Ground state density requirement

[dRic2]

Homework Statement

For any one-electron problem one can readily determine whether or not any given density is a possible ground state density. Using the known properties of solutions of the schrodinger equation, give a sufficient set of condition that any function must satisfy in order to guarantee that it is the ground state density of some potential.

Show that it is not possible to construct the density of the 2s state of the hydrogen atom (1 proton 1 electron) as the ground state of any smooth potential, i.e. one without delta functions.

Relevant Equations

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I've been think about it for hours but I'm really out of clue here... The only things I could think of are obvious or useless... Any help would be greatly appreciated.

 

[mfb]

Consider how the ground state looks like in different potentials. How much of a "wavelength" fits in there for the ground state? Can it be more?

 

[dRic2]

Thank you for the reply. I'm sorry but I'm not understanding: what does How much of a wavelength fits in there mean?

 

[mfb]

Check the ground state for some systems and you should see some pattern. Compare it to the second energy level for example.

 

[dRic2]

I noticed that usually gs densities are "bell-shaped" with only one "hill". Excited states have more that one "hill" (as in the case of 2s state of H). This, if true, is obviously related to the second derivative of the density, thus to the wave function, but I've yet to put it in a mathematical way. Am I on the right track ?

 

[mfb]

That's a good approach. What happens between the hills of the excited states? Is there some special point you can find?

 

[dRic2]

Yes, those point satisfy $n(r)=0$. Thus also $ψ(r)=0$. I think I can call those point "nodes". It seems to me that the number nodes is equal to "level" of the excited states.

Btw now I see what you meant by How much of a wavelength fits in there. I can infer from what I've found so far that the number of wavelength of a state equals the number of nodes thus the "level" of the excitation. The gs has only 1 wavelength (the largest one) corresponding to the value of lowest energy.

Now I have to show that $n_{\text{nodes}}=n$. I don't actually know how to prove it in a general way, but maybe I can exploit the fact that the Sch. eq (at least in 1 dimension) resembles a Strum-Liouville problem. Usually when solving SL problems I get polynomials (like for QHO or Hydrogen atom) for which it's known that the number of zeros equals the degree of the polynomial. Maybe there is a simpler way, what do you think ?

Thanks for all the help!

 

[mfb]

Btw now I see what you meant by How much of a wavelength fits in there. I can infer from what I've found so far that the number of wavelength of a state equals the number of nodes thus the "level" of the excitation. The gs has only 1 wavelength (the largest one) corresponding to the value of lowest energy.

There is a factor 2 between them, but yes, that's the right relation.

Now I have to show that $n_{\text{nodes}}=n$

Might need "+1" depending on the counting.
Not sure how to prove it mathematically.

 

[dRic2]

Ok, I did a quick search and if the potential has spherical symmetry (separations of variables in spherical coordinates holds) then you can prove it considering the properties of a Strum-Liouville problem, but it is not as easy as I thought. On the other hand the fact that GS can't have nodes can be shown in various methods, mainly with variational techniques. I found this paper which proposes a proof for a general non-velocity-dependent potential ( http://www.ecm.ub.es/nuclear/jordi/articles/varprinc2002.pdf )

Thanks again for all the help!

 

[vanhees71]

The textbook by Messiah on QM has very detailed proofs about 1D eigenvalue problems. There's also a nice elementary proof for the theorem about the nodes of eigenfunctions of self-adjoint operators.