% Probability current


# Probability current

In quantum mechanics, the **probability current** describes the movement
of the probability of finding a particle at given point in space.
In other words, it treats the particle as a heterogeneous fluid with density $|\psi|^2$.
Now, the probability of finding the particle within a volume $V$ is:

$$\begin{aligned}
    P = \int_{V} | \psi |^2 \dd[3]{\vec{r}}
\end{aligned}$$

As the system evolves in time, this probability may change, so we take
its derivative with respect to time $t$, and when necessary substitute
in the other side of the Schrödinger equation to get:

$$\begin{aligned}
    \pdv{P}{t}
    &= \int_{V} \psi \pdv{\psi^*}{t} + \psi^* \pdv{\psi}{t} \dd[3]{\vec{r}}
    = \frac{i}{\hbar} \int_{V} \psi (\hat{H} \psi^*) - \psi^* (\hat{H} \psi) \dd[3]{\vec{r}}
    \\
    &= \frac{i}{\hbar} \int_{V} \psi \Big( \!-\! \frac{\hbar^2}{2 m} \nabla^2 \psi^* + V(\vec{r}) \psi^* \Big)
    - \psi^* \Big( \!-\! \frac{\hbar^2}{2 m} \nabla^2 \psi + V(\vec{r}) \psi \Big) \dd[3]{\vec{r}}
    \\
    &= \frac{i \hbar}{2 m} \int_{V} - \psi \nabla^2 \psi^* + \psi^* \nabla^2 \psi \dd[3]{\vec{r}}
    = - \int_{V} \nabla \cdot \vec{J} \dd[3]{\vec{r}}
\end{aligned}$$

Where we have defined the probability current $\vec{J}$ as follows in
the $\vec{r}$-basis:

$$\begin{aligned}
    \vec{J}
    = \frac{i \hbar}{2 m} (\psi \nabla \psi^* - \psi^* \nabla \psi)
    = \mathrm{Re} \Big\{ \psi \frac{i \hbar}{m} \psi^* \Big\}
\end{aligned}$$

Let us rewrite this using the momentum operator
$\hat{p} = -i \hbar \nabla$ as follows, noting that $\hat{p} / m$ is
simply the velocity operator $\hat{v}$:

$$\begin{aligned}
    \boxed{
        \vec{J}
        = \frac{1}{2 m} ( \psi^* \hat{p} \psi - \psi \hat{p} \psi^*)
        = \mathrm{Re} \Big\{ \psi^* \frac{\hat{p}}{m} \psi \Big\}
        = \mathrm{Re} \{ \psi^* \hat{v} \psi \}
    }
\end{aligned}$$

Returning to the derivation of $\vec{J}$, we now have the following
equation:

$$\begin{aligned}
    \pdv{P}{t}
    = \int_{V} \pdv{|\psi|^2}{t} \dd[3]{\vec{r}}
    = - \int_{V} \nabla \cdot \vec{J} \dd[3]{\vec{r}}
\end{aligned}$$

By removing the integrals, we thus arrive at the **continuity equation**
for $\vec{J}$:

$$\begin{aligned}
    \boxed{
        \nabla \cdot \vec{J}
        = - \pdv{|\psi|^2}{t}
    }
\end{aligned}$$

This states that the total probability is conserved, and is reminiscent of charge
conservation in electromagnetism. In other words, the probability at a
point can only change by letting it "flow" towards or away from it. Thus
$\vec{J}$ represents the flow of probability, which is analogous to the
motion of a particle.

As a bonus, this still holds for a particle in an electromagnetic vector
potential $\vec{A}$, thanks to the gauge invariance of the Schrödinger
equation. We can thus extend the definition to a particle with charge
$q$ in an SI-unit field, neglecting spin:

$$\begin{aligned}
    \boxed{
        \vec{J}
        = \mathrm{Re} \Big\{ \psi^* \frac{\hat{p} - q \vec{A}}{m} \psi \Big\}
    }
\end{aligned}$$