## Physics Friday 57

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


(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).