Why is the gauge group of pseudo-fermion mapping referred to as $\mathrm{SU}(2)$ and not $\mathrm U(2)$?

Question

The representation of spin $\frac{1}{2}$ operators $\hat{S}^{a}$ by pseudo-fermions (also called Abrikosov fermions) is defined by the mapping

$$ \hat{S}^{a} = \frac{1}{2} \text{Tr}\big[ \hat{\psi}^{\dagger} \hat{\psi} \sigma^{a} \big], \quad a=x,y,z \tag1 $$

where $\sigma^{a}$ are Pauli matrices and $\hat{\psi}$ is a matrix consisting of fermion operators $\hat{f}_{\alpha}$ with spin $\alpha$

$$\hat{\psi} = \begin{pmatrix} \hat{f}_{\uparrow} & \hat{f}_{\downarrow} \\ \hat{f}^{\dagger}_{\downarrow} & -\hat{f}^{\dagger}_{\uparrow} \end{pmatrix}.$$

The mapping Eq. (1) has a gauge symmetry under a left multiplication of $\hat{\psi}$ with an arbitrary unitary matrix. Hence I would call the gauge group of a spin Hamiltonian on which the mapping is performed $\mathrm U(2)$. However, in the literature the gauge group is generally referred to as $\mathrm{SU}(2)$. Could someone explain the reason for this?

Answer 1

The issue is obscured by the notation / formulation that Wen uses. Fundamentally, your spin transformation must look like something of the form

$$ \begin{pmatrix} f_\uparrow \\ f_\downarrow^\dagger \end{pmatrix} \mapsto U \begin{pmatrix} f_\uparrow \\ f_\downarrow^\dagger \end{pmatrix} \hspace{2em} (1) $$

for an appropriately chosen unitary $U$. (It must be unitary to preserve fermion normalisation).

Left-multiplying $\Psi$ by $U$ implements exactly this for its left column, but not the right. Instead, it performs

$$ \begin{pmatrix} f_\downarrow \\ -f_\uparrow^\dagger \end{pmatrix} \mapsto U \begin{pmatrix} f_\downarrow \\ -f_\uparrow^\dagger \end{pmatrix} \hspace{2em} (2) $$

whereas appropriate manipulations of (1) would instead suggest the transformation law

$$ \begin{pmatrix} f_\downarrow \\ -f_\uparrow^\dagger \end{pmatrix} \mapsto RU^*R^{-1} \begin{pmatrix} f_\downarrow \\ -f_\uparrow^\dagger \end{pmatrix} \hspace{2em} (3) $$

where $R = \begin{pmatrix}&1\\-1&\end{pmatrix}$.

Now a generic $U(2)$ matrix can be written as

$$ U = \begin{pmatrix} a & b \\ -e^{i\theta}b^* & e^{i\theta} b^*\end{pmatrix} $$ for $a,b\in\mathbb{C},\ \theta\in\mathbb{R}$, $|a|^2 + |b|^2 = 1$.

It is then a quick calculation to show that the condition $RU^*R^{-1} = U$ is exactly equivalent to $\theta=0$, i.e., the condition $\det U = 1$.

TL;DR

It is a special property of only $SU(2)$ matrices that rotations of the fermion pair can be represented as left multiplication of a matrix, using $U(2)$ matrices here does something that breaks fundamental commutation relations.

However, it is also true that in principle one could have used a $U(2)$ matrix for equation (1). This preserves canonical commutators perfectly well - as Lie algebras, $\mathfrak{u}(2) \simeq \mathfrak{u}(1)\times \mathfrak{su}(2)$, with the Abelian part factoring out (indeed this ultimately ends up being a trivial phase for, e.g., the $f_{\uparrow}$'s). Note that as groups though, $U(1)\times SU(2)$ is a double cover of $U(2)$.