Are Local Hidden Variables Still Viable in Quantum Mechanics?

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In summary, Eric Reiter is proposing a "loading theory" of quantum mechanics in which quantization is a property of matter, not of light. His results suggest that the effect of a beamsplitter on gamma ray photons cannot be seen with low energy photons, where visible light does not have enough energy. This might explain why it has never been observed so far.
  • #1
vzn
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hi all in another recent thread there was an epic battle between
a local realist & a qm formalist. good theatre. but a stalemate in
the end.

I would like to post just a few key recent refs I was reminded of by that
thread, which might be useful for anyone interested in the area. from
your pt of view, either the LHV position is increasingly untenable for
anyone other than fanatics. or, another idea-- there is only a very
narrow slice of LHV theories remaining that have not been ruled out so far,
which have to have a higher degree of sophistication than some of the
"toy models" considered by bell.

more discussion on the "qm2" mailing list

http://groups.yahoo.com/group/qm2/

a. phd & nobel prize winner t'hooft is working on LHVs somewhat
recently which can be found in this paper. the idea is to use a set
of simple harmonic oscillators as the hidden systems. this would tend
to refute the idea that there is no point in working on LHVs, that the
whole matter is closed, that there is no possibility, etc. some other
researchers are already building on it

How Does God Play Dice? (Pre-)Determinism at the Planck Scale
http://www.arxiv.org/abs/hep-th/0104219

b. I think bell came up with a wonderful analysis of this problem but
in my opinion his form of the LHV is too strict. (or equivalently, you
can see it as ruling out all but a very restricted class of LHVs, which
have not been explored too much). let's look at the probability
distribution of the single hidden variable as we rotate analyzers in the
bell experiment. bells proof assumes this distribution does not change
as their difference angle changes. reasonable for a toy LHV model.

however there do exist LHV models such that the lambda probability
distribution can change as the analyzers change. the details are subtle
but mostly unexplored. therefore it makes a lot of sense to look
at LHV theories that are maximally compatible with the predictions of QM.

c. a very neat experiment purports to do a "efficient detection" of
a bell experiment for the 1st time. published in nature.
"experimental violation of a bells inequality with efficient detection"
by rowe et al, 2001. link to the paper & my analysis of this here.

http://groups.yahoo.com/group/qm2/message/9730

basically I agree its a
beautiful tour-de-force qm experiment which measures what it
purports to measure, but from the pt of view of bell rigor, the experiment
is very much lacking in rigor. the authors trapped two ions in an
atom trap. but the bell experiment consisted of a single laser and single
detector for BOTH ARMs of the experiment, which are traditionally at
least spatially separated!

d. it seems to me the ultimate hidden variable in QM is simply
"phi", the phase angle in the wavefn. it appears EVERYWHERE in
qm derivations yet the theory denies that it can be measured. suppose
there were some new theory that proposed that phi could somehow
be measured via some clever advancements (theoretical/experimental).
phi is clearly a hidden variable in QM--therefore "HVs" cannot
be so controversial. the major problem is figuring out how
to make a theory in which you have "locality", LHV.
 
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  • #2
oops, one other tidbit I forgot to mention. something remarkable
I ran across recently.

e. eric reiter claims on his site to be doing a classic
grangier,roger,aspect beamsplitter experiment with
gamma rays, using a slab of aluminum as a beamsplitter (or other variations).
the possible problem for this experiment is that I don't know of any
literature which allows for the possibility of using a beamsplitter
for gamma rays. so anyone can criticize his experiment on that
element.

I tend to think, barring contrary evidence, he may be
doing what he says he is doing. this is the most sophisticated
qm experiment by a non-university affiliated researcher Ive
ever seen. very impressive on that level alone.
he told me the whole effect can be demostrated
with a setup for less than $500. I would very much like to see
"professionals" analyze this experiment.

http://unquantum.com

conceptually, he looks at the histogram of timing differences between an
"idler" and a "signal" photon and finds a nonrandom peak at t=0.
in other words, the gamma ray photon goes "both ways" in the beamsplitter,
and energy is conserved on average but not in individual events.

to interpret his results reiter goes back to a "loading theory" of
plank in which quantization is a property of matter, not of light.
this is the basic thrust of semiclassical theories and
"stochastic electrodynamics".

now suppose this effect cannot be
seen with low energy photons, where visible light does not
have enough energy. that might explain why it has never
been observed so far. maybe the effect is proportional to photon
energy & is way too weak with visible light. although he does not
have a quantitative theory yet, reiter seems to be
proposing something along those lines.
 
  • #3
vzn said:
from
your pt of view, either the LHV position is increasingly untenable for
anyone other than fanatics. or, another idea-- there is only a very
narrow slice of LHV theories remaining that have not been ruled out so far,
which have to have a higher degree of sophistication than some of the
"toy models" considered by bell.

a. phd & nobel prize winner t'hooft is working on LHVs somewhat
recently which can be found in this paper. the idea is to use a set
of simple harmonic oscillators as the hidden systems. this would tend
to refute the idea that there is no point in working on LHVs, that the
whole matter is closed, that there is no possibility, etc. some other
researchers are already building on it

How Does God Play Dice? (Pre-)Determinism at the Planck Scale
http://www.arxiv.org/abs/hep-th/0104219

First, no one is saying that research should not be done in the area of EPR/Bell. However, I and others object to dismissing the existing experimental results, which are quite convincing.

Second, the existing experiments are not toys and they do not test toy models. They go straight to the heart of what EPR discussed. And this is in fact the heart and soul of all LHV theories: that the spin components are determined at the time the particles are created - and NOT at the time of the measurement.

Third, t'hooft says the following about LHV theories in the cited reference:

"The Einstein-Rosen-Podolsky paradox and the violation of the Bell inequalities. This is surely the most dicult aspect to be addressed, and a completely satisfactory response has not yet been given. ..."

So it is not like this isn't taken seriously by him - he recognizes that Bell is a significant obstacle in LHV development.
 
  • #4
vzn said:
oops, one other tidbit I forgot to mention. something remarkable
I ran across recently.

e. eric reiter claims on his site to be doing a classic
grangier,roger,aspect beamsplitter experiment with
gamma rays, using a slab of aluminum as a beamsplitter (or other variations).
the possible problem for this experiment is that I don't know of any
literature which allows for the possibility of using a beamsplitter
for gamma rays. so anyone can criticize his experiment on that
element.
...

This is not a comparable test to the Grangier experiments by a long shot. There is nothing to independently measure that a single photon is entering the apparatus.

By the way: Peer review is as open to Reiter as anyone. And you can see from the picture that he has a lot more that $500 worth of gear.
 
  • #5
vzn said:
c. a very neat experiment purports to do a "efficient detection" of
a bell experiment for the 1st time. published in nature.
"experimental violation of a bells inequality with efficient detection"
by rowe et al, 2001.

basically I agree its a
beautiful tour-de-force qm experiment which measures what it
purports to measure, but from the pt of view of bell rigor, the experiment
is very much lacking in rigor. the authors trapped two ions in an
atom trap. but the bell experiment consisted of a single laser and single
detector for BOTH ARMs of the experiment, which are traditionally at
least spatially separated!

This criticism of Rowe is misplaced. Aspect and others showed through time-varying analyzers with spatial separation that the Bell inequality was still violated. Therefore, one concludes that separation is not an issue.

