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Derivatives in spherical coordinates

  1. Apr 27, 2008 #1
    In quantum mechanics the momentum operator is a constant multiplied by the partial derivative d/dx. In spherical coordinates it's turning into something like that:
    constant*(1/r)(d^2/dr^2)r

    can anyone explain please how this result is obtained?
     
  2. jcsd
  3. Apr 27, 2008 #2

    Dick

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    That doesn't look like d/dx, it looks like the radial part of the laplacian. It's a second derivative. In general to convert derivatives between coordinate systems you use the chain rule for partial derivatives.
     
  4. Apr 27, 2008 #3
    the momentum operator is often written as -ihd/dx... but in that case x is the position vector. More generally, the momentum op is -ihDel (Del = the gradient operator, more easily generalized to different coordinate systems). (note also that i have written 'h' but mean "h-bar").
    As dick said, the operator you've shown doesn't look quite right... if something is only changing radially, then the grad in spherical is just d/dr (i believe).
     
  5. Apr 28, 2008 #4
    sorry, my mistake.
    I meant the radial part of the (momentum)^2 in spherical coordinates.

    If the Hamiltonian is:
    (Pr)^2/2m + L^2/2mr^2 +V(r)

    what's wrong in writing (Pr)^2 as
    -(hbar)^2*(d/dr)^2
    ?

    why is it written like that:
    (-(hbar)^2/r)*(d/dr)^2*r
    ?
     
  6. Apr 28, 2008 #5

    Dick

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  7. Apr 28, 2008 #6

    Dick

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    BTW there are much better ways to derive this using differential geometry, but it gives you an idea of WHY the answer isn't as simple as you think. No need to follow all of the details.
     
  8. Apr 28, 2008 #7
    wow... I'll take a look at that, thanks!
     
  9. Nov 24, 2008 #8
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