Question

The probability current density (probability density current) in one dimension is given by:

• A. \(\dfrac{i\hbar}{2m}\left(\psi\dfrac{\partial\psi^{*}}{\partial x}-\psi^{*}\dfrac{\partial\psi}{\partial x}\right)\)
• B. \(\dfrac{i\hbar}{2m}\left(\psi^{*}\dfrac{\partial\psi}{\partial x}-\psi\dfrac{\partial\psi^{*}}{\partial x}\right)\)
• C. \(-\dfrac{i\hbar}{2m}\left(\psi\dfrac{\partial\psi^{*}}{\partial x}-\psi^{*}\dfrac{\partial\psi}{\partial x}\right)\)
• D. \(-\dfrac{i\hbar}{2m}\left(\psi^{*}\dfrac{\partial\psi}{\partial x}-\psi\dfrac{\partial\psi^{*}}{\partial x}\right)\)

Hint:

Use the continuity equation with the Schrodinger equation; the potential cancels and the flux carries a leading negative sign.

Answer 1

The Correct Option is D

Write the current operator directly from the expectation value of momentum. The momentum operator is $\hat{p}=-i\hbar\,\partial/\partial x$, and the probability current is the real flux $J=\mathrm{Re}\!\left[\psi^{*}\dfrac{\hat{p}}{m}\psi\right]$.

Substituting the operator gives $J=\mathrm{Re}\!\left[-\dfrac{i\hbar}{m}\psi^{*}\dfrac{\partial\psi}{\partial x}\right]$. The real part of a complex number $z$ equals $\tfrac12(z+z^{*})$, so

\[J=\frac{1}{2}\left(-\frac{i\hbar}{m}\psi^{*}\frac{\partial\psi}{\partial x}+\frac{i\hbar}{m}\psi\frac{\partial\psi^{*}}{\partial x}\right).\]

Factoring out $-\dfrac{i\hbar}{2m}$ yields $J=-\dfrac{i\hbar}{2m}\left(\psi^{*}\dfrac{\partial\psi}{\partial x}-\psi\dfrac{\partial\psi^{*}}{\partial x}\right)$. This flux is real (as any measurable current must be) and reduces to $J=\dfrac{\hbar k}{m}|A|^{2}$ for a plane wave $\psi=Ae^{ikx}$, confirming its physical meaning as (density) times (velocity). Hence option (D) is correct.

\[\boxed{J=-\frac{i\hbar}{2m}\left(\psi^{*}\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^{*}}{\partial x}\right)}\]