# Time evolution of operators in QFT

$$\newcommand{\Ket}[1]{\left|#1\right>}$$ 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 $$$$i\hbar \frac{d}{dt} \Ket{\psi} = H \Ket{\psi}.$$$$

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?

• The paragraph in the middle is a direct quote from the lecture notes, right? When you cite sources directly, please a) link to the sources whenever possible and b) mark the quotes as blockquotes by adding a > before the quote in the editor Sep 7 at 15:27
• Not an exact quote, but thanks for the tip.
– Alex
Sep 7 at 15:34

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.