Add "Probability current"

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+# Probability current
+
+In quantum mechanics, the *probability current* expresses the movement
+of the probability of finding a particle. Or 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 given by:
+
+$$\begin{aligned}
+    P = \int_{V} | \psi |^2 \dd[3]{\vec{r}}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \pd{P}{t}
+    &= \int_{V} \psi \pd{\psi^*}{t} + \psi^* \pd{\psi}{t} \dd[3]{\vec{r}}
+    = \frac{i}{\hbar} \int_{V} \psi (\hat{H} \psi^*) - \psi^* (\hat{H} \psi) \dd[3]{\vec{r}}
+    \\
+    &= \frac{i}{\hbar} \int_{V} \psi \Big( \!-\! \frac{\hbar^2}{2 m} \nabla^2 \psi^* + V(\vec{r}) \psi^* \Big)
+    - \psi^* \Big( \!-\! \frac{\hbar^2}{2 m} \nabla^2 \psi + V(\vec{r}) \psi \Big) \dd[3]{\vec{r}}
+    \\
+    &= \frac{i \hbar}{2 m} \int_{V} - \psi \nabla^2 \psi^* + \psi^* \nabla^2 \psi \dd[3]{\vec{r}}
+    = - \int_{V} \nabla \cdot \vec{J} \dd[3]{\vec{r}}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \pd{P}{t}
+    = \int_{V} \pd{|\psi|^2}{t} \dd[3]{\vec{r}}
+    = - \int_{V} \nabla \cdot \vec{J} \dd[3]{\vec{r}}
+\end{aligned}$$
+
+By removing the integrals, we thus arrive at the *continuity equation*
+for $\vec{J}$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vec{J}
+        = - \pd{|\psi|^2}{t}
+    }
+\end{aligned}$$
+
+This states that 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:
+
+$$\begin{aligned}
+    \boxed{
+        \vec{J}
+        = \mathrm{Re} \Big\{ \psi^* \frac{\hat{p} - q \vec{A}}{m} \psi \Big\}
+    }
+\end{aligned}$$