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 $\endgroup$ Commented Jan 4 at 15:13
  • 2
    $\begingroup$ 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
    Commented Jan 4 at 17:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.