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How to calculate a commutator from hydrogen atom radial equation

  1. Sep 17, 2009 #1
    This is not homework, but is not general discussion, so not sure where this would go.

    In class we were deriving with the radial equations of a hydrogen atom, and in one of the equations was the commutator term:

    [tex]\left[ \frac{d}{d\rho}, \frac{1}{\rho}\right][/tex]


    my attempt was:
    [tex]\left[ \frac{d}{d\rho}, \frac{1}{\rho}\right][/tex]
    =
    [tex]\left(\frac{d}{d\rho} \cdot \frac{1}{\rho}\right) - - \left(\frac{1}{\rho} \cdot \frac{d}{d\rho}\right) = 2\left(\frac{d}{d\rho}\frac{1}{\rho}\right) = \frac{-2}{\rho^{2}}[/tex]


    I know above is incorrect, but how is this commutator calculated? For some reason when I try, I get the wrong result, but I'm confused on exactly how to do it.


    Also, I am confused on how to accept that the complex conjugate of [tex]\frac{d}{d\rho} = \frac{-d}{d\rho}[/tex].

    Can anyone help clear this up for me?
     
  2. jcsd
  3. Sep 17, 2009 #2

    gabbagabbahey

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    Gold Member

    When you operate on a wave-function with the commutator

    [tex]\left[ \frac{\partial}{\partial\rho}, \frac{1}{\rho}\right]=\frac{\partial}{\partial \rho}\frac{1}{\rho}-\frac{1}{\rho}\frac{\partial}{\partial \rho}[/tex]

    The first term [itex]\frac{\partial}{\partial \rho}\frac{1}{\rho}[/itex] tell you to first multiply the wavefunction by [itex]\frac{1}{\rho}[/itex] and then take the partial derivative of the result with respect to [itex]\rho[/itex]....that means, you are taking the derivative of a product of two functions, and so you need to use the product rule:

    [tex]\frac{\partial}{\partial \rho}\frac{1}{\rho}\psi=-\frac{1}{\rho^2}\psi+\frac{1}{\rho}\frac{\partial \psi}{\partial \rho}[/tex]

    So, in operator form,

    [tex]\frac{\partial}{\partial \rho}\frac{1}{\rho}=-\frac{1}{\rho^2}+\frac{1}{\rho}\frac{\partial }{\partial \rho}[/tex]
     
  4. Sep 17, 2009 #3

    turin

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    You are probably thinking of the Hermitian conjugate or adjoint. Basically, for differential operators this amounts to "integration by parts" (or what a friend of mine would call the "inverse product rule"):

    f(x)(dg/dx) = d(f(x)g(x))/dx - (df/dx)g(x)

    So, when there is a funciton f(x) on the left and a function g(x) on the right, the derivative can "act to the right" on g(x), like normal, or it can "act to the left" on f(x) with an extra minus sign. However, you must take care to deal with the "surface terms" - the extra "total derivative" that appears on the RHS.

    To relate this to complex conjugation:
    The complex conjugate of an inner product is equal to the inner product in which the bra and ket have switched places. So, for differential operators, this causes the derivative to "act in the opposite direction".
     
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