# How do I find the energy of this Hamiltonian?

So, I have a (confusing) Hamiltonian:

$$H = \int \mathrm{d}k\,\omega_{k} a^{\dagger}\bigl(\vec{k}\bigr)a\bigl(\vec{k}\bigr)$$

where $\omega_{k} = \sqrt{k^{2} + m^{2}}$ and the measure is $\mathrm{d}k = \frac{\mathrm{d}^{3}k}{(2 \pi)^{3} 2 \omega_{k}}$.

I want to find the energy of a single particle state $\lvert k\rangle = a^{\dagger}(k) \lvert 0\rangle$ and ask the energy of a multi-particle state, but I am not sure how to deal with that starting Hamiltonian.

A lot of the work for this problem (getting the Hamiltonian into this form) was already done, but I am not sure how to make the final jump to the energies. This style Hamiltonian doesn't seem conducive to such a solution...

Here's what happens when you apply $H$ to $|\mathbf k\rangle = a^\dagger_\mathbf k|0\rangle$ to find the energy of a single particle state: \begin{align} H|\mathbf k'\rangle &= \int \frac{d^3\mathbf l}{2(2\pi)^3}a^\dagger_\mathbf la_\mathbf la^\dagger_\mathbf k|0\rangle \\ &= \int \frac{d^3\mathbf l}{2(2\pi)^3}a^\dagger_\mathbf l([a_\mathbf l,a^\dagger_\mathbf k]-a^\dagger_\mathbf ka_\mathbf l)|0\rangle \\ &= \int \frac{d^3\mathbf l}{2(2\pi)^3}a^\dagger_\mathbf l((2\pi)^3 2\omega_\mathbf k\delta^{(3)}(\mathbf l - \mathbf k)-a^\dagger_\mathbf ka_\mathbf l)|0\rangle \\ &= \omega_\mathbf k a^\dagger_\mathbf k|0\rangle \\ &= \omega_\mathbf k|\mathbf k\rangle \end{align} The computation for a multi-particle state is simiar.