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Probability Current Free Particle

  1. Feb 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the probability current of a free particle.

    2. Relevant equations

    [itex]\Psi[/itex](x,t) = Aei(kx-[itex]\frac{(hbar)k^{2}t}{2m}[/itex])

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

    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. jcsd
  3. Feb 13, 2014 #2
    To answer the simplest question, [itex]\psi^*[/itex] is just the complex conjugate of ψ:

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

    [itex]c=Ae^{ix}[/itex], [itex]c^*=A^*e^{-ix}[/itex]

    Finally, [itex]cc^*=c^*c=|c|^2[/itex], which is a positive real number.

    The equation for a free particle in one dimension is

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

    Its complex conjugate is

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

    Note that by writing this as the wave function of the particle your starting with the assumption that [itex]\psi[/itex] 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 [itex]\psi[/itex] 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:
    [itex]J(x,t)=\frac{i\hbar}{2m}(\psi\frac{∂\psi^*}{∂x}-\psi^*\frac{∂\psi}{∂x})[/itex]

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