The evolution of observables in the Heisenberg picture

Question

Suppose that a system $S$ undergoes the following time-evolution:

$t_{1} \rightarrow t_{2}: |x_{a}\rangle \rightarrow \sum c_{i}|x_{i}\rangle$,

where $|x_{a}\rangle$ and each of $|x_{i}\rangle$ is a position eigenstate.

How is this process described in the Heisenberg picture? My trouble lies in the fact that $\sum c_{i}|x_{i}\rangle$ is (generally) not an eigenstate of the position operator. The operator (say, $O$) of which $\sum c_{i}|x_{i}\rangle$ is an eigenstate is (generally) not even an Hermitian operator, so the position operator would not be able to unitarily evolve into $O$. Then how is the state of $S$ at $t_{2}$ in the Schrodinger picture related to the (time-evolved) operator (observable) at $t_{2}$ in the Heisenberg picture?

Answer 1

In these types of problems, the phrase I always come back to is "expectation values don't care what picture we are in". This leads one to write \begin{align} \left ( \left < \psi | e^{\frac{i H t}{\hbar}} \right ) O \left ( e^{\frac{-i H t}{\hbar}} | \psi \right > \right ) = \left < \psi | \left ( e^{\frac{i H t}{\hbar}} O e^{\frac{-i H t}{\hbar}} \right ) | \psi \right >. \end{align} The left hand side has brackets which separate states from operators in the Schroedinger picture. The right hand side has brackets which separate states from operators in the Heisenberg picture. So the latter says that any operator (including the position operator) is evolved unitarily by $H$.

This notation helps in your example because $\sum_i c_i \left | x_i \right >$ will have to just be $e^{\frac{-i H t}{\hbar}} \left | x_a \right >$ for the evolution to be valid. The eigenstate of $x$ has therefore become an eigenstate of $e^{\frac{-i H t}{\hbar}} x e^{\frac{i H t}{\hbar}}$ which is Hermitian. You'd have to be given a Hamiltonian and coefficients $c_i$ which are incompatible with each other for this not to be the case.

Update

An arbitrary state $\left | \psi \right >$ is an eigenstate of the "observable" $\left | \psi \right > \left < \psi \right |$ (and many others). However, it may well be the case that these operators aren't easy to interpret. For a fixed observable (something we enjoy dealing with, say position) it is certainly the case that the system not being in an eigenstate of it is ubiquitous.

If we say that a state cannot be evolved unitarily from an eigenstate of position, that's equivalent to saying that it's not an eigenstate of $e^{\frac{-i H t}{\hbar}} x e^{\frac{i H t}{\hbar}}$ for any $t$. But there is nothing wrong with such a state. In the language of wavefunctions, it's just something that never looks like a delta function no matter how much we evolve forwards or backwards.

Maybe an analogy for an operator with a discrete spectrum will help. \begin{equation} \psi(x, 0) = \begin{cases} cx, & 0 < x < L/2 \\ c(L - x), & L/2 < x < L \\ 0, & \mathrm{otherwise} \end{cases} \end{equation} is a perfectly good initial state for the infinite square well. However, it never becomes an energy eigenstate.