Once a variable is ruled out as affecting application of a theory, it does not need to be addressed again in every permutation of that experiment. The simplest example of this concept is whether the results of experiments would be different during the day versus during the night, on Mondays versus Tuesday, in the lower hemisphere versus the upper hemisphere, etc. Failing to account for these permutations is not a scientific flaw, because the laws of physics have already been accepted as invariant to these. Ditto with the EPR "loopholes" which are largely closed at this time.
 
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  • #6
vzn said:
hi all in another recent thread there was an epic battle between
a local realist & a qm formalist. good theatre. but a stalemate in
the end.

If you make allusions to my discussion with nightlight, you might have misunderstood the debate (or I might !).
I was not defending QM for QM sake. I'm pretty open to theoretical challenges of QM. But the point was that nightlight was making an (in my honest opinion ; this was not some "point of view" matter) elementary mistake in what he claimed was *A QUANTUM THEORY PREDICTION*, in that he claimed that QED did NOT predict anticorrelation of photons in a setup a la Thorn.
Now, or I have a serious misunderstanding of quantum theory myself, or this prediction (contradicted by all people working in the field) is not a QED prediction.

I think I somehow proved that QED, in this particular case, did predict anti-correlation. Nightlight claimed that I could not use the same operator for the two detectors + electronics when there is and when there is not a beamsplitter in place, which I think is absurd because the beamsplitter didn't have anything to do with the "measurement apparatus", but just determined what state was presented to the measurement apparatus.

Now, if someone thinks that *I* made a serious mistake there, I'd like to know!

But the discussion was not about LHV versus QM. It was: "what is a correct prediction of QM".

cheers,
Patrick.
 
  • #7
vanesch.. a question for you .. I think you have referred
to the "many worlds" interpretation of qm at different
points. are you familiar at all with the "Ghirardi-Rimini-Weber"
model? at first I thought it was only a philosophical interpretation,
but I realized slightly to my shock
recently after skimming a pretty good ref on the
subject of bell thms (baggott, _the meaning of quantum theory_)
that GRW actually proposed a slight physical twist to
the collapse of the wavefn, with physically measurable parameters.

also acc to baggott, as I recall at this moment, he says they have
a LOCAL theory given only slight modifications to quantum mechanics.

from what I can tell
GRW seems to have highly influenced modern talk about
"decoherence" which is quite mainstream and
not at all as disreputable and stigmatized
as "LHV" investigation. in fact afaik, there are even now experiments
that "measure" "decoherence"... ?

if anyone understands the GRW model or knows of a nice intro
on it, please let me know
 
  • #8
I don't know GRW, but consider this : take original Bell's Ansatz for EPRB :

local hidden variable (whatever they are physically) in A : the result is

[tex] A(n_A,\lambda) [/tex] n_a is the direction of measurement in A, hence representing the measuring apparatus.

Interpreations :

1) Bell : [tex]\lambda[/tex] is a hidden variable (what's this ?) causing the result [tex] A(n_A,\lambda) [/tex] in A.. NB : the result can depend on the configuration of the measurement apparatus

2) MW : [tex] A(n_a,\lambda) [/tex] is interpreted as : the result of measurement given by the apparatus in confiuration n_a, IN THE UNIVERSE indexed by the extra coordinate [tex]\lambda[/tex]

So that in fact hidden variables and many-worlds are equivalent, they just are two different interpretations of the same physics (mathetmatical formula)
 
  • #9
See Ghiradi's introduction to spontaneous collapse theories at
http://plato.stanford.edu/entries/qm-collapse/

and Philip Pearle's lectures on the subject at

http://streamer.perimeterinstitute.ca/MediasiteLive30/LiveViewer/FrontEnd/Front.aspx?cid=2f1c7152-a146-47e9-ad22-38a861501270
 
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  • #10
kleinwolf said:
2) MW : [tex] A(n_a,\lambda) [/tex] is interpreted as : the result of measurement given by the apparatus in confiuration n_a, IN THE UNIVERSE indexed by the extra coordinate [tex]\lambda[/tex]

So that in fact hidden variables and many-worlds are equivalent, they just are two different interpretations of the same physics (mathetmatical formula)

Well, I may have misunderstood what you say here, but if you mean that "the choice of the branch" by Alice is the "hidden variable", then this is in a certain way true. However, this "hidden variable" can stay local, thanks to an original "trick" in MWI: the "measurement" at Bob's doesn't have to be decided at the moment when Alice "chooses her branch" !
Bob's "measurement" (from the point of view of Alice) only makes sense for Alice when she LEARNS ABOUT IT (through a "classical" communication channel). It is only at THAT MOMENT that the "hidden variable" (the chosen branch by Alice) determines Bob's outcome (from Alice's point of view).
But at that moment, there is no space-like separation anymore between the two "decision points". So this can happen "locally".
What Bell's theorem implies is that there cannot be a local hidden variable in the system which decides what will be Bob's outcome AT THE MOMENT WHEN ALICE FINDS HER OUTCOME. And MWI solves the riddle by *postponing* Bob's outcome (he's in a superposition from Alice's point of view) until she learns about Bob's "result", locally, at Alice's place.

cheers,
Patrick.
 
  • #11
vzn said:
...I realized slightly to my shock
recently after skimming a pretty good ref on the
subject of bell thms (baggott, _the meaning of quantum theory_)
that GRW actually proposed a slight physical twist to
the collapse of the wavefn, with physically measurable parameters.

also acc to baggott, as I recall at this moment, he says they have
a LOCAL theory given only slight modifications to quantum mechanics.

fyi, just to correct myself.
I was just rereading baggott. he said something about GRW
reproducing classical predictions which I misinterpreted. later he says
that GRW does not give a local theory for eg the bell experiments.
I must say however that there is a lot of literature on what is
called "dynamical collapse" by semimainstream physicists
which is a kind of hidden variable theory--
but generally not a local one.
 
  • #12
vanesch said:
Well, I may have misunderstood what you say here, but if you mean that "the choice of the branch" by Alice is the "hidden variable", then this is in a certain way true. However, this "hidden variable" can stay local, thanks to an original "trick" in MWI: the "measurement" at Bob's doesn't have to be decided at the moment when Alice "chooses her branch" !
Bob's "measurement" (from the point of view of Alice) only makes sense for Alice when she LEARNS ABOUT IT (through a "classical" communication channel). It is only at THAT MOMENT that the "hidden variable" (the chosen branch by Alice) determines Bob's outcome (from Alice's point of view).
But at that moment, there is no space-like separation anymore between the two "decision points". So this can happen "locally".
What Bell's theorem implies is that there cannot be a local hidden variable in the system which decides what will be Bob's outcome AT THE MOMENT WHEN ALICE FINDS HER OUTCOME. And MWI solves the riddle by *postponing* Bob's outcome (he's in a superposition from Alice's point of view) until she learns about Bob's "result", locally, at Alice's place.

cheers,
Patrick.

It's funny, you always speak about time, when the formla has absoutely nothing to do with it.

In fact time is one possible way out of Bell's intelligent trap (not admitted by everyone like everytime)

assume the result is time dependent : [tex] A(n_a,\lambda,t) [/tex]

then the correlation is : [tex] C(A,B,t)=\int A(n_a,\lambda,t)B(n_b,\lambda,t)d\lambda [/tex] assuming the result are simultanous (in the lab frame for example).

Then if you plug into CHSH, the correlation is measured in 4 different times :

[tex] CHSH=|C(A,B,t_1)-C(A,B',t_2)+C(A',B,t_3)+C(A',B',t_4)| [/tex]

since the correlation depends on t_i, and that those are all different, then you cannot factorize a la Bell, and you get CHSH<4.

