Proving Continuity Equation for Complex V

[Tales Roberto]

Hi. I have a new one!

Prove that if [tex] V \left(\stackrel{\rightarrow}{r} , t \right) [/tex] is complex the continuity equation becomes [tex] \frac{\partial}{\partial t}P \left(\stackrel{\rightarrow}{r},t \right)+\nabla \stackrel{\rightarrow}{j} \left(\stackrel{\rightarrow}{r},t \right) = \frac{2}{h} \int \right[Im \ V \left(\stackrel{\rightarrow}{r},t \right) \right] P \left(\stackrel{\rightarrow}{r},t \right) d\stackrel{\rightarrow}{r} [/tex]

so that the addition of an imaginary part of the potential describes the presence of sources if I am V > 0 or sinks if I am V < 0. Show that if the wave function is [tex]\Psi \stackrel{\rightarrow}{r},t \right)[/tex] is square integrable

[tex] \frac{\partial}{\partial t} \int P \left(\stackrel{\rightarrow}{r},t \right) d\stackrel{\rightarrow}{r}= \frac{2}{h} \int \right[Im \ V \left(\stackrel{\rightarrow}{r},t \right) \right] P \left(\stackrel{\rightarrow}{r},t \right) d\stackrel{\rightarrow}{r} [/tex]

This first is easy and i can prove it, now the second part! However i need to go now, later i return to continue!

 

[Jhero]

Sure, I can help you with the second part. Let's start by defining the wave function, \Psi \left(\stackrel{\rightarrow}{r},t \right), which represents the state of a quantum system at a particular point in space and time. It is related to the probability density, P \left(\stackrel{\rightarrow}{r},t \right), by the equation \Psi \left(\stackrel{\rightarrow}{r},t \right) = \sqrt{P \left(\stackrel{\rightarrow}{r},t \right)}. This means that the square of the wave function, |\Psi|^2, is equal to the probability density.

Now, if we take the time derivative of the integral of the probability density, we get:

\frac{\partial}{\partial t} \int P \left(\stackrel{\rightarrow}{r},t \right) d\stackrel{\rightarrow}{r} = \int \frac{\partial}{\partial t} P \left(\stackrel{\rightarrow}{r},t \right) d\stackrel{\rightarrow}{r}

Using the continuity equation, we can substitute in the expression for the time derivative of the probability density:

\frac{\partial}{\partial t} \int P \left(\stackrel{\rightarrow}{r},t \right) d\stackrel{\rightarrow}{r} = \int \left(-\nabla \stackrel{\rightarrow}{j} \left(\stackrel{\rightarrow}{r},t \right) \right) d\stackrel{\rightarrow}{r}

Now, let's focus on the term inside the integral. We can use the definition of the probability current density, \stackrel{\rightarrow}{j} = \frac{\hbar}{2mi} \left(\Psi^* \nabla \Psi - \Psi \nabla \Psi^* \right), to rewrite it as:

\begin{align*} -\nabla \stackrel{\rightarrow}{j} \left(\stackrel{\rightarrow}{r},t \right) &= -\nabla \left(\frac{\hbar}{2mi} \left(\Psi^* \nabla \Psi - \Psi \nabla \Psi^* \right) \right)\\ &= \frac{\hbar}{2mi} \left(\nabla \Psi^* \n