Degenerate perturbation theory help

[thoughtgaze]

So in time-independent degenerate perturbation theory we say that we can construct a set of wavefunctions that diagonalize the perturbation Hamiltonian (H') from the degenerate subspaces of the unperturbed Hamiltonian (H_o). Since the original eigenstates are degenerate, combinations of them are still eigenstates of H_o.

This means we have formed mutual eigenstates of both H_o and H'.

Now my question:

Doesn't this mean that the correction is EXACT in the new set of eigenstates that diagonalize both H_o and H' ? Where am I going wrong? In "Introductory Quantum Mechanics" 4th ed. by Liboff, he seems to say that there are higher order corrections.

Thank you.

EDIT:
Also, how does diagonalizing help the initial issue at hand (that the coefficients to the first order correction of the eigenstates are infinite when summing over a degenerate subspace)? Making the off-diagonal elements of H' equal to 0 simply makes the coefficients 0/0 (which is worse if you ask me lol). Is there some limit that needs to be taken?

I can see how this might work if the newly formed wavefunctions were exact, because then you needn't worry about any coefficients at all except the ones given by the secular equation.

 

[Bill_K]

We can exactly diagonalize H' across a degenerate eigenspace of H₀. However H' will still have off diagonal elements connecting one eigenspace of H₀ with another, so H₀ + H' is not completely diagonalized.

 

[thoughtgaze]

Okay, thanks for the reply.

So the corrections for the degenerate states ARE really solved exactly though right? I mean, they're not really corrections, but they are just written such that they are eigenstates of both H_o and H'. True over the whole space, H' is still not diagonalized, but say I want to find the energy correction just for one of the degenerate states.

 

[Bill_K]

No, it's still just an approximate eigenfunction and an approximate eigenvalue, due to the remaining off-diagonal elements. Let ψ be the wavefunction, and suppose you block-diagonalize across a subspace S. Before you diagonalize, Hψ = (sum of terms involving all eigenfunctions of H0). After you block-diagonalize, the terms in S are combined, but still Hψ = Eψ + (sum of terms involving all eigenfunctions of H0 not in S). So ψ is only an approximate eigenvector of H.

 

[thoughtgaze]

Sorry, still confused.

"...but still Hψ = Eψ + (sum of terms involving all eigenfunctions of H0 not in S). So ψ is only an approximate eigenvector of H. "

This is the part I don't quite understand. By definition it seems, the new combinations of subspace S (formed from degenerate subspace of H_o) are eigenstates of the total Hamiltonian.

defining {ψ_n} as the set of states that simultaneously diagonalizes H_o and H' (within the subspace) we have...

[H_o + H']ψ_n = [E_o + E'_n]ψ_n

I guess the question here is what does E'_n really represent? You and Liboff (so I assume you're probably right) seem to say that E'_n is only an approximation, and yet it IS an exact eigenvalue. What more would need to occur for it to be the exact energy of the system in that state?

 

[SpectraCat]

Remember that degenerate states are completely mixed by any perturbation. So the matrix solution to orthogonalize the degenerate states would converge on the exact solution in the limit of infinitesimally small perturbations. However, for any real perturbation, the non-degenerate states will also be mixed into some extent, which is why you need the complete treatment described in your text.

 

[thoughtgaze]

Well...

If I were to tell you there exists some wavefunction such that when operated on by a Hamiltonian gives that wavefunction multiplied by a constant i.e.

H[itex]\psi[/itex] = C[itex]\psi[/itex]

What do you call C?
The exact energy or only an approximation to the energy?

What you both seem to be saying here is that, C is not always the true energy of the system described by the wavefunction.

P.S.
If somehow someone still can't clarify this for me, can someone point me to some reading material that probably could?

 

[vanhees71]

To determine a quantum state completely, you have to measure a complete set of compatible observables. Then preparing the system in this state means to have a simultaneous eigenvector of the corresponding self-adjoint observable operators. Since the set of observables is complete by definition this determines the state uniquely, i.e., the corresponding eigenspace is one dimensional.

No, if you measure only the energy, it can be that the corresponding eigenspace is not one dimensional, but you have an eigenspace spanned by eigenvectors [itex]|E,\alpha \rangle[/itex], where [itex]\alpha[/itex] labels the different eigenvectors of the Hamiltonian with eigenvalue, [itex]E[/itex].

So if you measure only the energy of the system, you don't have complete information about the system, and thus you should describe the state of the system by a statistical operator. An objective estimate, according to information theory, is to use the state of maximum entropy, constraint by the known information. It turns then out that your statistical operator then should read

[tex]R=\frac{1}{Z} \sum_{\alpha} | E,\alpha \rangle \langle E,\alpha | \quad \text{with} \quad Z=\mathrm{Tr} \sum_{\alpha} | E,\alpha \rangle \langle E,\alpha |.[/tex]

That's the state with the least prejudice according to the information, i.e., in this case the energy of the system, you really have about it. This statistical operator defines the microcanonical ensemble of statistical physics.

 

[thoughtgaze]

vanhees71...

I think I understand what you are saying, but I don't understand how that answers my question. (so maybe I don't understand)

Oh well. Thanks for trying to explain it everyone. I'll search elsewhere for the answer.

