# Is the probability current an observable?

Is the probability current in Quantum Mechanics an observable? If so, how can it me measured (directly or indirectly)?

• It's not even a proper operator, how could it be an observable? – ACuriousMind Mar 6 '16 at 22:34
• @ACuriousMind so the rule is that for something to be an observable(to be something that is measurable) it must have an operator(Self-ajdoint one)? – TheQuantumMan Mar 6 '16 at 22:36
• You seem to be using "observable" to mean something other than by definition self-adjoint operator. In quantum mechanics, the word observable always means "self-adjoint operator". You could devise measurements that e.g. reconstruct the probability density or the current from many repeated measurement, but that doesn't make it an observable. – ACuriousMind Mar 6 '16 at 22:40
• you might also read the first sentence of the wikipedia article on observables: en.wikipedia.org/wiki/Observable :) – famfop Mar 6 '16 at 22:44
• @TheQuantumMan if you mean to ask whether value of probability current density can be measured in an experiment, you have to lose "an" before "observable". This indefinite article is triggering people to answer a different question about noun "observable" from quantum theory. – Ján Lalinský Apr 30 at 20:30

The probability current can be expressed in terms of an operator. Furthermore the electric current is charge times probability current, so measuring the probability current for a charged particle is as simple as measuring the electrical current and dividing by the charge.

The following Hermitian operator is the current operator $$\hat{j}(r)=\frac{1}{2m}[|r\rangle\langle r|\hat{p}+\hat{p}|r\rangle\langle r|]$$ so that the probability current in state $|\psi\rangle$ is the usual expression $$\langle\psi|\hat{j}(r)|\psi\rangle=\frac{1}{2m}[\psi^*(r)\{-i\hbar\nabla\psi(r)\}+\psi(r)\{-i\hbar\nabla\psi(r)\}^*]=\frac{\hbar}{2mi}[\psi^*\nabla\psi-\psi\nabla\psi^*]$$ When there are gauge fields around ones needs to be more careful about the momentum but the idea is still the same. For more details see this article.

The probability current is just the EM current divided by the charge. It is the charge-current that is the observable. The proportionality is no longer true in relativistic quantum mechanics, as the example of the Klein-Gordon equation shows.

No. It is not an observable. An observable is something that can be measured in a single measurement, not just inferred statistically from a series of measurements over time. Probability currents never fit that description.

It is important to distinguish between your question "is a probability current observable" for which the clear answer is "no", and the quite similar question that is closely related:

"Is a probability current real (i.e. something that has a physical existence)?"

The second question is up for debate and it isn't easy to think of experiments that determine if something that is not observable is real or not.

• If indeed the probability current cannot be expressed as an operator, then you're conclusions would follow. But this answer is lacking a proof of the former. (Hint: the explicit expression in James Roland's answer implies that such a proof does not exist.) – Emilio Pisanty Apr 30 at 19:56
• Even in the example in James Roland's answer, the measured observable is electrical current. The probability current is only inferred even though the inference is straight forward. The fact that something can be expressed as an operator does not in and of itself mean that it can be directly observed. Only things which are directly observed as "observables." – ohwilleke Apr 30 at 20:09