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## What does the probabilistic interpretation of QM claim?

I collect here info from another thread, to have a more focussed discussion.
 Quote by A. Neumaier The probability interpretation says that _if_ you can set up an experiment that measures a self-adjoint operator for a system in state psi then the probability of observing the k-th eigenvalue is psi^* P_k psi, where P_k is the projector to the k-th eigenspace. It says _nothing_ at all about which particular operators are observable in this sense. Everything beyond that is interpretation, and hence (at the current state of affairs) a matter of philosophy. In particular, which operators can be measured is not part of the probability interpretation but a matter of theoretical and experimental developments. Regarding what is arbitrarily _precisely_ measurable, there is a no go theorem by Wigner (I can give references if you want to check that) that states that _only_ quantities commuting with all additive conserved quantities are precisely measurable. The position operator is not among these. Nobody comparing QM with experiments is making use of this particular assumption.
''this particular assumption'' refers to the assumption that |psi(x_1,...,x_n|^2 is the probability density of observing simultaneously particle k at position x_k (k=1:N).
 Quote by A. Neumaier It is stated in the beginning as an interpretation aid without proof, and never taken up again in the context of real measurements where the claim would have to be justified. It is very common to make this sort of idealized assumption to get started; but once the formalism is established, this assumption is never used again. For example, Landau & Lifshitz begin in Section 2 of their Vol. 3 with such a statement, but immediately replace it in (2.1) and (3.10) by the more correct version about the interpretation of the expectation value = Psi^* K Psi, where K is an arbitrary observable (linear integral operator) depending on the form and values of the measurement. From then on, only the latter interpretation is used; never the fictitious, idealized introductory remark. And it cannot be different, since quantum mechanics is used in many situations where the state vectors used in the formalism have no interpretation as a function of position - the whole of quantum information theory and the whole of quantum optics belonging to this category.
 Quote by Avodyne For nonrelativistic particles, absolutely everyone comparing QM with experiments does make use of the wave-mechanics interpretation of $|\psi(x)|^2$ as a probability density.
Please show me a comparison with experiment that does this.

Nonrelativistic particles have no different interpretation than relativistic ones.

Particle detectors respond to the momentum of a particle, not to its position.
Scattering experiments are interpreted in the momentum picture. Nobody is interested in the position of particle tracks, only in their momentum (which tells about masses).
 Quote by meopemuk Yes, this is true. QM does not talk about the specifics of observations and measuring devices. For example, P_k can be a projection on the k-th eigenvalue of the position operator.
It could be this _only_ if you can prepare an experiment that realizes such a P_k. But this is a matter of experimental technique and not one about the interpretation of quantum mechanics. But there are no such operators since the spectrum of position is continuous.
 Quote by meopemuk Yes, this is true. QM does not talk about the specifics of observations and measuring devices. For example, P_k can be a projection on the k-th eigenvalue of the position operator. Then psi^* P_k psi is the probability (density) for measuring position k in the state described by psi. QM tacitly assumes that some ideal precise measuring device can be constructed, which does exactly that
No. This unrealistic assumption is needed _only_when one wants to insists on a probability density interpretation of |psi(x)|^2. And for the position representation of an N-particle state, one would need an even more ideal precise measuring device that can measure the simultaneous presence of N particles in N different, arbitrarily small regions
covering the size of an uranium atom (N=92), say.

This is ridiculous - such measurement devices are impossible!

Whereas the form in which I stated the probability interpretation assumes nothing. it makes claims only for those projectors that are actually realizable. it is therefore much more realistic.

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 there are no such operators since the spectrum of position is continuous.
Quantum uncertaqinty rules out continuity in position.

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 Quote by _PJ_ Quantum uncertainty rules out continuity in position.
Wigner's theorem (p.298 in: Wheeler & Zurek, Quantum theory and measurement, Princeton 1983) even rules out projective measurements of a particle being in a given region, since the corresponding projector does not commute with all additive conserved quantities.

## What does the probabilistic interpretation of QM claim?

 Quote by A. Neumaier Particle detectors respond to the momentum of a particle, not to its position. Scattering experiments are interpreted in the momentum picture. Nobody is interested in the position of particle tracks, only in their momentum (which tells about masses).
Would you then agree that $$|\psi(p)|^2$$ is the probability (density) of finding the particle with momentum p? Of course, $$\psi(p)$$ is the wave function in the momentum representation.

Eugene.

 Quote by Avodyne For nonrelativistic particles, absolutely everyone comparing QM with experiments does make use of the wave-mechanics interpretation of as a probability density.
 Quote by A. Neumaier Please show me a comparison with experiment that does this.
The double-slit experiment is a good example. See Feynman's Lectures on physics.

Eugene.

 Recognitions: Science Advisor I also reproduce what I wrote on this point in the other thread: Experimenters have been recording particle tracks in position space with cloud chambers, bubble chambers, spark chambers, and drift chambers for many decades. The experiments are typically done in a strong magnetic field, which allows for measuring the momentum of charged particles by measuring the curvature of a track in position space. Modern experiments also have calorimeters at the boundaries of detectors that measure energy deposited; this does give a direct measurement of a particle's energy, but not its momentum. For some recent pictures of particle tracks in position space from the LHC see http://public.web.cern.ch/press/pres.../PR15.10E.html
 I certainly believe that the superposition principle even applies when you can consider a quantum object in a definite state (Quantum Enigma, pg. 191; Entanglement, pg. 79; Absolutely Small, 371-372). So if it is in a definite state (this is obviously determined through what you see) then you can also consider all the other states it COULD be in as potentials (as you cannot see those other states). Of course, we can only talk about what we see. Perhaps those potentials ARE really real too, we just can't see them? But such questions are pointless as what we work with is from what we see. The Mental Universe -> http://henry.pha.jhu.edu/The.mental.universe.pdf
 When the particles in these chambers pass the medium, they ionise their surroundings. This interaction itself causes a collapse of the wafefunction. The resultant ionised particles of media are what is detected, the momentum can be calculated as can the energy of the particle, but since there is delay between the initial ionisation and the detectoin of the ionised media particles, the initial particle is no longer in that state and cannot be correlated with the measurement of its energy (velocity).

