Applying a Unitary Transformation to a Rashba Spin-Orbit Coupled System for EDSR

I have been trying to follow the derivation in [Ref. 1] so as to better understand the theory of Electric Dipole Spin Resonance (EDSR) in materials which experience Spin-Orbit Coupling (SOC). However, I’ve been unable to re-produce some of the equations in the paper and I was hoping that someone could point out where I’ve gone wrong in my derivation.

First, the authors give us the full Hamiltonian (Eq. 1) and then break it down into its components in Eq. 2. I’ve re-written the equations below to save time.

$$H = frac{p^2}{2m_e}+frac{1}{2}m_eomega^2x^2+alpha_Rsigma^yp+alpha_Dsigma^xp+frac{g_emu_BB}{2}sigma^n$$
$$alpha = sqrt{alpha_R^2 + alpha_D^2}$$
$$ phi = arctan(alpha_R/alpha_D)$$
$$H = H_0 + H_1$$
$$H_0 = frac{p^2}{2m_e} + frac{1}{2}m_eomega^2x^2+alphasigma^ap$$
$$H_{1} = frac{g_emu_BB}{2}[cos(theta-phi)sigma^a+sin(theta-phi)sigma^b]$$

I have assumed that all terms have their normal quantum mechanical meaning. That is:

• $p$, is the momentum operator of the form $-ihbarfrac{d}{dx}$ (assuming a 1D space)
• $sigma^x,sigma^y$ are the standard Pauli matrices
• $sigma^c = textbf{c}cdottextbf{$sigma$}$ where $textbf{c = a, b, c}$ and $textbf{a}=(cos(phi),sin(phi))$

Next, the authors carry out a unitary transformation to diagonalize the Hamiltonian $H_0$. The rules for a unitary transformation are explained on Wikipedia and I’ll include it as a reference [Ref. 2]. The Unitary operator in the paper takes the following form:

$$U = Exp[{i(m_ealpha/hbar)xsigma^a}] = Exp[i(m_ealpha/hbar)x(cos(phi)sigma^x+sin(phi)sigma^y)]$$
$$U^dagger = Exp[-i(m_ealpha/hbar)x(cos(phi)sigma^x+sin(phi)sigma^y)]$$

Because of how unitary transformations work, and the rules regarding Pauli matrix multiplication, I don’t expect the first two terms of $H_0$ to change. Only the third term is expected to change and so we can write out the equation for the third term:

$$Exp[i(m_ealpha/hbar)x(cos(phi)sigma^x+sin(phi)]*(-ihbaralpha(cos(phi)sigma^x+sin(phi)sigma^y)frac{d}{dx})*Exp[-i(m_ealpha/hbar)x(cos(phi)sigma^x+sin(phi)]$$

I tried doing this calculation by hand at first but I made several errors so I decided to check my work via Mathematica. I used the following code to check my math:

MatrixForm[
 FullSimplify[
  MatrixExp[
    I*((Subscript[m, e]*[Alpha]*x)/[HBar])*(Cos[[Phi]]*
        Subscript[[Sigma], x] + 
       Sin[[Phi]]*Subscript[[Sigma], y])] . 
       (([Alpha]*(Cos[[Phi]]*Subscript[[Sigma], x] + 
         Sin[[Phi]]*Subscript[[Sigma], y])) . 
          ((-I)*[HBar]*
       D[MatrixExp[(-I)*((Subscript[m, e]*[Alpha]*
              x)/[HBar])*(Cos[[Phi]]*Subscript[[Sigma], x] + 
            Sin[[Phi]]*Subscript[[Sigma], y])], x]))]]

However, the result that this produces is:
$$U(-ihbaralpha(cos(phi)sigma^x+sin(phi)sigma^y)frac{d}{dx})U^{dagger} = -m_ealpha^2mathbb{1}$$
(Where $mathbb{1}$ is the identity matrix)

This is close, but not exactly what is listed in the paper. In the paper the transformed Hamiltonian should have the form:
$$(frac{p^2}{2m_e} + frac{1}{2}m_eomega^2x^2-frac{1}{2}m_ealpha^2)mathbb{1}$$

Where does the factor of 1/2 come from? Is it possible that there is an uncorrected error in the original article? Or did I do something wrong when I input the equations into Mathematica? Is there an easier way to carry out this transformation?

What the authors describe next makes sense to me. The solution to transformed Hamiltonian ($frac{p^2}{2m_e} + frac{1}{2}m_eomega^2x^2+frac{1}{2}m_ealpha^2$) is that of a harmonic oscillator with a lowered ‘floor.’ Basically, all one needs to do is use the standard solutions to the quantum harmonic oscillator ($|{psi_n}rangle$) and adjust the energy so that $E_n = n+frac{1}{2}hbaromega – frac{1}{2}m_ealpha^2$. However, the next step doesn’t make as much sense to me. To transform the solution from our transformed Hamiltonian space back to the original Hamiltonian space I thought all we need to do is $U^{dagger}|{psi_n}rangle$ but the answer they give in the paper doesn’t seem to match this:

$$|Psi_{nuparrow}rangle = Exp[-i(m_ealpha/hbar)x]psi_n(x)|uparrow_arangle$$
$$|Psi_{ndownarrow}rangle = Exp[i(m_ealpha/hbar)x]psi_n(x)|downarrow_arangle$$

Where, $|uparrow_arangle = (sqrt{2}/2)(Exp[-iphi/2],space Exp[iphi/2])^T$ and $|downarrow_arangle = (sqrt{2}/2)(Exp[-iphi/2],space -Exp[iphi/2])^T$. I don’t really see how one goes from $U^{dagger}|{psi_n}rangle$ to the above answer so I would really appreciate some help with filling in the missing steps.

Thank you for time!