Asymptotic properties of Hydrogen atom wave function

  • Thread starter andrewkirk
  • Start date
  • #1
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,836
1,418

Main Question or Discussion Point

I am working through an explanation of the wave function of the Hydrogen atom.

I have got as far as deriving the version of Schrodinger's equation for the one-dimensional problem in which only the radial coordinate can vary:

##[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial^2 r}+\frac{\hbar^2 l (l+1)}{2mr^2}+V(r)]U_{El}(r)=EU_{El}(r)##

It is assumed that ##V(r)=-\frac{e^2}{r}##.

The presentation I am working through says:

'Find the asymptotic behavior of ##U_{El}(r)## as ##r\rightarrow 0##......
Solution: At ##r\rightarrow 0##, the first and second terms in the Hamiltonian will dominate, so [the above equation] becomes:

##\frac{\partial^2}{\partial^2 r}U_{El}(r)=\frac{l(l+1)}{r^2}EU_{El}(r)##'

I don't see why this follows. Certainly the second term in the square brackets will dominate the third term, which is only divided by ##r##, not ##r^2##, and the right-hand side of the equation, but why would it not also dominate the first term, which is not divided by ##r## at all?

Is it possible to derive the second equation above in a more convincing way?

Thank you.
 

Answers and Replies

  • #2
ChrisVer
Gold Member
3,337
440
Suppose it did not exist, you'd have a static thing (U would be constant or something).

The derivative will give you how things change...
 
  • Like
Likes 1 person
  • #3
Bill_K
Science Advisor
Insights Author
4,155
195
This is a standard technique in differential equations. To derive an approximate solution valid in a particular region (r → 0 in this case), assume a general form for the solution. Examine the magnitudes of each term in the equation, and keep only the leading ones.

So we assume u ~ rn and compare magnitudes of the four terms:

u'' ~ rn-2
u/r2 ~ rn-2
u/r ~ rn-1
Eu ~ rn

As r → 0, the first two terms have equal magnitude and dominate the others, so those are the ones we keep.

One could ask instead for an approximate solution for r → ∞. In that case, assume u ~ e-ar. Plugging it in, you'll find that the first and fourth terms dominate the others.
 
Last edited:
  • Like
Likes 1 person
  • #4
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,836
1,418
Thank you very much. That has completely cleared it up for me.

If I may, I would like to ask a supplementary question on the same topic area, although this time from a different text - Shankar.

Shankar (p341-2 of 2nd edition of 'Principles of Quantum Mechanics') discusses what constraints must be placed on ##U_{El}## in order for the Hamiltonian operator (part in square brackets in 1st equation - which he abbreviates to ##D_l(r)##) to be Hermitian with respect to U, ie for ##<D_l(r)U_{El}|=<U_{El}|D_l(r)## to hold.

He derives an equivalent condition that, for any two functions ##U_1## and ##U_2## obeying these constraints, we must have:

##[U_1^*\frac{dU_2}{dr}-U_2\frac{dU_1^*}{dr}]_0=0## [12.6.9]

He then states that 'this condition is satisfied if ##lim_{r\rightarrow 0}U=c## where '##c## is constant'.

I don't understand what this statement means - constant with respect to what? Obviously it is constant with respect to r, because it is a limit at a particular value of r, so that can't be his meaning. He might mean constant with respect to time but, given that we are talking about eigenvectors of the time-independent Schrodinger equation, the time dimension doesn't seem relevant. The only other interpretation I can think of is that he means:

##\exists c## such that ##\forall U: lim_{r\rightarrow 0}U(r)=c##.

But if that's what he means, I can't see how he proves that the above condition 12.6.9 follows from it.

Can anybody suggest what Shankar might be trying to say here, and how it can be proven?

Thank you very much.
 

Related Threads on Asymptotic properties of Hydrogen atom wave function

Replies
4
Views
989
Replies
7
Views
4K
Replies
5
Views
7K
Replies
2
Views
5K
  • Last Post
Replies
4
Views
2K
Replies
29
Views
8K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
13
Views
2K
Replies
3
Views
5K
  • Last Post
Replies
2
Views
974
Top