Revision history for comment 1005

Edited by Steve Flammia Dec 18 2017 20:54 UTC

The following observation resolves in the affirmative a decade-old open conjecture from this paper, except for $d=3$.

The Conjecture asks if any unitary 2-design must have cardinality at least $d^4 - d^2$, a value which is achievable by a Clifford group. This is true for any group unitary 2-design as follows. The cardinality of the group design has to be divisible by $d$, because a 2-design is also a 1-design, hence an irrep, and the order of the irrep has to divide the order of the group. Also $d(d\pm1)/2$ must divide the cardinality because the symmetric and antisymmetric reps must also be irreps to be a 2-design (by Theorem 6). The lcm $L$ of those must also be a divisor and is $d^3-d$ if $d$ is even and $\frac{d^3-d}{4}$ if $d$ is odd. The existing lower bound (Theorem 5) of $d^4-2d^2+2$ can be strengthened to the next greatest integer divisible by $L$, or equivalently, the least integer $k$ for which

$$ kL \ge d^4-2d^2+2. \qquad (1) $$

For the even case, it is easy to check that $k=d-1$ never yields solutions to (1), but $k=d$ is always a solution and yields the required bound. For the odd case, $k=4d-1$ only yields a solution to (1) for $d=3$ and $k=4d$ is always a solution and gives the desired bound. QED.

Although this resolves the conjecture for group 2-designs in dimensions $d\not=3$, it leaves open the possibility of exact 2-designs without any group symmetry that might have a slightly smaller cardinality.

Edited by Steve Flammia Dec 18 2017 20:52 UTC

The following observation resolves in the affirmative a decade-old open conjecture from this paper, except for $d=3$.

The Conjecture asks if any unitary 2-design must have cardinality at least $d^4 - d^2$, a value which is achievable by a Clifford group. This is true for any group unitary 2-design as follows. The cardinality of the group design has to be divisible by $d$, because a 2-design is also a 1-design, hence an irrep, and the order of the irrep has to divide the order of the group. Also $d(d\pm1)/2$ must divide the cardinality because the symmetric and antisymmetric reps must also be irreps to be a 2-design (by Theorem 6). The lcm $L$ of those must also be a divisor and is $d^3-d$ if $d$ is even and $\frac{d^3-d}{4}$ if $d$ is odd. The existing lower bound (Theorem 5) of $d^4-2d^2+2$ can be strengthened to the next greatest integer divisible by $L$, or equivalently, the least integer $k$ for which
$$ kL \ge d^4-2d^2+2\,. \quad (*) $$
For the even case, it is easy to check that $k=d-1$ never yields solutions to (*), but $k=d$ is always a solution and yields the required bound. For the odd case, $k=4d-1$ only yields a solution for $d=3$ and $k=4d$ is always a solution and gives the desired bound. QED.

Although this resolves the conjecture for group 2-designs in dimensions $d\not=3$, it leaves open the possibility of exact 2-designs without any group symmetry that might have a slightly smaller cardinality.

Edited by Steve Flammia Dec 18 2017 20:52 UTC

The following observation resolves in the affirmative a decade-old open conjecture from this paper, except for $d=3$.

The Conjecture asks if any unitary 2-design must have cardinality at least $d^4 - d^2$, a value which is achievable by a Clifford group. This is true for any group unitary 2-design as follows. The cardinality of the group design has to be divisible by $d$, because a 2-design is also a 1-design, hence an irrep, and the order of the irrep has to divide the order of the group. Also $d(d\pm1)/2$ must divide the cardinality because the symmetric and antisymmetric reps must also be irreps to be a 2-design (by Theorem 6). The lcm $L$ of those must also be a divisor and is $d^3-d$ if $d$ is even and $\frac{d^3-d}{4}$ if $d$ is odd. The existing lower bound (Theorem 5) of $d^4-2d^2+2$ can be strengthened to the next greatest integer divisible by $L$, or equivalently, the least integer $k$ for which
$$kL \ge d^4-2d^2+2\,. \quad (*)$$
For the even case, it is easy to check that $k=d-1$ never yields solutions to (*), but $k=d$ is always a solution and yields the required bound. For the odd case, $k=4d-1$ only yields a solution for $d=3$ and $k=4d$ is always a solution and gives the desired bound. QED.

Although this resolves the conjecture for group 2-designs in dimensions $d\not=3$, it leaves open the possibility of exact 2-designs without any group symmetry that might have a slightly smaller cardinality.

Steve Flammia commented on Evenly distributed unitaries: on the structure of unitary designs Dec 17 2017 20:25 UTC

The following observation resolves in the affirmative a decade-old open conjecture from this paper.

The Conjecture asks if any unitary 2-design must have cardinality at least $d^4 - d^2$, a value which is achievable by a Clifford group. This is true for any group unitary 2-design as follows. The cardinality of the group design has to be divisible by $d$, because a 2-design is also a 1-design, hence an irrep, and the order of the irrep has to divide the order of the group. Also $d(d\pm1)/2$ must divide the cardinality because the symmetric and antisymmetric reps must also be irreps to be a 2-design (by Theorem 6). The polynomial gcd of those must also be a divisor and is $d^3-d$. The existing lower bound (Theorem 5) of $d^4-2d^2+2$ can be strengthened to the next greatest integer divisible by $d^3-d$. The polynomial remainder between them is $2-d^2$, and adding this difference gives the exact value of $d^4 - d^2$ from the Conjecture. QED.

Although this resolves the conjecture for group 2-designs, it leaves open the possibility of exact 2-designs without any group symmetry that might have a slightly smaller cardinality.