Note that every "visible" variable allow this, hence every variable still present in the correlation, as an extra-parameter more than the angles of measurement.
 
  • #13
fyi, one fairly straightfwd LHV theory apparently not covered
by the bell impossibility/"no-go" thm:

let the hidden variable determine whether the particle
is detected or not, in conjunction with local polarizer angle.

from what I can tell (working from memory)
it is true that bell thms are fairly general and would tend
to rule out a LHV such that whether the particle is detected
at a remote arm is fixed by a hidden variable at the time of
generation.

however, if one examines a hidden variable that interacts
with the polarizer angle, the bell thm cannot rule it out. I noticed
that one can get different distributions of hidden variable
lambdas at each end, depending on polarizer orientation, a
situation not covered by his thms. (bell has a later thm that
shows that if the lambda distribution is the same at each end,
independent of polarizer angle, his thm still holds.)

hence, a perfect LHV may be possible. gisin has a paper with a very
similar idea.

A local hidden variable model of quantum correlation exploiting the detection loophole
http://arxiv.org/abs/quant-ph/9905018

I tend to disagree with them that the above model necessarily involves
a detector efficiency issue. you can model the dynamics in two ways:
a) lack of detection is due to detector inefficiency
b) lack of detection is not due to detector inefficiency. ie model holds
even with 100% efficient detector

the point is that (a) and (b) have IDENTICAL mathematics yet (b)
is NOT MEASURABLE. ie the difference between (a) and (b) cannot
be discriminated by experiment.

it is counterintuitive in the sense that it is not
deterministic in this way: not every "particle" or "wave"
launched at the source can be detected, no matter how the
measuring system is altered. yet it is deterministic in the sense
of having cause & effect, but just not being able to associate an
effect with every cause (not a 1-1 correspondence). it proposes
"unavoidable information loss" so to speak.

note this loophole is emphatically NOT the same as the efficiency
of detection loophole, although conceptually they seem almost
identical.

this "lambda and polarizer determine detection"
loophole involves the idea that even with a 100% efficient
detector, hidden variables control whether the particle is
detected.

it actually fits into the copenhagen interpretation in
a remarkable sense: it argues there
is a physically real parameter that influences
detection, but that there is no way to measure it.

it also has a strong resemblance to the Heisenberg uncertainty
principle, which says we cannot sharpen complementary
measurements past a certain point.
 
  • #14
vzn said:
fyi, one fairly straightfwd LHV theory apparently not covered
by the bell impossibility/"no-go" thm:

let the hidden variable determine whether the particle
is detected or not, in conjunction with local polarizer angle.

This variation has no effect on Bell, it still applies. The question is whether Malus' law [tex]i_{source}cos^2\theta[/tex] applies as QM predicts and has been amply verified. If that law applies, then Bell applies and the detection HV is simply mixed in with the ensemble that becomes [tex]\lambda[/tex]. This then leads to the contradiction.

Even in the scenarios in which the detectors are alleged to see an "unfair" sample - and perhaps you are referring to this element in your detection variation - there are serious problems: the "bias" (presumably a result of a local HV) must sometimes be positive and sometimes be negative, and must vary according to a very exacting formula intended to yield the observed results. (Experiments in which fair sampling is not an issue do not show any different results anyway.)

In addition, this hypothetical local HV would need to apply ONLY to photon pairs in the singlet state! Because nothing like this has ever been otherwise observed in classical optics, which agrees with the predictions of QM.
 
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  • #15
kleinwolf said:
It's funny, you always speak about time, when the formla has absoutely nothing to do with it.

The formula only has a meaning when you think that the process that "made the measurement happen", happened about simultaneously at Bob's and at Alice's, because the factorization (the locality condition) only applies at that point.
This means that somehow, Bob DID a measurement, and Alice DID a measurement, and they HAVE a result.
MWI "weasels out" by having Bob, and his measurement, in a "suspended state" (entanglement) until Alice learns about it, which is a local interaction at Alice's (I'm giving here purely the Alice point of view). So there's (from Alice's point of view) no "probability" to be associated with Bob's "measurement" at all. The only two real measurements are done by Alice:
She first does her measurement on the photon, and then she does her second measurement on the messenger from Bob.
Bob never measured anything, he's just part of the outer world with respect to Alice.
And as the only probabilities generated are now locally, at Alice's, indeed, Bell's theorem doesn't apply here because everything happens locally at Alice's.
The only thing she could eventually deduce is that Bob got some *information* from her first decision, by calculating if her second measurement contains information from her first CHOICE (not measurement). She will then find out that no such information transfer took place.

cheers,
Patrick.
 
  • #16
Patrick,

The truth is that I understand almost nothing from the text. I rather believe in the principle : a formula is worth 1000 words.

If you compute from basic quantum axioms the probabilities of a local measurement made by Alice only, you get that a parameter local at Bob's place from the definition of the probability in QM, gets then local in A...however there is no indication if there is a continuous displacement of that parameter. This is summarized by the equation of the prob. of let say + in A by : ([tex] \Psi[/tex] is the singlet state)

[tex] p(+_A)=|<+_A\phi|\Psi>|^2 [/tex]

here [tex]\phi[/tex] is located in B

[tex] \Rightarrow p(+_A)=\frac{1}{4}(1-\cos(\theta_A-2\phi) )[/tex]

here [tex]\phi[/tex] is located in A

If you now have an interpretation of this quantum formalism then I think you're lucky...personally I think the formulas are self-explaining, and that phrasal intepretations are just making the clearness of the mathematical language opaque.
 
  • #17
kleinwolf said:
If you now have an interpretation of this quantum formalism then I think you're lucky...personally I think the formulas are self-explaining, and that phrasal intepretations are just making the clearness of the mathematical language opaque.

Ok, so let's do it formula-wise:



Let us take as initial state a singlet state of 2
1/2 spin particles:
|psi_start> = 1/sqrt(2) (|z+>|z-> - |z->|z+>)

We know that if the axis n makes an angle b with the
z-axis, then we can write:
|z+> = cos(b/2) |n+> + sin(b/2)|n->
|z-> = -sin(b/2)|n+> + cos(b/2)|n->

Rewriting psi_start with the n-axis for the second
particle, we have:

|psi_start> = 1/sqrt(2) {
-sin(b/2)|z+>|n+> + cos(b/2)|z+>|n->
- cos(b/2)|z->|n+> - sin(b/2)|z->|n-> }

Note that the association of the first and second
ket of each tensor product could just as well be obtained
by an association of indices, with each ket living individually
in its own Hilbert space.

Let us also introduce "initial states" for Alice and Bob:
|A0> and |B0>.

So our starting state is |psi_start>|A0>|B0>

Now, Alice "measures" the z-component and Bob "measures" the
n-component (where Bob decides about the b-angle).
This "measurement" is just a LOCAL interaction, where, for
Alice, this corresponds to the following time evolution
operator:

U_A: |z+>|A0> -> |noz> |A+>
|z->|A0> -> |noz> |A->

(|noz> corresponds to an absorbed particle)

This U_A operator acts only on the Alice space and the first particle
space.

