Consider a single particle wavefunction (in the position basis) . Then for a volume *V*, the probability *P* that the particle will be measured to be in *V* is

.

The time derivative of this is:

.

Now, we recall that , where ^{*} indicates the complex conjugate. Thus, via the product rule,

So

Now, consider the time dependent Schrödinger equation:

Solving for the time derivative, we get

And taking the complex conjugate of that:

.

(Note that the potential is real).

Thus

,

and

Adding these, we get

(the potential energy terms cancel).

So

.

Here, we need to use some vector calculus; namely, the product rule for the divergence operator : for scalar valued function *φ* and vector field , the divergence of their product is given by

Now, if our vector field is itself the gradient of a scalar function *ψ*, then

Swapping *φ* and *ψ*,

And taking the difference, we find

.

Putting in our wavefunction and it’s conjugate for *φ* and *ψ*,

So

.

Let us call the vector-valued function that is the argument of the divergence in the integrand :

Then

Now, recalling that the probability *P* is the integral over *V* of the probability density

So in terms of the probability density

and as this holds for all *V*, the integrand must vanish:

Now, this should look familiar to some of you. For any conserved quantity *ρ* with a flux given by the function , and no sources or sinks, the quantity and flux obey the continuity equation

.

[For example, if *ρ* is electric charge density, then the conservation of electric charge gives

,

where is the electric current density.]

Now, as the probability density for the particle must always integrate over all space to unity, we similarly expect it to be a conserved quantity. Thus the above result is our continuity equation for quantum probability, and so as defined above is our probability current (or probability flux). It has units of probability/(area × time), or probability density times velocity.

Note that the continuity equation tells us that for a stationary state, the divergence of the probability current must be zero. However, this does not mean that the current itself must be zero. Consider the three-dimensional plane wave .

Then

So

So

.

Note that as is the particle’s velocity, the probability current of the plane wave, a stationary state, is the amplitude squared times the particle velocity.

Lastly, consider a wavefunction which has the same complex phase for all locations at any given time; that is , where is a real-valued function. Then , and , and so we see the probability current is zero for such wavefunctions (one example of which are solutions to the particle in a one-dimensional box).