In the Schrodinger picture, states are time dependent and operators time-independent. So expected values look like: $\langle s_1,t|\hat{A}|s_1,t\rangle$.
If we go over to the Heisenberg picture the states are time-independent and the operators time dependent: $\langle s_1|\hat{A}(t)|s_1\rangle$.
My question is what happens if we make the ket $|s_1\rangle$ dependent on an operator. For example, take $|s_1\rangle = a_{p_1}^\dagger|0\rangle$ where $a_{p_1}^\dagger$ creates particles with momentum $p_1$ in the Schrodinger picture.
When we move to the Heisenberg picture, does the creation operator $a_{p_1}^\dagger$ become time dependent? And if so, does the Heisenberg ket $|s_1\rangle$ also become time dependent since it is defined in terms of the creation operator? I thought kets in the Heisenberg picture were supposed to be time-independent.
To provide a little bit of context, this question arose while I was reading my QFT textbook on S-matrix elements. I know what is meant by the Heisenberg and Schrodinger picture in ordinary single particle quantum mechanics, but I am getting confused in QFT because of the question asked above.
Best Answer
Let $t_0$ be the reference time, at which the Schrodinger and Heisenberg pictures are the same: $$ | \psi \rangle_H = | \psi(t_0) \rangle_S, $$ $$ \mathcal{O}_H(t_0) = \mathcal{O}_S $$ where $|\psi\rangle$ is a generic state, $\mathcal{O}$ a generic operator, and the subscripts $S$ and $H$ denote respectively the Schroedinger and Heisenberg pictures. Suppose also that we can write $$|\psi\rangle = c^\dagger |0 \rangle$$ for some creation operator $c^\dagger$. In what picture should we read this equation?
Of course you also ask how does the creation operator evolve in time. Again, in the Schroedinger picture it does not. In the Heisenberg picture you have the usual Heisenberg time evolution of an operator: $$ c_H^\dagger(t) = e^{i \mathcal{H} (t-t_0)} c_H^\dagger(t_0) e^{-i \mathcal{H} (t-t_0)} = e^{i \mathcal{H} (t-t_0)} c_S^\dagger e^{-i \mathcal{H} (t-t_0)}$$