Re: Superposed observers (was No new Einstein)
- From: rof@xxxxxxxxxxxx
- Date: Tue, 4 Oct 2005 04:33:50 +0000 (UTC)
"I.Vecchi" <vecchi@xxxxxxxxxxxxx> writes:
>rof@xxxxxxxxxxxx wrote:
>> "I.Vecchi" <vecchi@xxxxxxxxxxxxx> writes:
>>
>> >rof@xxxxxxxxxxxx ha scritto:
>>
>..
>>
>> >No difference here. Coin-tossing and Geiger counters do the same
>> >quantum job.
>>
>> When a coin is tossed, the side on which it will land is mostly
>> determined by the initial conditions.
>The point is that the initial conditions are not known to the observer.
>If they are known, then the process is no longer random. And any
>chaotic process will quickly blow initial quantum scale
>indeterminacies up to macroscopic scale ([2]).
>Others may believe, unlike me, that physical instances of probability
>come in two kinds, those with a quantum soul and those without.
So, you consider that probabilities obtained using quantum mechanics
are no different to probabilities which are obtained simply as a
consequence of using incomplete information. Yet you still insist
on talking as though there are many worlds, with one corresponding
to each possible outcome. Why?
If you had never heard of quantum mechanics, but had played with
dice, would you have decided to start talking about people
"splitting" when they see the result of a die roll?
>> However, it isn't normalised. If I set alpha=1, for example,
>> then my calculator gives the squared norm of the state vector
>> to be (0.5*((1/sqrt(2))+1))^2+(0.5*((1/sqrt(2))-1))^2+(1/sqrt(2))^2=1.25
>>
>> Now the probability that the photon is absorbed should be 1/4, so
>> you can solve the normalisation problem and the absorption
>> problem by making the final state
>> 0.5[(alpha/sqrt(2))-1)i|2> - (alpha/sqrt(2))+1)i|1>] + i|absorbed>/2
>> instead.
>>
>> Is this what you intended?
>Indeed, as you point out sqrt(2) should be replaced by 2 , but since
>this affects only the fraction that is absorbed it is irrelevant to the
>final result at the detector.
It is, but I wanted to make sure that it was a typo and not
a state vector which was to be normalised later.
>> Okay, so let me just check that you agree with the following
>> points:
>>
>> 1. My calculation, which indicates that the probability that the
>> detector will click 1/8 of the time (and not 3/8) is correct,
>> according quantum mechanics as presented in standard quantum mechanics
>> textbooks.
>>
>> Yes/No?
>When I give an answer I choose the format.
But I managed to get you to respond to each point in turn rather
than dismiss them all with a single short comment.
>I would be grateful if you could provide references to the textbooks
>you mention.
A reasonable discussion is to be found in "Quantum Computation and
Quantum Information" by Nielson and Chuang. More elementary
discussions are found in the second edition of Gottfried and
in Basdevant and Dalibard. There's a brief introduction to
the subject in wikipedia:
http://en.wikipedia.org/wiki/Quantum_entanglement
>Do they provide an explicit definition of entanglement?
Yes, they all do.
>Do they discuss
>whether it is a property of the wave-function or of the measurement
>process?
No; that's something which only you discuss.
Almost everybody else is in agreement that the Hilbert space of a
composite system is the tensor product of the Hilbert spaces of
the subsystems. A state |psi> of the product space H=H_a x H_b
is called entangled when it cannot be written as a product
|psi>=|psi_a>|Psi_b>. This is a definition of entanglement.
Definitions can't be incorrect. By definition, whether a state
is entangled is determined by the state and the factorisation
H_a x H_b of the Hilbert space.
You might disagree with the idea that the tensor product of
the Hilbert spaces is the right space to use for the
composite system. There are a number of reasons for using
the tensor product. The simplest is to notice that the
space of square-integrable functions on the configuration
space of a composite system is the tensor product of the
spaces of square-integrable functions on the configuration
spaces of the subsystems. Nielson gives a few other reasons,
but includes it as an axiom of quantum mechanics in his
system.
>Based on my experience, entanglement is often introduced
>handwavingly and the conceptual issues involved are ignored. Your usage
>of the entanglement notation in this case leads to conclusions that
>in my opinion are testably erroneous.
The problem is that although you have criticised the standard
procedure, you haven't suggested any alternative. What
mathematical structure do you use, which the tensor product
is an approximation to? Under what conditions is the
tensor product a good approximation?
