B Does the EPR experiment imply QM is incomplete?

[vanhees71]

Only if you add a collapse postulate. In the MWI there isn't only one outcome; all outcomes occur. And the MWI is perfectly consistent with decoherence--in fact, I believe decoherence was originally developed in order to explain why the different branches in the MWI don't interfere.

There is no collapse. It's decoherence, which is very hard to avoid to the dismay of quantum-computer engineers.

 

[stevendaryl]

Quantum mechanics in its minimal interpretation (usually physicists working on physics and not philosophy follow the "shutup and calculation interpretation", which is just the minimal interpretation adding ignorance of unnecessary philosophical quibbles irrelevant to physics) is applied with great success to all kinds of mesoscopic and macroscopic systems. Condensed-matter physics is all about this. There is no contradiction between QT and experiment nor is there any inconsistency within QT in its application to macroscopic systems. To the contrary, condensed-matter physics is one of the success stories of QT with evergrowing numbers of scientists involved in it and leading even to applications in everyday life like the laptop I'm hacking this posting right now.

I don't have any disagreement with any of that. The issue is not its usefulness but its completeness and consistency. Yes, ad hoc procedures can deal with inconsistencies.

The inconsistency is that the claim that a measurement device always returns an eigenvalue of the observable being measured contradicts the claim that that measurement device obeys quantum mechanics. The resolution is to have rules of thumb for when you treat the measurement device as a measuring device and when you treat it as a quantum-mechanical system. You have rules of thumb for resolving the contradictions.

 

[vanhees71]

Where are, in your opinion, inconsistencies?

 

[stevendaryl]

Where are, in your opinion, inconsistencies?

The Born rule predicts that after a measurement, a measuring device is in a definite pointer state. Unitary evolution predicts that it's not.

 

[stevendaryl]

The Born rule predicts that after a measurement, a measuring device is in a definite pointer state. Unitary evolution predicts that it's not.

In the particularly simple case of measuring the spin of a Fermion, let's assume that the state of the device + environment + whatever can be described using a basis: $|P,j\rangle$, where $P$ is the "pointer state" variable, which can take on values "U" (measured spin-up) or "D" (measured spin-down) or "0" (no measurement yet), and where $j$ describes all other variables that we're not interested in.

If we assume that in the initial state, the Fermion was in the superposition $\alpha|u\rangle |u\rangle + \beta |d\rangle$, and that the device had pointer state $0$, then at a later time, the state of the composite system will be described by a density matrix of the form:

$\rho_{jkk'uu} |U, k\rangle \langle U, k'\rangle + \rho_{jkk'ud} |U, k\rangle \langle D, k'\rangle + \rho_{jkk'du} |D, k\rangle \langle U, k'\rangle + \rho_{jkk'ud} |U, k\rangle \langle D, k'\rangle$

In contrast, the Born rule would predict that the coefficients for the "cross" terms, $\rho_{jkk'ud}$ and $\rho_{jkk'du}$, would be zero.

Those are different predictions. It's not a matter of interpretation. Those are different states.

 

[PeterDonis]

There is no collapse.

Then there isn't one outcome. So your statement in post #190, which I quoted...

it explains, why there's only one outcome.

...doesn't make sense if there is no collapse. So which is it?

 

[zonde]

Well, the difference between a macroscopic object, such as a measuring device, and a microscopic object, such as an electron, is that for any given macroscopic state (what we would intuitively, pre-quantum mechanics, think of a state, such as "the readout shows the number 32" or "the pointer points to the left" or "the left light is on") there are many, many microscopic states that correspond to it.

I don't have the mathematical sophistication to accurately describe the situation using quantum mechanics, but perhaps it's something like the following:

The complete system perhapse can be described by three variables: $|s, S, j\rangle$, where $s$ is the observable corresponding to the system being measured (an electron's spin, maybe), $S$ is the corresponding value of the "pointer variable", and $j$ represents all the other degrees of freedom.

To make it both simple and definite, we will assume that there are two possible values for $s$:$u$ and $d$, and three possible values for $S$: $0, U, D$. There are many (possibly infinitely many) values for the other degrees of freedom, $j$.