In the same way, there is an operator corresponding to Bob:

U_B(b): |n+> |B0> -> |noz> |B+>
|n-> |B0> -> |noz> |B->

After Alice and Bob's respective LOCAL interactions, (after
applying the time evolution operator U_A x U_B(b) ), we have:

|state1> = 1/sqrt(2) |noz>|noz> {
-sin(b/2)|A+>|B+> + cos(b/2)|A+>|B->
- cos(b/2)|A->|B+> - sin(b/2)|A->|B-> }

Again, the association in tensor products can be replaced by
indices which have been inherited ; the evolutions take place
individually, in each factor space.

Now let us assume that we take the viewpoint of Alice as an
observer. This means that at this point, Alice will only observe
one of her two branches, with probability given by the Born rule.
She can be in the branch A+ with a probability given by the sum of the
absolute square of all amplitudes of A+:
((-sin(b/2))^2 + (cos(b/2))^2)/2 = 1/2
or she can be in the branch A- with a probability given by:
((-cos(b/2))^2+(-sin(b/2))^2)/2 = 1/2

Let us, for sake of example, suppose that our Alice
as an observer is in the A+ branch; which we will indicate by *:

|state1> = 1/sqrt(2) |noz>|noz> {
-sin(b/2)|A+*>|B+> + cos(b/2)|A+*>|B->
- cos(b/2)|A->|B+> - sin(b/2)|A->|B-> }

Next, there is a transmission of information from Bob to Alice,
because Bob traveled to Alice's:
this is implemented by the following time evolution operator:
(which can now act LOCALLY upon the Bob and Alice, because both
are now in the same spot at the same time:

U_T : |A+>|B+> -> |A++>|B++>
|A+>|B-> -> |A+->|B-+>
|A->|B+> -> |A-+>|B+->
|A->|B-> -> |A-->|B-->

(note that A-+ means that Alice first had locally - and then saw
Bob with a +, while B+- means that Bob first had locally a + and
then saw Alice with a -)

So now we have:

|state2> = 1/sqrt(2) |noz>|noz> {
-sin(b/2)|A++*>|B++> + cos(b/2)|A+-*>|B-+>
- cos(b/2)|A-+>|B+-> - sin(b/2)|A-->|B--> }

But this is again something that is an observation from Alice's point
of view (because it affects her body state).
So we have to apply again Born's rule, which has now to choose
between the A++ and the A+- state, with relative probabilities:

sin(b/2)^2 and cos(b/2)^2

So let us suppose that the first case wins:

We now have:

|state2> = 1/sqrt(2) |noz>|noz> {
-sin(b/2)|A++*>|B++> + cos(b/2)|A+->|B-+>
- cos(b/2)|A-+>|B+-> - sin(b/2)|A-->|B--> }

Alice observed first a z+ state, with probability of 1/2.

Next she observed Bob, when he came to see her,
to be in a state B+ with (relative)
probability (sin(b/2))^2

Knowing what Bob did, she deduces that this means that Bob
"measured" an n+ state. But that's a deduction, which does not
need to imply that Bob somehow "was" in an n+ state and so that
something happened at Bob's.

We can also tell the story from Bob's point of view. It is quite
different, but the states are the same.

After the "measurement" interactions, we have the state:

|state1> = 1/sqrt(2) |noz>|noz> {
-sin(b/2)|A+>|B+> + cos(b/2)|A+>|B->
- cos(b/2)|A->|B+> - sin(b/2)|A->|B-> }

Because this changed Bob's body state, and Bob is now our observer,
we have that Bob is in the B+ state with probability 1/2 and
in the B- state with probability 1/2.
Let's assume that our observer Bob is associated with the B- state,
which we indicate with #:

|state1> = 1/sqrt(2) |noz>|noz> {
-sin(b/2)|A+>|B+> + cos(b/2)|A+>|B-#>
- cos(b/2)|A->|B+> - sin(b/2)|A->|B-#> }

Now, Bob goes and sees Alice ; the same state2 as before is there:

|state2> = 1/sqrt(2) |noz>|noz> {
-sin(b/2)|A++>|B++> + cos(b/2)|A+->|B-+#>
- cos(b/2)|A-+>|B+-> - sin(b/2)|A-->|B--#> }

But because we see things from our Bob observer point of view,
Bob will observe only one of his body states B-+ or B--, according
to the relative Born rule: cos(b/2)^2 versus sin(b/2)^2.

Let us suppose that, with a probability cos(b/2)^2, our Bob observer
is associated with B-+:

|state2> = 1/sqrt(2) |noz>|noz> {
-sin(b/2)|A++>|B++> + cos(b/2)|A+->|B-+#>
- cos(b/2)|A-+>|B+-> - sin(b/2)|A-->|B--> }

We could eventually consider both adventures together:

|state2> = 1/sqrt(2) |noz>|noz> {
-sin(b/2)|A++*>|B++> + cos(b/2)|A+->|B-+#>
- cos(b/2)|A-+>|B+-> - sin(b/2)|A-->|B--> }

Alice "observer" ended up in the A++ body state, and in that branch, observes a Bob body state which is in a B++ state, and whose body will
act (like talking) accordingly ; so for Alice observer, it is clear that
"Bob" saw an n+ state.

For Bob "observer", he ended up in the B-+ body state. In that branch,
he observes Alice in a body state +-, so it is clear for him that
Alice observed + and that he observed - (the Alice body in his branch
will agree with that).

cheers,
Patrick.
 
  • #18
You always remain local by an a priori position :

Now, Alice "measures" the z-component and Bob "measures" the
n-component (where Bob decides about the b-angle).
This "measurement" is just a LOCAL interaction, where, for
Alice, this corresponds to the following time evolution
operator:

1) you speak about the measurement operator : [tex] (\vec{\sigma}\cdot\vec{n}_A)\otimes(\vec{\sigma}\cdot\vec{n}_B) [/tex] which is not local, because it disturbs both systems A and B.

The local measurement operator for A is [tex] (\vec{\sigma}\cdot\vec{n}_A)\otimes(I_2) [/tex]. This measurement is LOCAL in A in the sense : it does not disturb B.

The correlation between A and B is then given by a superposition of both type of measurements.

2) Then there is no indication of how many time the collapse in a measurement takes, but it is a unitary operation of course, so you abusively speak about time evolution.

Following Copenhagen QM :

a) there is no indication if the collapse is composed by steps (continuous or discrete) between the initial state and the final wavefunctions.

Let's put the hypothesis the process is composed by steps :

b) These steps can have another physical dimension than time

b) If this process is continuous for example, then it can be non unitary between the initial and the final points. Expressed mathematically, the collapse of the WF could be put as the Ansatz :

[tex] \psi(\epsilon=0)=|\Psi\rangle [/tex] singlet state
[tex] \psi(\epsilon=1)=|+_A\rangle [/tex]

We can naturally have [tex] ||\psi(0<\epsilon<1)||^2\neq 1 [/tex].


3)

a) if I'm not mistaken what you think is : p(+_A)=p(-_A)=p(+_B)=p(-_B)=1/2.
if A and B measure along the same direction.

b) What quantum mecahnics gives you locally is, in common with my previous message :

[tex] p(-_A)=\frac{1}{4}(1+\cos(\theta_A-2\phi)) [/tex]
[tex] p(+_B)=\frac{1}{4}(1+\cos(\theta_B-2\phi)) [/tex]
[tex] p(-_B)=\frac{1}{4}(1-\cos(\theta_B-2\phi)) [/tex]

hence :

[tex] p(+_A)+p(-_A)+p(+_B)+p(-_B)=1 [/tex]

with [tex] \max_{\phi} p(+_A)=1/2 [/tex]

so that the sum of all the single event is one, not only one side of the system...because QM always considers the wholeness of the system , since it is described by a single wf.

