# Feynman Hellman Theorem: dependence of E on $\ell$ Hydrogen

1. Nov 13, 2013

### MisterX

The theorem states
$\frac{\partial E}{\partial \lambda} = \langle \psi \mid \frac{\partial H}{\partial \lambda} \mid \psi \rangle$

Where $\mid \psi \rangle$ is an eigenket of H.

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

But, how do we make sense of $\frac{\partial n}{\partial \ell}$. Don't we normally (when $\ell$ is not varied continuously) think of $n$ as being somewhat independent of $\ell$?

2. Nov 14, 2013

### DrDu

Could you provide a link to the wikipedia article?

3. Nov 14, 2013

### MisterX

4. Nov 14, 2013

### Jano L.

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 $\partial n'/\partial l' =1$.

5. Nov 15, 2013

### DrDu

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.