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What if V(x) is complex?

  1. Sep 10, 2008 #1
    I see how one can calculate

    ∂/∂t P(x,t)= (∂ψ*)/∂t ψ+ψ*∂ψ/∂t

    by plugging in the general schrodinger equation and its complex conjugate but I have read that to do this V(x) must be real

    Why does the potential energy V(x) have to be real though?

    How would you find ∂/∂t P(x,t) if V(x) were complex?

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Sep 10, 2008 #2

    reilly

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    Look up the "Optical Potential". Just as is the case in, say, microwave propagation, the imaginary component of the potential describes absorption or loss -- much like resistance in circuit theory.
    Regards,
    Reilly Atkinson
     
  4. Sep 11, 2008 #3
    Thank you for the reply! I'm not sure I understand exactly how this works though, I have not yet taken complex analysis. Why does V(x) have to be real for this to work?

    Thanks again
     
  5. Sep 19, 2008 #4
    The equation you wrote above for P(x,t) is just a consequence of normal calculus, where for most intents and purposes you can treat i as an ordinary constant. (Complex differentiability is a slightly different kettle of fish, but here you aren't differentiating wrt a complex variable, so you don't have to worry about it :smile:) It's nothing to do with the Schroedinger equation- all you need to assume is that [tex]P(x,t)=\Psi\Psi^*[/tex] and use the chain rule.

    The schroedinger equation is the equation that a function psi has to satisfy in order to describe a real system. The reason the potential term usually has to be real is because for a potential V(x) that doesn't depend on time (as yours doesn't seem to!) then you can obtain the time-independent Schroedinger equation:
    [tex](-\frac{\hbar^2}{2m}\nabla^2+V(x))\Psi=E\Psi[/tex]
    Here E is an energy-which must be real, as it's a physical, measurable quantity. So whatever you get in front of psi when you apply the big operator in brackets (the hamiltonian) must be a real constant- hence, V must be real. More simply, V is a potential energy function- doesn't it make sense that it should be real?

    I wrote "usually" above... I've never heard of an optical potential before Reilly's post- and after a quick wiki/google I'm still none the wiser! I think I recognise what he's alluding to though- I'm sure he'll correct me if I point you in completely the wrong direction here!
    Say you have some electromagnetic wave of the form
    [tex]E=E_0 e^{i(kx-\omega t)}[/tex]
    If your wavenumber k is imaginary, then your wave is a product of a complex exponential (from the real part of k) and a decaying exponential (from the imaginary part of k, using i^2 =-1). So the amplitude of the field decays exponentially, but the field still oscillates.
    Now, how this translates to QM I have no idea. The constant E in the TISE above arises as a constant of separation (if you're familiar with the method of solving partial differential equations by separation of variables?). So, just from the maths, it could be complex. But it would have to have a negative imaginary part, or the resulting solution would explode exponentially with time, not decay (separable solutions are of the form [tex]\psi(x) e^{-iEt}[/tex], so you'd get an increasing real part because of that minus sign). Reilly's a lot more knowledgable than I am, so if he says there's a situation in QM that's more like the EM wave I described above, I'll believe him! But I don't know what it is.
    Anyway, I hope the first parts of my answer helped at least!
     
  6. Sep 21, 2008 #5
    Thank you so much for your help, I see now how V(x) must be real as a potential energy function (except in cases of particle absorbtion and decay which I now know is what this problem is about). Again thank you for taking the time to help me with this problem.
     
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