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I was reading a paper that claimed the following Hamiltonian had $C_4$ rotational symmetry,

$$ \hat{H} = \sum_{k} c^\dagger_k h_s(\bf{k}) \sigma_s c_k$$

where the Bloch hamiltonian is given by

\begin{equation}h_s(\bf{k})\sigma_s = t \sin(k_x)\sigma_x + t \sin(k_y)\sigma_y + (m - \cos(k_x) - \cos(k_y))\sigma_z \end{equation}

I can't seem to verify this claim. The momentum should transform under a $\pi/2$ rotation as $k_x \rightarrow -k_y$ and $k_y \rightarrow k_x$, while the Pauli matrices transform as (I think) $\sigma_i \rightarrow -\sigma_i$ by using the transformation $(\sigma_i)' = R\sigma_iR^{-1}$. With that, we'd have

\begin{equation} \hat{H}'= t\sin(k_y)\sigma_x - t\sin(k_x)\sigma_y -(m-\cos(k_x) - \cos(k_y))\sigma_z \end{equation}

The above is not invariant, so there is probably a very basic error I'm making in transforming, if someone could point out what I'm missing I'd be grateful.

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  • $\begingroup$ you have to find a $R$ such that $R\sigma_{x}R^{-1}=\sigma_{x}$ and $R\sigma_{y}R^{-1}=-\sigma_{y}$ the thing is these Pauli matrices does not represent actual spin they represent pseudo spin so you $R$ is not the usual rep of rotation matrix. $\endgroup$ – physshyp Nov 9 '18 at 22:12
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    $\begingroup$ You get to choose how the Pauli matrices transform under the rotation. These Pauli matrices don't represent spin, they represent sites of the unit cell. When they say this Hamiltonian has rotational symmetry, they mean there exists SOME unitary such that $U^\dagger H(k_x,k_y)U=H(-k_y,k_x)$. The actual form of the unitary will depend on what the internal degrees of freedom represent physically. $\endgroup$ – Jahan Claes Nov 9 '18 at 23:25
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A rotation about $\hat{n}$ by an angle $\phi$ acts on spins according to $R(\hat{n},\phi) = e^{\frac{\vec{s}\cdot\hat{n}\phi}{i\hbar}}$, where $\vec{s} = \frac{\hbar}{2} \vec{\sigma}$ where, so that, equivalently, $R(\hat{n},\phi)=e^{-i\vec{\sigma}\cdot\hat{n}\frac{\phi}{2}}$. The factor of $\frac{1}{2}$ should resolve your contradiction.

For rotations of wave functions it is the same after substituting $\vec{s} \to \vec{l}=-i\hbar \vec{r} \times \vec{\nabla}$.

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