What is the origin and the theoretical proof of the probability current in quantum mechanics, particularly quantum scattering theory?

Question

In quantum scattering theory, the probability current is commonly introduced to calculate differential cross sections but I cannot find its origin. Please someone has an explanation of this point?

Answer 1

The so called continuity relation is really the charge conservation law, which follows from the Schrodinger Lagrangian by the Noether theorem. Thus $e|\psi|^2$ is the charge density and needs not be postulated. It is the Lagrangian that is postulated and from it follow the Schrödinger equation and conservation of charge-current, energy-momentum, orbital angular and spin momentum.

I interpret Born's rule, as saying that this and other expressions of conserved quantities have a probabilistic meaning. This is a postulate and cannot be derived as far as I know.

Answer 2

Probability is a conserved quantity (the integral of probability density is unity). Therefore it satisfies a conservation law $${\partial \over \partial t}\rho=-\nabla\cdot\mathbf j $$ where $\mathbf j$ is the probability current or probability flux. The expected flow of probability must correspond to a flow of mass density, that is it corresponds to momentum density. The Hermitian operator with this property is the current density operator, $$\mathbf J(x):\psi \rightarrow - {i\over 2m}(\nabla - \overleftarrow \nabla)\psi \bigr |_x $$ This notation is used to make clear that the argument is from the current density operator, not from the wave function. One must be clear that probability current is related to momentum density, not directly to the momentum operator which would be found by integrating over $x$. It is perhaps clearer in ket notation $$\mathbf J(x)|\psi\rangle = - {i\over 2m}|x\rangle(\nabla - \overleftarrow \nabla)\langle x|\psi \rangle $$

Answer 3

The following paper answers my question, link.

The starting point is conservation probability using the continuity equation of charge conservation which reads

$$ \frac{d}{dt}\rho_e(r)=-\nabla.j_e(r)$$

If we assume that our system contains just one electron we can write $$\rho=e|\psi|^2,$$ where $\psi$ is the system wave function. Therefore, from the first equation, the rhs defines locally the probability current operator j(r) and the lhs could be evaluated by using the TDSE ($i\hbar\psi=H\psi$).

$$ \frac{d}{dt}\rho_e(r)= e\frac{d}{dt}(\psi*\psi)$$ which leads to $$e\frac{d}{dt}(\psi*\psi)=\frac{e}{i\hbar}[\psi^*(H\psi)-(H^*\psi^*)\psi]$$ Here, a critical point, if we take just the kinetic energy, we assume that the H Hamiltonian $H=\frac{\vec{p}^2}{2m}$ we find easily the following expression $$ \frac{d}{dt}\rho_e=\vec{\nabla}.\{\frac{ie\hbar}{2m}(\psi^*(\nabla \psi)-(\psi^*\nabla) \psi ) \}=-\vec{\nabla}.j$$ thus, we can write $$j(r)=-\frac{ie\hbar}{2m}(\psi^*(\nabla \psi)-(\psi^*\nabla) \psi )$$

In order to write the probability current as an operator that acts on $\psi$, I can use the expression given by Charles Francis. All this is ok, but I still have confusion to replace the H operator just by the kinetic energy and that makes the case for free particle but what about a general Hamiltonian contains other types of energy?