Hamiltonian of Klein-Gordon Field The Hamiltonian of the Klein-Gordon Field may be written $$H=\int\frac{d^{3}p}{(2\pi)^{3}}\frac{1}{2\omega_{\mathbf{p}}}\omega_{\mathbf{p}}\left(a^{\dagger}(p)a(p)+\frac{1}{2}(2\pi)^{3}2\omega_{\mathbf{p}}\delta^{(3)}(0) \right)$$ where $a(p)$ and $a^{\dagger}(p)$ are the usual creation and annihilation operators and $\omega_{\mathbf{p}}=\sqrt{|\mathbf{p}|^{2}+m^{2}}$ is the oscillator frequency. 
Act on a singly excited state $|q\rangle=a^{\dagger}(q)|0\rangle$ with this Hamiltonian. We find $$H|q\rangle=\left(\omega_{\mathbf{q}}+E_{0}\right)|q\rangle$$ where the (infinite) ground state energy $$E_{0}=V\int\frac{d^{3}p}{(2\pi)^{3}}\frac{\omega_{\mathbf{p}}}{2}$$ and the volume has been included using $$(2\pi)^{3}\delta^{(3)}(0)=\lim_{L \to \infty}\int_{-\frac{L}{2}}^{\frac{L}{2}}d^{3}xe^{i\mathbf{p}\cdot\mathbf{x}}|_{\mathbf{p}=0}=\lim_{L \to \infty}\int_{-\frac{L}{2}}^{\frac{L}{2}}d^{3}x=V$$
I am a bit confused because:


*

*In the ground state energy, the oscillator zero point energy seems to be contributing to an energy density (hence the volume factor).

*In the excitation energy, the oscillator excitation energy is not being treated as an energy density.
This appears to be a contradiction, since the oscillators should either contribute to the total energy or the energy density of the field, not both - can anybody explain this?  
 A: "In the ground state energy, the oscillator zero point energy seems to be contributing to an energy density (hence the volume factor)" 
It is not. For a single oscillator the zero point energy is finite and hence there is no problem with it. But for klien-gordon field, which is a collection of infinite number of oscillators, have infinite zero point energy. Therefore To deal with it we consider energy density instead which is $E_0 /V$. This also answered your (2) confusion.
A: I'm not sure I understand exactly what you are asking. One intuitive way to see the calculation of the energy is 
$$E (\textrm{state } | q \rangle) = \sum\limits_{\textrm{Oscillators } \mathbf{p}} (n_{\mathbf{p}}+1/2) \omega_{\mathbf{p}} =  \sum\limits_{\textrm{Oscillators } \mathbf{p} } n_{\mathbf{p}}\omega_{\mathbf{p}} + \sum\limits_{\textrm{Oscillators } \mathbf{p} } \frac{1}{2}\omega_{\mathbf{p}} =\omega_{\mathbf{q}} + E_0  $$
where we use that the excitation number $ n_{\mathbf{p}}$ is always zero, except when $\mathbf{p} = \mathbf{q}$. Does it answer the question? 