This gives a simpler way than yours, since for example let say : [tex]\theta_A=\theta_B=0 [/tex]. (A and B measure along the same direction)

Let's assume : hypothesis : [tex] p(-_A)=0 [/tex].

From this we are locally sure that the result is +_A.

If you take the above probabilities, then you see this implies : [tex] p(+_B)=0 [/tex].

Hence the result is -_B in B.

As summary : taking the basic axioms of Copenhagen QM allows you to explain only with a single variable [tex] \phi [/tex], how the measurement on 1 side affects the other. NB: [tex]\phi[/tex] is just some eigenvector of the identity operator.
 
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  • #19
kleinwolf said:
1) you speak about the measurement operator : [tex] (\vec{\sigma}\cdot\vec{n}_A)\otimes(\vec{\sigma}\cdot\vec{n}_B) [/tex] which is not local, because it disturbs both systems A and B.

Ah, well, then apply first the A "measurement" operator (U_A x 1), and later apply the B "measurement" operator. That comes down to the same thing.
Note that these operators are just the time evolution operator corresponding to the interaction of Alice (Bob) with some apparatus and the system under study, which must have a Hamiltonian associated with it. It is an abuse of language to call it a "measurement" operator ; it is just a time evolution like any other time evolution.

The local measurement operator for A is [tex] (\vec{\sigma}\cdot\vec{n}_A)\otimes(I_2) [/tex]. This measurement is LOCAL in A in the sense : it does not disturb B.

Exactly. However, it is not [tex]
\vec{\sigma}[/tex] ; it is much more complicated and describes the typical action of a measurement device (but expressed as a physical system, with its interaction hamiltonian), and Bob's (Alice's) body state with all its atoms and so on.

2) Then there is no indication of how many time the collapse in a measurement takes, but it is a unitary operation of course, so you abusively speak about time evolution.

Well, there is a certain interaction time needed for Bob's body and his measurement apparatus to interact with the particle ; it is this complicated interaction that is symbolised by the time evolution operator U_B, in a very complex Hilbert space (containing all of Bob's body states).
But somehow we can assume that after a reasonable amount of time (say, a few seconds) Bob's body is in a certain state which corresponds to "Bob saw the red LED of his apparatus flash".
As I cannot write down of course explicitly the hamiltonian (with all interactions in his retina, his nerve cells and so on on the molecular level: go figure!) I just write it down symbolically with U_A(t, t+a few seconds).


Following Copenhagen QM :

a) there is no indication if the collapse is composed by steps (continuous or discrete) between the initial state and the final wavefunctions.

But I'm giving you an MWI QM picture, which DOES suppose that there is no such "collapse" but only unitary interactions!

Let's put the hypothesis the process is composed by steps :

b) These steps can have another physical dimension than time

Well, the MWI viewpoint is that it is just a time evolution like any other (of which the time derivative is the hamiltonian).

b) If this process is continuous for example, then it can be non unitary between the initial and the final points. Expressed mathematically, the collapse of the WF could be put as the Ansatz :

[tex] \psi(\epsilon=0)=|\Psi\rangle [/tex] singlet state
[tex] \psi(\epsilon=1)=|+_A\rangle [/tex]

We can naturally have [tex] ||\psi(0<\epsilon<1)||^2\neq 1 [/tex].

Yes, but that's explicitly non-unitary (you cannot have it unitary). And now you will have to tell me what physical processes happen according to this evolution equation, and what physical processes happen according to the unitary time evolution. This is the dilemma of the Copenhagen interpretation.



a) if I'm not mistaken what you think is : p(+_A)=p(-_A)=p(+_B)=p(-_B)=1/2.
if A and B measure along the same direction.

No, in all cases ! Even if they don't measure according to the same direction.
That's true, no matter what "approach" (MWI, Copenhagen...) you take, no matter who measures first etc...
It is essential, in that entanglement does not allow for information transfer.
Alice will see half of her electrons spin up, and the other half: spin down ; and for Bob, the same ; no matter what angles they happen to chose.

b) What quantum mecahnics gives you locally is, in common with my previous message :

[tex] p(-_A)=\frac{1}{4}(1+\cos(\theta_A-2\phi)) [/tex]
[tex] p(+_B)=\frac{1}{4}(1+\cos(\theta_B-2\phi)) [/tex]
[tex] p(-_B)=\frac{1}{4}(1-\cos(\theta_B-2\phi)) [/tex]

hence :

[tex] p(+_A)+p(-_A)+p(+_B)+p(-_B)=1 [/tex]

with [tex] \max_{\phi} p(+_A)=1/2 [/tex]

Well, that's wrong then. At least if you mean by p(-_A) the probability that Alice will measure spin down (unconditionally). This is 1/2. No matter what Bob does, and no matter which axis Alice chooses.
We should first clear out this.


so that the sum of all the single event is one, not only one side of the system...because QM always considers the wholeness of the system , since it is described by a single wf.

Well, it is clear that p(-_A) + p(+_A) = 1, because Alice can only obtain 2 possible answers: spin up or spin down, right ?
So an event will have given spin down at Alice's or it will have given spin up at Alice's. The sum of these probabilities must be equal to 1!

Let's clear out this problem first...

cheers,
Patrick.
 
  • #20
vanesch said:
Ah, well, then apply first the A "measurement" operator (U_A x 1), and later apply the B "measurement" operator. That comes down to the same thing.

That does NOT comes to the same thing...But to explain you the BASICS of QM I think it really becomes a hard work: Let start with the singlet state [tex] \Psi [/tex] :

Apply AxI, suppose you get + in A, hence the final state is [tex] \psi_1=|+_A>|\phi> [/tex].

[tex] |\phi> [/tex] is any unitary vector of R^2.

Apply IxB on the state obtained..suppose B got -. Then the final state is

[tex]\psi_2=|\phi>|-_B> [/tex].

However applying AxB gives you the final state [tex] \psi_3=|+_A>|-_B> [/tex]

if you suppose you get the same results of measurement (+ in A and - in B)

You immediatly sees this are not the same final states...

This is exactly the difference between local operators AxI, IxB and the non-local one AxB.

Note that these operators are just the time evolution operator corresponding to the interaction of Alice (Bob) with some apparatus and the system under study, which must have a Hamiltonian associated with it. It is an abuse of language to call it a "measurement" operator ; it is just a time evolution like any other time evolution.

I thought the time evolution operator should contain time, but I learn everyday like everyone else.



Exactly. However, it is not [tex]
\vec{\sigma}[/tex] ; it is much more complicated and describes the typical action of a measurement device (but expressed as a physical system, with its interaction hamiltonian), and Bob's (Alice's) body state with all its atoms and so on.

It's known that calculable theoretical physics will never be able to reproduce the whole reality, by chance to give work to some physicist...However if you want to do numerical large simulation it's up to you, but those will still remain just in the box and not real.



Yes, but that's explicitly non-unitary (you cannot have it unitary). And now you will have to tell me what physical processes happen according to this evolution equation, and what physical processes happen according to the unitary time evolution. This is the dilemma of the Copenhagen interpretation.

Well there are two processes you mixed up : the time evolution of the qbit system in a magnetic field for example, because you need the energy (Schroedinger equation)...and the wave function collapse (which is not an evolution) when you try to find out in which state your qbit is..

You mixed up because in the case of qubit the energy is a multiple of the spin operator (with the gyromegnetic factor aso..)

