The paper Qubit stabilizer states are complex projective 3-designs states in the final paragraph that "any orbit of a unitary t-design is a complex projective t-design." Using this fact one can take the simple proof that the Clifford group is a t-design and turn it into a simple proof that the set all stabilizer states is a complex projective t-design.

It definitely seems intuitive that the orbit of an orthogonal t-design would be a spherical t-design and the orbit of a unitary t-design would be a complex projective t-design. But what is a good proof of this?


1 Answer 1


One of the equivalent definitions of a unitary t-design $\{U_i\} \subset \mathbb{U}(d)$ is that $$ \frac{1}{n}\sum_{i=1}^n (U_i^{\otimes t})M(U_i^{\otimes t})^\dagger = \int_{\mathbb{U}(d)} (U^{\otimes t})M(U^{\otimes t})^\dagger {\rm d}\mu(U) $$ for any matrix $M$, where $\mu$ is the Haar measure on the group of unitaries $\mathbb{U}(d) \subset \mathbb{C}^{d \times d}$.

If we take $M = |v\rangle\langle v|^{\otimes t}$ then it's easy to see that $\{|v_i\rangle\ = U_i|v\rangle\}$ satisfies $$ \frac{1}{n}\sum_{i=1}^n \big(|v_i\rangle\langle v_i| \big)^{\otimes t} = \int_{\mathbb{C}P^{d-1}} \big(|\phi\rangle\langle \phi|\big)^{\otimes t} d\mu(\phi) $$ which is equivalent to the definition of projective t-design.

  • $\begingroup$ Ah I see so the reference is specifically talking about unitary designs that are groups, otherwise the idea of taking an orbit doesn't really make sense $\endgroup$ Jan 4 at 15:13
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    $\begingroup$ Not necessary. $\{U_i\}$ can be just a set of unitaries. It's orbit is the set $\{U_i | v \rangle \}$ for some state $| v \rangle$. $\endgroup$
    – Danylo Y
    Jan 4 at 17:24

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