# Probability Current

1. Jul 21, 2007

### ehrenfest

I am confused about the the probability density equation:

$$\vec j = \frac{\hbar}{2mi}\left(\Psi^* \vec \nabla \Psi - \Psi \vec \nabla \Psi^*\right)$$

Psi is not a ket on the right side, correct?
If not how can perform a del on it and get a vector on the right side?

Can someone give me a concret example of what you would plug in to this expression:
$$\Psi^* \vec \nabla \Psi$$

2. Jul 21, 2007

### Gokul43201

Staff Emeritus
Psi is not a ket, it is a position-space wavefunction, given by the inner product: $\psi _n(x) = \langle x|n \rangle$

E.g., in the 1D infinite square well of width L, $\psi_n(x)=\sqrt{2/L}~sin(n \pi x/L)$

The gradient of a scalar field is a vector field.