You cannot distinguish between a unitary evolution and a non-unitary one in the sense : you only have access to the initial and the final points, which have the same norm...but you don't know what happens inbetween.




No, in all cases ! Even if they don't measure according to the same direction.
That's true, no matter what "approach" (MWI, Copenhagen...) you take, no matter who measures first etc...
I told you : in the case you measure along the same direction, not in all cases.

It is essential, in that entanglement does not allow for information transfer.
Alice will see half of her electrons spin up, and the other half: spin down ; and for Bob, the same ; no matter what angles they happen to chose.

Are you always using a prioris like this ?


Well, that's wrong then. At least if you mean by p(-_A) the probability that Alice will measure spin down (unconditionally). This is 1/2. No matter what Bob does, and no matter which axis Alice chooses.
We should first clear out this.

If you want this to be 1/2 then you can forget QM. Just use the basic axioms...you always use your intuition instead of computing...that's misleading.

For example the exercise is : compute the probability of getting +_A with the local operator AxI on the singlet state. You will see it's not 1/2 but :

[tex] p(+_A)=|<+_A|<\phi|\Psi>|^2 [/tex] you know this ? This is the definition of the prob. of an outcome of a measurement in A in the z_direction. Put [tex] |+_A>=\left(\begin{array}{c} 1\\0\end{array}\right)[/tex] eigenvector of [tex]\sigma_z[/tex] and [tex] |\phi>=\left(\begin{array}{c}\cos(\phi)\\\sin(\phi)\end{array}\right)[/tex] an eigenvector of I.

then :

[tex] p(+_A)=\frac{1}{2}|\left(\begin{array}{c} 1\\0\end{array}\right)\otimes\left(\begin{array}{c}\cos(\phi)\\\sin(\phi)\end{array}\right)|\left(\begin{array}{c} 1\\0\end{array}\right)\otimes\left(\begin{array}{c} 0\\1\end{array}\right)-\left(\begin{array}{c} 0\\1\end{array}\right)\otimes\left(\begin{array}{c} 1\\0\end{array}\right)|^2[/tex]

[tex] p(+_A)=\frac{1}{2}\sin(\phi)^2[/tex].

NB :[tex]\phi[/tex] IS NOT THE DIRECTION OF MEASUREMENT IN A !

So you see : the prob. is not always 1/2...it CAN BE, but on average over the variable [tex]\phi[/tex] it's 1/4...If you allow the direction in A to change then you get what I wrote in the previous message.

Well, it is clear that p(-_A) + p(+_A) = 1, because Alice can only obtain 2 possible answers: spin up or spin down, right ?

No, globally there are 4 events : +_A, -_A,+_B,-_B...why do you split the world in 2 ?

So an event will have given spin down at Alice's or it will have given spin up at Alice's. The sum of these probabilities must be equal to 1!

It's always the same problem : you follow your intuition and hence don't see the globality of the system. What do you do with B ? You again split the world in 2, so that you REDUCE THE UNIVERSE OF PROBABILITIES...but you don't explain why you reduce...and what's the point of reducing it..and why you're allowed to do it.

Another way to explain the problem is : you have a GLOBAL system at A and B...Let's say you choose randomly one side (they are equivalent if you measure along the same directions) with prob. p(A)=1/2.

Then you have partial probabailites p(+)=1/2 for + p(-)=1/2 for -..hence the total probability is :

p(+_A)=p(A)*p(+)=1/2*1/2=1/4.

This is the same for all single events, hence the events in A and B are equiprobable...but the sum of ALL of them should be 1...

cheers,
Patrick.[/QUOTE]
 
  • #21
kleinwolf said:
That does NOT comes to the same thing...But to explain you the BASICS of QM I think it really becomes a hard work: Let start with the singlet state [tex] \Psi [/tex] :

Apply AxI, suppose you get + in A, hence the final state is [tex] \psi_1=|+_A>|\phi> [/tex].

[tex] |\phi> [/tex] is any unitary vector of R^2.

Apply IxB on the state obtained..suppose B got -. Then the final state is

[tex]\psi_2=|\phi>|-_B> [/tex].

Yes, and apply AxI and then IxB and that's the same as applying A x B.

However, you make a mistake here in using Born's PROJECTORS while I was talking about unitary evolution. Unitary evolution is a linear operator, so when you apply it to a sum, it is the sum of its application to the terms.


I thought the time evolution operator should contain time, but I learn everyday like everyone else.

It DOES contain time ! The time of the measurement interaction (a few seconds) ; you can assume that after that, the states do not evolve anymore (which is also part of the time evolution operator).


Well there are two processes you mixed up : the time evolution of the qbit system in a magnetic field for example, because you need the energy (Schroedinger equation)...and the wave function collapse (which is not an evolution) when you try to find out in which state your qbit is..

In MWI there is NO collapse ! That was the whole point of the explanation. It is only in Copenhagen QM that you have collapse. But the probabilities you can calculate come out the same.

For example the exercise is : compute the probability of getting +_A with the local operator AxI on the singlet state. You will see it's not 1/2 but :

[tex] p(+_A)=|<+_A|<\phi|\Psi>|^2 [/tex] you know this ?

No, I don't know this. What you write is the probability to find the state A+ AND phi. Not the probability to find A+.
If you want to find the probability to find A+, you should SUM this over all orthogonal vectors in the phi space, and not just ONE, because then you calculate a JOINT probability.
So you should do the calculation with (cos phi, sin phi) and also with (-sin phi, cos phi), and then sum them.

And then you get 1/2. No matter what phi is.

This is the definition of the prob. of an outcome of a measurement in A in the z_direction. Put [tex] |+_A>=\left(\begin{array}{c} 1\\0\end{array}\right)[/tex] eigenvector of [tex]\sigma_z[/tex] and [tex] |\phi>=\left(\begin{array}{c}\cos(\phi)\\\sin(\phi)\end{array}\right)[/tex] an eigenvector of I.

As I told you, this is NOT the probability of an outcome of a measurement in A in the z-direction ; it is the probability of the event:
"outcome of a measurement A in the z-direction AND a phi direction in B"
because that is what your bra <A+|<phi| describes.

cheers,
Patrick.
 
  • #22
vanesch said:
Yes, and apply AxI and then IxB and that's the same as applying A x B.

yes, it's clear that (AxI)(IxB)=AxB

But the final states are not the same and the probabilites of the final state are not the same neither (the final states were in the previous message)


It DOES contain time ! The time of the measurement interaction (a few seconds) ; you can assume that after that, the states do not evolve anymore (which is also part of the time evolution operator).

Sorry, but I see no continuous parameter like t in your evolution operator.




No, I don't know this. What you write is the probability to find the state A+ AND phi. Not the probability to find A+.

If you want to find the probability to find A+, you should SUM this over all orthogonal vectors in the phi space, and not just ONE, because then you calculate a JOINT probability.
So you should do the calculation with (cos phi, sin phi) and also with (-sin phi, cos phi), and then sum them.

The problem is : you measure once...so you get only 1 phi

As I told you, this is NOT the probability of an outcome of a measurement in A in the z-direction ; it is the probability of the event:
"outcome of a measurement A in the z-direction AND a phi direction in B"
because that is what your bra <A+|<phi| describes.

If you speak about the direction of the measurement operator, then there is no direction of measurement in B, because the operator is I, in other words : there exist no direction of measurement of the spin that gives you the identity operator.

Well we can debate on this hours : I just ask you :

1) what measures the operator AxI what is the interpretation of it...
2) what are the eigenvectors ?
3) what is [tex] |\phi\rangle [/tex] ?

