1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$

Your Answer

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

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