So far, you have only presented me with phrases such as "Entanglement is
observer-dependent" and "Entanglement is a property of the measurement
process". I cannot use these phrases to do calculations, but I
can certainly use the tensor-product system to do calculations.
>> 2. Your assertion is that quantum mechanics, as presented in
>> textbooks, and the current treatment of entanglement in
>> particular, is incorrect and needs to be modified according
>> to your prescription.
>>
>> Yes/No?
>Again, please provide references to textbooks containing an explicit
>definition of entanglement and a discussion thereof. Otherwise we are
>talking about nothing.
Let's use wikipedia's page about entanglement as a standard
reference because it doesn't involve a trip to the library.
There you'll find the standard talk about tensor products and
the standard definition of entanglement along with links to
pages about EPR and so on. You have claimed many times that
this tensor-product procedure is merely a "rule of thumb",
suggesting that you know of some mathematical procedure
which isn't a rule of thumb. Can you say precisely what that
procedure is?
>> 3. Your proposed modification of the notion of entanglement
>> asserts that, as long as the observer doesn't receive any
>> information from the system, "entanglement is an empty concept
>> and all photons can be summed and subtracted, if you can track their
>> phases." This has the following consequence for the two-slit
>> interference experiment: If a detector is placed over one of the slits,
>> so that it records which slit the particle passed through, *but*
>> the observer doesn't extract information from the detector, then
>> the particles will form an interference pattern. The standard
>> answer - that no interference pattern will be formed because
>> the particle's state is entangled with the detector, is wrong,
>> according to you, because "entanglement is an empty concept".
>>
>> Yes/No?
>Not really. Roughly speaking , interaction with the detector will
>fudge the interference pattern anyways, but the "marks" it leaves are
>just ripples on the wave function as long as they are not interpreted
>by measurement.
Can you say it without speaking roughly?
>> 4. Presuming that you agree with point 1 above, standard quantum
>> mechanics with the usual notion of entanglement would predict
>> that your experiment can not detect superpositions. Your modification
>> to the notion of entanglement must be made in order to deduce that
>> the experiment can detect superpositions.
>>
>> Yes/No?
>Again, as far as entanglement is concerned , I am not sure what you
>mean by "Standard quantum mechanics",unless you provide references. As
>I said previously, entanglement is widely misunderstood.
>Misunderstandings breed mistakes.
By standard quantum mechanics I mean quantum mechanics in which
the tensor product is used to represent states of the composite
system. If I use this, I find that the probability of the
detector clicking is 1/8, and not 3/8. That is, I find that
your experiment cannot detect superpositions. My question
was: Do you agree that if I use the tensor-product system
and treat the photon as being entangled (in the tensor-product
sense) with the absorber, then the calculated probability
of the detector clicking is 1/8?
More specifically, is it necessary, in order to deduce that
your experiment can detect superpositions, to reject the
tensor-product procedure for composite systems and replace
it with another procedure? What is that other procedure?
>> >> ... essentially it comes down to your claim in quant-ph/0007117
>> >> that it is possible to distinguish between cases (a) and (b) above.
>> >>
>> >> As I said, I believe there's a calculational error in that paper and
>> >> that in both cases the detector clicks 1/8 of the time, so it
>> >> can't be used to detect superpositions.
>>
>> >I am confident that the calculations are correct.
>>
>> Confidence can be measured by wager.
>> I think it's very unlikely
>> that you would be able to get a paper published in, say,
>> Foundations of Physics, in which you present this experiment
>> and claim that it can detect macroscopic superpositions. How
>> much would you be prepared to bet?
>I bet only for fun and on outcomes that I regard as intrinsically
>interesting.
Oh, go on. I'll give you great odds: If you can get a paper
published in Found. Phys. within a year, claiming that this
experiment can detect macroscopic superpositions, I'll give
you EUR 1,000. If you can't, you give me EUR 100. I think
that would be fun; don't you?
>> >In order to
>> >facilitate verification however, I will provide a rationale for the
>> >result.
>> >The key point is very simple. The shutter on the upper photon induces a
>> >random phase change, which destroys any interference pattern between
>> >the upper and the lower photon. This is quite intutive and is actually
>> >proven at the bottom of page 5 of [1].
>>
>> It's certainly not intuitive for me.
>> I can't find a proof of it
>> at the bottom of page 5. I just find assertions that it's true.
