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Feynman Hellman Theorem: dependence of E on [itex]\ell[/itex] Hydrogen

  1. Nov 13, 2013 #1
    The theorem states
    [itex]\frac{\partial E}{\partial \lambda} = \langle \psi \mid \frac{\partial H}{\partial \lambda} \mid \psi \rangle[/itex]

    Where [itex]\mid \psi \rangle[/itex] is an eigenket of H.

    An example (given on Wikipedia) is to find [itex]\langle \psi \mid \frac{1}{r^2} \mid \psi \rangle [/itex] for a Hydrogen eigenstate using this method with [itex]\lambda = \ell[/itex]. It is straightforward to differentiate H with respect to [itex]\ell[/itex]. However the common expression for energy only depends [itex]n[/itex]. In the Wikipedia article there is
    [itex]\frac{\partial E}{\partial \ell} = \frac{\partial E}{\partial n}\frac{\partial n}{\partial \ell}[/itex].

    But, how do we make sense of [itex]\frac{\partial n}{\partial \ell}[/itex]. Don't we normally (when [itex]\ell[/itex] is not varied continuously) think of [itex]n[/itex] as being somewhat independent of [itex]\ell[/itex]?
     
  2. jcsd
  3. Nov 14, 2013 #2

    DrDu

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    Could you provide a link to the wikipedia article?
     
  4. Nov 14, 2013 #3
  5. Nov 14, 2013 #4

    Jano L.

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    Yes, in the basic theory of hydrogen atom, we usually do not think of ##n## as of a function of ##l##, because usually we consider all possible combinations of ##n,l##, and for given ##l##, we have many possible ##n = l +1, l+ 2, ...##, so obviusly ##n## is not a function of ##l##.

    However, if we restrict the set of the eigenfunctions, we can regard ##n## as a function of ##l##.

    In the mentioned application of the Hellmann-Feynman theorem, the argument runs probably as follows.

    We have special operator ##\hat{H}_l## which depends on ##l## only. From this it follows that each its eigenfunctions ##\phi_{nl}## is a function of ##l##. The number ##n## with possible values ## l+1, l+2, ...## is introduced as a whole number indexing various different eigenfunctions for the given ##l##.

    In the calculation of ##\langle \frac{1}{r^2}\rangle## for ##\phi_{nl}##, we deal with one eigenfunction ##\phi_{nl}## for given numbers ##n,l##. Given these two numbers, we can find exactly one natural number ##c > 0## such that ##n = l + c##. Now, from the set of all eigenfunctions ##\phi_{n'l'}##, let us choose the subset ##\phi_{l'+c,l'}##, in other words, we have ##n' = l'+ c##. Then we have ##\partial n'/\partial l' =1##.
     
  6. Nov 15, 2013 #5

    DrDu

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    Another possibility would be that the degeneracy of eigenvalues with different l but same n is broken once non-integer values of l are considered. It is necessary to consider non-integer l to be able to take the derivatives. I fear a clean calculation of the expectation value of 1/r^2 via the HF theorem is much more complicated than a direct calculation.
    In Wikipedia there is also a reference for this problem. Maybe you can check it out.

    Edit: Thinking about it, this will probably amount to the same what Jano L. said, i.e. dn/dl=1 as n=l+c also holds for non-integer values of l.
     
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