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

Question

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!

Answer 1

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.

Answer 2

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 $|0\rangle|0\rangle$

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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.

Answer 3

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.

Answer 4

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).