Let $ P $ be the Pauli group and $ Cl $ (the Clifford group) be the normalizer of $ P $ in the unitary group $ U_d $ . Consider the representation given by acting the unitary group $ U_d $ on the Lie algebra $ \mathfrak{su}_d $ by conjugation (essentially the adjoint representation) $$ \text{ad}: U_d \to \text{End}(\mathfrak{su}_d) $$ Since the Clifford group is a subgroup of $ U_d $ we can restrict $ \text{ad} $ to get a representation $$ \text{ad}:Cl \to \text{End}(\mathfrak{su}_d) $$ Is this representation of the Clifford group an irreducible representation?
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asked Mar 20, 2022 at 17:09
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2$\begingroup$ I think you can argue from the fact that the adjoint representation of U is irreducible that the adjoint representation of the Clifford group is also irreducible, since the Clifford group is a 2-design. $\endgroup$– Jahan ClaesCommented Mar 20, 2022 at 19:30
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$\begingroup$ Ya that sounds right but I was curious to see if someone had an easier/different perspective/ or how they would work out the details of the 2-design argument $\endgroup$– Ian Gershon TeixeiraCommented Mar 21, 2022 at 1:54
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The 2-design proof is exactly about proving that this is an irrep since these are equivalent statements. I think that this is about the simplest proof that you can find.
Let me elaborate. Let $\tau^{(2)}(U):= U\otimes U$ be a representation of $U(d)$. Then, a subgroup $G\subset U(d)$ is a unitary 2-design if and only if $\tau^{(2)}|_{G}$ decomposes into the same irreps as $\tau^{(2)}$. This is because the 2-design definition reads $$ \int_G U^{\otimes 2} (\cdot) (U^{\otimes 2})^\dagger \,dU = \int_{U(d)} U^{\otimes 2} (\cdot) (U^{\otimes 2})^\dagger \,dU. $$ These are both projectors onto the commutant of $\tau^{(2)}$ and $\tau^{(2)}|_{G}$, respectively. Since the dimension of the commutant is the sum of squared multiplicities of the irreps and $G$ is a subgroup, the commutant can only be the same if the irreps of $\tau^{(2)}$ are still irreducible under $G$.
Note that this implies the same statement for $(U \mapsto U\otimes\overline U) \simeq (U\mapsto U(\cdot)U^\dagger)$.
It is easy to see that the latter representation decomposes as $1 \oplus \mathrm{ad}$. Clearly, the trivial irrep is irreducible for $G=\mathrm{Cl}_n$, so $\mathrm{Cl}_n$ is a unitary 2-design if and only if $\mathrm{ad}|_{\mathrm{Cl}_n}$ is irreducible.
The usual proof uses that $\mathrm{Cl}_n$ acts transitively on traceless Pauli matrices. This already implies the correct dimension of the commutant, see Lemma 2 in [Zhu]. Alternatively, on can evaluate the above integral on the Pauli basis, see e.g. Sec. IV.C in [Gross et al.].
