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Complex conjugate derivative

  1. Dec 26, 2009 #1
    Hi.

    Sometimes in my quantum mechanics course we encounter derivatives such as [itex]\frac{d}{dz}z^*[/itex], i.e. the derivative of the complex conjugate of the complex variable z wrt z. We are told that this is just zero, even though I know that the complex conjugate is not an analytic function ... Has anyone else encountered this? I asked my teachers but they didn't know what was going on. Does this have anything to do with the fact that for an analytic function [itex]f(z)[/itex], the derivative wrt [itex]z^*[/itex] is zero, i.e. [itex]\frac{d}{dz^*}f(z)=0[/itex]?
     
  2. jcsd
  3. Dec 26, 2009 #2

    LCKurtz

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    The derivative of the complex conjugate isn't zero. It doesn't exist. Consider

    [tex]\frac{f(z)-f(z_0)}{z-z_0}[/tex]

    for [itex]f(z) = \overline z[/itex] and [itex]z_0 = 0[/itex]:

    [tex]\frac{\overline z - \overline 0}{z - 0} = \frac{x-iy}{x+iy}[/tex]

    Now [itex]z\rightarrow 0[/itex] won't give a limit at all as you can see by letting either x or y approach 0 first.
     
  4. Dec 26, 2009 #3
    Yeah, that's what I'm saying. The complex conjugate is not analytic (holomorphic). I'm just asking if anyone else have encountered the "trick" where you say that the derivative is just zero.

    I've seen this used in quantum mechanics in connection with the variational principle. You expand the state vector on a basis and then you take the expectation value of the Hamiltonian in this state. Now you want the combination of coefficients that will give the lowest energy. To do this you differentiate the expectation value wrt the expansion coefficients, and when you encounter a conjugated coefficient, you just say that the derivative is zero. This will turn into a matrix equation in the expansion coefficients.

    So the method works, I'm just asking why you can put the derivative of the conjugated expansion coefficients to zero.
     
    Last edited: Dec 26, 2009
  5. Dec 29, 2009 #4
    I think you're dealing with something similar to this: [tex]\frac{\partial f}{\partial \overline{z} } = \frac{1}{2} \left( \frac{\partial f}{\partial x } +i \frac{\partial f}{\partial y } \right)[/tex] and so a function [tex]f[/tex] is analytic iff [tex]\frac{\partial f}{\partial \overline{z} }=0[/tex] . This is sometimes taken to mean that [tex]f[/tex] does not depend on [tex]\overline{z}[/tex] which would give a certain meaning to your equation (assuming of course that there is some other function [tex]f[/tex] involved).
     
  6. Dec 30, 2009 #5
    This is used in mathematics too. But only advanced mathematics. Instead of considering f a function of two variables x and y (where z = x+iy) we can consider f a function of two variables z and [itex]z^*[/itex] (or, as mathematicians say, z and [itex]\overline{z}[/itex] ). So now we do partial derivatives with respect to these two variables. Please note that partial derivative [itex]\partial/\partial z[/itex] is NOT THE SAME as ordinary derivative [itex]d/dz[/itex]. The Cauchy-Riemann equations become [itex]\partial f/\partial \overline{z} = 0[/itex], very nice.

    Should physicists who don't know the mathematics avoid using it? Try telling that to physicists! It has no effect!
     
  7. Dec 30, 2009 #6
    That's a good point. I was thinking about analyticity in terms of the total derivative.
     
    Last edited: Dec 30, 2009
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