Quantum Mechanics – Identity in the Continuity Equation of a Wave Function

conservation-lawshomework-and-exercisesquantum mechanicsschroedinger equation

In my scrips about theoretical quantum mechanics I got stuck at the continuity equation for a wave function. Specifically at the term for the Probability current $\vec{j}$.
$$
\nabla\vec{j}=\frac{\hbar}{2mi}\nabla(\Psi\nabla\Psi^*-\Psi^*\nabla\Psi)=\frac{\hbar}{m}\mathrm{Im}(\Psi^*\nabla\Psi)
$$

where $\nabla=\nabla_{\vec{r}}$ and $\Psi=\Psi(\vec{r},t)$
Now I don't know how I can come from the middle term to the right term. Till now I've got
$$
\frac{\hbar}{2mi}\nabla(\Psi\nabla\Psi^*-\Psi^*\nabla\Psi)=\frac{\hbar}{2mi}(\nabla\Psi\nabla\Psi^*+\Psi\nabla^2\Psi^*-\nabla\Psi^*\nabla\Psi-\Psi^*\nabla^2\Psi)
$$

But I don't know how to work with the $\mathrm {Im}()$ now. I hope someone can give me a tip/help me.

Best Answer

The equation you've written is not correct, for more than one reason.

  1. The left hand side is a divergence and as such is a scalar. The right hand side, however, is the imaginary part of a (function times a) gradient, which is a vector. Therefore, you cannot equate the two, no matter how hard you try!

  2. I also think the equation for the divergence of $\mathbf{j}$ has the wrong sign, there should be an overall negative sign in front of it.

Once both of these have been fixed, we can try to get an "answer". From the identity: $$\nabla \cdot \mathbf{j} = -\frac{\hbar}{2mi}\nabla \cdot\Bigg(\Psi \nabla \Psi^* - \Psi^* \nabla \Psi \Bigg),$$ one can infer that the probability current is given by:

$$\mathbf{j} = -\frac{\hbar}{2mi} \Bigg(\Psi \nabla \Psi^* - \Psi^* \nabla \Psi \Bigg).$$

Now, it is this quantity that is proportional to the imaginary part of $\Psi^* \nabla \Psi$, by definition.

If you have a complex number $z = a+i b$, then you should be able to show quite simply that the real and imaginary parts ($a$ and $b$ respectively) are given by:

\begin{align} a = \text{Re}(z)&= \frac{z + z^*}{2},\\ b = \text{Im}(z) &= \frac{z - z^*}{2i}.\\ \end{align}

Using $z = \Psi^* \nabla \Psi$, you should easily be able to show that $$\mathbf{j} = \frac{\hbar}{m} \text{Im}(\Psi^*\nabla\Psi).$$