In case of a time dependent Hamiltonian of the sort $$H=\frac{p^2}{2m}+\frac{1}{2}m \omega(t) x^2$$ I have solved for the time evolution operator using the Schrodinger equation and got $U(t,0)$. If, I am working in the Schrodinger picture, my ket at some later time $t$ would be given as $|\phi(t)\rangle=U|\phi(0)\rangle$. Also, operators like the creation and annihilation operators $a^\dagger$ and $a$ have explicit time dependence and at any later time $t$ they can be written as $a(t)=S(t,0)a(0)S^\dagger(t,0)$. Now, in a paper I am reading they treat $S=U$. How is this a valid assumption and if it is valid, which picture am I exactly working in, Schrodinger, Heisenberg or interaction? Also, while computing expectation values like $\langle \phi(t)|x(t)|\phi(t)\rangle$, what rules of transformation I must use now for the establishing a relation between $x(0)$ and $x(t)$ and similarly for $|\phi(t)\rangle$?


It seems like you have meddled with all different pictures of QM at the same time. Your first approach is perfectly correct where you find the time evolution operator and evolve your state using it. You are now working in the Schrödinger picture where kets evolve with time and operators are stationary. Next you mention that the operators are time dependent, which is an incorrect statement to make if you are in the Schrödinger picture.

$ a(t)=U^{†}(t,0)a(0)U(t,0) $ is the expression giving the time evolution of operators in the Heisenberg picture. Hence in the paper they are definitely working in the Heisenberg picture. You have not mentioned what S(t,0) is and I do not know why you are confused about this.

Your expectation value $ ⟨ϕ(t)|x(t)|ϕ(t)⟩ $ has time dependence on both the operators and the state, which can only mean that you are now working in the interaction picture. However, I think you are just basically confused between the Schrödinger and Heisenberg pictures because of the time dependence of the Hamiltonian. Note that in the Schrödinger picture, the kets evolve with time and operators do not. In the Heisenberg picture, the operators evolve in time and kets do not. This could seem counterintuitive but notice that the expectation values (which is what you are concerned with) is exactly the same irrespecive of picture you use to do in your calculations.

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  • $\begingroup$ The operator $a$ depends explicitly on $t$ as $\omega(t)$ comes in the definition of $a$ and $S$ is the bogoliubov transformation that has been made. $\endgroup$ – Naman Agarwal Jun 26 '18 at 9:29

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