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Schrodinger equation in the spherical coordinates

  1. Nov 28, 2006 #1
    If using spherical coordinates (r, theta, phi) , what is the meaning of the canonical momentum of theta, phi? What are their definitions and mathematical form?

    In solving the Hydrogen problem, one has not take into consideration P_theta and P_phi at all.

    Quantum River
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  3. Nov 28, 2006 #2
    Canonical momenta:

    [tex]p_i \equiv \frac{\partial L}{\partial \dot{q}_i}.[/tex]

    So you have to get the Lagrangian out of the Hamiltonian, which is given by

    [tex]L(q_i, \dot{q}_i) = \sum_i p_i q_i - H(p_i, q_i),[/tex]

    where you substitute for [itex]\dot{q}_i[/itex] wherever you find a [itex]p_i.[/itex]

    Remember that the canonical momenta in Q.M. are promoted to derivative operators, so that

    [tex]p_\theta \rightarrow -i\hbar \frac{\partial}{\partial \theta}.[/tex]

    This has to be because the canonical commutation relations hold for co-ordinates and their canonical momenta (since in classical mechanics, the corresponding relations hold for the Poisson bracket).

    So these momenta do in fact appear in the Hamiltonian for the hydrogen atom: they usually appear as squared terms in the [itex]\nabla ^2[/itex] term (i.e. the kinetic energy).
  4. Nov 29, 2006 #3


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    [tex] p_{r}=:\frac{\partial L}{\partial \dot{r}} [/tex]

    [tex] p_{\theta}=:\frac{\partial L}{\partial \dot{\theta}} [/tex]

    [tex] p_{\phi}=:\frac{\partial L}{\partial \dot{\phi}} [/tex]

    The form depends on the form of the lagrangian, of course. It's important the lagrangian be nondegenerate, else the quantization will be difficult.

  5. Nov 29, 2006 #4
    Physically, the canonical momenta for spherical coordinates turns out to be angular momentum components.
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