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.


2 Answers 2


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?

  • In the Schroedinger picture we have: $$ |\psi(t) \rangle_S = c^\dagger_S | 0(t) \rangle_S = c_H^\dagger(t_0) | 0(t) \rangle_S$$ where $c_S^\dagger$ is the time-independent creation operator in the Schroedinger picture, and $|0\rangle_S$ is a time-dependent vacuum state (or some ground state or the system, or any other state really).
  • In the Heisenberg picture we have $$ |\psi(t)\rangle_H \equiv |\psi(t_0)\rangle_S \equiv |\psi\rangle_H = c^\dagger_H(t_0) | 0\rangle_H$$ note that in this case you are always "asking" for the state at the reference time $t_0$, so no time-dependence at all.

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: $$ c_H^\dagger(t) = e^{i \mathcal{H} (t-t_0)} c_H^\dagger(t_0) e^{-i \mathcal{H} (t-t_0)} = e^{i \mathcal{H} (t-t_0)} c_S^\dagger e^{-i \mathcal{H} (t-t_0)}$$


In your particular situation, no. Because your initial state is $|s\rangle$, as what you defined. It would be the invariant state in the Heisenberg picture.

Remember that the time dependent observable values $O(t)$ should be an invariant physical quantity in any physical pictures. Suppose the initial state is $|\psi\rangle$. Then in Schroedinger picture, we have final state as $|\psi(t)\rangle=e^{-iHt}|\psi\rangle$, so the observable is $$O(t) = \langle \psi(t)| O | \psi(t)\rangle = \langle \psi| e^{iHt} O e^{-iHt}|\psi\rangle$$ and in the Heisenberg picture, $$O(t) = \langle \psi| O(t) |\psi\rangle = \langle \psi| e^{iHt} O e^{-iHt}|\psi\rangle$$ In your example, $a_{p_1}^\dagger$ is not related to any observable, so your won't use the time dependent form.


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