Operator relations¶

In this section, the relation

\[\begin{equation} Se^XS^{-1}=e^{SXS^{-1}} \end{equation}\]

for operators \(S\) and \(X\) is derived. This identity expresses how operator exponentials transform under similarity transformations.

First, recall the power series definition of the exponential of an operator:

\[\begin{equation} e^X=\sum_{n=0}^\infty\dfrac{X^n}{n!}. \end{equation}\]

Using this expansion, the left-hand side can be written as

\[\begin{equation} Se^XS^{-1}=S\left(\sum_{n=0}^\infty\dfrac{X^n}{n!}\right)S^{-1}. \end{equation}\]

Since \(S\) is a linear operator, it can be distributed over the sum, allowing \(S\) and \(S^{-1}\) to be moved inside:

\[\begin{equation} \label{appendix:powerseries1} Se^XS^{-1}=\sum_{n=0}^\infty\dfrac{SX^nS^{-1}}{n!} \end{equation}\]

The problem is thus reduced to evaluating the transformed powers \(SX^nS^{-1}\).

To proceed, recall the identity

\[\begin{equation} S(AB)S^{-1}=\left(SAS^{-1}\right)\left(SBS^{-1}\right), \end{equation}\]

which shows how products of operators transform under conjugation.

Applying this relation to powers of \(X\), for \(n=2\) one obtains

\[\begin{equation} SX^2S^{-1}=\left(SXS^{-1}\right)\left(SXS^{-1}\right), \end{equation}\]

and for \(n=3\)

\[\begin{equation} SX^3S^{-1}=\left(SXS^{-1}\right)\left(SXS^{-1}\right)\left(SXS^{-1}\right). \end{equation}\]

This pattern generalizes to arbitrary \(n\), yielding

\[\begin{equation} SX^nS^{-1}=\left(SXS^{-1}\right)^n, \end{equation}\]

so that each term in the series transforms into a corresponding power of the conjugated operator.

Substituting this result back into \(\eqref{appendix:powerseries1}\) gives

\[\begin{equation} \label{appendix:powerseries_final} \begin{aligned} Se^XS^{-1}&=\sum_{n=0}^\infty\dfrac{\left(SXS^{-1}\right)^n}{n!}=e^{SXS^{-1}} \end{aligned} \end{equation}\]

This establishes the desired relation, showing that exponentiation and similarity transformation commute.