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.