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

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

• A QFT is just one type of quantum system, it has the usual Schrödinger equation and the energy of a state is the expectation value of the Hamiltonian. – Heidar Jan 19 at 10:19
• Does this answer your question? Is there any theorem that suggests that QM+SR has to be an operator theory? – AccidentalFourierTransform Jan 19 at 12:34
• Please try to give us an idea of what research you have done and what you couldn't find out/understand is when asking such a question. This helps answerers understand what you're actually looking for. For instance, there are different levels of abstraction at which one can call an equation "the Schrödinger equation", and there are always the "time-independent" and the "time-dependent" versions. So it's unclear what you really mean by "Schrödinger equation" here. – ACuriousMind Jan 19 at 19:24

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.

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.