[NOTES/QFT-04005] S Matrix in Interaction Picture qft-lsn-04005

In the interaction picture the total Hamiltonian is split into two
parts \(H= H_0+ H{'}\).
Let \(u_n\) denote the eigenfunctions of \(H_0\) with eigenvalues
\(E_n\)
\begin{equation}\label{EQ01} H_0 u_n({\mathbf x})= E_n u_n(\mathbf x) \end{equation}
In the interaction picture the field operator obey EOM with \(H_0\) as
the Hamiltonian.
\begin{equation}\label{EQ02} i\hbar\dd[\psi(\mathbf x,t)]{t} = H_0 \psi(\mathbf x,t) \end{equation}
Taking the expansion of the field in terms of \(u_n(x)\) as
\begin{equation}\label{EQ03} \psi(\mathbf x,t) = \sum_n a_n u_n(\mathbf x) e^{-iE_nt/\hbar}, \qquad \psi^\dagger(\mathbf x,t) = \sum_n a_n^\dagger u_n^*(\mathbf x) e^{iE_nt/\hbar}, \end{equation}
We note that the operators \(a_n\) will be independent of time. Verify This.

The field operators obey equal time commutation relations.

\begin{eqnarray} [\psi(\mathbf x,t),\psi^\dagger(\mathbf y,t)] = \delta(x-y) \end{eqnarray} The operators \(a_m,a^\dagger_n\) obey commutation relations \begin{equation} [a_m, a^\dagger_n] = \delta_{mn} \end{equation} Using these commutators, it is straight to verify \begin{eqnarray} \psi(x)=\sum_n u_n(x,t) a_n, \quad \psi^\dagger(x)=\sum_n u_n^*(x,t) a^\dagger_n, \\{} [a_n, \psi^\dagger(\mathbf x,t)] = u_n^*, \qquad [\psi(\mathbf x,t), a_n^\dagger] = u_n. \end{eqnarray}

The states corresponding to \(\nu_1,\nu_2,...\) particle in levels
\(m_1, m_2,...\) are defined by
\begin{equation} \ket{\nu_1, \nu_2, ...}= \prod_m\frac{(a_{k}^\dagger)^{\nu_k}}{\sqrt{\nu_k!}}\,\ket{0}. \end{equation}