# Degenerate perturbation theory question

1. Mar 3, 2005

### mattlorig

Can anybody explain what Griffiths means when he talks about "good eigenstates" in degenerate time-independent purturbation theory?

Mathematically, I know he is just talking about the eigen-vectors of the W matrix (where Wij = <pis_i|H'|psi_j>). But what do the eigen-vectors physically represent? Are they the 1st order approximations of the of the new eigen-states of H = H0 + H'?

2. Mar 3, 2005

### dextercioby

No.They are just that:eigen-vectors of the perturabation matrix:$$W^{(n)}_{i,i'}=:\langle n i|\hat{W}|n'i'\rangle$$...Once you determine the first-order corrections to the energy levels,then u can worry about eigenvectors of the perturbation.

Daniel.

3. Mar 3, 2005

### Galileo

I`m not entirely sure, because I think Griffiths does a horrible job of explaining it.

I guess the point is that you can take any linear combination of eigenstates of H corresponding to a degenerate eigenvalue and it will still be an eigenstate of H with the same eigenvalue. So your choice of basis is not unique.
You want $\langle \psi_i^0 | H' | \psi_j^0 \rangle$ to be zero if $i\not=j$, then the result will be the same as using nondegenerate pertubation theory.
Therefore the good linear combination to take is the one in which your choice of basisvectors is such that they are eigenvectors of H'. In that case the off-diagonal matrix elements vanish.
So you have to choose your basis as to diagonalize the pertubation.

4. Mar 3, 2005

### Dr Transport

You first calculate the matrix elements $$\langle \psi_{i} ^{0} | H' |\psi_{j} ^{0} \rangle$$ for all $$i,j$$ degenerate pairs. After finding the matrix of these values, find the eigen-values of this matrix and use them to find the combinations of the orginal basis functions that diagonalize the matrix so that $$\langle \psi_{i} ^{0} | H' |\psi_{j} ^{0} \rangle = 0$$ for $$i \neq j$$.

5. Mar 4, 2005

### mattlorig

DO I HAVE THIS RIGHT?
OK, I think I understand this a bit better now. The eigenvectors obtained from the W-matrix simply represent a change of basis. The original basis consisted of functions which were eigenstates of H0 and had an eigenvalue of E0. The new basis consists of functions which are eigenstates of H' and have eigenvalues E1 (which, may or may not be distinct). The eigenstates of H' are what griffiths calls "good states".

ANOTHER QUESTION
griffiths goes on to talk about a cute way of finding the "good states" without constructing the W-matrix:
Find an operator that commutes with H', and which has distinct eigenvalues for each of its eigenstates. These eigenstates will automatically be the "good states".

To be frank, this doesn't seem any easy at all since you have to:
1) come up with an operator A
2) show that A commutes with H'
3) find the functions which are simultaneously eigenstates of A and H'

I was hoping griffiths would shed a bit more light on this subject when he talked about the relativistic correction to the Hydrogen Energies. But, rather than showing that p^4 commutes with L^2 and Lz, he simly says that it does because it's spherically symetric. (which still isn't obvious to me). Furthermore, the eigenstates of L^2 and Lz are Y_lm. How does griffiths make the jump from Y_lm to Psi_nlm? (e.i., how does he know that know that Psi is an eigenfunction of H'?) Lastly, since we are dealing with two operators, what is A in this case? L^2 + Lz? I suppose that would work, since it would give distinct eigenvalues for each eigenstate:
l(l+1)hbar^2 + mhbar

LAST QUESTION:
Forgive my stupidity, but how do you guys write your equations nicely on the posts? LaTex? (I admittedly know nothing about LaTex...perhaps I should investigate).

6. Mar 4, 2005

### dextercioby

Last:Yes,Latex it is that we use for writing formulas.
Penultimate.Yes,it is straightforward to show that $\hat{L}^{2},\hat{L}_{z},\hat{p}^{4}$ form a csco,therefore have common eigenvectors.
No,the addition of $\hat{L}^{2}$ and $\hat{L}_{z}$ has no physical meaning.

Daniel.

7. Mar 4, 2005

### mattlorig

I guess I'm still not clear on what the operator A would be. It can't be just L^2 or Lz, since the eigenstates of those operators do not produce distinct eigenvalues:
|211> --> L^2 = 2 hbar^2
|210> --> L^2 = 2 hbar^2
|210> --> Lz = 0
|200> --> Lz = 0
Although Lz + L^2 has no physical meaning, at least it would give distince eigenvalues for all |nlm>.

I also still don't see how we make the jump from Y_lm to Psi_nlm. I know the operators that commute share eigenstates. But how did we know Psi_nlm was an eigenstates of H'?

Lastly, it may be straightforward to show the p^4 commutes with Lz and L^2. But I tried doing the calculation this morning, and it takes a LOT of work (I didn't get the correct answer either, but I'm not about to try it again).

8. Mar 4, 2005

### dextercioby

Remember the free particle Hamiltonian..?It's $\frac{\hat{p}^{2}}{2m}$.Add to it the Coulomb term...Now consider the relativistic correction.Add to it the free particle Hamiltonian (see above) and the Colomb term.Denote the first Hamiltonian through $\hat{H}_{0}$ and the second through $\hat{H}$...Consider the 2 Hamiltonians and the angular momentum operators.Show they form a csco...
Think about whether $\frac{\hat{p}^{2}}{2m}$ commutes with the ang.momentum ops.

Daniel.

9. Mar 15, 2005

### mattlorig

what's a "csco"?
my best guess is "blank blank commuting operators"

10. Mar 15, 2005

### mattlorig

oh, I just got it: complete set of commuting observables

11. Jun 2, 2009

### HoangFat

Omg !
I found another question about degenerate perturbation, in some book I found I only contain the first approx, but the second approx ?
A electric field affect to a rigid rotator (2 different charge at 2 poles of the rigid rotator)