How Do You Find the Probability Current of a Free Particle?

[petera88]

Homework Statement

Find the probability current of a free particle.

Homework Equations

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

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

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*?

 

[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.