To say that the pointer accurately measures the z-component of spin is to say something like the following:

• If you start in the state $|u, 0, j\rangle$, and you allow the system to evolve, then you will end up most likely in a superposition of the form
  • $\sum_k c_{ujk} |u, U, k\rangle$
• If you start in the state $|d, 0, j\rangle$, and you allow the system to evolve, then you will end up most likely in a superposition of the form
  • $\sum_k c_{djk} |d, D, k\rangle$
• It follows from the linearity of the evolution operator that if you start in a superposition of the form $\alpha |u, 0, j\rangle + \beta |d, 0, j\rangle$, then you will end up in a superposition of the form $\alpha \sum_k c_{ujk} |u, U, k\rangle + \beta \sum_k c_{djk} |d, D, k\rangle$

By "end up", I mean applying the evolution operator $e^{-iHt}$.

You say: "It follows from the linearity of the evolution operator". But how can I check that evolution operator is indeed linear? I suppose that for evolution operator to be linear Hamiltonian should be linear as well, right? But is this the case when we have configuration that can evolve by avalanche type process? Total potential energy for system of several particles interacting by coulomb potentials would have cross terms. Is this not a problem if we assume that Hamiltonian is linear?

 

[stevendaryl]

You say: "It follows from the linearity of the evolution operator". But how can I check that evolution operator is indeed linear?

That's what QM says about it.

I suppose that for evolution operator to be linear Hamiltonian should be linear as well, right? But is this the case when we have configuration that can evolve by avalanche type process? Total potential energy for system of several particles interacting by coulomb potentials would have cross terms. Is this not a problem if we assume that Hamiltonian is linear?

Saying that evolution is linear means that if $\psi_1(x,t)$ and $\psi_2(x,t)$ are solutions to Schrodinger's equation, then so is a linear combination:

$\alpha \psi_1(x,t) + \beta \psi_2(x,t)$

Quantum mechanics assumes that evolution is linear.

An example of a nonlinear theory might be something like this:

$\frac{-\hbar^2}{2m} \frac{d^2 \Psi}{dx^2} + V(x) \Psi + (\Psi)^3 = i \hbar \frac{d\Psi}{dt}$

The presence of second or higher powers of $\Psi$ would make it nonlinear.

Actually, field theory does consider differential equations that look like that, but in those cases, $\Psi$ is not the wave function, it's a field operator. There is something corresponding to the wave function in quantum field theory, and its evolution is still linear.

 

[vanhees71]

Then there isn't one outcome. So your statement in post #190, which I quoted...
...doesn't make sense if there is no collapse. So which is it?

There's one outcome also in the minimal interpretation. That there is a unique outcome of a measurement is part of the definition of the word "measurement" here. If the measurement doesn't deliver one outcome (including a systematic and statistical error estimate) it's not a valid measurement by definition, and any experimentalist's paper sending such a result to a serious science journal will be rejected by peer review.

 

[Mentz114]

There's one outcome also in the minimal interpretation. That there is a unique outcome of a measurement is part of the definition of the word "measurement" here. If the measurement doesn't deliver one outcome (including a systematic and statistical error estimate) it's not a valid measurement by definition, and any experimentalist's paper sending such a result to a serious science journal will be rejected by peer review.

Agree. The definition of 'measurement' is particularly important. We could rewrite the Born rule as 'a successful measurement is one where the apparatus indicates correctly the eigenstate of the system'. It would save some confusion in my opinion.

 

[vanhees71]

An apparatus never "indicates the eigenstate of the system". I don't even know what "the eigenstate of the system" means. What a good measurement apparatus does is precisely what its name suggests, it measures the values of observables defined by (an equivalence class of) measurement procedures. In the same sense a quantum state is (an equivalence class) of preparation procedures.

 

[Mentz114]

An apparatus never "indicates the eigenstate of the system". I don't even know what "the eigenstate of the system" means. What a good measurement apparatus does is precisely what its name suggests, it measures the values of observables defined by (an equivalence class of) measurement procedures. In the same sense a quantum state is (an equivalence class) of preparation procedures.

