Obtain, with the aid of the time-dependent Schrödinger equation, the conservation equation

$$\frac{\partial}{\partial t} \rho(\mathbf{x}, t)+\nabla \cdot \mathbf{j}(\mathbf{x}, t)=0$$

where $\rho(\mathbf{x}, t)$ is the probability density and $\mathbf{j}(\mathbf{x}, t)$ is the probability current. What have you assumed about the potential energy of the system?

Show that if the potential $U(\mathbf{x}, t)$ is complex the conservation equation becomes

$$\frac{\partial}{\partial t} \rho(\mathbf{x}, t)+\nabla \cdot \mathbf{j}(\mathbf{x}, t)=\frac{2}{\hbar} \rho(\mathbf{x}, t) \operatorname{Im} U(\mathbf{x}, t)$$

Take the potential to be time-independent. Show, with the aid of the divergence theorem, that

$$\frac{d}{d t} \int_{\mathbb{R}^{3}} \rho(\mathbf{x}, t) d V=\frac{2}{\hbar} \int_{\mathbb{R}^{3}} \rho(\mathbf{x}, t) \operatorname{Im} U(\mathbf{x}) d V$$

Assuming the wavefunction $\psi(\mathbf{x}, 0)$ is normalised to unity, show that if $\rho(\mathbf{x}, t)$ is expanded about $t=0$ so that $\rho(\mathbf{x}, t)=\rho_{0}(\mathbf{x})+t \rho_{1}(\mathbf{x})+\cdots$, then

$$\int_{\mathbb{R}^{3}} \rho(\mathbf{x}, t) d V=1+\frac{2 t}{\hbar} \int_{\mathbb{R}^{3}} \rho_{0}(\mathbf{x}) \operatorname{Im} U(\mathbf{x}) d V+\cdots$$

As time increases, how does the quantity on the left of this equation behave if $\operatorname{Im} U(\mathbf{x})<0$ ?