What does it mean to expand a Hamiltonian using perturbation theory?
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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: H0H0, the unperturbed Hamiltonian, and H1H1, 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!
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: H0H0, the unperturbed Hamiltonian, and H1H1, 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!
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definition in terms of first principle physics ... Then we can always expand the eigenfunctions of the full Hamiltonian in terms of .... But first-order perturbation theory tells us
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