Time evolution of operators in the Heisenberg picture

Question

In the Schrodinger picture, a state $|{\psi_{S}(t)}\rangle$ at a time $t$ is given by applying the time-evolution operator $\hat{U}(t)=e^{-\frac{i\hat{H}t}{\hbar}}$ to the state $|{\psi_{S}(0)}\rangle$ at time $t=0$, such that $$|{\psi_{S}(t)}\rangle = \hat{U}(t)|{\psi_{S}(0)}\rangle.$$ $U(t)$ is a unitary operator. Consider a general operator $\hat{O}_{S}$ in the Schrodinger picture that does not depend on time. The representation of this operator in the Heisenberg picture does evolve with time. However, the representation of the state $|{\psi}\rangle$ in the Heisenberg picture will be constant with time. Therefore we have $\hat{O}_{H}(t)$ and $\hat{O}_{H}(0)$, but a constant $|{\psi_{H}}\rangle$. The state $|{\psi_{H}}\rangle$ is given by $$|{\psi_{H}}\rangle = |{\psi_{S}(0)}\rangle = \hat{U}^\dagger(t)|{\psi_{S}(t)}\rangle.$$ This is a unitary transformation from one Hilbert space to another, where both describe the same physical system. Intuitively, we can interpret the transformation in the following way: to go from the Schrodinger picture at time $t$ to the Heisenberg picture at time $t$, we need to "de-evolve" the Schrodinger-picture system, "deleting" the time evolution, and thereby achieving a picture where the state vector is a time constant. This interpretation is consistent with the operator transformation: $$\hat{O}_{H}(t) = \hat{U}^\dagger(t)\hat{O}_{S}\hat{U}(t).$$ The action of the $\hat{O}$ operator in the Heisenberg picture is equivalent to "evolving" the system "up to" the Schrodinger picture, applying $\hat{O}$ in the Schrodinger picture, and then "de-evolving" back to the Heisenberg picture (I understand that this unitary transformation operator does not signify time evolution, but it is identical to the time-evolution operator, which makes sense intuitively).

At time $0$, when no evolution has yet occurred, we get that $\hat{U}(t) = \hat{I}$ is the identity operator (substitute $t=0$). This means that $\hat{O}_{H}(0) = \hat{O}_{S}$ i.e. the Heisenberg and Schrodinger pictures are equivalent at $t=0$ as expected. However, if we substitute this result in $\hat{O}_{H}(t) = \hat{U}^\dagger(t)\hat{O}_{S}\hat{U}(t)$ we get $$\hat{O}_{H}(t) = \hat{U}^\dagger(t)\hat{O}_{H}(0)\hat{U}(t),$$ which is the transformation describing the time-evolution of oprator $\hat{O}$ in the Heisenberg picture. However, this reads as follows: to get the effect of $\hat{O}_{H}(t)$ in the Heisenberg-picture Hilbert space at time $t$ we evolver the system (by $\hat{U}(t)$) up to the Heisenberg-picture Hilbert space at time $0$. Then we apply $\hat{O}_{H}(0)$ and de-evolve the system back to the space at time $t$.

This suggests that operators in the Heisenberg picture evolve according to $\hat{U}^\dagger$. But this operator was initially defined to describe the time evolution of the system in the Schrodinger picture. So it would seem as if operators in the Heisenberg picture are evolving backwards? Is there an intuitive explanation behind this?

Answer 1

The crux of the matter is this, the Schrödinger and Heisenberg pictures are both valid because they both yield the same observable results. That is to say, the amplitude $$\langle \psi\lvert U^\dagger(t) H U(t) \lvert\psi\rangle \tag{1}$$ is the same whether you choose the "use" the Heisenberg picture or the Schrödinger picture.

If we start from the Schrödinger picture, i.e., states evolve according to $U(t)\lvert \psi \rangle$, then to obtain the Heisenberg picture we do not consider "how the operator $H$ should evolve [via some Schrödinger picture interpretation]". Rather, we consider how should we evolve the operator $H$ such that the amplitude at time $t$ is still given by (1).