# Time evolution of the time-dependent eigenvalue corresponding to an operator and its state vector in the interaction picture

I am new in studying the three pictures in quantum mechanics and I have a specific doubt regarding the time evolution of the time-dependent eigenvalue corresponding to an operator and its state vector. How to get an expression for the time derivative of an eigenvalue in terms of its state vector and operator?

Let us take,

$$A(t)$$ $$|\psi_n(t)$$> = $$a_n(t)$$ $$|\psi_n(t)$$>, A is the operator and $$a_n$$'s are its eigenvalues corresponding to |$$\psi_n$$>.

$$\therefore$$ $$a_n$$ = <$$\psi_n$$|A|$$\psi_n$$>

$$\implies$$ $$\dot a_n$$ = <$$\dot \psi_n$$|A|$$\psi_n$$> + <$$\psi_n$$|$$\dot A$$|$$\psi_n$$> + <$$\psi_n$$|A|$$\dot \psi_n$$>, where I've used $$\dot a_n$$ for $$\frac {d}{dt}a_n$$ to make the expressions handy.

Now <$$\dot \psi_n$$|A|$$\psi_n$$>$$^\dagger$$ = <$$\psi_n$$|A|$$\dot \psi_n$$>

$$\therefore$$ $$\dot a_n$$ = <$$\psi_n$$|$$\dot A$$|$$\psi_n$$> + 2Re(<$$\psi_n$$|A|$$\dot \psi_n$$>) ... (1)

Also from interaction picture of quantum mechanics, we know that

$$\dot A$$ = -$$\frac {i}{\hbar}[A_I(t), H_{0,S}]$$ and |$$\dot \psi_n$$> = -$$\frac {i}{\hbar}H_{1,I}(t)$$ $$|\psi_n$$> (Source: Wikipedia article)

Now I want to know,

(i) Am I conceptually right and proceeding correctly?

(ii) If so, then if I plug the expressions of $$\dot A$$ and |$$\dot \psi_n$$> in the equation (1), the expression for $$\dot a_n$$ becomes more complicated and I can't understand how to proceed further to get the most simplified expression for $$\dot a_n$$.

(iii) What is the most simplified version for the expression of $$\dot a_n$$ in general?