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Let $X\subset\mathbb{C}^d$ be a (finite, non-empty) set of unit vectors. A standard way to define $X$ being a spherical $t$-design, is to impose it saturates the Welch bounds for all $k\le t$. Following the notation in https://arxiv.org/abs/quant-ph/0502031, this means $$\frac{1}{|X|^2} \sum_{x,y\in X}|\langle x|y\rangle|^{2k} = \frac{1}{\binom{d+k-1}{k} }.\tag1$$ Analogous definitions are used in https://arxiv.org/abs/1510.02767 and other related papers. I think this generalises to weighted designs by simply replacing the LHS with $\sum_{x,y\in X}p_x p_y \langle x|y\rangle|^{2k}$ for some probability distribution $(p_x)_x$.

However, I'm pretty sure I've seen an alternative definition used in the literature based on the identity $$S_k \equiv \sum_{x\in X} p_x |x\rangle\!\langle x|^{\otimes k} = \frac{\Pi_{\rm sym} }{\binom{d+k-1}{k}}, \qquad k=1,...,t,\tag2$$ where $\Pi_{\rm sym}$ is the projection on the completely symmetric subspace of $(\mathbb{C}^d)^{\otimes k}$.

It's pretty easy to see that (2) implies (1), as taking the inner product of the operator with itself, we get $$\sum_{x,y\in X}p_x p_y |\langle x|y\rangle|^{2k} = \operatorname{tr}(S_k^2) \equiv \langle S_k,S_k\rangle = \frac{\operatorname{tr}(\Pi_{\rm sym})}{\binom{d+k-1}{k}^2}= \frac{1}{\binom{d+k-1}{k} }.$$ How is the other direction proved? Equivalently, what's a reference showing this?

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Figured out after posting that the reference I was thinking about, which was also the same I mentioned in this other question, was (Scott 2006), where they discuss the statement at hand at the beginning of page 4.

Namely, define the operator $$S_t \equiv \sum_{x\in X}p_x |x\rangle\!\langle x|^{\otimes t}.$$ Then, as mentioned in the question, $\operatorname{tr}(S_t)=1$ and $\operatorname{tr}(S_t^2)=\sum_{x,y\in X} p_x p_y |\langle x,y\rangle|^{2t}$. We can then use the general fact that, for any positive semidefinite operator $Q\ge0$, $$\operatorname{tr}(Q)^2\le \operatorname{rank}(Q)\operatorname{tr}(Q^2).$$ Applied to $S_t$, this inequality translates to $$\operatorname{rank}(S_t) \ge \binom{d+t-1}{t},$$ where we used the assumption $\sum_{x,y\in X} p_x p_y |\langle x,y\rangle|^{2t}=\binom{d+t-1}{t}^{-1}$.

But by its definition the span of $S_t$ is contained in the fully symmetric subspace of $(\mathbb{C}^d)^{\otimes t}$, which has dimension $\binom{d+t-1}{t}$. It follows that $\operatorname{rank}(S_t)\le \binom{d+t-1}{t}$, and therefore $\operatorname{rank}(S_t)= \binom{d+t-1}{t}$.

But this means that $\operatorname{tr}(Q)^2\le \operatorname{rank}(Q) \operatorname{tr}(Q^2)$ saturates when $Q=S_t$, and this is in general true iff $Q$ is a multiple of the identity. We conclude that $S_t$ is a multiple of the identity (identity in the symmetric subspace, which means $S_t$ is a multiple of the projection onto this subspace), and from $\operatorname{tr}(S_t)=1$ we conclude that $$S_t = \frac{\Pi_{\rm sym}}{\binom{d+t-1}{t}}.$$

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