Is every quantum measurement reducible to measurements of position and time?

I am currently studying Path Integrals and was unable to resolve the following problem. In the famous book Quantum Mechanics and Path Integrals, written by Feynman and Hibbs, it says (at the beginning of Chapter 5 Measurements an Operators, on page 96):

So far we have described quantum-mechanical systems as if we intended to measure only the coordinates of position and time. Indeed, all measurements of quantum mechanical systems could be made to reduce eventually to position and time measurements (e.g., the position of a needle on a meter or time of flight of a particle). Because of this possibility a theory formulated in terms of position measurements is complete enough to describe all phenomena.

To me this seems to be a highly non trivial statement (is it even true?) and I was unable to find any satisfying elaboration on this in the literature.

I would be thankful for any answer to resolve this question and any reference to the literature!

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Hm. Nice question. Initially I thought "well what's with spin", but of course the stern gerlach experiment is an example of how to convert spin to position. –  Lagerbaer May 17 '13 at 23:29
I don't see how it applies to spin measurements. Maybe the assumption is that you can impose an external field, and then do something like Stern-Gerlach? –  Ben Crowell May 17 '13 at 23:29
@Lagerbaer: Beat me to it! –  Ben Crowell May 17 '13 at 23:30
Make your grad student do whatever measurement is needed. Afterwards, he writes down the result and mails it to you. Then measure the position of the ink on the paper. I'm not being sarcastic: this scheme suffices to reduce any possible measurement into a position measurement, although not the most efficient way possible. One could also think of an instrument that displays the measurement on an analog voltmeter, and this is what your text means by "position of a needle". –  Dan Stahlke May 18 '13 at 15:40
Lagerbaer, Stahlke, SMeznaric, and Mitchison are all essentially right. In particular, the statement by Feynman and Hibbs is not as mathematical sophisticated as you might suspect. The key idea is that we can make the so-called "Heisenberg cut" at multiple scales between initial amplification and our physical observation. en.wikipedia.org/wiki/Heisenberg_cut All are observationally indistinguishable and, furthermore, all the cuts above some scale will tend to have macroscopic objects in well-defined positions. –  Jess Riedel May 18 '13 at 17:35

One point to consider, although not a definitive answer, is the following. The validity of the pilot-wave theory (Bohmian mechanics) relies on the truth of Feynman & Hibbs' postulate (F&H). This is because the pilot-wave theory only makes predictions about the positions of all particles, which along with the unobservable wave function constitute a complete description of reality. In order for Bohmian mechanics to be consistent with non-relativistic quantum mechanics (QM), all measurements must therefore be reducible to position measurements. The rationale is that the outcome of any measurement $-$ momentum, spin, or otherwise $-$ is ultimately decided by the time-dependent positions of a macroscopic number of atoms or electrons, which belong to a pointer or electrical circuit in the experimental apparatus. I believe that this is also the argument that Feynman & Hibbs are making here.

So apparently, a counterexample to F&H would also be an experimental phenomenon that cannot be explained by Bohmian mechanics. Although most people don't believe in Bohm's theory, there is still a grudging consensus that it completely reproduces the predictions of ordinary QM. Demonstrating otherwise would be quite a noteworthy result. This suggests to me that no one has yet managed to think of a counterexample to F&H, although of course it is not a proof that no such counterexample exists.

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It might be true for those systems whose only degrees of freedom are time and position. However, there are other internal components such as spin that do not directly reduce to position or time. So this is one example where this statement fails.

Having said that, I would not discount it entirely. Spin can certainly be measured by observing trajectories of particles in the magnetic field and perhaps that is what Ferynman meant. But if that's the case then we might also say that all measurements are reducible to the measurement of the electromagnetic field intensities (which is particularly true for our species since at the end of the day we read off all measurements with our eyes; so we could treat all measurement devices as a tool that transforms the measured components into distinct visible spectrum electromagnetic fields). I personally prefer the more direct line of thought of what is being measured by the device, in which case we've got to accept that Feynman was perhaps not as infallible as we would like to believe.

