Answer to Question #238394 in Quantum Mechanics for Matimu

Question #238394

. Consider the SchrΓΆdinger equation for a particle in the presence of the potential of the form 𝑉(π‘₯) = π‘ˆ(π‘₯) + π‘–π‘Š(π‘₯) where π‘ˆ and π‘Š are real functions of π‘₯. What form does the conservation equation take?


1
Expert's answer
2021-09-17T09:22:52-0400

let us consider the SchrΓΆdinger equation with potential 𝑉(π‘₯)=π‘ˆ(π‘₯)+π‘–π‘Š(π‘₯)𝑉(π‘₯) = π‘ˆ(π‘₯) + π‘–π‘Š(π‘₯):


iβ„βˆ‚Οˆβˆ‚t=βˆ’β„22mβˆ‡2ψ+(π‘ˆ(π‘₯)+π‘–π‘Š(π‘₯))ψ(1)i\hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2\psi+( π‘ˆ(π‘₯) + π‘–π‘Š(π‘₯))\psi\quad (1)

The conjugate equation

βˆ’iβ„βˆ‚Οˆβˆ—βˆ‚t=βˆ’β„22mβˆ‡2Οˆβˆ—+(π‘ˆ(π‘₯)βˆ’π‘–π‘Š(π‘₯))Οˆβˆ—(2)-i\hbar \frac{\partial \psi*}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2\psi^*+( π‘ˆ(π‘₯) - π‘–π‘Š(π‘₯))\psi^*\quad (2)

After multiplying Eqn 1 by Οˆβˆ—\psi^*, Eqn 2 by ψ\psi and subtracting, we get

iℏ(βˆ‚Οˆβˆ‚tΟˆβˆ—+βˆ‚Οˆβˆ—βˆ‚tψ)=i\hbar\left( \frac{\partial \psi}{\partial t}\psi^*+\frac{\partial \psi^*}{\partial t}\psi\right)=

βˆ’β„22m(Οˆβˆ—βˆ‡2Οˆβˆ’Οˆβˆ‡2Οˆβˆ—)+2iW(x)ΟˆΟˆβˆ—-\frac{\hbar^2}{2m}\left(\psi^*\nabla^2\psi-\psi\nabla^2\psi^*\right)+2iW(x)\psi\psi^*

or

βˆ‚Οβˆ‚t+βˆ‡jβˆ’Ξ³Ο=0\frac{\partial \rho}{\partial t}+\nabla {\bf j}-\gamma\rho=0

Here

ρ=ΟˆΟˆβˆ—\rho=\psi\psi^*

j=ℏ2mi(Οˆβˆ—βˆ‡Οˆβˆ’Οˆβˆ‡Οˆβˆ—){\bf j}=\frac{\hbar}{2mi}\left(\psi^*\nabla\psi-\psi\nabla\psi^*\right)

Ξ³=2W(x)/ℏ\gamma=2W(x)/\hbar


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