# Why is the orbit of a unitary t design a complex projective t design?

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?

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.
• Not necessary. $\{U_i\}$ can be just a set of unitaries. It's orbit is the set $\{U_i | v \rangle \}$ for some state $| v \rangle$. Jan 4 at 17:24