Probability Current

The probability current, also known as the probability flux, of a wave function at a certain point describes the rate of flow at which probability passes through that point, analogous to the way electrical current describes the rate of flow at which electrical charge passes through a point in a medium. Probability currents are used, for example, when calculating reflection and transmission coefficients for particles encountering potential steps or potential barriers. Here, we'll derive the formula that gives the probability current of a wave function, $\psi(\mathbf{r}, t)$, where $\mathbf{r}$ is the position vector of the particle in question and $t$ is time.

Conservation of Probability

To start, let's derive the conservation law for probability. We know that the probability, $P(R, t)$, that the particle is in the three-dimensional region $R$ at time $t$ is given by:

\begin{equation}
P(R, t) = \int_R |\psi(\mathbf{r}, t)|^2 dR
\end{equation}

The rate over time at which the amount of probability in the region $R$ changes is thus given by:

\begin{equation}
\frac{\partial}{\partial t}P(R, t) = \frac{\partial}{\partial t} \int_R |\psi(\mathbf{r}, t)|^2 dR
\end{equation}

As you can imagine, probability needs to be conserved and so, the probability current vector, $\mathbf{j}$, can be related to $\frac{\partial}{\partial t}P(R, t)$ like so:

\begin{equation}
\frac{\partial}{\partial t}P(R, t) + \int_S \mathbf{j} dS = 0
\end{equation}

where $S$ is the surface of the region $R$.

\begin{equation}
\frac{\partial}{\partial t} \int_R |\psi(\mathbf{r}, t)|^2 dR + \int_S \mathbf{j} dS = 0
\end{equation}

The above equation is the conservation law for quantum mechanical probability.

Divergence Theorem

To proceed any further, we'll need to use the divergence theorem, so let's derive that here.

\begin{equation}
\int_R \nabla \mathbf{j} dR = \int_{R_z} \int_{R_y} \int_{R_x} \left[ \frac{\partial}{\partial x}j_x + \frac{\partial}{\partial y}j_y + \frac{\partial}{\partial z}j_z \right] dxdydz
\end{equation}

where $j_x$, $j_y$, and $j_z$ are respectively the $x$, $y$, and $z$ components of the flux, $j$, vector and $R_x$, $R_y$, and $R_z$ are respectively the $x$, $y$, and $z$ boundaries of the region $R$.

\begin{equation}
\int_R \nabla \mathbf{j} dR = \int_{R_z} \int_{R_y} [j_x]_{R_x} dydz + \int_{R_z} \int_{R_x} [j_y]_{R_y} dxdz + \int_{R_y} \int_{R_x} [j_z]_{R_z} dxdy
\end{equation}
\begin{equation}
\int_R \nabla \mathbf{j} dR = \int_S \mathbf{j} dS
\end{equation}

The above equation is the mathematical statement of the divergence theorem.

Solving for the Probability Current

Using the divergence theorem, we can rewrite equation $(4)$ as:

\begin{equation}
\frac{\partial}{\partial t} \int_R |\psi(\mathbf{r}, t)|^2 dR + \int_R \nabla \mathbf{j} dR = 0
\end{equation}

Since $R$ is arbitrary, we can remove the definite integration from the above equation by differentiation.

\begin{equation}
\frac{\partial}{\partial t} \psi(\mathbf{r}, t) \psi^*(\mathbf{r}, t) + \nabla \mathbf{j} = 0
\end{equation}

The time-dependent Schrodinger equation states:

$$i \hbar \frac{\partial}{\partial t} \psi(\mathbf{r}, t) = -\frac{\hbar^2}{2m}\nabla^2 \psi(\mathbf{r}, t) + V(\mathbf{r}, t) \psi(\mathbf{r}, t)$$

If we take the complex conjugate of that, we get:

$$-i \hbar \frac{\partial}{\partial t} \psi^*(\mathbf{r}, t) = -\frac{\hbar^2}{2m}\nabla^2 \psi^*(\mathbf{r}, t) + V(\mathbf{r}, t) \psi^*(\mathbf{r}, t)$$

Using the above two equations, equation $(9)$ can be rewritten as:

\begin{equation}
-\psi(\mathbf{r}, t)\frac{i\hbar}{2m}\nabla^2 \psi^*(\mathbf{r}, t) + \psi^*(\mathbf{r}, t) \frac{i\hbar}{2m}\nabla^2 \psi(\mathbf{r}, t) + \nabla \mathbf{j} = 0
\end{equation}
\begin{equation}
\frac{i\hbar}{2m}\left[ -\psi(\mathbf{r}, t) \nabla^2 \psi^*(\mathbf{r}, t) + \psi^*(\mathbf{r}, t) \nabla^2 \psi(\mathbf{r}, t)\right] + \nabla \mathbf{j} = 0
\end{equation}
\begin{equation}
\frac{i\hbar}{2m}\nabla \left[ -\psi(\mathbf{r}, t) \nabla \psi^*(\mathbf{r}, t) + \psi^*(\mathbf{r}, t) \nabla \psi(\mathbf{r}, t)\right] + \nabla \mathbf{j} = 0
\end{equation}
\begin{equation}
\mathbf{j} = \frac{i\hbar}{2m} \left[\psi(\mathbf{r}, t) \nabla \psi^*(\mathbf{r}, t) - \psi^*(\mathbf{r}, t) \nabla \psi(\mathbf{r}, t)\right] + \mathbf{C}
\end{equation}

Applying the condition that $\mathbf{j}=0$ at position $\mathbf{r}$ when $\psi(\mathbf{r}, t) = 0$ for all $t$, we find that $\mathbf{C}=0$. Thus, the probability current, $\mathbf{j}$, is derived to be:

\begin{equation}
\mathbf{j} = \frac{i\hbar}{2m} \left[\psi(\mathbf{r}, t) \nabla \psi^*(\mathbf{r}, t) - \psi^*(\mathbf{r}, t) \nabla \psi(\mathbf{r}, t)\right]
\end{equation}