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In the basic quantum mechanics lectures, we learn that we can measure any observable. That means mathematically, all Hermitian operators correspond to a physically measurable quantity. In strong contrast, it seems that experimentally, we only ever have access to the position of a quantum system. To measure its other properties, one has to find a way to entangle the other property with the position and then measure the position. A few examples:


The spin of an electron

To measure the spin of an electron, we send it into a Stern-Gerlach apparatus and let the inhomogeneous magnetic field change its trajectory depending on its spin projection. Then we detect its position on a CCD chip. $$\left( \alpha \left| \uparrow \right\rangle + \beta \left| \downarrow \right\rangle \right) \otimes \left| x_{in} \right\rangle \to \alpha \left| \uparrow, x_{out, \uparrow} \right\rangle + \beta \left| \downarrow, x_{out, \downarrow} \right\rangle$$

The energy of a photon

To measure the energy/momentum/wavelength of a photon you send it into a grating or prisma and see where it comes out. Here, the energy gets entangled with the position, like $$\left( \int s(\nu) \left| \nu \right\rangle d\nu\right) \otimes \left| x_{in} \right\rangle \to \int s(\nu) \left| \nu, x_{out, \nu} \right\rangle d\nu.$$

The momentum of a charged particle

Similar to a photon spectrometer, we can spatially separate different momenta through the Lorentz force. A particle with charge $q$ and a momentum $p$ is deflected with a radius $R = \frac{p}{qB}$ when it enters a homogeneous magnetic field $B$. Again, we create entanglement between the momentum and the position of the particle, then measure its position. $$\left( \int s(p) \left| p \right\rangle dp\right) \otimes \left| x_{in} \right\rangle \to \int s(p) \left| p, x_{out, p} \right\rangle dp$$

If we think about what a measurement means, it becomes clear: A measurement is an interaction of the particle with a larger system of many degrees of freedom. For the interaction to happen, the particle must spatially be close to the detector. Or in other words – interaction takes place in position space.

This all seems like it doesn't make any difference for how we use quantum mechanics, because after all we can always infer the desired property from the position measurement. But it does make a big difference when it comes to formulating a physical theory based on observations. If one tries to come up with a novel theory, which should be consistent with all measurements ever done, it only has to explain how we end up measuring particles at particular places. All the other stuff, like energy, spin, momentum, etc. are just useful concepts, but not necessarily part of the physical reality.

Is this reduction to position measurements in conflict with the statement in the introduction that any Hermitian operator corresponds to a physical measurement?
Does this concept have a name? I didn't know what to search for. The closest thing I found is Plato's allegory of the cave.

Is there a way to discriminate between a quantum mechanics formulation which allows only position measurements and one which also allows to directly measure other observables? As pointed out in Mark Mitchison's answer on Is every quantum measurement reducible to measurements of position and time?, the validity of Bohm's pilot wave approach to quantum mechanics relies on the assumption that indeed position measurements are sufficient to completely describe quantum mechanics. This could help identifying differences – if Bohm's theory makes different predictions, it could be because it doesn't allow all theoretically possible measurements.

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  • $\begingroup$ @TobiasFünke Thanks for the links! In particular the first one is helpful, as it quotes a literature resource. I will extend the question to specifically ask for cases which can't be explained with Bohmian mechanics, as pointed out in Mark Mitchison's answer. $\endgroup$
    – A. P.
    Feb 11, 2023 at 10:40
  • $\begingroup$ I'm not a real particle physicist, but as I heard, energy of a photon can be also measured by a bolometer or a CCD. They don't operate on position, but on excitation of the electrons. More energy - more excitation. $\endgroup$
    – user46147
    Feb 11, 2023 at 10:49
  • $\begingroup$ @user46147 I also don't know the exact operation principle of these. In the case of a CCD I could imagine that high-energy photons lift a certain number of electrons into the conduction band, which are later found in the capacitor of an analog-to-digital converter. I will read up on those. $\endgroup$
    – A. P.
    Feb 11, 2023 at 11:13
  • $\begingroup$ Bohm's theory is not a local hidden variable theory. There is $\psi$ and its effect is non-local. $\endgroup$ Feb 11, 2023 at 17:59

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This is an event from a bubble chamber experiment, where proton $K^{-}$ interactions are measured :

K-proton

(look for "muon" in the ("do you want to see" link)

The quantum mechanical position is the vertex of the event out of which four charged track appear. ( in this answer I give details)

If you mean by position measurement the ionization of the hydrogen that gives the tracks, we only measured a few points on the track and fitted it to get the momentum, given a table for the ionization of particles that gave the mass. This does not make "position" measurement dominant, except in a philosophical framework, not appropriate for this site.

From the measurements of thousands of such events the standard model of particle physics was built up.

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    $\begingroup$ Thanks for the example! Could you please elaborate why position measurement is not the dominant one? To me it looks like a long sequence of position measurements, from which other properties like momentum and charge are inferred. $\endgroup$
    – A. P.
    Feb 11, 2023 at 20:13
  • $\begingroup$ As I said, it ends up into philosophy, our existence could be a position measurement, as we exist in 3 dimensions and time. Space time is important the way the sounds we make are important but would you call them dominant in a discussion? The alphabet is important, but is it dominant in a book? $\endgroup$
    – anna v
    Feb 11, 2023 at 20:20

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