# Heisenberg picture with creation annihilation operators

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.

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.

They key distinction is that it's observables, not operators, that evolve in the Heisenberg picture. This might sound like semantics at first, given that in most introductory presentations "(Hermitian) operator" is often used as a synonim of "observable". However, operators are just mathematical objects, without an intrinsic physical meaning, and they can be used in the definition of observables, states, or in other ways. Therefore, there isn't an intrinsic rule on how operators evolve in the Schrödinger or Heisenberg picture, it depends on the role they play.

In the OP, the creation operator is used to define a state, so it doesn't evolve in the Heisenberg picture. On the other hand, if the creation operator defines an observable (for example, when you expand the field observables in normal modes), then it evolves as $$a^{\dagger}_p \mapsto a^{\dagger}_{p,\mathrm{H}}(t):= e^{iHt} a^{\dagger}_p e^{-iHt}$$.

For completeness, we can look at the evolution of the creation operator in the Schrödinger picture, when it is used to define a state as in the OP (this was mentioned in another post, but with a small mistake). The state evolves as $$$$a^{\dagger}_p|0\rangle \mapsto e^{-iHt} a^{\dagger}_p|0\rangle$$$$ (note, this is not the same as $$a^{\dagger}_p|0(t)\rangle$$). We can insert $$\mathbb{I} = e^{iHt} e^{-iHt}$$ between the operator and the vector to get $$$$a^{\dagger}_p|0\rangle \mapsto a^{\dagger}_{p,\mathrm{S}}(t) |0(t)\rangle_{\mathrm{S}},$$$$ where \begin{align} a^{\dagger}_{p,\mathrm{S}}(t) &:= e^{-iHt} a^{\dagger}_p e^{iHt}, \\ |0(t)\rangle_{\mathrm{S}} &:= e^{-iHt} |0\rangle. \end{align} Note that the "Schrödinger evolution" of the operator $$a^{\dagger}_{p,\mathrm{S}}(t)$$ is not the same as the time evolution of an observable in the Heisenberg picture. In this case, they are related by $$a^{\dagger}_{p,\mathrm{H}}(t) = a^{\dagger}_{p,\mathrm{S}}(-t)$$. Don't take this as a rule though, it really depends what you are using the operators for!

Finally, it's worth mentioning that, similarly to how operators are not always observables, vectors are not necessarily states, so they might evolve or not in either picture. For example, if you calculate transition amplitides $$\langle \phi | \psi \rangle$$, the vector $$|\phi\rangle$$ plays the role of an observable, so it will evolve in the Heisenberg picture and remain invariant in the Schrödinger picture (while $$|\psi\rangle$$ does the opposite). This is necessary to preserve the transition amplitude $$\langle \phi | e^{-iHt}|\psi \rangle$$ in all pictures.

If this seems confusing, remember that (as mentioned in another post), the key is to reproduce the relevant physical quantities, which are the time-evolving expectation values $$$$\langle O(t)\rangle = \langle \psi | e^{iHt} O e^{-iHt} | \psi \rangle.$$$$ Different pictures are simply obtained by attaching the evolution operators to one or another part of the expression.

• I disagree completely with the statement that "The key distinction is that it's observables, not operators, that evolve in the Heisenberg picture". This is simply incorrect. All operators evolve in the Heisenberg picture, not doing so will give incorrect predictions. Commented Apr 14, 2023 at 9:01
• Sorry but that's just wrong, as the example in the OP shows. If you use the creation operator to define a state, and then evolve it as if it were an observable in the Heisenberg picture, you get a time-evolving state. But states are constant in the Heinseberg picture, so you clearly get a contradiction. For another example, take the density operator representation of a state: $\rho=|\psi\rangle\langle \psi |$. This is clearly constant in the Heisenberg picture (because $\psi$ is constant). Again, you have an operator that doesn't evolve in the Heisenberg picture. Commented May 2, 2023 at 15:23
• I'll rephrase my complaint. I'm complaining about using the word observable in a non-standard way, since this is usually restricted to the hermitian operators. If you don't add clarification that you're not using the standard definition in this context, then it reads that you're suggesting only the hermitian operators need evolve, not the non hermitian ones. As far as I understand the point you are trying to make is that operators representing states don't evolve, which is fine except as written you haven't actually explained this in your answer Commented May 3, 2023 at 9:59
• I have a hard time understanding the complaint, since the whole post is about the fact that operators (Hermitian or not) are not always associated to observables. I'm not referring to any "non-standard" definition of observable: the whole discussion applies to observables defined as Hermitian operators (except the very last bit, where the observable is associated with an eigenvector, which however I would claim is pretty standard too). The OP asks how operators evolve when they don't define observables, which is what the answer is about. I changed the first paragraph for further clarity. Commented May 4, 2023 at 11:32