Can local conservation be verified experimentally?

[Demystifier]

This question was raised but not answered in a thread which is now permanently closed. Consider the local conservation of charge $\partial_{\mu}j^{\mu}=0$. In quantum field theory it is valid as an operator identity, but operators as such do not have a direct operational (experimental) interpretation. Operationally, it is valid in the average statistical sense $\langle\psi|\partial_{\mu}j^{\mu}|\psi\rangle=0$. But is it valid on the individual level? Is it possible, in principle, to perform an experiment which will show it's true in every single measurement?

I think it's not. To do that one would need to simultaneously measure 4 observables $\partial_{0}j^{0}$, $\partial_{1}j^{1}$, $\partial_{2}j^{2}$ and $\partial_{3}j^{3}$, but those operators do not all mutually commute, so such a measurement is impossible. Or do I miss something?

A possible way out is not to measure the 4 observables, but only their sum as a single observable. But it looks like a cheating to me. That would be like saying that a quantum field obeys the classical deterministic equation of motion $(\Box +m^2)\phi =0$ on an individual level because the left-hand side can be measured as a single observable.

 

[gentzen]

Is it possible, in principle, to perform an experiment which will show it's true in every single measurement?

Shouldn't the question be the other way around? Obviously, no finite number of experiments will ever be able to show that something will be true in every instance. To me, the relevant question instead seems to be whether there is an experiment that contradicts conservation on the individual level.

 

[vanhees71]

I'd say that the local conservation laws are never be measurable at all, not even within classical field theory, because the conserved currents are only defined up to a "pseudo-gauge". E.g., if you have a conserved charge-current, $j^{\mu}$, where $\partial_{\mu} j^{\mu}=0$, then you can as well use another current, $j^{\prime \mu}=j^{\mu}+\partial_{\nu} \chi^{\mu \nu}$ with an antisymmetric tensor $\chi^{\mu \nu}=-\chi^{\nu \mu}$. The same holds for the energy-momentum-stress tensor and the angular-momentum-density tensor.

What is observable are the corresponding total conserved charges (or energy, momentum, angular momentum, center-of-momentum velocity), which are given by
$$Q=\int_{\mathbb{R}^3} \mathrm{d}^3 x j^{\mu}(t,\vec{x}),$$
which also are scalars if (and only if!) the local conservation law, $\partial_{\mu} j^{\mu}=0$, holds.

For details about the energy-momentum and angular-momentum tensors within classical field theory, see
Sect. 3.2 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Demystifier]

Shouldn't the question be the other way around? Obviously, no finite number of experiments will ever be able to show that something will be true in every instance. To me, the relevant question instead seems to be whether there is an experiment that contradicts conservation on the individual level.

By "true in every measurement" I meant true whenever you perform the measurement. If you perform N measurements, then the question is whether it will be true in those N measurements, not whether it would be true in other hypothetic unperformed measurements. But that's not important here, the point is local law vs. global law.

 

[Demystifier]

I'd say that the local conservation laws are never be measurable at all, not even within classical field theory, because the conserved currents are only defined up to a "pseudo-gauge". E.g., if you have a conserved charge-current, $j^{\mu}$, where $\partial_{\mu} j^{\mu}=0$, then you can as well use another current, $j^{\prime \mu}=j^{\mu}+\partial_{\nu} \chi^{\mu \nu}$ with an antisymmetric tensor $\chi^{\mu \nu}=-\chi^{\nu \mu}$. The same holds for the energy-momentum-stress tensor and the angular-momentum-density tensor.

Well, this means that local current is not measurable directly. But if one measures a local classical field as a function of space and time, and then computes the current from the field by the standard formula, then one gets a unique answer in classical physics. The point is that in quantum physics, if you measure the local field as a function of time, then the field obtained this way does not obey the classical equation of motion.

 

[samalkhaiat]

Well, this means that local current is not measurable directly.

What is wrong with ammeter?

The point is that in quantum physics, if you measure the local field as a function of time, then the field obtained this way does not obey the classical equation of motion.

