# Homework Help: Probability Current Free Particle

1. Feb 12, 2014

### petera88

1. The problem statement, all variables and given/known data

Find the probability current of a free particle.

2. Relevant equations

$\Psi$(x,t) = Aei(kx-$\frac{(hbar)k^{2}t}{2m}$)

J(x,t) = $\frac{ihbar}{2m}(ψψ*' - ψ*ψ')$

3. The attempt at a solution

I figured it was just take the derivative of the time dependent wave function and plug it in. This is my first experience with quantum mechanics so I find myself getting caught up on working with the wave function. My question it if it's real, then the psi and psi* are the same and it would equal 0. This isn't the answer of course. How do I work with the complex psi? What's the difference in the wave function for psi and psi*?

2. Feb 13, 2014

### smithhs

To answer the simplest question, $\psi^*$ is just the complex conjugate of ψ:

For some complex number $c=a+bi$, $c^*=a-bi$.
For a real number $a$, $a=a^*$
In the case of complex exponential functions:

$c=Ae^{ix}$, $c^*=A^*e^{-ix}$

Finally, $cc^*=c^*c=|c|^2$, which is a positive real number.

The equation for a free particle in one dimension is

$\psi(x,t) = Ae^{ikx}e^{-i\frac{\hbar k^2}{2m} t}$

Its complex conjugate is

$\psi^*(x,t) = A^*e^{-ikx}e^{i\frac{\hbar k^2}{2m} t}$

Note that by writing this as the wave function of the particle your starting with the assumption that $\psi$ is complex; it almost always IS complex, with a few exceptions (such as the energy eigenstates the particle in a box. When a particle's wave function is real there is 0 probability current; the probability distribution of the particle does not evolve in time (though its wave function still does). You work with complex $\psi$ like any other function, just making sure to due the complex arithmetic correctly; the derivitives all behave the same way as a real valued function.

So, back to the problem:

The probability current is defined as:
$J(x,t)=\frac{i\hbar}{2m}(\psi\frac{∂\psi^*}{∂x}-\psi^*\frac{∂\psi}{∂x})$

Just take derivitives like you would normally; I will say that the final answer is not zero.