Quantum fields in infinite volume This problem might sound a bit trivial at face value but to me is sorta hard to grasp if not paradoxical!
We know that the energy gap between the quantum state of $n$ particles and $n+1$ particles is finite. Given the fact that quantum field are beyond any notion of waves or particles (and they are there to solve the wave-particle duality problem), how is it possible to visualize that: to excite a quantum field to another higher state, one needs only a finite amount of energy regardless of the volume!?
In fact in both cases of finite and infinite volume (consequently standing and propagating waves) the energy gaps are independent of the volume! How is it possible to excite an open-ended field at all points of space with a finite amount of energy? Or how is it possible to excite a standing quantum field (like the primitive instance of a quantized string in the 3 man paper by Jordan) by adding a fixed amount of energy that doesn't depend on the volume (of the black body or string or whatever capacitor)?
 A: This question has nothing to do with quantum mechanics. The same "paradox" happens in classical field theory, where things are easier to visualize. Consider for example the electromagnetic field in $\mathbb R^3$. How can we excite this field with finite energy? Well, we make an excitation that is localized in space.
Recall that the energy of the electromagnetic field is
$$
E=\int_{\mathbb R^3}\boldsymbol E^2+\boldsymbol B^2
$$
If $\boldsymbol E,\boldsymbol B$ are non-zero in a compact region (or if they decay sufficiently fast at infinity), then $E$ is finite, even if the total integration region is infinite.
In QFT the energy is given by an identical expression,
$$
E=\int_{\mathbb R^3}\mathcal H(\varphi)
$$
where $\mathcal H$ is the Hamiltonian density, a function of all the fields of the theory. Just like in the classical case, a finite-energy field configuration has fields of compact support (or that vanish sufficiently fast at infinity).
Please do keep in mind that not every field configuration satisfies $E<\infty$. There are important situations where this is violated. For example, solitons have finite energy and are topologically protected, because their decay involves configurations that have infinite energy. So keep in mind that not every field configuration has finite energy, but those that don't are not physical and cannot be realized in the real world. A more down-to-earth example of a configuration with infinite energy is the case of plane waves, which are constant in space (so they do not decay). Again, this field configuration is not physical and cannot be realized in the real world. There are no plane waves in nature.
A: If I understand your question properly, you are concerned that (1) the Hamiltonian is given by a Hamiltonian density which we integrate over all space, and (2) that an energy eigenstate (in an infinite system) is a plane wave which exists everywhere with the same amplitude, differing from point to point only by a phase factor.  Therefore, either the Hamiltonian density acting on this state gives zero or something nonzero, corresponding to an excitation with either zero energy or infinite energy.
The answer to this is basically the same as the answer to a similar answer in elementary QM: In an infinite, free system, energy eigenstates are non-normalizable and hence unphysical.
We talk about momentum eigenstates $a^\dagger_\mathbf p|0\rangle$ in essentially the same way as we talk about them in 1D scattering problems from our first course in QM.  They're useful tools, but all physical states will be of the form
$$|\psi\rangle = \int\frac{d^3p}{(2\pi\hbar)^3} f(\mathbf p) a^\dagger_{\mathbf p} |0\rangle$$
for some nicely-behaved function $f$.  Expressing this state in terms of the spatial creation operators (i.e. the field operators) would yield
$$|\psi \rangle = \int d^3 x \ \tilde f(\mathbf x) \phi^\dagger(\mathbf x)|0\rangle$$
with $\tilde f$ the Fourier transform of $f$.   As a result, all physical states will have some finite spread in momentum as well as some degree of spatial localization, as $\tilde f$ should be square-integrable.  If you apply the Hamiltonian operator to this state, you will find that
$$H|\psi\rangle = \int\frac{d^3p}{(2\pi\hbar)^2} f(\mathbf p) \omega_\mathbf p a^\dagger_\mathbf p|0\rangle$$
$$\implies \frac{\langle \psi|H|\psi\rangle}{\langle \psi|\psi\rangle} = \frac{\int \frac{d^3p}{(2\pi\hbar)^3} |f(\mathbf p)|^2 \omega_p}{\int\frac{d^3p}{(2\pi\hbar)^3} |f(\mathbf p)|^2} \geq \omega_0 = mc^2$$
