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I'm trying to compute the following integral: $$\int U^{\otimes t}\left|x_1,\cdots,x_t\middle\rangle\middle\langle x'_1,\cdots,x_n'\right|\left(U^\dagger\right)^{\otimes t}\,\mathrm{d}\mu(U)$$ Where $\mu$ is the Haar-measure. There are two easy special cases:

  1. If $x_i=x'_i$ for every $i$, then the integral is equivalent to the one in this answer, which means that it is equal to $\begin{pmatrix}d+t-1\\t\end{pmatrix}^{-1}P_{\text{Sym}^t\left(\mathbb{C}^d\right)}$, with $P_{\text{Sym}^t\left(\mathbb{C}^d\right)}$ being the projector on the symmetric subspace $\left(\mathbb{C}^d\right)^{\otimes t}$.
  2. If $t=1$, then it is $\frac1d\delta_{x_1,x_1'}I$.

However, I'm not sure how to proceed in the general case. I wanted to use this answer, but I'm not sure about what are the irreps of $U\mapsto U^{\otimes t}$. For instance, in the case of $t=2$, if we take $x_1=x_1'$ and $x_2\neq x_2'$, I feel like the irreps are a sub-symmetric space and another one. More generally, I feel like I have to "group" the $x_i$ and $x_i'$ that are equal into sub-symmetric groups, but I'm unsure about how to do it since I'm not familiar with representation theory.

How to proceed in this case? Is there a simple argument that I'm missing? I've heard about Schur-Weyl duality, but do not understand it at the moment.

EDIT: possible duplicate of Random quantum states and Schur-Weyl duality ?

Tristan Nemoz
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1 Answers1

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Consider the permutation matrices $\Pi_\pi$: $$ \Pi_\pi|x_1,\cdots,x_t\rangle = |x_{\pi^{-1}(1)},\cdots,x_{\pi^{-1}(t)}\rangle, $$ where $\pi \in S_t$ is a permutation of $\{1,\dots,t\}$.

It's easy to see that the map $$ \Phi(M) = \int U^{\otimes t}M\left(U^\dagger\right)^{\otimes t}\,\mathrm{d}\mu(U) $$ acts as identity for $M=\Pi_\pi$.

It's a consequence of Schur-Weyl duality that this map is actually the orthogonal projection of $M$ onto the subspace spanned by all matrices $\Pi_\pi$.

The problem is that matrices $\Pi_\pi$ are non-orthogonal, so in general it's not that easy to compute the projection onto the subspace they generate. The related answers cover the approaches.

It's easy for $t=2$ though. There are only two permutation matrices, the identity $I$ and the swap $U_{sw}$: $U_{sw}|x,y\rangle = |y,x\rangle$. You can orthogonalize them by taking $$ U' = U_{sw} - {\rm Tr}(U_{sw})I/d^2=U_{sw}-I/d, $$ $$ U''=U'/||U'||_{HS}=(U_{sw}-I/d)/\sqrt{d^2-1}. $$ Then $$ \Phi(M) = {\rm Tr}(M)I/d^2 + {\rm Tr}(M(U'')^\dagger)U''. $$

Another way is to notice that $$ P_{\rm sym} = \frac{1}{2}(I + U_{sw}), $$ $$ P_{\rm asym} = \frac{1}{2}(I - U_{sw}), $$ where $P_{\rm sym}$, $P_{\rm asym}$ are projectors on the symmetric and asymmetric subspaces respectively. Clearly, $$ {\rm span}\langle P_{\rm sym}, P_{\rm asym} \rangle = {\rm span}\langle I, U_{sw} \rangle. $$
But $P_{\rm sym}$, $P_{\rm asym}$ are orthogonal to each other. So we can write $$ \Phi(M) = {\rm Tr}(MP_{\rm sym})P_{\rm sym}\frac{2}{d(d+1)} + {\rm Tr}(MP_{\rm asym})P_{\rm asym}\frac{2}{d(d-1)}, $$ where coefficients come from $$||P_{\rm sym}||_{HS}^2 = {\rm Tr}(P_{\rm sym})=\frac{d(d+1)}{2}, $$ $$ ||P_{\rm asym}||_{HS}^2 = {\rm Tr}(P_{\rm asym})=\frac{d(d-1)}{2}.$$

In general, ${\rm span}\langle \Pi_\pi\rangle$ equals to ${\rm span}\langle P_\lambda\rangle$, where $P_\lambda$ are projections onto invariant subspaces (common eigenspaces) of the representation $\pi \rightarrow \Pi_\pi$ of $S_t$.

Danylo Y
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