I don’t know the meaning of “measure” in your sentence. In order for a local field (operator-valued distribution) to become an operator-valued $C^{\infty}$ function of time, you need to smear it with a good function $f(\mathbf{x}) \in \mathscr{S}^{3}(\mathbb{R}^{3})$. That is, for real KG field, $$\varphi (t , f) \equiv \int_{\mathbb{R}^{3}} d^{3}\mathbf{x} \ f (\mathbf{x}) \varphi (t , \mathbf{x}) .$$ In this case, it is easy to show that the KG equation, $\left( \partial^{2} + m^{2} \right) \varphi (x) = 0$, becomes $$\left( \partial_{t}^{2} + m^{2}\right) \varphi (t , f) = \varphi (t , \nabla^{2}f) .$$

In QFT, current conservation has a remarkable consequence for any matrix element of $j^{\mu}(x)$. If you take the matrix element of $\partial \cdot j = 0$ between arbitrary states $|a \rangle$ and $|b \rangle$ and then use translation invariance, you obtain $$0 = \langle b |\big[ iP^{\mu} , j_{\mu}(x) \big]|a \rangle = iq^{\mu} \langle b|j_{\mu}(x)|a\rangle ,$$ where $q^{\mu} \equiv p^{\mu}(a) - p^{\mu}(b)$ is the “momentum transfer” vector. The equation $$q^{\mu}\langle b | j_{\mu}(x)|a \rangle = 0,$$ is the simplest example of a “Ward-Takahashi identity”, a relation that must be satisfied by any conserved operator and plays a vital role in proving the renormalizability of the theory.

 

[Demystifier]

What is wrong with ammeter?

It measures the flux of the current $I=\int d{\bf S}\cdot {\bf j}$ through the wire, not the local current ${\bf j}$ itself. The flux is invariant under the redefinitions of the local current that @vanhees71 mentioned, provided that his (otherwise arbitrary) addition to the current vanishes at the boundary of the wire.

 

[Demystifier]

I don’t know the meaning of “measure” in your sentence.

I mean as in the well-known fact that one cannot simultaneously measure field and its canonical momentum. To put it in a form you like, if you smear both the field and its canonical momentum, the two well-defined operators don't commute so cannot both be measured at the same time.

 

[samalkhaiat]

It measures the flux of the current $I=\int d{\bf S}\cdot {\bf j}$ through the wire, not the local current ${\bf j}$ itself.

Thank you for that valuable piece of information If you cannot infer the value of $\mathbf{j}$ from the reading of the ammeter, you can measure the electric field which, in many situations, allows you to determine the electric current by “Ohm’s law” $\mathbf{j} = \sigma \mathbf{E}$. Look, if you cannot measure $\rho$ and $\mathbf{j}$, you will not be able to solve Maxwell’s equations for the electromagnetic propagation in a medium containing charges and currents. Jackson and Landau texts explain this stuff for you.

 

[samalkhaiat]

I mean as in the well-known fact that one cannot simultaneously measure field and its canonical momentum.

Again, thanks for that elementary fact. You said: “…., if you measure the local field as a function of time, …” and I still don’t understand the meaning of “measure” in THAT sentence. So, what does it mean to “measure” say the KG field operator? And how you do it?

 

[Demystifier]

if you cannot measure $\rho$ and $\mathbf{j}$,

I didn't say that they cannot be measured. I said that they cannot be measured directly. If you define density and current in terms of more fundamental classical quantities (e.g. in terms of charged classical fields or in terms of classical charged particle positions and their velocities), measure those more fundamental quantities, and compute the density and current from the fundamental quantities, then by this procedure you measure the density and current indirectly.

 

[Demystifier]

Again, thanks for that elementary fact. You said: “…., if you measure the local field as a function of time, …” and I still don’t understand the meaning of “measure” in THAT sentence. So, what does it mean to “measure” say the KG field operator? And how you do it?

So you ask me how to measure something as a function of time? Only a theoretical physicist can ask such a question. You measure it at time $t$, then you measure it again at time $t+\delta t$, then again at time $t+2\delta t$, and so on. The $\delta t$ is finite but small.

Or perhaps you asked me how to measure the Klein-Gordon field even ones, at one time? I don't know how to do it in practice, but in principle it should be measurable as long as it is a self-adjoint operator. Von Neumann taught is in 1932 how to write down the interaction Hamiltonian that can describe the dynamics of measurement of an arbitrary observable, but writing down a Hamiltonian on the paper is not the same as realizing this Hamiltonian in the laboratory.

 

[vanhees71]

A neutral Klein-Gordan field describes, e.g., neutral pions. As any particle they are defined as asumptotic free one-particle Fock states, and you can measure the cross sections of their production, scattering, etc. I don't think that you can measure more "field-like" states like coherent states of this field.

Even for the electromagnetic field, what you can measure are probability distributions for detecting photons. Here, of course, we usually observe coherent fields, which behave like classical electromagnetic waves or classical static fields, etc.