2.4 \(\mathbb{Z}_2\times \mathbb{Z}_2\) projective representation

Here we introduce a systematic method in studying the projective phase, using the same example of qubit. For simplicity, we concentrate on the following \(\mathbb{Z}_2\times \mathbb{Z}_2\) subgroup inside \(SO(3)\): \[\begin{equation} \tag{2.22} \{\mathrm{Id}, R_x(\pi), R_y(\pi), R_z(\pi)\} \subset SO(3). \end{equation}\] Note that \(R_x(\pi) = R_y(\pi)R_z(\pi) = R_z(\pi)R_y(\pi)\), so this subset is closed under the multiplication forming the group \(\mathbb{Z}_2\times \mathbb{Z}_2\). To be concise, we rename the generators of this group as \(a := R_z(\pi)\) and \(b := R_y(\pi)\), then \(R_x(\pi) = ab = ba\). The explicit actions of these generators on the coordinates are \[\begin{align} \tag{2.23} a \curvearrowright (\theta,\phi) &\mapsto (\theta,\phi+\pi)\\ b \curvearrowright (\theta,\phi) &\mapsto (\pi-\theta,\pi-\phi) \end{align}\] From this and (2.9), we can read the action of \(a\) and \(b\) on the states as \[\begin{align} \tag{2.24} a \curvearrowright [\, \cos(\theta/2)\ket{0} + \sin(\theta/2) e^{\mathrm{i}\phi}\ket{1} \,] \mapsto & [\, \cos(\theta/2)\ket{0} - \sin(\theta/2) e^{\mathrm{i}\phi}\ket{1} \,]\\ b \curvearrowright [\, \cos(\theta/2)\ket{0} + \sin(\theta/2) e^{\mathrm{i}\phi} \ket{1}\,] \mapsto & [\, \sin(\theta/2)\ket{0} - \cos(\theta/2) e^{-\mathrm{i}\phi}\ket{1} \,]\\ \sim & [\, \sin(\theta/2)e^{\mathrm{i}\phi}\ket{0} - \cos(\theta/2) \ket{1} \,]. \end{align}\] The last line is the equivalence between the states by the overall phase \(e^{\mathrm{i}\phi}\). These transformations can be easily realized by the unitary matrices \[\begin{equation} \tag{2.25} U_a = \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \sigma_3,\quad U_b' = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} = \mathrm{i}\sigma_2. \end{equation}\] However, the latter matrix squares to \(-\mathbf{I}_2\), although \(b^2=1\). To fix this, one can instead use \[\begin{equation} \tag{2.26} U_b = \begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \\ \end{pmatrix} =\sigma_2. \end{equation}\] However, an issue remains: although \(ab = ba = R_x(\pi)\), we have \(U_a U_b = - U_b U_a\). This discrepancy between the multiplication law among group elements and their corresponding unitaries is the projective phase. If we set \(U_{ab} = U_b U_a = \mathrm{i}\sigma _1\), the multiplication law among \(U_g\), \(g\in \mathbb{Z}_2\times\mathbb{Z}_2\) is \[\begin{equation} \tag{2.27} U_g U_{g'} = e^{\mathrm{i}\alpha(g,g')}U_{gg'}, \end{equation}\] where in this example the projective phase is \[\begin{equation} \tag{2.28} \alpha(a,b) = \alpha(ab,ab) = \pi,\quad \alpha(g,g') = 0 \,\, \text{(otherwise)}. \end{equation}\]

Recall that we have the room to redefine each \(U_g\) by a phase. Can we use this freedom to eliminate the projective phase? Let us say that we redefine \(U_g\) by phase \(\beta(g)\): \[\begin{equation} \tag{2.29} U_g \mapsto e^{\mathrm{i}\beta(g)}U_g. \end{equation}\] However, such phase rotation cannot eliminate the commutator \([U_a,U_b]\), which has to be zero if \(\alpha = 0\).

When a projective phase cannot be removed by the phase rotation of the unitaries, the phase is called (cohomologically) nontrivial. The phase \(\alpha\) in (2.28) is nontrivial.

One significance of the nontrivial projective phase is that it prohibits a one-dimensional representatoin: in the \(\mathbb{Z}_2\times\mathbb{Z}_2\) example, since \([U_a,U_b]\neq 0\) (even though the group \(\mathbb{Z}_2\times \mathbb{Z}_2\) is abelian), the such operators cannot act on a one-dimensional space. Note that for a linear representation one always has the trivial representation; with a nontrivial projective phase, there is no such thing.

This in turn put a restriction on the possible Hamiltonian. Assume that the Hamiltonian on the qubit preserves the \(\mathbb{Z}_2\times \mathbb{Z}_2\) symmetry. If the two energy were distinct, each state should consist a one-dimensional representation with the projective phase, which contradicts with the above observation. Hence the projective action of \(\mathbb{Z}_2\times \mathbb{Z}_2\) is enough to ensure that the two states are degenerate.

Now, we also see that the entire group \(SO(3)\) acting on the qubit also has a nontrivial projective phase. This is because other elements of \(SO(3)\) does not help to resolve the projective phase that \(\mathbb{Z}_2\times \mathbb{Z}_2\) has.