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 Quote by meopemuk Would you then agree that $$|\psi(p)|^2$$ is the probability (density) of finding the particle with momentum p? Of course, $$\psi(p)$$ is the wave function in the momentum representation.
Definitely yes! Momentum is measurable, and (in direction, often also in magnitude) measured in every scattering experiment. And this doesn't contradict Wigner's theorem since momentum is one of the additively conserved quantities.

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 Quote by meopemuk The double-slit experiment is a good example. See Feynman's Lectures on physics.
It cannot, since photons (usually used in the double slit experiment) have no Schroedinger position picture.

If it doesn't in the most important case, there is no reason to believe it should in other cases.

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 Quote by Avodyne I also reproduce what I wrote on this point in the other thread:
 Quote by Avodyne Experimenters have been recording particle tracks in position space with cloud chambers, bubble chambers, spark chambers, and drift chambers for many decades. The experiments are typically done in a strong magnetic field, which allows for measuring the momentum of charged particles by measuring the curvature of a track in position space.
Yes; one measures the momentum by finding the curvature of the track. Nobody is interested in the position of the particle. Just as the photon in a double slit experiment is a spherical wave which leaves a detector click at a random detector position, so
the particle is a spherical wave which leaves a track in the detector (according to Mott's famous analysis) in a random initial direction - the scattering direction and the particle energy (which together give the momentum) form the observables, not the position.

 Recognitions: Homework Help Science Advisor I don't think its really correct to say that no one is interested in particle tracks. Although I'm not a particle experimentalist, I certainly think they're interesting to look at! More seriously, detailed reconstruction of particle tracks is important for determining interaction vertices. For example, interesting unstable particles may decay after travelling some macroscopic distance, thus leading to an offset in the positions of the decay products. Deciding whether certain decay products came from a "primary vertex" or from some "secondary vertex" is critical for determining the nature of the unstable particles. For example, it helps in deciding what invariant mass to compute. At least that's my understanding from chatting with my experimentalist friends. See here: http://lhcb-public.web.cern.ch/lhcb-public/ for an example. The April 21, 2010 entry.

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 Quote by A. Neumaier Yes; one measures the momentum by finding the curvature of the track. Nobody is interested in the position of the particle.
Whether or not anyone is interested in the position of the particle is irrelevant. The point is that position of the particle has been measured. You had claimed that "Particle detectors respond to the momentum of a particle, not to its position." This is wrong. Particle detectors at the LHC respond to, and record, the position of a particle at a sequence of times, and then infer the momentum from the path of that the particle took.

 Quote by A. Neumaier Definitely yes! Momentum is measurable, and (in direction, often also in magnitude) measured in every scattering experiment. And this doesn't contradict Wigner's theorem since momentum is one of the additively conserved quantities.
So, you are not against probabilistic interpretation per se. You are against probabilistic interpretation of position measurements.

But experimentalists do measure particle positions all the time. They use rulers, photographic plates, bubble chambers and other not-so-sophisticated devices to do so. Then they can specify a certain volume in space and count how many times particle passed through that volume (e.g., using a Geiger counter). So, experimentalists can calculate the probability of the particle being in that volume. Probabilistic interpretation of position measurements exists experimentally. However, for some reason, you do not allow us to do the same in theory.

I would appreciate if you give an original reference to the "Wigner theorem", which, as you say, does not allow us to measure position and interpret it probabilistically. This claim sounds unbelievable to me.

Eugene.

 Quote by A. Neumaier It cannot, since photons (usually used in the double slit experiment) have no Schroedinger position picture.
When a photon hits the photographic plate or a CCD detector it leaves a clear trace, whose position is well-defined (at least in our macroscopic world). So, it seems that there is no problem in measuring photon position with the precision of few micrometers, or so. We must have a theory, which would explain these kinds of measurements. Quantum mechanics is exactly this kind of theory, and the probabilistic interpretation of measurements of position (and all other obervables) is the cornerstone of quantum mechanics. If you deny that, then you invite a major revision of the entire foundation of quantum mechanics.

I don't know what would happen if we tried to determine photon's position with the precision of, say, less than 1 Angstrom. Perhaps, in this case we would meet some difficulties that you are referring to. However, there are no experimental devices, which can measure photon's position so precisely. So, the issue of the absence of a photon's position operator is an academic issue, in my opinion.

You do not deny the existence of photon's momentum eigenstates. And I can always form linear combinations of these eigenstates (with factors like exp(ipx)) which would behave *almost* like position eigenstates. For example, the linear combinations with different x and x' will be (almost) orthogonal. A space translation of an x-combination will move it to the x+a-combination. So, all properties characteristic to position will be approximately satisfied.
This should be sufficient for defining position measurements and related probabilities at least in our macroscopic world with micrometer-or-so-precision measurements.

All these problems are absent for massive particles, like electrons. So, to keep our discussion simple, let us focus on the double-slit experiment with electrons.

Eugene.

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 Quote by Physics Monkey I don't think its really correct to say that no one is interested in particle tracks. Although I'm not a particle experimentalist, I certainly think they're interesting to look at! More seriously, detailed reconstruction of particle tracks is important for determining interaction vertices.
Yes. particle tracks _are_ important; but because they allow one to measure the momentum of a particle. But particle position is irrelevant, and doesn't exist on the quantum field level.

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