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I am not a physicist, and not sure whether I want the adjective, or the noun, observable here.

Example 1) If we view the mass and velocity of a classical particle as observables, we calculate the kinetic energy. Is that observable? an observable?

Example 2) In electrodynamics I read that charge and current densities are observable, and that we can calculate the scalar and vector potentials in the Lorenz gauge, but that those are not observable.

If the answer in (1) is yes, how do I reconcile that with (2)?

And generally, Is a quantity calculated from observables, observable?

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3 Answers 3

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Is a quantity calculated from observables, observable?

A quantity calculated only from observables is observable.

Example 1) If we view the mass and velocity of a classical particle as observables, we calculate the kinetic energy. Is that observable? an observable?

Yes. Note that although this is observable, it will also be relative to the frame of the velocity measurement device.

Example 2) In electrodynamics I read that charge and current densities are observable, and that we can calculate the scalar and vector potentials in the Lorenz gauge, but that those are not observable.

Correct, the potentials are not observable.

If the answer in (1) is yes, how do I reconcile that with (2)?

The calculation in 2 depends on the gauge. The gauge is not observable. So the calculation in 2 is not based only on observables.

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  • $\begingroup$ "A quantity calculated only from observables is observable." - this is true in classical physics, but Quantum Mechanics complicates it - I wrote a separate answer below. $\endgroup$ Commented Jun 5, 2023 at 17:19
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    $\begingroup$ @NadavHar'El yes, QM complicates everything. But although it is more complicated, it is still a true statement. I.e. what combination of observables are possible is restricted, but if a combination is observable then calculations are also observable. $\endgroup$
    – Dale
    Commented Jun 5, 2023 at 22:07
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    $\begingroup$ Might want to elaborate on what gauges are, since they're not obvious, and OP says they are not a physicist. $\endgroup$
    – Allure
    Commented Jun 6, 2023 at 2:48
  • $\begingroup$ @BobTerrell you specifically mentioned the Lorenz gauge in your question. So I assumed that you understood what a gauge is. Do you need an explanation? It seems off topic, but I can do a brief one if it is actually needed $\endgroup$
    – Dale
    Commented Jun 6, 2023 at 16:21
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I think you're hitting on a subtle point that is being missed here.

Classically, any function of observables such as charge density or velocity is of course an observable itself, but in the trivial sense of "I know that $\rho={\rm some\,\, number}$, so I know that $f(\rho)={\rm some\,\,other\,\,number}$". This is, I think, the spirit of what you're getting at with your first point.

However, your second point is different in an important way. When you, for example, use Gauss' law to obtain an electric field from your charge density: $$\nabla\cdot\boldsymbol{E}=4\pi\rho,$$ you're not simply writing down a function of $\rho$ and evaluating it, you're relating two ontologically distinct entities of the world. So the fact that, in practice, you calculate the value of an object in your theory from observable quantities does not necessarily imply that that object is an observable.

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  • $\begingroup$ What makes the kinetic energy ontologically distinct from the electric field? Or perhaps more precisely I should ask what makes energy ontologically non-distinct from mass and velocity, such that it's an observable which can be calculated with a function, but the electric field is ontologically distinct from charge density such that it is not? $\endgroup$
    – R.M.
    Commented Jun 6, 2023 at 15:01
  • $\begingroup$ @R.M. The formal solution to $\nabla\cdot\mathbf{f}=4\pi\rho$ is the field $\mathbf{f}_{\rho}$, which is an observable since it only depends on $\rho$ (well...and boundary conditions). The "law of physics" is the claim that it's a general fact that $\mathbf{f}$ is numerically identical to the electric field $\mathbf{E}$ in the presence of $\rho$. Unlike $\mathbf{f}$, the field $\mathbf{E}$ is ontologically distinct from $\rho$ in that it is a separate feature of the world. We constantly talk about it as such ("the electric field looks like...", "the electric field evolves...", etc.). $\endgroup$ Commented Jun 6, 2023 at 17:00
  • $\begingroup$ ...(cont.) Energy, on the other hand, seems to me like it belongs to the other class. We aren't relating e.g. $f(m,v)=mv^2/2$ to an independent substance called "kinetic energy" through a physical law, but rather defining kinetic energy as $f(m,v)$ itself. $\endgroup$ Commented Jun 6, 2023 at 17:01
  • $\begingroup$ The charge density just fixes the divergence of your electric field, which is not enough to fully determine it. For example in the trivial case $\rho = 0$ their could be any number of plane waves with any polarisations, intensities, frequencies and propagation directions, all contributing to $\mathbf{E}$ but all divergence-free. So with $\rho$ you still don't have enough info to infer $\mathbf{E}$, no matter which position you take on the ontology/epistemic status of the electric field. This makes me feel the ontology is a red herring. $\endgroup$
    – Dast
    Commented Jun 7, 2023 at 9:16
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    $\begingroup$ @Dast Okay, sure. But you can easily replace $\rho$ with "all the observed data required to fix $\mathbf{E}$" or even consider a generic law like $L(Q, D)=0$ for data $D$ and differential equation $L$. That's not the point I'm trying to make. I'm merely trying to draw a distinction between the solution to the equation (which is trivially an observable as it's entirely fixed by the data) and the actual physical object that that equation is designed to predict the value of (which may or may not be an observable). That's what I interpreted the source of the OP's confusion to be anyway. $\endgroup$ Commented Jun 7, 2023 at 10:17
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And generally, Is a quantity calculated from observables, observable?

