- #1
malcomson
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Doing a problem and I'm not entirely sure what the wording is referring to.
I won't get into details of the question unless people want to know (not really relevant) but I'm looking at the quantum ferromagnet and applying the Holstein-Primakoff transformation to find a new form of the Hamiltonian.
"Show that, to order S, the Hamiltonian is Quadratic in the boson annihilation and creation operators a and a-dagger"
basically I'm a bit unsure as to what they're asking - should the equation have an a^2 and (a-dagger)^2 term multiplied by S or does it mean something else?
Earlier on in the question it mentions we're looking at the limit of large S, and there is an expansion in 1/S which is multiplied by S so does the question mean - "If we ignore all terms in the expansion of order less than S, - show the Hamiltonian is quadratic in the boson operators"
I've done some working and if you ignore those constant terms in the expansion you do get a quadratic Hamiltonian in the boson operators, but the quadratic term is not of order S.
The quadratic term is a constant that comes from elsewhere in the equation so I'm not sure if I should count it (it would have been discarded if it were in the expansion so why keep it now).
Apologies if I'm unclear - I'm not entirely sure how to make it better though
I won't get into details of the question unless people want to know (not really relevant) but I'm looking at the quantum ferromagnet and applying the Holstein-Primakoff transformation to find a new form of the Hamiltonian.
"Show that, to order S, the Hamiltonian is Quadratic in the boson annihilation and creation operators a and a-dagger"
basically I'm a bit unsure as to what they're asking - should the equation have an a^2 and (a-dagger)^2 term multiplied by S or does it mean something else?
Earlier on in the question it mentions we're looking at the limit of large S, and there is an expansion in 1/S which is multiplied by S so does the question mean - "If we ignore all terms in the expansion of order less than S, - show the Hamiltonian is quadratic in the boson operators"
I've done some working and if you ignore those constant terms in the expansion you do get a quadratic Hamiltonian in the boson operators, but the quadratic term is not of order S.
The quadratic term is a constant that comes from elsewhere in the equation so I'm not sure if I should count it (it would have been discarded if it were in the expansion so why keep it now).
Apologies if I'm unclear - I'm not entirely sure how to make it better though