# Probability current density

1. Dec 6, 2007

### stunner5000pt

1. The problem statement, all variables and given/known data
This isnt quite a question per se. But in my review im supposed to know how to derive a condition
$$\psi (\infty,t) = \psi^*(-\infty,t) = 0$$

In addition i'm supposed to show that

$$-\frac{d}{dt} \int_{V\infty} \rho(r,t) dV = \oint_{S\infty} J(r,t)\cdot dS' =0$$

where $$\rho(r,t) = |\Psi|^2$$
and J is the probability current.

2. The attempt at a solution
Im not sure how that condition can be used.
Can it used to how that the probability goes to zero as we approach infinity?

Integration of the probability current over an infinite space gives zero? Do i simply integrate the probability current by parts and use the condition given??

Also one question... is it possible to derive the momentum operator?? Can i do the same for the position operator?

Could i use $$<p> = m \frac{d}{dt} <r>$$
where $$<r> = \int r |\Psi|^2 d^3 r$$

When doing the time derivative of the integrand, am i allowed to assume that dr/dt = 0?? After all the operators are not supposed to vary with time... right?
Thanks for the help!

Last edited: Dec 6, 2007
2. Dec 7, 2007

### dwintz02

You are expected to know that probability is conserved (remember this when solving). Bring the d/dt into the integral (remember the chain rule) and substitute in from the time-DEpendent Schrodinger equation and integrate by parts to show the first part. When you integrate by parts, what will 'kill' your extra boundary term?

Edit: The integration of probability over all space does not give zero (remember normalization) but the time-derivative of probability equals zero (probability is conserved, aka the particle must be somewhere).

Last edited: Dec 7, 2007