How to derive the electric dipole spin resonance (EDSR) Hamiltonian in a slanting Zeeman field?

Background

I have been reading through the Supplementary Notes from [1] and I am having some trouble understanding part of the derivation. To begin with, the paper describes a spin qubit within a 2D electron gas. The qubit is manipulated via a time-dependent electric field which generates an alternating potential within the dot. The complete Hamiltonian that the electron experiences within the dot is made up of three terms:

$$H = H_{QD} + H_{M} + H_{SO}$$

Where the differnt parts of the Hamiltonian are (from left to right) the Quantum Dot potential term, the magnetic field term, and the spin-orbit coupling term. Each of these has a fairly simple/intuitive form:

$H_{M} = -g mu_{B} vec{B_{M}} cdot vec{sigma}/2$, where $vec{sigma} = (sigma_x , sigma_y, sigma_z)$ are the Pauli spin matricies, g is the g-factor and $mu_B$ is the Bohr magneton. The magnetic field, $B_M$, has the form $vec{B_M} = (B_0 + delta B_0 – |b_{sl}|z) hat{x} – |b_{SL}|x hat{z}$.

$$H_{SO} = alpha(p_{x’} sigma_{y’} – p_{y’} sigma_{x’}) + beta(-p_{x’} sigma_{x’} + p_{y’} sigma_{y’})$$

Where $alpha$ and $beta$ are the Rashba and Dresselhaus spin-orbit coefficients respectively, while $p_{x’,y’}$ and $sigma_{x’,y’}$ are the momentum and spin operators in the $x’$ and $y’$ directions.

Now here is where I get a bit confused in the derivation. The authors then say "the spin-orbit coupling can be conveniently accounted up to first order in $alpha$ and $beta$ by the canonical transformation: $tilde{H} = U^{dagger}HU$ where $U = Exp[i vec{n} cdot vec{sigma}/2]$." The authors then explain that $H_{QD}$ is unaffected (which I assume is because it is not a matrix quantity) but that $H_M + H_{SO}$ is transformed into:

$$tilde{H_M} + tilde{H_{SO}} = -g mu_B (vec{B_M} + vec{B_{SO}}) cdot vec{sigma}/2$$

Where:

$$vec{B_{SO}} = vec{n} otimes vec{B_{SO}}$$
$$n_{x’} = frac{2m^{*}}{hbar}(alpha y’ + beta x’)$$
$$n_{y’} = frac{-2m^{*}}{hbar}(alpha x’ + beta y’)$$
$$n_z =0$$

Question

What I don’t understand is how the exponentiated matrix $U = Exp[i vec{n} cdot vec{sigma}/2]$ is applied to the Hamiltonian $H = H_{QD} + H_{M} + H_{SO}$ to obtain $tilde{H_M} + tilde{H_{SO}} = -g mu_B (vec{B_M} + vec{B_{SO}}) cdot vec{sigma}/2$. I understand how this works for the 1D case but I don’t see how the dot-product is carried through here and how it works with the other Paul spin-matrices.

• How does one go about this derivation?
• What is the significance of the dot-product in the exponential transformation matrix?
• Why (mathematically) is $H_{QD}$ unaffected by the $U^{dagger}HU$ transformation but $H_{M} + H_{SO}$ are?

Thank you for taking the time to read this question. I appreciate any and all help with understanding this derivation including links, book recommendations and (of course) written derivations. Thank you for your time.