Time evolution of operators in QFT

Question

$\newcommand{\Ket}[1]{\left|#1\right>}$ I began to study QFT using David Tong's notes. This is what the notes says (http://www.damtp.cam.ac.uk/user/tong/qft.html page 21).

I work in the Schrödinger picture so my operators $\phi(\boldsymbol{x})$ and $\pi(\boldsymbol{x})$ do not have a time dependence. All time dependence in the Schrödinger picture is encapsulated in the states $\Ket{\psi}$ which evolve according to the Schrödinger equation \begin{equation} i\hbar \frac{d}{dt} \Ket{\psi} = H \Ket{\psi}. \end{equation}

The notation $\Ket{\psi} $ is deceiving since the wave function in quantum field theory is a functional of every possible configuration of the field $\phi$.

What does it mean that the time dependence of the operators are encapsulated by the functional $\Ket{\psi}$? If I solve the Schrödinger equation I can find $\Ket{\psi(t)}$, but then how do I relate this to my fields?

Answer 1

Tong is merely referring to the usual relation between the Schrödinger picture and the Heisenberg picture. In the Heisenberg picture you have operators $O(t)$ and states $\lvert \psi\rangle$, in the Schrödinger picture you have operators $O$ and states $\lvert \psi(t)\rangle$.

The $\lvert \psi(t)\rangle$ "encapsulates" the time-dependence of the operators in the sense that the same time-evolution operator $U(t)$ yields the time-dependent states via $\lvert \psi(t)\rangle = U(t)\lvert \psi(0)\rangle$ and the time-dependent operators via $O(t) = U(t)O(0)U(t)^\dagger$, so both $\lvert \psi(t)\rangle$ and $O(t)$ "encapsulate" the same information about time evolution.

That in this case the operator $O$ additionally depends on position and you write $\phi(\vec x,t)$ in the Heisenberg picture and $\phi(\vec x)$ in the Schrödinger picture changes nothing.