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In (Dankert et al. 2009), the authors define a unitary t-design as a finite set of unitaries $\{U_k\}_{k=1}^K\subset \mathbf U(D)$ such that for all polynomials $P_{(t,t)}(U)$ of "degree at most $t$ in the matrix elements of $U$ and at most $t$ in the complex conjugates of those matrix elements" we have $$\frac1 K \sum_{k=1}^K P_{(t,t)}(U_k) = \int_{\mathbf U(D)} dU\, P_{(t,t)}(U).$$ Shortly thereafter, they define the "$\mu$-twirl superoperator" as $$\mathbb{E}_\mu(\Lambda) \equiv \int_{\mathbf U(D)} d\mu(U)\, \mathcal E_{U^\dagger}\circ \Lambda\circ\mathcal E_U, \qquad \mathcal E_U(X)\equiv UXU^\dagger,$$ that is, more explicitly, $$\mathbb{E}_\mu(\Lambda):X\mapsto \int_{\mathbf U(D)}d\mu(U)\, U^\dagger\Lambda(UXU^\dagger)U.$$

The authors then write that, if I understand correctly, unitary 2-designs are characterised by the property that for any quantum map $\Lambda$, $$\mathbb{E}_\mu(\Lambda) = \mathbb{E}_{\rm Haar}(\Lambda),$$ where $\mu$ is the discrete uniform measure over the finite set of unitaries making up the 2-design.

To justify this, they consider maps of the form $X\mapsto AXB$, work out explicitly what the equality of the twirls corresponds to (Eq. (6) in the paper) and argue that one gets the defining relation of 2-designs considering $A,X,B$ of the form $|i\rangle\!\langle j|$.

I don't quite see why this is the case. How do we get an arbitrary polynomial $P_{(t,t)}$ from the expressions with the twirls?

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  • $\begingroup$ another relevant reference discussing characterisation of k-designs via twirling is arxiv.org/abs/1510.02769 $\endgroup$
    – glS
    Jan 9, 2023 at 13:06

1 Answer 1

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Eq. 6 is equivalent to $$ \frac{1}{K}\sum_k \langle i|U_k^\dagger A U_kXU_k^\dagger BU_k |j\rangle = \int_{\mathbf U(D)} \langle i|U^\dagger A UXU^\dagger BU |j\rangle, $$
for all $i,j \in [D]$.

If we take $A,B,X$ also in the form $|i\rangle\langle j|$, then the expression $\langle i|U^\dagger A UXU^\dagger BU |j\rangle$ becomes $u_1^*u_2u_3^*u_4$, where $u_i$ are some matrix entries of $U$. They can be any four entries, and thus by linearity we can form any homogeneous polynomial $P$ of total degree $(2,2)$. So that the defining equality
$$ \frac1 K \sum_{k=1}^K P(U_k) = \int_{\mathbf U(D)} dU\, P(U) $$ holds for such $P$.

Note that $u_{11}u_{11}^* + \dots + u_{D1}u_{D1}^* = 1$ for any unitary. Thus $u_{11}u_{11}^*u_xu_y^* + \dots + u_{D1}u_{D1}^*u_xu_y^* = u_xu_y^*$, and we can deduce that defining equality also holds for monomials of degree $(1,1)$.

I'm not sure how to get other (unbalanced) monomials, but it looks like the average over them must be 0.

Anyway, if the defining equality holds for all homogeneous polynomials of total degree $(2,2)$, then it's easy to see that we can go backwards, i.e. prove the twirling equality because it's linear over $A,B,X$.

Update
We actually can't get all monomials from the twirling equality. That equality doesn't depend on the phases of $U_k$. But $\sum_k P(U_k)$ for the monomial $P(U)=u_{11}$ does.

I can't present a 2-desing that shows the problem, but it's known that the set generated by clock and shift matrices $$ \{ X^kZ^l, ~~ k,l \in [D]\}, $$ where $$ X = \sum_i |i+1\rangle\langle i|,~~ Z = \sum_j e^{2\pi j/D} |j\rangle\langle j| $$ forms a unitary basis, and thus a 1-design (as defined in https://arxiv.org/abs/0809.3813v1). One can see that $\sum_{kl} P(X^kZ^l) = D \neq 0$ for $P(U)=u_{11}$.

Also, the definition in https://arxiv.org/abs/quant-ph/0611002 states that polynomials must be homogeneous of degree $(2,2)$.

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  • $\begingroup$ thanks. I also got to writing a similar expression, but was taken aback by the fact that terms from the twirling have the form $A_{ik}X_{\ell m} B_{np} U_{k\ell} U_{pq} \bar U_{ij} \bar U_{nm} |j\rangle\!\langle q|$, whereas a general poly I think is $a_{ijk\ell mnpq} U_{ij} U_{k\ell} \bar U_{mn} \bar U_{pq}$, so it seems like there's less freedom in choosing polynomials from twirling. It's like the difference between all (1,1) polys and the set of expressions $UXU^\dagger$ for all $X$. If I understand, you're saying that the difference disappears due to the defining property of unitaries? $\endgroup$
    – glS
    Dec 2, 2022 at 17:06
  • $\begingroup$ ah, maybe I get it now. If the averaging property holds for all expressions of the form $UXU^\dagger$, then it holds for the linear combinations, and we can thus build up arbitrary polynomial expressions by taking suitable linear combinations. And same holds for the case of 2-designs. $\endgroup$
    – glS
    Dec 2, 2022 at 17:15
  • $\begingroup$ not arbitrary, monomials should have the same number of real and conjugate variables. I don't think we can get the monomial $u_{11}$ from twirling. $\endgroup$
    – Danylo Y
    Dec 2, 2022 at 17:26
  • $\begingroup$ right, didn't consider the "degree at most $t$" part. So general polynomials should include lowest degree terms. So you're saying it's not actually true that we can characterise 2-designs via equality of the twirled maps, because identifying these we can't ensure equality of lower-order polynomials in $U$? Or does identity on higher-order polynomials somehow ensure identity of lower order ones at least on average? $\endgroup$
    – glS
    Dec 2, 2022 at 17:33
  • $\begingroup$ I think the twirling equality is equivalent to defining equality only on the subset of polynomials $-$ that contain monomials or degree $(2,2)$ and $(1,1)$ but no others in their linear expansion. $\endgroup$
    – Danylo Y
    Dec 2, 2022 at 17:44

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