Why can unitary 2-designs be characterised via twirling superoperators?

Question

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?

Answer 1

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 $(2,2)$.