In Quantum Mechanics, there are 3 pictures: Schrodinger, Heisenberg and Dirac (Interaction) picture. In Schrodinger picture, the states evolve in time and operators are constant in time, and vice versa in Heisenberg picture.
Now consider, a hermitian operator $\hat{A}$ with eigenvectors $\vert i\rangle$ and eigenvalues $A_{i}$. We can expand the operator as
$\hat{A} = \sum_{i} A_{i} \vert i\rangle \langle i\vert$ (1)
Which picture do we assume to perform this calculation ? If this is completely general (independent of picture), then why is the density operator defined as
$\hat{\rho} = \vert \psi(t)\rangle \langle \psi(t)\vert$
constant in time in the Heisenberg picture ? The definition of the density operator is in the form of Equation (1), yet every other operator other than density operator evolves in time in the Heisenberg picture. Furthermore, by the virtue of that expansion in Equation (1); if an operator evolves in time, the eigenvectors must also evolve in time as there is no other element from which an operator can get its time dependence ! What is the problem in my line of reasoning ?
Best Answer
You might well have skewed expectations about the logic behind time dependence in these pictures. I skip self-explanatory operator carets and use lower case for eigenvalues. Assume $A_S$ and the hamiltonian H have no intrinsic time dependence for simplicity, and $|\psi\rangle_H= |\psi(0)\rangle_S$, with $A_S=A_H(0)$ and $\rho_H= \rho_S(0)$, and recall $$ \langle A(t)\rangle = _S\!\!\langle \psi(t)|A_S| \psi(t)\rangle_S \\ = _H\!\!\langle \psi|e^{iHt}A_S e^{-iHt} |\psi\rangle_H = _H\!\!\langle \psi| A_H(t)|\psi\rangle_H \\ A_H(t)= e^{iHt}A_S e^{-iHt}, \qquad \rho_H= e^{iHt}\rho_S(t) e^{-iHt} , \\ \bbox[yellow]{ \langle A(t)\rangle = \operatorname{Tr}\rho_S(t) A_S= \operatorname{Tr}\rho_H A_H(t)}~~. $$
Recall the eigenstates of $A_S$ are not eigenstates of H. Your (1) follows by constructing a resolution of the identity on the right of your eigenvalue equation; in the Schroedinger picture, $$ A_s|i\rangle= a_i |i\rangle ~, $$ so that $$ A_H(t) e^{-iHt}|i\rangle = a_i e^{-iHt}|i\rangle,\leadsto \\ A_H(t) |i(t)\rangle = a_i |i(t)\rangle, $$ where I defined $|i(t)\rangle\equiv e^{-iHt}|i\rangle $, an infinite superposition of $|i\rangle$ s with time-dependent coefficients.
Consequently, $$ A_H(t)= \sum_i a_i |i(t)\rangle \langle i(t)|~~. \tag{1'} $$
So, indeed, your (1) is general and applies to both pictures, since the eigenvalue equation applies to both pictures.
Your expectations about time dependence were off.