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---
title: "Probability current"
firstLetter: "P"
publishDate: 2021-02-22
categories:
- Quantum mechanics
- Physics
date: 2021-02-22T21:37:26+01:00
draft: false
markup: pandoc
---
# 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:
$$\begin{aligned}
\boxed{
\vec{J}
= \mathrm{Re} \Big\{ \psi^* \frac{\hat{p} - q \vec{A}}{m} \psi \Big\}
}
\end{aligned}$$
|