The point is that : you cannot say what "an outcome [...] "phi direction in B"" because you don't measure "in DIRECTIONS" in B !

I remind you : "The results of measurement are the eigenVALUES of the measurement operators" in QM.

So if we measure with the 1 operator in B, we know with certainty the result in B : it's 1...so multiplied with the result in A, it leaves it the same (identity)

The problem is : What is the interpretation of the final state [tex] |\phi\rangle [/tex]

I already asked this question in the thread :

https://www.physicsforums.com/showthread.php?t=68693 [/URL]
 
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  • #23
kleinwolf said:
Sorry, but I see no continuous parameter like t in your evolution operator.
Well, it is there, I only didn't write it explicitly. I took their "asymptotic" states where the time doesn't matter anymore.


The problem is : you measure once...so you get only 1 phi



If you speak about the direction of the measurement operator, then there is no direction of measurement in B, because the operator is I, in other words : there exist no direction of measurement of the spin that gives you the identity operator.

I'm affraid that that is a mistake.
You may know that the prescription |<phi|psi>|^2 as the probability is a simplified version of the real probability prescription, which is:
|<psi | P_i | psi>|^2, and P_i is the projection operator on the eigenspace that corresponds with the outcome of which we are calculating the probability.
In the case that there is no degeneracy left in the outcome (the measurement result specifies COMPLETELY the state), then of course both prescriptions are equivalent. But if the outcome is degenerate (meaning that the outcome does not completely specify the state), then there is an ambiguity in the choice of phi (and that's where you are making a mistake).
We are in this case: the outcome at A alone does not completely specify the state: there are different states possible with the same outcome at A for sure. So we have to project on an eigenspace which is more than one-dimensional, and you just pick out ONE direction. The projection on that direction will then of course be shorter than the one on the entire subspace.

An analogy with Euclidean geometry is possible: imagine a vector 1/sqrt(62) (3,2,7), and the "eigenspace" on which we have to project is the xy plane.
Clearly the projection on the xy plane is 1/sqrt(62) (3,2,0) with length Sqrt(13/62).
However, if you arbitrary choose ONE projection direction, say, the x-axis, then of course we'll find something shorter: sqrt(9/62). But that's not the length of the projection on the relevant eigenspace.
In the same way, you specify ONE phi vector (like my x-axis), arbitrarily. But that is not the length of the projection of the eigenspace.

Well we can debate on this hours : I just ask you :

1) what measures the operator AxI what is the interpretation of it...
2) what are the eigenvectors ?
3) what is [tex] |\phi\rangle [/tex] ?

A x I is indeed the hermitean operator which corresponds to the measure at A.
It has 4 eigenvectors:
(1,0) x (1,0) and (1,0) x (0,1) correspond to + at A
(0,1) x (1,0) and (0,1) x (0,1) correspond to - at A

The probability to have + at A is the length of the total state vector projected upon the eigenspace spanned by the two eigenvectors given in the first line. The projector corresponding to it, P_+A has then the following action on a general ket (a,b)x(c,d):
P_+A (a,b)x(c,d) = (a,0) x (c,d).

The point is that : you cannot say what "an outcome [...] "phi direction in B"" because you don't measure "in DIRECTIONS" in B !

You do, by specifying a specific eigenvector in the eigenspace of A x I.
If you want to talk only about an eigenvalue of A x I, you should use the entire eigenspace, and not an arbitrarily selected vector in that eigenspace.

I remind you : "The results of measurement are the eigenVALUES of the measurement operators" in QM.

So if we measure with the 1 operator in B, we know with certainty the result in B : it's 1...so multiplied with the result in A, it leaves it the same (identity)

The problem is : What is the interpretation of the final state [tex] |\phi\rangle [/tex]

Well, the final state after only a measurement at A, assuming we have outcome +, is then simply P_+A applied to the state.

cheers,
Patrick.
 
  • #24
Hm, your interpretation is

But if the outcome is degenerate (meaning that the outcome does not completely specify the state), then there is an ambiguity in the choice of phi (and that's where you are making a mistake).
We are in this case: the outcome at A alone does not completely specify the state: there are different states possible with the same outcome at A for sure. So we have to project on an eigenspace which is more than one-dimensional, and you just pick out ONE direction. The projection on that direction will then of course be shorter than the one on the entire subspace.

Let me show geometrically that you restrict yourself and not me :

1) Your way : you project a vector on a plane and normalize the result. You agree this will give you ONE vector.

2) Presented way : you project the vector on ANY normalized vector in the plane parametrized by phi...You can adjust phi so that this vector corresponds to yours, but there are infinitely many other possibilities.


You immediatly see : the projection on the plane is a special case of the projection on an arbitrary vector in the plane.

A x I is indeed the hermitean operator which corresponds to the measure at A.
It has 4 eigenvectors:
(1,0) x (1,0) and (1,0) x (0,1) correspond to + at A
(0,1) x (1,0) and (0,1) x (0,1) correspond to - at A

once again you restrict yourself to special vectors (1,0) and (0,1) for the eigenvector of I, but they are arbitrary (cos(phi),sin(phi)) and (cos(alpha),sin(alpha))...phi and alpha are free.

That's what you say after : you have to consider the whole eigenspace since the only eigenvalue of I is degenerate.

(cos(phi),sin(phi)) is not a direction of measurement, it's the endstate of the wavefunction at B's place...
 
  • #25
kleinwolf said:
Hm, your interpretation is



Let me show geometrically that you restrict yourself and not me :

1) Your way : you project a vector on a plane and normalize the result. You agree this will give you ONE vector.

2) Presented way : you project the vector on ANY normalized vector in the plane parametrized by phi...You can adjust phi so that this vector corresponds to yours, but there are infinitely many other possibilities.

This is not "my interpretation" ; it is standard quantum theory.

*) Cohen-Tannoudji (French version) vol I: p 217 4th postulate in the case of a degenerate spectrum:
P(a_n) = sum_i=1^gn |<u_n^i | psi>|^2
where gn is the degree of degeneracy and u_n^i are an orthonormal basis of the Eigenspace corresponding to the eigenvalue a_n

*) von Neumann: p 247: projections as propositions.
Quite opaque but thorough treatment of what I've been saying here.

*) Messiah vol I p 298 :
|u> and ideal measurement yielding result D ; state after that is P_D |u>

...

You immediatly see : the projection on the plane is a special case of the projection on an arbitrary vector in the plane.

Yes, that is because the projection in the plane is the component of the initial state with the right eigenvalue, and it is this component which survives in the Copenhagen view.

once again you restrict yourself to special vectors (1,0) and (0,1) for the eigenvector of I, but they are arbitrary (cos(phi),sin(phi)) and (cos(alpha),sin(alpha))...phi and alpha are free.

You are right ; only, you have to take a COMPLETE BASIS in that eigenspace. So if you really insist on making life difficult, then you can work with the two basis vectors (cos(alpha), sin(alpha)) and (-sin(alpha),cos(alpha)).
In any case, you will have to take the SUM of both contributions (that's a basic axiom in QM, see my citations), and of course because the norm of a vector is invariant under change of orthogonal basis, the result is the same.
(cos(phi),sin(phi)) is not a direction of measurement, it's the endstate of the wavefunction at B's place...

You're not free to choose: the endstate of the wavefunction at B is what you obtain after PROJECTION onto the eigenspace.

cheers,
Patrick
 
  • #26
vanesch said:
This is not "my interpretation" ; it is standard quantum theory.

