Eigenkets in Interaction Picture

Question

Let us consider a system. In Schrodinger picture, its Hamiltonian $H$ is given by $H = H_0 + V(t)$, where $H_0$ is the unperturbed Hamiltonian and $V(t)$ is the time-dependent perturbation.

In interaction picture, the state ket $|\psi, t \rangle_I$ of this system is defined as $$|\psi, t \rangle_I = e^{iH_0(t-t_0)/\hbar} \, |\psi, t \rangle, $$ where $|\psi, t \rangle$ is the state ket of the system in Schrodinger picture.

In interaction picture, an operator $\hat{A}_I$ is defined as $$\hat{A}_I = e^{iH_0(t-t_0)/\hbar} \,\hat{A} \, e^{-iH_0(t-t_0)/\hbar},$$ where $\hat{A}$ is the operator in Schrodinger picture.

My questions are as follows.

1. What is the eigenvalue equation for $\hat{A}_I(t)$?

2. If the eigenkets are time dependent, then how do they evolve in time?

3. If an eigenket of $\hat{A}_I$ is $|i, t\rangle_I$, then how is it related to the eigenket $|i\rangle$ of $\hat{A}$ in Schrodinger picture?

Answer 1

In the interaction picture, the evolution of any state is dictated by $$i\hbar\frac{d}{dt}|\psi,t\rangle_I=V_I(t)|\psi, t\rangle_I\tag{1}$$ where $$|\psi,t\rangle_I=\exp(\frac{iH_0(t-t_0)}{\hbar})|\psi,t\rangle_S$$ and $$V_I(t)=\exp(\frac{iH_0(t-t_0)}{\hbar})V\exp(-\frac{iH_0(t-t_0)}{\hbar})$$ is the interaction hamiltonian in the interaction picture.

The solution to (1) is given by $$|\psi,t\rangle_I=U(t,t_0)|\psi,t_0\rangle_I$$ If both $H_0,V(t)$ are hermitian and the states are taken to be normalized on both sides then $U$ is unitary. It is also trivial to verify that,$$U(t_1,t_3)=U(t_1,t_2)U(t_2,t_3)\hspace{1cm} \text{for} \hspace{1cm} t_1\geq t_2\geq t_3$$ Substituting this into the equation (1) we get the equation satisfied by $U(t,t_0)$ is $$i\frac{\partial}{\partial t}U(t,t_0)=V_I(t)U(t,t_0).\tag{2}$$ If we can solve (2), we will plug the solution in (1), to obtain $|\psi,t\rangle_I$ ant any later time $t$. The solution to (2) in this equation with the initial condition $U(t_0,t_0)=\mathbb{1}$ given in terms of the following Dyson series, $$U(t,t_0)=T[\exp{(-i\int\limits_{t_0}^{t}dt^\prime V_I(t^\prime)})].$$

Therefore, an eigenstate of $A_I(t)$ will also evolve according to (1).

Answer 2

The state kets and operators in the interaction picture are defined via a unitary transformation:

$$ |\psi(t)\rangle_I = e^{iH_0t} |\psi(t)\rangle_S, \quad A_I(t) = e^{iH_0t} A_S \, e^{-iH_0t} $$ where for simplicity I have assumed $t_0=0$.

The eigenvalue equation in the Schrodinger picture is given by \begin{eqnarray} && \qquad A_S |a\rangle_S = a |a\rangle_S \\ && \Rightarrow A_I(t) \left(e^{iH_0t}|a\rangle_S\right) = a \left(e^{iH_0t}|a\rangle_S\right) \end{eqnarray} which is an eigenvalue equaiton for $A_I(t)$.

The eigen ket of $A_I(t)$ is given by $$e^{iH_0t}|a\rangle_S \equiv |a, t \rangle_I.$$

Then the eigenvalue equation becomes $$ A_I(t) |a, t \rangle_I = a |a, t \rangle_I. $$