# Rewrite hamiltonian from integral over momenta to integral over energies

How do I rewrite a Hamiltonian written as some integral over momenta, e.g., $$H = \int\frac{d^d \vec k}{(2\pi)^d}\omega(\vec k) a_{\vec k}^\dagger a_{\vec k}$$ 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: $$\omega(k)=\omega,\\ a_{\omega,+} = a_{k}, \text{ if } k>0,\\ a_{\omega,-} = a_{k}, \text{ if } k<0.$$
• Right, but does the density of states have to appear as function $f(\omega)$? When I do a change of variables, I can choose any function there. I suspect there is some choice for the commutation relations that's usually made? Dec 7, 2020 at 10:57
• When you change the variables, the only function you can get there is the Jacobian (in the case of a unique function), which is the density of states: $\int dk ... = \int \frac{dk}{d\omega} d\omega$. Dec 7, 2020 at 11:02