 

[SpectraCat]

Well...

If I were to tell you there exists some wavefunction such that when operated on by a Hamiltonian gives that wavefunction multiplied by a constant i.e.

H[itex]\psi[/itex] = C[itex]\psi[/itex]

What do you call C?
The exact energy or only an approximation to the energy?

What you both seem to be saying here is that, C is not always the true energy of the system described by the wavefunction.

In your example, C would be the exact energy corresponding to the Hamiltonian that was used to calculate it. That is how it is defined. However, in perturbation theory, you typically have TWO Hamiltonians to consider. There is the reference Hamiltonian ... the degenerate wavefunctions are eigenfunctions of this Hamiltonian ... i.e. they all give the same energy when the *reference* Hamiltonin is applied to them. You also have the FULL Hamiltonian, which is the reference Hamiltonian, plus a perturbation term. The eigenfunctions of the reference Hamiltonian that you start with are obviously not eigenfunctions of the full Hamiltonian. You use perturbation theory to calculate *approximations* to the eigenfunctions of the full Hamiltonian, based on how the perturbation acts to couple the reference eigenfunctions. When you have degenerate eigenfunctions of the reference Hamiltonian .. the perturbation will mix those states COMPLETELY (c.f. the two state system in your textbook). Therefore, in order to make the math easier in the long run, it makes sense to orthogonalize the degenerate states before calculating the effects of the perturbation ... having the reference wavefunctions be orthogonal GREATLY simplifies the mathematical treatment. For example, say that the ground state of the reference Hamiltonian is doubly degenerate, and label those states |1> and |2> ... you create the linear combinations:

[itex]|\psi_1>=|1> + |2>[/itex] and [itex]|\psi_2>=|1> - |2>[/itex]

Those states are still degenerate with respect to the reference Hamiltonian, but now they are orthogonal, and thus their overlap integrals will always be zero in the derivation of the perturbation corrections.

[EDIT] For maximum clarity, I should point out that the procedure above is to create the correct set of zero-order degenerate states by orthogonalizing the degenerate eigenstates of the reference Hamiltonian taking the perturbation into account (I neglected to mention that explicitly in the above description). This is done by solving the secular equation for the perturbation to the Hamiltonian to obtain the appropriate expansion coefficients as described in your book.

Note that the while zero-order solutions thus obtained are still eigenfunctions of the reference Hamiltonian, they are NOT eigenfunctions of the perturbed Hamiltonian. The energies you obtain are only zero-order corrections to the energy ... they typically need to be improved by considering coupling to other, non-degenerate states. Perturbation theory is approximate .. it only gives exact solutions (i.e. the true eigenfunctions of the full Hamiltonian) in the limit of infinite order.

 

[SpectraCat]

@thoughtgaze

I just noticed an important error in your original post. The following statement:

This means we have formed mutual eigenstates of both H_o and H'.

Is incorrect. They are not mutual eigenstates of the two Hamiltonians. If they were, then we could get an exact solution for the ground state of a helium atom considering only the coupling of two 1s-electrons due to the electron-electron repulsion, without having to consider the coupling to possible excited states.

The solutions to the secular determinant considering only the degenerate states ARE eigenfunctions of something, it is true ... that is mathematically guaranteed by the way the solutions are found. What you need to realize is that they are eigenfunctions of a non-physically restricted solution space, and thus are (likely) poor approximations to the solutions of the full Hamiltonian. Note that this is completely equivalent to restricting the basis set size in variational calculation by the Raleigh-Ritz method.

 

[thoughtgaze]

@thoughtgaze
What you need to realize is that they are eigenfunctions of a non-physically restricted solution space, and thus are (likely) poor approximations to the solutions of the full Hamiltonian. Note that this is completely equivalent to restricting the basis set size in variational calculation by the Raleigh-Ritz method.

Okay. I think this is the answer I wanted. Thank you.

Perhaps this is the source of my confusion.

Let me see if I get this straight. Let H_o be the unperturbed Hamiltonian and H' be the perturbation Hamiltonian. Also, let {[itex]\psi[/itex]_n} be the set of states in a degenerate subspace of H_o and {[itex]\overline{\psi}[/itex]_n} be the new degenerate states that diagonalize H' in that same subspace.

Then,
[H_o + H'][itex]\overline{\psi}[/itex]_n = [E_o + E'_n][itex]\overline{\psi}[/itex]_n

more importantly I want to focus on,

H'[itex]\overline{\psi}[/itex]_n = E'_n[itex]\overline{\psi}[/itex]_n

You are telling me that E'_n , although an eigenvalue corresponding to the H' hamiltonian acting on the [itex]\overline{\psi}[/itex]_n state, (and this is the weird part for me) [itex]\overline{\psi}[/itex]_n nevertheless is not an eigenfunction of H' but instead "eigenfunctions of a non-physically restricted solution space."
I'll look into the variational calculation by the Raleigh Ritz method that you have mentioned.

 

[thoughtgaze]

For the record, now it is blatantly obvious.

Thanks. I see what you were saying now Bill and Spectra.

Also, thanks Vanhees for an informative reply.