>> I would go so far as to say that it's not true, and that you
>> have made a mistake here. Can you present a proof of your assertion
>> using equations?
>It's the calculation of Pb(D) at the bottom of page 5, where the
>photon's probability at the detector is obtained by averaging the
>phases over [0, 2pi], where they are equidistributed,.
Sorry; I didn't make myself clear. What I can't find a proof of
is the assertion that "The shutter on the upper photon induces a
random phase change". I see no reason why this should be true.
In fact, since you rely on this (unproved) statement to reach
the conclusion that you can do something which is impossible
(as I explain below, again), I claim that this statement
is actually false. That is, I claim that the photon doesn't
simply undergo a random phase change when passing by the
absorber. The proof that I was asking for is a proof of
the statement that the photon undergoes a random phase
change when passing by the absorber.
>> >At the upper beam-splitter you
>> >have an incoming upper photon with probability 1/4 and and incoming
>> >lower photon with probability 1/2. Since they do not interfere, after
>> >being halved passing through the beam splitter , they will yield a
>> >1/4+1/8=3/8 probabilty at the detector. In the non-superposed shutters
>> >situation the phases stay put and you get cancellation at the upper
>> >beam-splitter, so the result is 1/8.
>>
>> This relies on the assertion that the photon undergoes a random phase
>> change when passing by the absorber.
>Yes. More in general, on the assetion that interaction with the
>absorber will destroy phase coherence with the other instances of the
>photon.
So if it turns out that the photon doesn't undergo a random phase
change, then you will agree that this experiment can't detect
superpositions?
>> >Your calculation is based on an interpretation of entanglement that I
>> >consider erroneous, although it may provide a handy rule of thumb in
>> >trivial situations.
What's the alternative?
>> >I have tackled the issue of entanglement in my post [2] (cf. [3],[4]).
>> >Indeed the whole procedure I am proposing depends on the fact that
>> >entanglement is an observer-dependent property of the measurement
>> >process, not of the wave function, i.e. as long as evolution is
>> >unitary, entanglement has no object.
>> >In the setting we are discussing that implies that prior to measurement
>> >all photons can be summed and subtracted, as long as their phases can
>> >be tracked.
>>
>> I can't find anything in [2] or [3] which actually presents any
>> concrete mathematical alternative to entanglement, while [4]
>> seems to merely introduce a notion of "generalized entanglement",
>> without actually claiming that any of the standard uses of
>> entanglement give incorrect answers, which is what you are claiming.
>Standard uses ... missing references ....common misconceptions ... .
>I think that what I have written about entanglement is sufficient to
>motivate my argument in the current setting.
Perhaps you have written something that I haven't seen where
you introduce a mathematical formalism and demonstrate that
as some parameter becomes very small your formalism is
approximated by the tensor-product system. Have you?
>> >> Now, from Alice's point of view, if she measures along the x-axis,
>> >> then, since she adopts the Copenhagen interpretation (which gives
>> >> the same experimental predictions as anybody's favourite interpretation,
>> >> and which she is entitled to use), she considers that Bob is receiving
>> >> particles which are _either_ |up_x> or |down_x> and not a superposition
>> >> of the two. Hence, if she measures along the x-axis, then she can
>> >> say with certainty: "Bob's results, whatever they are, will be
>> >> consistent with the results that he would get if he was receiving
>> >> a beam of particles, each of which is in either |up_x> or |down_x>.
>> >> That is, Bob's beam absorber is _either_ in the state |SA> or
>> >> in the state |SB>."
>> >>
>> >> On the other hand, if Alice measures along the z-axis, then
>> >> Bob (as far as Alice is concerned) is receiving particles
>> >> with a well-defined value of z-spin. That is, Bob will receive
>> >> particles whose states are either |up_x>+|down_x> or
>> >> |up_x>-|down_x>. Consequently, Bob's absorber will end up
>> >> in the states |SA>+|SB> or |SA>-|SB>.
>> >>
>> >> Notice the crucial point - Alice can *control* whether Bob's
>> >> absorber is in a superposition or not.
>>
>> >This is meaningless.
>> >In your setting Alice cannot control whether the system is in a
>> >superposition or not in Bob's perspective.
>>
>> She doesn't have to control Bob's perspective. My point is
>> that after she measures the state of her particle, she assigns
>> a definite quantum state to Bob's particle.
