Should Hamiltonians in quantum field theory be linear operators? The usual structure of quantum mechanics imposes that Hamiltonians are linear operators. I am not sure if this really holds in quantum field theory. Do non-linear Hamiltonian operators really make sense in quantum field theory?
 A: A common confusion about quantum field theory is thinking that the fields, $\psi(x)$, $\phi(x)$, etc. generalize the wavefunction of quantum mechanics. If you think this and then notice that QFT Hamiltonians usually look like $\int dx \, -(\nabla \phi(x))^2 + m^2 \phi(x)^2$ then it is natural to think that the Hamiltonian is no longer a linear operator on the wavefunction.
But the fields do not generalize the wavefunction. Instead they generalize the position operator. In quantum mechanics the Hilbert space of the theory has a basis made up of position eigenstates, in quantum field theory the basis is made up of field eigenstates, where $\phi(x) |\varphi\rangle = \varphi(x) |\varphi \rangle$.
A general state is then a linear combination of these states. In QM a general state is $|\Psi\rangle = \int dx \Psi(x) |x\rangle$. In QFT a general state is $|\Psi\rangle = \int \mathcal{D}\varphi\, \Psi[\varphi] |\varphi\rangle$. The only differences are that 
 the "wavefunction" $\Psi$ takes functions and returns numbers instead of taking positions and returning numbers, and that instead of integrating over positions with $dx$ we are somehow integrating over all possible fields $\mathcal{D}\varphi$.
Finally, note that the field operators act linearly on the basis of field-eigenstates. Since the Hamiltonian is a sum of products of field operators, the Hamiltonian also acts linearly on these field eigenstates, so the Hamiltonian is still a linear operator on the Hilbert space.
A: Quantum field theory is a subset of quantum mechanics, and its fundamental rules still hold:


*

*Observables, including the Hamiltonian, are linear self-adjoint operators.

*Probabilities are given by the squared absolute values of inner products.

*The Schrödinger equation $i \hbar \frac{d|\psi\rangle}{dt} = H |\psi\rangle$ holds.


And so on. The only thing that changes is that you now have infinite degrees of freedom, so the Hamiltonian is an integral over space of local operators.
