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,
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).
Adding these, we get
(the potential energy terms cancel).
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,
Let us call the vector-valued function that is the argument of the divergence in the integrand :
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 .
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).