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In [1] it is mentioned that the probability flux

\begin{equation}\label{eqn:fluxAndMomentum:20}

\Bj(\Bx, t) = -\frac{i\Hbar}{2 m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj },

\end{equation}

is related to the momentum expectation at a given time by the integral of the flux over all space

\begin{equation}\label{eqn:fluxAndMomentum:40}

\int d^3 x \Bj(\Bx, t) = \frac{\expectation{\Bp}_t}{m}.

\end{equation}

That wasn’t obvious to me at a glance, however, this can be seen by recasting the integral in bra-ket form. Let

\begin{equation}\label{eqn:fluxAndMomentum:60}

\psi(\Bx, t) = \braket{\Bx}{\psi(t)},

\end{equation}

and note that the momentum portions of the flux can be written as

\begin{equation}\label{eqn:fluxAndMomentum:80}

-i \Hbar \spacegrad \psi(\Bx, t) = \bra{\Bx} \Bp \ket{\psi(t)}.

\end{equation}

The current is therefore

\begin{equation}\label{eqn:fluxAndMomentum:100}

\begin{aligned}

\Bj(\Bx, t)

&= \frac{1}{2 m}

\lr{

\psi^\conj \bra{\Bx} \Bp \ket{\psi(t)}

+\psi {\bra{\Bx} \Bp \ket{\psi(t)} }^\conj

} \\

&= \frac{1}{2 m}

\lr{

{\braket{\Bx}{\psi(t)}}^\conj \bra{\Bx} \Bp \ket{\psi(t)}

+ \braket{\Bx}{\psi(t)} {\bra{\Bx} \Bp \ket{\psi(t)} }^\conj

} \\

&= \frac{1}{2 m}

\lr{

\braket{\psi(t)}{\Bx} \bra{\Bx} \Bp \ket{\psi(t)}

+

\bra{\psi(t)} \Bp \ket{\Bx} \braket{\Bx}{\psi(t)}

}.

\end{aligned}

\end{equation}

Integrating and noting that the spatial identity is \( 1 = \int d^3 x \ket{\Bx}\bra{\Bx} \), we have

\begin{equation}\label{eqn:fluxAndMomentum:n}

\int d^3 x \Bj(\Bx, t)

=

\bra{\psi(t)} \Bp \ket{\psi(t)},

\end{equation}

This is just the expectation of \( \Bp \) with respect to a specific time-instance state.

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.