Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is probability current in quantum mechanics? Why define such a thing?

I mean the meaning of probability current. I know the formula for it but I just don't get the idea of a flow of probability over time. Does it imply that the probability of finding a particle along $dx$ changes over time such that it is not fixed. In other words, at $t=0$, the probability can be $0$, then $t=1$, probability can be $0.5$.

I am quite new to quantum mechanics, so I am still trying to understand this new world.

share|cite|improve this question

The total probability of all mutually exclusive alternatives must always be 100%, so it is conserved. The conservation law in the spacetime tend to be "local", so just like for the charge conservation, we may derive the conservation of the probability in Schrödinger's equation from the local continuity equation $$ \frac{\partial \rho}{\partial t} + \mathbf \nabla \cdot \mathbf j = 0 $$ So the probability $\rho$ in a tiny box decreases exactly by the amount that may be calculated as the flux of the probability current through the six faces of the little box (through its boundary) via Gauss' theorem. It's useful to know the form of the vector $\vec j$ that allows us to satisfy the continuity equation above.

But the probability current has a much more direct interpretation. Imagine that you have a photographic plate of a sort that is guaranteed to absorb a particle that hits the plate (i.e. that makes any reflection impossible). Then the probability per unit time that the particle will really be absorbed by the surface $\Sigma$ (and create a "dot" somewhere on this surface) is given by $$ \frac{dP_{\rm absorb}}{dt} = \int_\Sigma d\vec S \cdot \vec \jmath $$ with the right sign convention for the vectors. Even if one calculates the simple question what the density of dots will be on a plate in the double slit experiment, the probability current is directly relevant for that. Somewhat sloppily, one might imagine that the distribution of the dots matches $\rho$. But it's much more accurate that it matches $j_n$, the normal component (to the plate) of the probability current. These two functions of $x,y$ are only proportional to each other assuming that the speed of the particle is "constant" everywhere. If it's not, $j_n$ gives the more correct representation of the "density of dots" than $\rho$.

share|cite|improve this answer

To add to @Luboš Motl's great answer, I just want to mention the connection to the electric current and electric charge density from electrodynamics, which you may be more familiar with.

Note that probability is unitless, so probability density has units of 1/Volume and probability current has units of 1/area*time. These are the same units as electric charge density and electric current up to a factor of electric charge. Indeed, if you multiply these quantities by some electric charge Q, you get totally sensible electric charge densities and electric currents that satisfy the continuity equation for electrodynamics (which has the same form as the one posted above, except that $\rho$ and $\textbf{j}$ are interpreted as the relevant quantities in electrodynamics, not quantum mechanics.

What's more, if you calculate $p$ and $\textbf{j}$ for the electron in a hydrogen atom (part of any first course in quantum mechanics) and multiply them by $e$ the fundamental electric charge, you get the associated electric charge density and electric current for hydrogen. If this doesn't seem noteworthy to you, try using the bohr radius and the electric current you just calculated to calculate the magnetic dipole moment of hydrogen: $\mu = I A$. It comes out pretty close to the actual value! So this interpretation of probability flow as creating a "probabilistic electric current" is actually meaningful, in that it gives a heuristic approximation to the real magnetic dipole moment of hydrogen.

Last comment: it was actually by trying to address such problems in electrodynamics that Schroedinger first developed his famous equation. The original wavefunction actually had units of $\sqrt{\frac{Q}{P}}$ where Q is charge and P is probability. It was only later that he realized the greater significance of his result and dropped the charge factor.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.