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I'll admit this is a matter of opinion, but I think you have missed Feynman's point. The question is where do you draw the boundary between quantum and classical, or more precisely, how you fix the boundary conditions of the quantum path integral. Depending on this choice you can say that the "outcome" was a set of positions, a set of field strengths, a set of spin orientations, whatever. As long as all of these results are consistent with the same physics between "observations", the choice is irrelevant - QM does not give a prescription for where exactly the observation takes place. –  Mark Mitchison May 18 '13 at 17:56
You make a good point about electromagnetic field intensities, but it is easy to shift the emphasis again to positions. After all, "reading off measurements with our eyes" involves conformation changes of certain proteins within retinal cells. These in turn stimulate an action potential as different ionic species are pumped across the membranes of neurones, and you can keep going like this all the way until you hit the question of what is consciousness anyway. All of these involves changes in positional degrees of freedom. Of course, in practical terms this view hugely complicates matters. –  Mark Mitchison May 18 '13 at 18:01
But on matters of theoretical principle, it clearly should be possible to phrase the outcome of any measurement process in terms of final positions of particles, just by pushing the quantum/classical boundary back a few steps further. Often it only matters that one can do something in principle for a particular theoretical manipulation to be valid. –  Mark Mitchison May 18 '13 at 18:01
Also see the comments of Dan Stahlke and Jess Riedel for a slightly better explanation of what I was trying to say :) –  Mark Mitchison May 18 '13 at 18:07

I don't know whether this will satisfy you, but when I came across that line in the same book, I resolved it by trying to think of an example where I cannot reduce a measurement to position or time measurements. It will be interesting if you share any example which you think is irreducible.

Examples vary from trivial ones like kinetic energy to non-trivial like measurement of spin, like someone mentioned above.

That seems to be why there are no other fundamental operators than position and momentum in non-relativistic quantum mechanics (don't count Pauli spin operators as they come out of "nowhere", Pauli's theory was phenomenological and the proper explanation lies in relativistic quantum mechanics).

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There is a difference between the mathematical treatment of Quantum Mechanics, and the pratical job of an experimenter.

Quantum mechanics says that the outcome of a measure is a particular eigenvalue of an operator:

$$Particle \, Position : X^i(t)|\psi \rangle = x^i(t)|\psi \rangle$$ $$Particle \, Spin : S_z\psi \rangle = s_z|\psi \rangle$$ $$Fields : A_i(x,t)|\psi \rangle = a_i(x,t)|\psi \rangle$$

For the experimenter, the only practical test that he can do is :

$$Yes \,\, or \,\, No$$

So, let's take the exemple of a test experience of violation of Bell inequalities. You will have 2 particles and 4 possible states :

$$|0\rangle|0\rangle, |0\rangle,|1\rangle, |1\rangle|0\rangle, |1\rangle|1\rangle$$

Can a experimenter "measure" these states ? No.

So the experimenter is doing the following thing: he makes the entanglement of each state with an optical path, so you will have now :

$$|0\rangle|0\rangle|NORTH\rangle, |0\rangle|1\rangle|EAST\rangle$$ $$|1\rangle|0\rangle|SOUTH\rangle, |1\rangle|1\rangle|WEST\rangle$$

So now, the experimenter does not need to mesure the states, he could "measure" optical paths instead.

But can the experimenter measure optical paths? No.

But he can put a counter for each optical path. In some sense, the optical path and the counter are some subset of space-time with an intersection.

When the experimenter got a $Yes$ answer ,for instance a NORTH counter click, he knows that the optical path is NORTH, and then he knows that the state was $|0\rangle|0\rangle$

So, practically, an experimenter knows only $Yes \,or \,No$, $True \,or\, False$, he cannot measure directly position, spin, fields, etc, but he can measure indirectly with entanglements. So, I think that Feynmann and Hibbs assertion is not enough precise.

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