**Canonical transformation**

A canonical transformations of the Hamiltonian is given by
\begin{equation}
H' = e^{S} H e^{-S},
\end{equation}
with $S$ anti-hermitian. The idea is to eliminate certain terms in the Hamiltonian by making a basis change. We trade old degrees of freedom by new ones in the hope that the new Hamiltonian becomes easier to solve. The specific choice of $S$ depends on the application. The canonical transformation with
\begin{equation}
[S,H_0] = -\lambda V,
\end{equation}
is called the **Schrieffer-Wolff transformation**. More applications can be found here.

**Schrieffer-Wolff transformation**

The idea behind this scheme is to facilitate the projection of the Hamiltonian into some low-energy subspace up to a specific order in $\lambda$. Consider the case where the first-order result is zero:
\begin{equation}
H_{eff} = P_g H_0 P_g + P_g V P_g = E_g P_g,
\end{equation}
where $P_g$ is the projection operator onto the (degenerate) ground state of the unperturbed system. An example is Anderson's superexchange mechanism for the Hubbard model in the limit of weak site hopping. In this model, tunnelling between sites only occurs via higher-order virtual processes.

So we want to eliminate $\lambda V$ from the Hamilonian, which will give a representation where the higher-order processes are manifest in the Hamiltonian. To better keep track of the order in $\lambda$, we take $H' = e^{\lambda S} H e^{-\lambda S}$. By choosing $S$ so that $[S,H_0] = V$, you can get rid of all terms at first order in $\lambda$ in the transformed Hamiltonian:
\begin{equation}
H' = H_0 + \frac{\lambda^2}{2} [ V, S ] + \mathcal O(\lambda^3).
\end{equation}
Every term in $H'$ except for $H_0$ represents a higher-order process. Up to second order, the effective Hamiltonian then becomes
\begin{equation}
H_{eff} = P_g H' P_g = E_g P_g + \frac{\lambda^2}{2} P_g [V, S] P_g.
\end{equation}

**Example**

As an example, consider a degenerate ground state with $E_g=0$ and let $P_e$ be the projection onto an excited subspace with unperturbed energy $E_e$. Furthermore, assume that $V$ only connects off-diagonal elements between the ground state and this excited state:
\begin{equation}
V = P_g V P_e + P_e V P_g.
\end{equation}
Now take
\begin{equation}
S = \frac{P_g V P_e - P_e V P_g}{E_e}.
\end{equation}
and note that $S^\dagger = -S$ and that
\begin{equation}
[S,H_0] = \frac{1}{E_e} [P_g V P_e - P_e V P_g,H_0] = P_g V P_e + P_e V P_g = V.
\end{equation}
The transformed Hamiltonian becomes
\begin{align}
H' & = H_0 + \frac{\lambda^2}{2E_e} [ V, P_g V P_e - P_e V P_g ] \\
& = H_0 + \frac{\lambda^2}{E_e} \left( P_e V P_g V P_e - P_g V P_e V P_g \right).
\end{align}
Note the interpretation of both second-order terms. The term $P_g V P_e V P_g$ represents a two-step tunnelling process from the ground state to an excited state and back. The effective low-energy Hamiltonian becomes
\begin{equation}
H_{eff} = P_g H' P_g = -\frac{\lambda^2}{E_e} P_g V P_e V P_g.
\end{equation}
In the superexchange model of Anderson I mentioned before, this effective Hamiltonian represents an antiferromagnetic exchange interaction.

Let $\rho_i$ be a set of right eigenvectors of $\mathcal L$, ie $\mathcal L\rho_i = \lambda_i \rho_i$. Then, taking $\tilde \rho_i = U\rho_iU^\dagger$ we have :
\begin{align}
\tilde{\mathcal L}\tilde \rho_i &= (U\mathcal LU^\dagger )(U\rho_iU^\dagger) \\
&= U(\mathcal L\rho_i)U^\dagger\\
&= \lambda_i U\rho_i U^\dagger\\
&= \lambda_i \tilde\rho_i
\end{align}
therefore we can take $\mu_i = \lambda_i$.

## Best Answer

For the right A take the CC of the exponent and for the Pauli spin matrix take adjoint. Hope it works.