Other people answered the more specific questions, I want to comment on this general one. In Quantum Mechanics, you can have two observables but your are not able to observe both at the same time, so in particular cannot generally observe some calculation based on both values.

A canonical example is position and momentum (velocity) of a particle: Each one of the two is observable, but you cannot observe both at the same time (this is known as the Heisenberg Uncertainty Principle). So you cannot, for example, observe the value of the silly vector "position + velocity", because you cannot measure the two parts of this calculation at the same time.

The way this is formulated in quantum mechanics is that observables are Hermitian operators (you can think of them like matrices) which operate on the space of states. The observations are the eigenvalues of these operators. Two observables can be measured at the same time only if the two operators commute (i.e. $AB$ = $BA$). As an example, the operators $P_x$, $P_y$, and $P_z$, giving the x coordinate, y coordinate, and Z coordinate of the momentum of a particle, commute, so you can measure all of them at the same time, and, for example, calculate $|P| = \sqrt{P_x^2 + P_y^2 + P_z^2}$.

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    $\begingroup$ There is a common misconception here. The only actual "observables" in quantum mechanics that are in any way "precise" are energy, momentum, angular momentum and charges. These are the "quanta" that we are measuring in a single quantum event. Why are these the fundamental quantities? Because they are the only conserved quantities in physics. Position, on the other hand, is not a conserved quantity, neither is timing. What the theory calls "observables" or "measurements" are actually statistical distributions of these quanta, but those are derived properties, not primary physical ones. $\endgroup$ Commented Jun 6, 2023 at 1:32
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    $\begingroup$ FlatterMann, what you said is true, but I don't understand how it is relevant to the question. Sure, if something is not conserved, every time you observe it can be different. Moreover, if you're not in an eigenstate of this specific observable, then an observation will yield a "random" result from some distribution. But we still call this an observation. And in practice, we can truly observe something like, say, the time of decay of a radioactive atom, even if the time we observe looks "random", it's still an observation. $\endgroup$ Commented Jun 6, 2023 at 7:16
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    $\begingroup$ It is relevant to the question because what we call "measurements" in quantum mechanics are purely theoretical predictions, not laboratory reality. Just as one can not actually measure a classical probability distribution, it's not possible to measure a von Neumann observable. We can, at most, get frequentist approximations, most of which are completely irrelevant, anyway, The precision tests of quantum mechanics are spectra and scattering cross sections. The general "observable" of the theory is a nice mathematical tool, but it is not how we actually do physics. $\endgroup$ Commented Jun 6, 2023 at 9:17
  • $\begingroup$ "you cannot, for example, observe the value of the silly vector "position + velocity"" I'm actually not totally sure what you mean by this, could you elaborate? $|\langle \psi | (\hat{X} + \hat{P}/m) | \psi \rangle|^2$ can be calculated and, in principle, measured. $\endgroup$
    – Rococo
    Commented Jun 16, 2023 at 2:33

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