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"Don't panic!"

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*instantaneous eigenstates*in the Heisenberg picture. Does it simply mean that if vectors in the corresponding Hilbert space are eigenstates of some operator, then they won't necessarily be so for all times ##t##, the eigenstates of the operator will change as it evolves in time?

As far as I understand, in the Heisenberg picture the states are time-independent and instead it is the operators that evolve in time - does this mean that their eigenstates also evolve in time or are how I described above?

The reason I ask is that I've been reading a text in which they consider the position operator ##\hat{x}## in both the Schrödinger picture and the Heisenberg picture. They start off in the Heisenberg picture and state that an

*instantaneous eigenstate*of the position operator ##\hat{x}(t)## at some time ##t_{i}## is defined by $$\hat{x}(t_{i})\lvert x,t_{i}\rangle =x_{i}\lvert x,t_{i}\rangle$$ and take care to point out that the eigenstate ##\lvert x,t\rangle## is simply parametrised by time ##t## and is not dynamical - it simply labels the time at which the position is measured. They then move on to consider the same operator in the Schrödinger picture, denoting it as ##\hat{x}=\hat{x}(0)## to explicitly show that it is not time dependent in this picture. Clearly the eigenvalues of the operator should be the same in both pictures, and so $$\hat{x}\lvert x\rangle =x\lvert x\rangle$$ where ##\lvert x\rangle## is a corresponding eigenstate of the position operator in the Schrödinger picture. They then note that the Heisenberg position operator is related to its Schrödinger counterpart by $$\hat{x}(t_{i})=e^{i\hat{H}t_{i}}\hat{x}e^{-i\hat{H}t_{i}}$$ and proceeds as follows $$ \hat{x}(t_{i})\lvert x,t_{i}\rangle =e^{i\hat{H}t_{i}}\hat{x}e^{-i\hat{H}t_{i}}\lvert x,t_{i}\rangle \quad\Rightarrow\quad \hat{x}\left(e^{-i\hat{H}t_{i}}\lvert x,t_{i}\rangle\right)=x_{i}\left(e^{-i\hat{H}t_{i}}\lvert x,t_{i}\rangle\right)$$ where we have operated from the left on both sides by ##e^{-i\hat{H}t_{i}}##. Upon comparing this with $$\hat{x}\lvert x\rangle =x\lvert x\rangle$$, the above implies that the Schrödinger and Heisenberg eigenstates are related by $$\lvert x_{i}\rangle =e^{-i\hat{H}t_{i}}\lvert x,t_{i}\rangle\quad\Rightarrow\quad \lvert x,t_{i}\rangle =e^{i\hat{H}t_{i}}\lvert x_{i}\rangle$$

I get that the eigenstates of an operator in the Schrödinger picture should be time independent since the operator itself is time independent and so it's eigenstates should remain constant in time. However, I'm unsure as to how this concept translates into the Heisenberg picture?