

Correct !

 

[Zafa Pi]

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?

 

[Zafa Pi]

In responding to zonde:

I'm certain that you can't explain me anything about probability theory, because I happen to be a professional scientist,

Wow!

 

[secur]

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?

I'm pretty sure they would have responded "What the deuce does Alice have to do with physics?"

 

[Zafa Pi]

I'm pretty sure they would have responded "What the deuce does Alice have to do with physics?"

Ok, then how about Marie and Pierre (Curies). Anyway what's your answer to my question?

 

[Demystifier]

Someone - it may have been [USER=61953]@Demystifier - corrected me in some other thread months ago, when I made the same comment about anoxia. Actually, depending how the air got into that room corner, you might have died then, if it all rushed over there very quickly.[/USER]

It hasn't been me.

 

[A. Neumaier]

Anything that can happen does happen" is, in a sense, the simplest scenario

But a completely nonpredictive one, hence it is no science.

William of Ockham was also known as William of Occam. Adeptly applying his razor you should use Occam.

His razor was made for shaving scientific statements, not names.

 

[Zafa Pi]

His razor was made for shaving scientific statements, not names.

My bad.

 

[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.