# Probability current

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.14.

Using the Schrödinger equation we can derive an interesting quantity called the probability current. Using the probabilistic interpretation of the wave function, the probability of a particle being between ${x=a}$ and ${x=b}$ is

$\displaystyle P_{ab}=\int_{a}^{b}\left|\Psi^*\Psi\right|dx \ \ \ \ \ (1)$

The rate of change of this probability can then be expressed in terms of spatial derivatives using the Schrödinger equation:

 $\displaystyle \frac{dP_{ab}}{dt}$ $\displaystyle =$ $\displaystyle \int_{a}^{b}\left[\frac{\partial\Psi^*}{\partial t}\Psi+\frac{\partial\Psi}{\partial t}\Psi^*\right]dx\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{a}^{b}\left\{ -\frac{i\hbar}{2m}\frac{\partial^{2}\Psi^*}{\partial x^{2}}-\frac{1}{i\hbar}V\Psi^*\right\} \Psi dx\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle$ $\displaystyle +\int_{a}^{b}\left\{ \frac{i\hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+\frac{1}{i\hbar}V\Psi\right\} \Psi^*dx\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2m}\int_{a}^{b}\left[\frac{\partial^{2}\Psi}{\partial x^{2}}\Psi^*-\frac{\partial^{2}\Psi^*}{\partial x^{2}}\Psi\right]dx \ \ \ \ \ (5)$

We can now apply integration by parts to each term.

 $\displaystyle \int_{a}^{b}\frac{\partial^{2}\Psi}{\partial x^{2}}\Psi^*dx$ $\displaystyle =$ $\displaystyle \left.\frac{\partial\Psi}{\partial x}\Psi^*\right|_{a}^{b}-\int_{a}^{b}\frac{\partial\Psi}{\partial x}\frac{\partial\Psi^*}{\partial x}dx\ \ \ \ \ (6)$ $\displaystyle -\int_{a}^{b}\frac{\partial^{2}\Psi^*}{\partial x^{2}}\Psi dx$ $\displaystyle =$ $\displaystyle \left.-\frac{\partial\Psi^*}{\partial x}\Psi\right|_{a}^{b}+\int_{a}^{b}\frac{\partial\Psi}{\partial x}\frac{\partial\Psi^*}{\partial x}dx \ \ \ \ \ (7)$

Adding these terms together, we get

$\displaystyle \frac{dP_{ab}}{dt}=\frac{i\hbar}{2m}\left[\left.\frac{\partial\Psi}{\partial x}\Psi^*\right|_{a}^{b}\left.-\frac{\partial\Psi^*}{\partial x}\Psi\right|_{a}^{b}\right] \ \ \ \ \ (8)$

If we define the probability current as

$\displaystyle J(x,t)\equiv\frac{i\hbar}{2m}\left(\frac{\partial\Psi^*}{\partial x}\Psi-\frac{\partial\Psi}{\partial x}\Psi^*\right) \ \ \ \ \ (9)$

we can write the rate of change of probability as

$\displaystyle \frac{dP_{ab}}{dt}=J(a,t)-J(b,t) \ \ \ \ \ (10)$

As an example, if the wave function is given by

$\displaystyle \Psi(x,t)=\left(\frac{2am}{\pi\hbar}\right)^{1/4}e^{-a\left[\left(mx^{2}/\hbar\right)+it\right]} \ \ \ \ \ (11)$

(we’ve taken the constant in front so that it normalizes the wave function), then

$\displaystyle \frac{\partial\Psi}{\partial x}=-\frac{1}{\pi^{1/4}}\left(\frac{2am}{\hbar}\right)^{5/4}xe^{-a\left[\left(mx^{2}/\hbar\right)+it\right]} \ \ \ \ \ (12)$

So for the probability current we get

 $\displaystyle J(x,t)$ $\displaystyle =$ $\displaystyle -\frac{i\hbar}{2m}\frac{1}{\pi^{1/4}}\left(\frac{2am}{\hbar}\right)^{5/4}xe^{-a\left[\left(mx^{2}/\hbar\right)-it\right]}\left(\frac{2am}{\pi\hbar}\right)^{1/4}e^{-a\left[\left(mx^{2}/\hbar\right)+it\right]}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle$ $\displaystyle +\frac{i\hbar}{2m}\frac{1}{\pi^{1/4}}\left(\frac{2am}{\hbar}\right)^{5/4}xe^{-a\left[\left(mx^{2}/\hbar\right)+it\right]}\left(\frac{2am}{\pi\hbar}\right)^{1/4}e^{-a\left[\left(mx^{2}/\hbar\right)-it\right]}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (15)$

A bit of an anti-climax after all that mathematics.

## 6 thoughts on “Probability current”

1. chang

Great solution. For (b), you can also convince yourself that Psi*del(Psi*)/del(x) = psi \times d(psi)/dx (sorry for the terrible typesetting, but Psi here is the capital psi and psi is the small psi, del for partial derivative and d for definite derivative) and you get psi \times d(psi)/dx – psi\times d(psi)/dx = 0 inside the parenthesis for J.