>When she measures the state of her particle she splits , in Bob's
>perspective, into different instances measuring different states.
>It should be clear that in an epistemic model there can be no such
>thing as a "definite quantum state". Any state is relative to an
>observer.
Fine, but Alice is using quantum mechanics, and her predictions
will be correct. Bob's perspective is irrelevant. Bob's only
role is to perform the experiment on the incoming particle.
He could be replaced by a machine, or a well-trained chimpanzee.
Bob makes no predictions so he doesn't need to know about
quantum mechanics and doesn't need to have a perspective.
>> When Bob shows her
>> his results long after the experiment is over, his results
>> will be consistent with the state that she assigned.
>When she assigned, she split. Is that clear?
No; that's a pseudo-religious assertion about the existence of
parallel worlds, and it's irrelevant because it doesn't
have any bearing on the experimental results that will be obtained.
The point is that Alice can make a measurement on her particle
and then assign a state to Bob's particle. You responded to
this by asserting that Bob is a many-worlds enthusiast who has
daydreams about Alice splitting into parallel worlds. Well,
Bob may very well be such a person, but fortunately it
doesn't matter because he doesn't have to be competent and
doesn't have to do any calculations.
It is only Alice who assigns a quantum state. She measures
the spin of her particle along the z-axis (for example),
and finds (say) spin up. Then she assigns the quantum state
|z_down> to Bob's particle.
You may very well interrupt here and talk about what Bob
is thinking about, but it doesn't matter what Bob is
thinking about because Bob doesn't do any calculations. The fact
that Bob might be somebody who likes to think about parallel
worlds is irrelevant. Alice assigns a quantum state to
Bob's particle and Alice makes a prediction (for example,
that whatever experiment is performed on Bob's particle,
the result of the experiment will be consistent with
the state of the particle being |z_down>). Alice's
prediction will be correct, or quantum mechanics fails.
>> If she
>> measures her particle along the z-axis, she will say "Bob's
>> particle has a definite z-spin".
>>
>> When she meets Bob, and sees what he did with that particle,
>> she will say "Oh yes, that particle - it had a definite
>> spin along the z-axis. I know that because I measured the
>> z-spin of its partner particle." Bob's results will be
>> consistent with his particle having a definite z-spin,
>> or else quantum mechanics fails. If quantum mechanics
>> doesn't fail, then Bob's results *must* be consistent with the
>> quantum state that Alice assigns to Bob's particle. This
>> is because Alice is using quantum mechanics, and if quantum
>> mechanics doesn't fail, then Alice's predictions must be
>> correct.
>>
>> Similarly, Alice can measure the spin of her particle along
>> the x-axis. Then she will say that Bob's particle has a
>> definite value of spin along the x-axis - that is, that
>> it is in a superposition of z-spin states. Bob's results
>> will then be consistent with his particle having a definite
>> value of spin along the x-axis.
>Do you realise that when you use the word "can" you are referring to
>Alice's free will?
Yes, I do. It is because Alice has free will that she can signal
faster than light. If she didn't have free will then one could
hypothesise that some event in her past light cone determined
both her choice and Bob's result, and caused the illusion of
faster than light communication when only subluminal effects
were involved.
>And that such a process will put her in a
>superposition in Bob's perspective?
You are focussing on Bob as though his perspective were somehow
relevant. It is not. Bob can be replaced by a machine. You
seem to be suggesting that if we pay enough attention to
Bob and his perspective and talk about Alice splitting
for long enough, then Alice's predictions will become incorrect.
They will not.
>> If there is any possible experiment that Bob can do
>> to detect whether his particle has a well-defined z-spin,
>> rather than a well-defined x-spin, then Alice can signal
>> to him faster than light. From Alice's point of view,
>> Alice can control whether Bob's particle has a well-defined
>> z-spin or a well-defined x-spin.
>No, she can't. You neglect the fact that Bob's input in my experiment
>is determined all the instances of Alice, not just the one he will end
>up talking to.
Imagine the following simpler experiment. Imagine that Alice
prepares a particle in a specific spin state and gives the
particle to Bob. She prepares the particle by taking a
large collection of particles and measuring the spin of
each one until she finds one with the right spin, and then
gives that particle to Bob. Suppose she decides that she
will give Bob a particle in the state |z_down>.
Now I claim that in this simpler case, Alice can control the
spin state of Bob's particle. Let me know if you disagree with
this.
