Do the probability density and the probability current density have a unit

Probability, as such, has no units — it is simply a dimensionless number.

A probability density, however, measures probability over a unit of space (or time, or phase space, or whatever), and thus its unit is the inverse of the unit you're using to measure the space the density is distributed over.

For example, if you have a probability density over a one-dimensional space measured in meters, then the unit of the probability density is 1 / meter. If the probability density was distributed over a two-dimensional space, you'd measure it in units of 1 / meter², and if it was a density over time, you could measure it in units of 1 / second, etc.

Similarly, a probability current is a measure of the probability (which, again, is dimensionless) passing through the boundary of an area per unit of time. Thus, if the boundary is measured in units of X, the probability current is measured in units of 1 / X / second (or whichever unit of time you're using). Of course, in one-dimensional space, a boundary is simply a point, and thus also dimensionless, so the probability flux in 1D space simply has units of 1 / second.

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To get a visual intuition for this, you can simply think of probability as analogous to a physical fluid, whose amount you can measure in moles or kilograms or (assuming incompressibility) liters. Probability density is then simply analogous to the amount of the fluid over a given volume / area / length, which you can measure in conventional units of density, e.g. moles per cubic meter, and probability current is just analogous to the flow of the fluid through an area / interval / point, measured e.g. in moles per square meter per second. The only difference between actual probability and the fluid analogy is that probability itself, unlike fluid amount, is unitless, so the unit of probability density becomes (e.g.) one per cubic meter, and that of probability current becomes one per square meter per second.

In fact, this analogy has a clear physical meaning. On one hand, if you have a gas consisting of a large number of particles, each randomly distributed according to the same probability density, then the density of the gas is proprotional to the probability density of the distribution (and the constant of proportionality is simply the total amount of gas).

Conversely, if you have a volume of gas with a given density distribution, and you randomly choose one of the particles comprising the gas, the probability of the chosen particle being in a given volume of space is given by a probability distribution whose density is proportional to the density of the gas (with the constant of proportionality being, again, the total amount of gas).

Thus, in a very real sense, a probability density can be interpreted as just a physical particle density divided by the total amount of particles. This holds even if we're looking at just one particle — if the position of the particle is uncertain, so that we have, say, a probability of 0.05 of finding it within a given volume, we can indeed meaningfully say that, on average, that volume contains 0.05 particles.

Of course, a similar interpretation can equally well be given also for, say, probability distributions over time, just with particles replaced by events. Again, the basic idea is that, if we have lots of independent and identically distributed events, then the average number of events over an interval of time is simply the probability of an event occurring over that interval, times the total number of independent events. Thus, the "event density" is simply proportional to the probability density, but its unit may be multiplied by whatever unit we're using to measure the amount of events.

They are both densities, that is they have the form

$$ frac{text{base quantity}}{text{"volume"}}$$

• The base quantity has whatever units it normally has. Amperes for current, or (dimensionless) for probability.

• The "volume" (which is in quotes because (a) it's not necessarily 3d as in a linear charge density and (b) can exist in abstract space with coordinate dimensions other than length as with a distribution in momentum space or Fourier space) probably has units too: some power of the dimension of a single coordinate in that space. And so on. You are expected to be able to deduce the correct units for each case.

So, current density has units of $mathrm{A/m^2}$ because it is an areal density. Probability density in a 3D position space has units of $mathrm{m^{-3}}$. The probability density in momentum space as measured in a quasi-elastic scattering experiment, has units of $mathrm{{GeV}^{-1}}$ because we only probe one dimension of dependence (and being nuclear physicists we have obviously employed natural units).

I want to add up to the probability current density, which is:

$$j = frac{hbar}{2mi}left[psi^*frac{dpsi}{dx}-psifrac{dpsi^*}{dx}right]$$

With the general form of wave function:

$$u(x)=Ae^{ikx}+Be^{-ikx}$$

$A$ and $B$ are dimensionless constants, and $k$ is the wave number.

Eventually, $k$ comes out of the derivatives, and the current density becomes:

$$frac{hbar k}{m}(A^2-B^2)$$

Where $$hbar k$$ is the linear momentum, $$frac{hbar k}{m}$$ becomes the velocity. Hence, the unit is meter per second.