Consider a finite subset $X\subset\mathbb{CP}^{d-1}$ of $d$-dimensional pure states. Following e.g. (Roy and Scott 2007), we say that $X$ is a complex projective $t$-design if $$\frac1{|X|}\sum_{x\in X} \mathbb{P}_x = \frac{\Pi_{\rm sym}^{(t)}}{\binom{d+t-1}{t}}, \qquad \mathbb{P}_x\equiv |x\rangle\!\langle x|,$$ where $\Pi_{\rm sym}^{(t)}$ is the projection onto the totally symmetric subspace of $(\mathbb{C}^d)^{\otimes t}$. Among other things, it is well-known that MUBs and SIC-POVMs are 2-designs, as per quant-ph/0502031, whereas stabiliser states are 3-designs, as per 1510.02767 and 1510.02619.

Perhaps the prototypical example of an MUB for a single qubit is given by the set of vectors $$X_{\rm MUB} =\{|0\rangle,|1\rangle, |0\rangle+|1\rangle,|0\rangle-|1\rangle,|0\rangle+i|1\rangle,|0\rangle-i|1\rangle\}.$$ It is not too hard to see that this is a 2-design, which more explicitly means that $$\frac16\sum_{x\in X_{\rm MUB}} \mathbb{P}_x\otimes \mathbb{P}_x = \frac{\Pi_{\rm sym}^{(t)}}{\binom{2+2-1}{2} } = \frac32 (I\otimes I+W),$$ where I used the explicit formula for the projection on the symmetric bipartite subspace, and $W$ is the swap operator.

Which brings me to my question. By playing numerically with these things, I found that $X_{\rm MUB}$ seems to be not only a 2-design, but a 3-design. That is, we have the stronger identity $$\frac16 \sum_{x\in X_{\rm MUB}} \mathbb{P}_x^{\otimes 3} = \frac14 \Pi_{\rm sym}^{(3)}.$$ Verifying this is equally pretty straightforward (unless I'm misunderstanding something here, which is possible). You can use for example the following Mathematica snippet

mubProjections = {
  {1, 0}, {0, 1}, {1, 1}/Sqrt@2, {1, -1}/Sqrt@2, {1, I}/Sqrt@2, {1, -I}/Sqrt@2
} // Map[KroneckerProduct[#, Conjugate@#] &];
1/6 KroneckerProduct[#, #, #] & /@ mubProjections // Total // MatrixForm

and you can recognise the structure of the projection onto the symmetric subspace, which in this case looks like $$\Pi_{\rm sym}^{(3)} = \mathbb{P}_{000} + \mathbb{P}_{111} + \mathbb{P}\left(\frac{|100\rangle+|010\rangle+|001\rangle}{\sqrt3}\right) + \mathbb{P}\left(\frac{|101\rangle+|011\rangle+|110\rangle}{\sqrt3}\right).$$

Interestingly doing the same with 4 tensor products reveals that this MUB is not a 4-design, although it's not too far from being one.

Reading around, I didn't find any mention of MUBs being also 3-designs, though I also haven't found mentions of them not being so. By contrast, stabiliser states are known to be 3-designs but not 4-designs, as per the references above. I haven't checked other MUBs yet, so I'm not sure whether them being 3-designs holds in general (I'd expect not). Still, is there a good way to see why this MUB is a 3-design? Are there other MUBs that are so, and if so, is there a characterisation for the class of MUBs that are 3 designs?


2 Answers 2


There is a bound for the size $n$ of a complex projective $t$-design (see Eq. 2.5 of Roy and Scott paper): $$ n \ge \binom{d-1+⌊t/2⌋}{⌊t/2⌋}\binom{d-1+⌈t/2⌉}{⌈t/2⌉}. $$ So, for a 3-design, $n\ge \binom{d}{1}\binom{d+1}{2} = d^2(d+1)/2 > d(d+1)$ for $d > 2$.

As for the qubit case, a complex projective $t$-design is the same thing as a spherical $t$-design on the sphere $S^2$ in $\mathbb{R}^3$ (it's via the Bloch sphere). MUB vectors in this case are just vertices of an octahedron. It's known that they form a 3-design (see e.g. Hardin and Sloane 1996).

  • $\begingroup$ ah, very nice! So it can't be true in higher dimensions as per the inequality. And I suppose all single-qubit MUBs are equivalent to the standard one? (seems like that has to be the case) So this property holds iff we're talking about MUBs with $d=2$ $\endgroup$
    – glS
    Commented Jan 25, 2023 at 21:28
  • $\begingroup$ Yeah, all 1-qubit MUBs must correspond to octahedrons (we know all the distances between points). And vice versa. $\endgroup$
    – Danylo Y
    Commented Jan 25, 2023 at 21:40

Another straightforward approach is to consider the alternative definition of (complex projective) $t$-designs in terms of frame potentials. Namely, as also mentioned e.g. in this other post, a set $X\subseteq\mathbb{CP}^{d-1}$ is a $t$-design iff, for all $k\le t$, $$\frac{1}{|X|^2} \sum_{x,y\in X}|\langle x|y\rangle|^{2k} = \frac{1}{\binom{d+k-1}{k} }.\tag1$$ Let's thus compute the LHS for our case. MUBs have $|X|=d(d+1)$ elements ($d+1$ bases, and $d$ element in each basis), and the inner products between different MUB elements are either $\langle x|y\rangle=0$ if $x,y$ are elements of the same basis, or $|\langle x|y\rangle|=1/\sqrt d$ if they belong to different bases. Therefore we have $$\frac{1}{|X|^2} \sum_{x,y\in X}|\langle x|y\rangle|^{2k} = \frac{1}{|X|^2}|X| \sum_{y\in X} | \langle x|y\rangle|^{2k} = \frac{1}{d(d+1)}\left(1+\frac{d^2}{d^k}\right) = \frac{d^{k-1}+d}{(d+1)d^k}.$$ For small $k$ this gives: $$ k=1\to \frac{1+d}{(d+1)d} = \frac1d = \frac{1}{\binom{d+1-1}{1}}, \\ k=2\to \frac{d+d}{(d+1)d^2} = \frac{2}{(d+1)d} = \frac{1}{\binom{d+2-1}{2}}, \\ k=3\to \frac{d^2+d}{(d+1)d^3} = \frac{1}{d^2}. $$ Note how in the $k=3$ case we get $1/d^2$, which is in general different from $\binom{d+3-1}{3}^{-1}=\binom{d+2}{3}^{-1}$. Except it can be equal in some cases, as $d^2=\binom{d+2}{3}$ amounts to the equation $d^2-3d+2=0$, which has solutions $d=2,1$. We conclude that MUBs are in fact 3-designs precisely when we are in two dimensions.

For larger $k$, the identity is not satisfied for any dimension $d$, hence MUBs are at most a 3-design.


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