Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

On UC Davis chemwiki website, the Hamiltonian for quadrupolar coupling in NMR is analyzed. (The details of this aren't important.)

It is said in the analysis that:

The expansion of the Hamiltonian is done using perturbation theory.

It then proceeds to "expand the Hamiltonian" according to first and second order terms.

I'm an undergraduate reading up on quantum physics for a summer research position, and I've spent the last day or so immersing myself in perturbation theory. It's my understanding that the basic premise of the theory is that there are two (and only two) parts to the Hamiltonian: $H^0$, the unperturbed Hamiltonian, and $H^1$, the perturbing one. What does it mean to expand the Hamiltonian itself?

EDIT: It appears the premise of this question is mistaken. As far as I can tell, perturbation theory itself doesn't result in first- and second-order Hamiltonians. Instead, NMR researchers have just happened to call a pair of Hamiltonians "first-order" and "second-order". Since both of these are perturbing Hamiltonians, I got confused at the overlap in terminology.

Thanks for helping me figure this out!

share|cite|improve this question
This is explained in any reasonable book on QM. See, for instance, Griffiths' book, ch. 6. – Danu Jul 14 '14 at 17:57
Hmmm. That reference is indeed quite confusing, and it's not very clear what they are doing. Do you have additional references that parallel this treatment? It doesn't look like they are doing a perturbation expansion, it looks to me (right off the bat) like a full expansion of the change from the Principal Axis System into the lab frame. – Emilio Pisanty Jul 14 '14 at 18:17
@Danu I just read over Griffith's chapter 6. It's a great reference but nowhere does it answer this question. – NcAdams Jul 14 '14 at 18:32
@NcAdams My sincere apologies: I was under the impression that it was the 'standard expansion' you were asking after, which is explained there. – Danu Jul 14 '14 at 18:47
I think what they mean is that the perturbed Hamiltonian is actually itself very complicated, and so they will "perturb the perturbed Hamiltonian." This is a little tricky and not a standard intro QM thing. In other words, when they talk about "perturbing the hamiltonian," they do not mean that they are going to compute perturbations to the energy spectrum using QM perturbation theory [though they will ultimately do this later on]. What they mean is that the hamiltonian $H_Q$ that represents the "quadrupolar correction" is very complicated and depends on the exact shape of the nucleus,... – Andrew Jul 14 '14 at 20:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.