Physics Asked on June 8, 2021
How do I rewrite a Hamiltonian written as some integral over momenta, e.g.,
$$ H = intfrac{d^d vec k}{(2pi)^d}omega(vec k) a_{vec k}^dagger a_{vec k} $$
in terms of energy eigenmodes?
That is, what’s the function $f$, the cutoff $omega_c$ and what are the modes $a_omega$ in the following formula:
$$ H = int_0^{omega_c}f(omega) a_omega^dagger a_omega.$$
Maybe this can’t be done in most cases — the problem is that $omega(vec k)=omega_0$ is not invertible. So in addition to the integral over $omega$ there needs to be a sum or an integral over all the states with the same frequency. But I’m mostly interested in cases where it works.
I know about substitution of variables, but I’m unsure how the operators transform. Can $f(omega)$ be anything and $a_omega$ still fulfil canonical commutation relations? Mostly thinking about bosons, but shouldn’t be very different for fermions I assume.
Converting momentum summations to integrals density of states is a rather basic procedure, which can be found in many textbooks. What requires special care here is the degeneracy of the spectrum - how this one is treated depends very much on the specific problem at hand. For example, in 1D one usually resorts to introducing operators for forward and backward going particles: $$ omega(k)=omega, a_{omega,+} = a_{k}, text{ if } k>0, a_{omega,-} = a_{k}, text{ if } k<0. $$
Correct answer by Roger Vadim on June 8, 2021
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