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

Step 1: Start from the continuity equation of probability. If \(P=\psi^{*}\psi\) is the probability density, conservation demands \(\dfrac{\partial P}{\partial t}+\dfrac{\partial J}{\partial x}=0\), where \(J\) is the probability current density.

Step 2: Differentiate \(P\) and use the time-dependent Schrodinger equation \(i\hbar\dfrac{\partial\psi}{\partial t}=-\dfrac{\hbar^{2}}{2m}\dfrac{\partial^{2}\psi}{\partial x^{2}}+V\psi\) and its complex conjugate. The potential terms cancel, giving

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

Step 3: Matching this with \(\dfrac{\partial P}{\partial t}=-\dfrac{\partial J}{\partial x}\) identifies

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

Step 4: This is the standard probability current density, matching option (D).

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