What is the 'QFT equivalent' of the Schroedinger equation?

Question

How do you describe the energy of a certain system in quantum field theory?

Answer 1

Exactly the same Schrodinger equation as in quantum mechanics, only with a different Hamiltonian.

The QFT Hamiltonian is $$ \hat{\mathcal{H}} = \int d^3 x \, \hat{T}^{00}, $$

where $\hat{T}_{\mu \nu}$ is the quantization of the classical stress-energy tensor $T_{\mu \nu}$ (strictly speaking, it is not required to mathematically exist; only its integral over space is). The exact expression for $T_{\mu \nu}$ depends on the model, see wikipedia for details.

Answer 2

The Schrödinger equation can be quantized canonically to get the Schrödinger field. The latter is a quantum field theory that has the Schrödinger equation as its operator equation of motion. Here is an outline of how that works.

The Schrödinger equation is

$$H\psi(r,t) = i\frac{\partial}{\partial t} \psi(r,t).$$

where $H$ is the ''first quantized'' Hamiltonian (note that people do not like this terminology...).

The corresponding canonically quantized Hamiltonian is

$$ \hat{H} = \int dr \hat{\psi}^\dagger (r,t) H \hat{\psi}(r,t),$$

where $\hat{\psi}(r,t)$ is now a field operator with harmonic oscillator equal-time commutation relations at each point (see e.g. linked wikipedia article)

$$[\hat{\psi}(r,t), \hat{\psi}^\dagger(r',t)] = \delta(r-r').$$

One can also write this quantum field theory in terms of the energy eigenstates of the Schrödinger equation, which then involves the creation and annihilation operators $\hat{c}_m(E,t)$ associated with an eigenstate at energy $E$. This can be thought of as a diagonalization of the second quantized Hamiltonian, giving

$$\hat{H} = \sum_m \int dE E \hat{c}^\dagger_m(E,t)\hat{c}_m(E,t). $$

Deriving the latter Hamiltonian from the first one is left as an exercise for the reader.