The free particle: probability current

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.19.

The rate of change of probability of a particle in a given range of ${x}$ can be written as the difference in probability current at the two ends. The current is defined as

$\displaystyle J(x,t)\equiv\frac{i\hbar}{2m}\left(\frac{\partial\Psi^*}{\partial x}\Psi-\frac{\partial\Psi}{\partial x}\Psi^*\right) \ \ \ \ \ (1)$

For the free particle, a stationary state is given by

$\displaystyle \Psi(x,t)=Ae^{ikx}e^{-i\hbar k^{2}t/2m} \ \ \ \ \ (2)$

The probability current for this state is found by working out the derivative:

 $\displaystyle \frac{\partial\Psi}{\partial x}$ $\displaystyle =$ $\displaystyle ikAe^{ikx}e^{-i\hbar k^{2}t/2m}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle ik\Psi \ \ \ \ \ (4)$

So we get

 $\displaystyle J(x,t)$ $\displaystyle =$ $\displaystyle \frac{i\hbar k}{2m}\left(-i\left|\Psi\right|^{2}-i\left|\Psi\right|^{2}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar k}{m}\left|A\right|^{2} \ \ \ \ \ (6)$

(The complex exponentials cancel out in ${\left|\Psi\right|^{2}}$.) Since the current is positive, it ‘flows’ in the positive ${x}$ direction. Note that the current is independent of ${x}$, so the probability of a particle being found in any given range of ${x}$ is constant. (Actually, as we’ve seen, a free particle can’t exist in a single stationary state since such a state cannot be normalized.)

2 thoughts on “The free particle: probability current”

1. George

If I understood the theory well, a free particle cannot exist in a stationary state so we should probably have the integral over k form of the wave equation? Are we allowed to use stationary states as a mathematically useful trick which does not have any physical meaning, and getting right results? The integral form shouldn’t have given the same result?