*) Cohen-Tannoudji (French version) vol I: p 217 4th postulate in the case of a degenerate spectrum:
P(a_n) = sum_i=1^gn |<u_n^i | psi>|^2
where gn is the degree of degeneracy and u_n^i are an orthonormal basis of the Eigenspace corresponding to the eigenvalue a_n

*) von Neumann: p 247: projections as propositions.
Quite opaque but thorough treatment of what I've been saying here.

*) Messiah vol I p 298 :
|u> and ideal measurement yielding result D ; state after that is P_D |u>

...

Yes, that's why this position (till now nobody said (s)he agreed with it) with this parametrization of the eigenstates in a degenerate case is not accepted...but in fact it's not so important, because the standard calculation of probabilities is just a special case of this one...however this parameter can induce interesting results...see at the end.

You're not free to choose: the endstate of the wavefunction at B is what you obtain after PROJECTION onto the eigenspace.

That's exactly the point I don't understand...if you read the other thread I put as an url, u see that :

In the case of the I operator and a vector u, the probability that the endstate is u is 1. With the standard interpretation the problem is closed. However, if you compute the probabilities corresponding to projection on other eigenstates IN THIS eigenspace, the the probability is not 0...so does that mean those other endstates are allowed ?

Now why is the use of [tex]\phi[/tex] interesting : because in the case of experimental measurement of CHSH in Bell type experiment you never get CHSH=2.82. I think the most precise value obtained sofar is 2.37 (the wineland, monroe experiment already cited in this thread)...A large community attributes this to the precision of the apparatus. However, if you take the correct definition of the covariance and use the average over [tex]\phi[/tex] then you get CHSH=2.47...without use of any experimental a posteriori adjustment like precision or detection loophole.

That's just why I say that projection on the parametrization of the eigenendstates instead of the projection on the eigenspace can lead to interesting results.

As summary : suppose I have a law to follow...This law is a special case of another one...so I'm not punishable if I apply the more general law...is that right ? (of course it has to be checked that special cases of the general law do not contradict the first law...but this out of scope here)
 
  • #27
I would like to ask you something, a very simple "exercise" to see the difference between what you do and what I claim standard QM does.

Let us consider a hydrogen atom, and its stationary states |n,l,m> as usual.
Remember that the energy is only function of n (we just consider the coulomb potential), E(n).

Now let us imagine that the initial state is:

|psi0> = a |4,1,-1> + b |4,1,1> + c |4,2,-2> + d |5,2,0> + e |1,0,0>

What, according to you, is the probability of measuring E(4), E(5) and E(1) ?

cheers,
Patrick.
 
  • #28
DrChinese said:
This criticism of Rowe is misplaced. Aspect and others showed through time-varying analyzers with spatial separation that the Bell inequality was still violated. Therefore, one concludes that separation is not an issue.

This seems like a good conclusion. Also, just the idea
that crossed linear polarizers are analyzing, ie., the same
light (in given coincidence interval) should give cos^2 theta
correlation curve for mutual detection indicates that
separation should not be an important factor as long as
the emission characteristics of the light can be preserved
until it hits the polarizers.

I don't see any mystery (nonlocal 'influences') in these
experiments. BI's are violated because the lhv formulations
don't take into account that probabilities for the individual
results change when one side registers a detection and
initiates a coincidence interval.

Cos^2 theta probability of coincidental detection doesn't
depend on the *specific* value of the emission polarization,
per se. It depends on the light *incident on* the polarizers
during any particular coincidence interval having the same
physical characteristics.
 
Last edited:
  • #29
vanesch said:
I would like to ask you something, a very simple "exercise" to see the difference between what you do and what I claim standard QM does.

Let us consider a hydrogen atom, and its stationary states |n,l,m> as usual.
Remember that the energy is only function of n (we just consider the coulomb potential), E(n).

Now let us imagine that the initial state is:

|psi0> = a |4,1,-1> + b |4,1,1> + c |4,2,-2> + d |5,2,0> + e |1,0,0>

What, according to you, is the probability of measuring E(4), E(5) and E(1) ?

cheers,
Patrick.

Wow I'm not in atomic physics at all...this looks difficult


Well I'll modfify your question if you allow me this :

let put [tex]\Psi(intial)=a|4,1,-1>+b|4,1,1>+c|5,2,0>+e|1,0,0> [/tex]

Let say I want to know the probability of measuring E(4)...then

This special case gives : [tex] P(E=E(4))=a^2+b^2 [/tex]

I think this is what standard QM would give ? Because it does not specify and does not need the endstate !

But you agree that in the EPR experiment, you also have to know which endstate your system falls into : It's like if I say :

"I measure the energy at first, let say I got E(4) out. Then I want to know the prob. of a certain dipole allowed transition"

You agree that in that case you have to know the endstate precisely...to continue your work...


>>>>


That's why I will ask you another question :

Let say I prepare an initial state in the E=E(4) eigenstate let say :

[tex] \Psi(initial)=\cos(\beta)|4,1,-1\rangle+\sin(\beta)|4,1,1\rangle [/tex]

I measure the energy...my measure gives p(E=E(4))=1 out...but what is the final state ?

Standard QM affirms : the state stays the same, because it's an eigenstate of the energy. However I affirm, the endstate can be

[tex] \Psi(end)=\cos(\delta)|4,1,-1>+\sin(\delta)|4,0,0> [/tex]

The probability the final state is the latter is : [tex] \cos(\beta)^2\cos(\delta)^2 [/tex] which can be nonzero...

So that when computing the prob. of degenerate eigenvalues outcome, you should always add : to which endstate it finishes...even if standard QM affirms the sate remains the same...because there are other possible endstates.

So this should give a possible experimental verification of which version is correct : if the endstate remains the same, then by some "dipole selection rules" you should be able to see if the other states are completely forbidden, or still probable...

If you compare this to the singlet state local measurement : +_A corresponds to E(4) in this problem...the parametrization [tex]\delta[/tex] of the eigenvectors in the E(4) energy "restricted" eigenspace corresponds to the [tex]\phi[/tex] variable in the EPRB experiment
 

1. What is the difference between LHVs and qm?

LHVs (Local Hidden Variables) and qm (Quantum Mechanics) are two different theories used to explain the behavior of particles at the subatomic level. LHVs assume that particles have definite, hidden properties that determine their behavior, while qm states that particles do not have definite properties until they are measured.

2. Which theory is the most widely accepted among scientists?

Currently, qm is the most widely accepted theory among scientists. It has been extensively tested and has been found to accurately predict the behavior of particles at the subatomic level.

3. What are some newer ideas related to LHVs and qm?

Some newer ideas related to LHVs and qm include the concept of entanglement, which states that particles can become connected in such a way that their properties are dependent on each other, even when separated by large distances. Another idea is the concept of non-locality, which suggests that particles can influence each other instantaneously, without any physical connection.

4. Can LHVs and qm coexist or be merged together?

There have been attempts to merge LHVs and qm together, but so far, no successful unified theory has been developed. Some scientists believe that it may be possible to reconcile the two theories by incorporating elements of both, but this is still a subject of ongoing research and debate.

5. What implications do LHVs and qm have for our understanding of the universe?

The theories of LHVs and qm have revolutionized our understanding of the universe at the smallest scales. They have led to technological advancements, such as quantum computing, and have also challenged our traditional notions of cause and effect. These theories continue to be studied and have the potential to unlock even more mysteries of the universe.

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