# Proof of equivalence between Welch-bound-based and frame-potential-based definitions of t-designs

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?

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 $$S_t = \frac{\Pi_{\rm sym}}{\binom{d+t-1}{t}}.$$