Current in Single Particle Quantum Mechanics
January 14, 2014
For simplicity, I will only use one-dimension in this post, but this can be generalized to higher dimensions rather easily.
Many textbooks on Quantum Mechanics mention current density can be derived from the continuity equation and probability. The usual method for figuring this out is to assume you have some Hamiltonian where is the momentum and is the position. In this way the current density is written in terms of the wave function as
This then satisfies the continuity equation
with density . It should be noted that if you write the wave function as , then the current is just proportional to the gradient of the phase , giving the spatial change in phase a physical significance.
However, there are two lingering questions:
- Is this current density related to the Heisenberg operator which tracks the velocity of the system?
- If so, does it generalize to more arbitrary Hamiltonians?
To answer these questions, we consider the more arbitrary Hamiltonian
where is some potential and the kinetic energy is some polynomial
We are unworried about bounding the energy, so odd-order Kinetic energy terms are allowed (in the higher dimensional case, the Dirac-like Hamiltonians have linear terms in ). At this point, we can take our Heisenberg operator and find
where is the derivative of with respect to its argument. Now, we would like to obtain a current density from this quantity. We can certainly define the total current at a specific time as
where in the last line we go from the Heisenberg to Schroedinger picture. Now to get density, we need to use a complete set position states, so that
Now, acts as a derivative on position kets, so that one can verify that
However, there is an ambiguity here since we can write
This ambiguiuty in how to choose the derivatives leaves us with many way to define the current density. Fortunately, only one of these combinations satisfies the continuity equation. To figure out which one that is, let us reverse engineer the continuity equation to obtain a solution. The density is , and so using the Schroedinger’s equation, we have
Thus, the continuity equation must become
If we now assume that we have a current density that takes the form
and satisfies the continuity equation, , then we can equate operators to obtain
Anticipating the answer, we write the general form of as
Then we can take the left hand side Eq. \eqref{eq:diff-ops-cty} and write
On the other hand, we can calculate the right hand side of Eq. \eqref{eq:diff-ops-cty} to be
Equating the left and right sides, we can just read off that , and , so that .
Thus, we have
Returning all the way to when we were considering as an integral over position, this suggests that in Eq. \eqref{eq:T-delta}, we want to consider
Given the expression for total current Eq. \eqref{eq:total-current} and integrating the delta function by parts numerous times, we can replace with and with , and then the total current is just
which actually integrates the current density! Thus, we have shown that
and that
Indeed, does track the current of the problem and can even be written as the integral of a current density. Even for the more arbitrary Hamiltonian .