The system I refer to is the one being measured. Clearly I misunderstand everything about the Born rule and measurement. For instance what is written here https://en.wikipedia.org/wiki/Born_rule

The Born rule states that if an observable corresponding to a Hermitian operator A {\displaystyle A}

with discrete spectrum is measured in a system with normalized wave function | ψ ⟩ {\displaystyle |\psi \rangle }

(see bra–ket notation), then

• the measured result will be one of the eigenvalues λ {\displaystyle \lambda }
  
  of A {\displaystyle A}
  
  , and
• the probability of measuring a given eigenvalue λ i {\displaystyle \lambda _{i}}
  
  will equal ⟨ ψ | P i | ψ ⟩ {\displaystyle \langle \psi |P_{i}|\psi \rangle }
  
  , where P i {\displaystyle P_{i}}
  
  is the projection onto the eigenspace of A {\displaystyle A}
  
  corresponding to λ i {\displaystyle \lambda _{i}}
  
  .

 

[vanhees71]

Well, Wikipedia is not very accurate here... There's no "king's path" to understanding quantum physics, at least it's not provided by Wikipdia. You have to study the real thing as written in good textbooks like Dirac's.

 

[stevendaryl]

Well, Wikipedia is not very accurate here...

What is inaccurate about it?

 

[atyy]

Well, Wikipedia is not very accurate here... There's no "king's path" to understanding quantum physics, at least it's not provided by Wikipdia. You have to study the real thing as written in good textbooks like Dirac's.

Dirac's book has collapse.

 

[vanhees71]

Dirac's book first of all has a very good exposition of the formalism in Dirac's own bra-ket formulation. The collapse is not overemphasized, and of course, to read about interpretation (which you shouldn't be too worried about as a beginner anyway, because it's a side subject for specialists; the real success of QT as a physical theory simply needs the minimal statistical interpretation and not philosophical details which should be postponed to be studied by the interested student after s/he has mastered the math, because you cannot talk about QT without the math) should be studied from more modern textbooks. E.g., for the minimal statistical interpretation Ballentine, for Bohmian mechanics the writings of Dürr, consistent histories by Griffiths or Omnes. I've no clue of which value the socalled many-world interpretation should be, but I guess there Deutsch is the main proponent with the best books.

 

[vanhees71]

About Wikipedia

https://en.wikipedia.org/wiki/Born_rule

What is inaccurate about it?

Is this a joke? You'd have to rewrite the entire article from the beginning to the end to answer this question. To answer the opposite question is simple: What's accurate about it? Nothing!

 

[Mentz114]

About Wikipedia

https://en.wikipedia.org/wiki/Born_ruleIs this a joke? You'd have to rewrite the entire article from the beginning to the end to answer this question. To answer the opposite question is simple: What's accurate about it? Nothing!

From Jochen Rau, Statistical Physics and Thermodynamics (Oxford, 2017), page 10

A measureable physical quantity -an observable - is represented by a Hermitian operator. When measured the result will be one of the eigenvalues of the operator.

Maybe that is an oversimplification but it is what Wiki and most textbooks say.

 

[stevendaryl]

Is this a joke? You'd have to rewrite the entire article from the beginning to the end to answer this question. To answer the opposite question is simple: What's accurate about it? Nothing!

It summarizes the Born rule in about the same way that I've always heard it summarized by every physicist that's ever tried to explain it. So I have no idea what you are talking about.

The main points, which I think the Wikipedia gets across, are (1) a measurement of an observable produces an eigenvalue of the corresponding operator, and (2) the probability of getting an eigenvalue is the square of the corresponding amplitude (which Wikipedia gives in terms of projection operators).

That's almost the same as I would have described it. If it's completely inaccurate, then I don't see it. You need to spell it out.

 

[stevendaryl]

Googling for "Born Rule" gives as the first result after Wikipedia this article: http://www.math.ru.nl/~landsman/Born.pdf

It describes the rule this way:

Let $\hat{a}$ be a quantum-mechanical observable, mathematically represented by a self-adjoint operator on a Hilbert space $H$ with inner product denoted by $( , )$. For the simplest formulation of the Born rule, assume that $\hat{a}$ has non-degenerate discrete spectrum: this means that $\hat{a}$ has an orthonormal basis of eigenvectors $(e_i)$ with corresponding eigenvalues $\lambda_i$ , i.e. $\hat{a}\ e_i = λ_i\ e_i$ . A fundamental assumption underlying the Born rule is that a measurement of the observable a will produce one of its eigenvalues $\lambda_i$ as a result. In what follows, $\Psi \in H$ is a unit vector and hence a (pure) state in the usual sense. Then the Born rule states: If the system is in a sate $\Psi$, then the probability $P(a = \lambda_i | \Psi)$ that the eigenvalue $\lambda_i$ of $\hat{a}$ is found when $\hat{a}$ is measured is $P(a = \lambda_i | \Psi) = |(e_i , \Psi)|^ 2$.