Supposing for the moment that you don't disagree, let's make
it a little more complicated by changing the procedure which
Alice uses to control the state of Bob's particle. Suppose
instead that she takes a collection of pairs of particles,
where each pair is an EPR-style pair, so that each particle
will be found to have opposite spin to its partner particle.
She can measure one particle from each pair, say along the
z-axis. If she finds a particle which has spin up along
the z-axis, she assigns the state |z_down> to the partner
particle, and gives the partner particle to Bob.
It is my claim, which I take to be uncontroversial, that
in this case as well, Alice can control the spin state
of the particle which she gives to Bob. Let me know if
you think that this is wrong.
In these cases above, Alice could say with confidence
that whatever experiment Bob does with his particle, the
results of the experiment will be consistent with
the initial state of the particle being |z_down>.
Now let's make it a little more complicated again, by
supposing that Alice doesn't get to choose which
particle gets given to Bob. Say, for example, that there's
only one pair, and Alice measures the spin of one of
the particles (along the z-axis) and gives the other
one to Bob. Alice will get either spin up or spin down
for her particle. If she finds that her particle is
spin up, then she will assign the state |z_down> to
the Bob's particle. If she finds that her particle
is spin down, then she will assign the state |z_up>
to Bob's particle. Please let me know if you disagree
with this.
If you don't, then we notice that in this last case,
Alice can't control the spin state of Bob's particle.
However, she does assign a state to Bob's particle,
either |z_up> or |z_down>. If she assigns the state
|z_up>, then she can predict with confidence that
whatever experiment Bob does, the result will be
consistent with the initial state of the particle being |z_up>.
If she assigns the state |z_down>, then she can predict with
confidence that whatever experiment Bob does, the result will be
consistent with the initial state of the particle being |z_down>.
Please let me know if you disagree with this, bearing
in mind that Bob might be a machine.
If you do, then we notice that, no matter what result
Alice gets when she measures the z-spin of her particle, Alice
can say the following with certainty: "Whatever experiment
Bob does, the result will be consistent with the initial
state of Bob's particle being either |z_up> OR |z_down>."
If you agree with that, then notice that, if Alice had
measured her particle along the x-axis instead of the z-axis,
then she would say: "Whatever experiment Bob does, the result will
be consistent with the initial state of Bob's particle being either
|x_up> OR |x_down>."
Now Alice, with her free will, can choose whether to measure
along the x-axis or the z-axis. Hence she can choose whether
(in her perspective) Bob's particle has a definite x-spin
or a definite z-spin.
If there was a way for Bob to distinguish between a collection of
particles which have definite (but random) x-spins, and a
collection with definite but random z-spins, then Alice
could measure her particles along the z-axis (say) and then
assign definite z-spin states to Bob's particles, and
assert that if Bob does the experiment, his results
will be consistent with the particles having definite
values of z-spin (and not x-spin).
However, if Bob could do such an experiment, then he could receive
a signal from Alice. He does not need to do any calculations or
write a wavefunction which encodes Alice and her behaviour. All Bob
has to do is concern himself with the question of whether the
particles that arrive have a definite z-spin or a definite x-spin.
He does not need to concern himself with the question of where they
came from or whether there are parallel worlds and parallel Alices.
If there is an experiment which can distinguish between definite
x-spin particles and definite z-spin particles, he just has to do
that experiment and nothing else.
If your experiment worked, it would be such an experiment,
and would hence allow signalling faster than light.
>> Your interferometry experiment, if it worked, would allow
>> Bob to tell whether the incoming particle had a well-defined
>> z-spin or a well-defined x-spin.
>Your conclusion is erroneous for the same reasons that I explained in
>my previous post ([3]).
In that post, you said things like:
>The key point here is that using a Geiger counter, tossing a coin or
>resorting to one's free will have exactly the same effect on the
>procedure, i.e. they will create a superposition in the perspective of
>an external observer. From the point of view of an external observer
>impredictability and free will are indistinguishable/identical. When
>Alice decides which bit to send next, she is acting as a Geiger
>counter, creating uncertainty and therefore a superposition of herself
>in Bob's perspective.
Basically, what you said was that Bob thinks that Alice is in
a superposition. This may or may not be true, but it's certainly
irrelevant. In the end what I'm saying is that Alice can use
quantum mechanics to make predictions and her predictions will
be correct. You keep responding by saying "No, because
Bob thinks that Alice is in a superposition."
R.
.
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