Probability current density

[stunner5000pt]

Consider the infinite square well potentail over the range 0<x<a with energy eigenfunctions $Psi _{n} (x,t)$

Show that if the quantum state of a particle is descirbed by a single eigenfuncation i.e. the particle has a sharply defined eneryg, then the probability current density inside the well vanishes

ok Delta E = 0

$$\Psi (x,t) = \psi(x) \exp(\frac{-iEt}{m}$$

$$j= \frac{\hbar}{2im} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right)$$

ok let's say that $$P(x,t) = | \Psi(x,t)|^2 = \psi^*(x) \psi(x)$$

we also know that \frac{\partial}{\partial t} (\psi^* \psi)= -\frac{\hbar}{2im} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) [/tex]

and hence $$\frac{\partial}{\partial t} P(x,t) = \frac{\partial}{\partial t}( \psi^*(x) \psi(x) )= 0 = -\frac{\partial}{\partial t} j(x,t) = 0$$

so the position derivative of probability current wrt position is zero

but does that show that it vanishes??

thank you for your input

 

[Jhero]

!

Yes, that does show that the probability current density inside the well vanishes. This is because if the quantum state of a particle is described by a single eigenfunction, it means that the particle has a sharply defined energy. This also means that the wavefunction is time-independent, as shown in the equation \Psi (x,t) = \psi(x) \exp(\frac{-iEt}{\hbar}). Therefore, the time derivative of the probability density is equal to zero, and as a result, the probability current density inside the well also vanishes. This indicates that the particle is not moving and is confined within the well, in accordance with the infinite square well potential.