I would say that that's substantially the same as what Wikipedia says. Wikipedia uses the projection operator $P_i$ defined to be $|e_i\rangle \langle e_i|$, but that's equivalent.

 

[Mentz114]

@stevendaryl .
The point I was leading up to is about your 'soft contradiction'. It seems to me that (1) something is prepared in a superposition (2) it interacts with the apparatus to form a superposition with it (as I think you spelt out) (3) something happens (X) and the system and the apparatus are left in the same (or highly correlated) state which indicates one of the members of the superposition.

Isn't this just a (oldish) statement of the measurement paradox ? We don't know what X is but it has to be non-unitary. I don't see a contradiction anymore because the two options (in your contradiction) are not mutually exclusive but different stages in a process.

 

[stevendaryl]

Isn't this just a (oldish) statement of the measurement paradox ? We don't know what X is but it has to be non-unitary.

Right. If you allow for something non-unitary to be happening during a measurement, then there is no paradox. But unless that something is spelled out, then the formalism is not complete, because part of the dynamics (the nonunitary part) is not spelled out.

I don't see a contradiction anymore because the two options (in your contradiction) are not mutually exclusive but different stages in a process.

I wouldn't call them stages. If you analyze the measurement process from the point of view of the Born rule, then you get a different answer than if you analyze the same process considering it a quantum mechanical interaction like any other. One or the other analysis has to be wrong.

If the analysis that uses unitary evolution applied to the measurement device (and environment) is the one that's wrong, then to me, it shows that unitary evolution is incorrect, and is only approximately true (it's only true in the limit of small systems).

 

[Mentz114]

... (allow) something non-unitary to be happening during a measurement...

Surely this is tautological. Can you call something that is unitary a measurement ? My point is that measurement is always non-unitary and a untary interaction is not a measurement.

I think you've always asserted that measurement is not an ordinary (unitary ?) interaction. I agree but I don't see it as a problem.

I wouldn't call them stages. If you analyze the measurement process from the point of view of the Born rule, then you get a different answer than if you analyze the same process considering it a quantum mechanical interaction like any other. One or the other analysis has to be wrong.

What other point of view is there ? Projective measurement cannot be analysed otherwise, can it ? You probably will say evolution, but evolution is a process and the Born rule is a constraint which has an implicit definition of 'measurement'.

If the analysis that uses unitary evolution applied to the measurement device (and environment) is the one that's wrong, then to me, it shows that unitary evolution is incorrect, and is only approximately true (it's only true in the limit of small systems).

It is incorrect for dissipative processes. Even if only one photon escapes to infinity unitarity is gone.

If this looks like hair-splitting I apologise. I broadly agree with your points.

 

[stevendaryl]

Surely this is tautological. Can you call something that is unitary a measurement ?

I suppose you could define a measurement in such a way that it must be non-unitary, but then it's an open question whether a measurement is possible. According to both the Bohm and Many-Worlds interpretations, evolution is always unitary.

 

[stevendaryl]

I suppose you could define a measurement in such a way that it must be non-unitary, but then it's an open question whether a measurement is possible. According to both the Bohm and Many-Worlds interpretations, evolution is always unitary.

It's sort of like (and I'm not sure whether this analogy is deep, or not) defining a measurement in classical mechanics so that only irreversible changes can be measurements. Then presumably you could prove from Newton's laws that no measurements are possible.

 

[vanhees71]

That's already way better than Wikipedia, and that's how you indeed start to explain it when you start introducing QT, but it's not the final word. First of all, indeed the operators representing observables in QT have to be self-adjoint; Hermitean is not enough to guarantee the consistency of Born's rule. Another important point is that not the Hilbert-space vectors represent (pure) states but rays in Hilbert space (or more conveniently and equivalently by projection operators $|\Psi \rangle \langle \Psi|$, with $|\Psi \rangle$ normalized. Of course, the special case of entirely non-degenerate spectra to degenerate ones is also an important point but easily generalized.