# Integrate continuity equation in QM

From Shankar's QM book pg. 166:

The continuity equation for probability density in QM is $$\frac{\partial P(\vec{r},t)}{\partial t}=-\nabla \cdot \vec{j}(\vec{r},t),$$ where $$P=\psi^*\psi$$ is the probability density and $$\vec{j}$$ is the probability current density given by$$\vec{j}=\frac{\hbar}{2mi}(\psi^*\nabla\psi-\psi\nabla\psi^*).$$

Integrating over all space, we get $$\frac{d}{dt}\int P(\vec{r},t) d^3\vec{r}=-\int \nabla\cdot\vec{j}(\vec{r},t)d^3\vec{r}=-\int_{S_\infty}\vec{j}\cdot d\vec{S}$$ where the last integral is over the sphere at infinity.

It was said that for wave functions which are normalizable to unity, $$r^{3/2}\psi\rightarrow0$$ as $$r\rightarrow \infty$$ in order that $$\int\psi^*\psi r^2drd\Omega$$ is bounded, and the surface integral of $$\vec{j}$$ on $$S_\infty$$ vanishes.

Why is the author considering the expressions $$r^{3/2}\psi$$ and $$\int\psi^*\psi r^2drd\Omega$$? What do these expressions represent?

My understanding is that $$\psi\rightarrow0$$ as $$r\rightarrow \infty$$ for normalizable wavefunctions, hence $$\vec{j}$$ vanishes and the integral $$\int_{S_\infty} \vec{j} \cdot d\vec{S}$$ vanishes.

• This is integration in cylindrical coordinates: $dxdydz=r^2drd\thetad\phi$. May 5 at 6:59

The integral $$\int \psi^*\psi\ r^2\ dr\ d\Omega$$ is just the same as integral $$\int P(\mathbf{r},t) d^3\mathbf{r}$$ representing the total probability of finding the particle anywhere. Remember that $$P(\mathbf{r},t)=\psi^*\psi$$ is the probability density, and $$d^3\mathbf{r}=r^2\ dr\ d\Omega$$ is the volume element (a slice with thickness $$dr$$ and area $$r^2\ d\Omega$$).
The requirement $$\lim_{r\to\infty}r^{3/2}\psi=0$$ makes sure that in the integral $$\int \psi^*\psi\ r^2\ dr\ d\Omega$$ the integrand $$\psi^*\psi\ r^2$$ stays smaller than $$O(\frac{1}{r})$$, and thus the integral from $$r=0$$ to $$r=\infty$$ doesn't get infinitely large.