$\mathcal{H}$ is a $d$-dimensional complex vector space. $\mathcal{E}$ maps matrix on $\mathcal{H}^{\otimes m+k}$ to matrix on $\mathcal{H}^{\otimes m}$ through $$\mathcal{E}(X)=EXE^{+},$$ where $m,k$ are integers, $E$ is a $d^m\times d^{m+k}$ matrix and $E^+$ stands for the Hermitian conjugate of $E$

What do we know about the following mapping

$$\int_U {{U}^{+}}^{\otimes m}EU^{\otimes m+k}X{{U}^{+}}^{\otimes m+k}E^{+}U^{\otimes m} dU$$

where $U^{\otimes m+k}=U\otimes U\cdots\otimes U$, and the integral ranges over Haar measure $U$ over $\mathcal{H}$.

A particular interesting case is $m=1$.

  • $\begingroup$ The Weingarten function may be helpful in evaluating the Haar integral. See this MO post and relevant wikipedia articles. $\endgroup$
    – anecdote
    Oct 6, 2016 at 22:18
  • $\begingroup$ is $+$ actually the "Hermitian" conjugate instead of the complex conjugate? Is $U^{\otimes m }$ randomly taken from $U(d^m)$ or it's just a $u\otimes u \otimes u \cdots u$ where $u$ is randomly taken from $U(d)$. I don't understand the measure of the integral, because $U^{\otimes m}$ and $U^{\otimes m+k}$ live in $U(d^m)$ and $U( d^{m+k} )$ with different Haar measure. Please edit to clarify especially the dimensions of those matrices and which unitary group you take the Haar measure. $\endgroup$
    – anecdote
    Oct 6, 2016 at 22:36
  • $\begingroup$ @anecdote Please see the updated one. $\endgroup$
    – gondolf
    Oct 7, 2016 at 5:58

1 Answer 1


This is an extended version of my comment of Weingarten functions, a brute force computation that perhaps not very insightful.

I'll call $m+ k = n$ in the following and use $\dagger$ to denote Hermitian conjugate.

Let $\{e_i\}$ to be a basis in $\mathcal{H}$ then the basis in $\mathcal{H}^{\otimes m} \times \mathcal{H}^{\otimes n}$(where $E$ lives in) is \begin{equation} e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_m} \otimes e^{\dagger}_{j_1} \otimes e^{\dagger}_{j_2} \otimes \cdots \otimes e^{\dagger}_{j_n} \end{equation} The matrix $E$ can be decomposed as follows where repeated indices are assumed to be summed over \begin{equation} E = E_{i_1 i_2 \cdots i_m , j_1 j_2 \cdots j_n} e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_m} \otimes e^{\dagger}_{j_1} \otimes e^{\dagger}_{j_2} \otimes \cdots \otimes e^{\dagger}_{j_n} \end{equation} Similarly $U_{ij}$ are the components of $U$ under $e_i \otimes e_j^\dagger$ basis, then the integral becomes \begin{equation} \begin{aligned} \int_U U^{\dagger}_{\bar{i}_1 i_1} U^{\dagger}_{\bar{i}_2 i_2} \cdots U^{\dagger}_{\bar{i}_{n+m} i_{n+m} }U_{ j_1 \bar{j}_1 } U_{ j_2 \bar{j}_2 } \cdots U_{j_{n+m}\bar{j}_{n+m} } \\ E_{i_1 i_2 \cdots i_m,j_1 j_2 \cdots j_n} X_{\bar{j}_1 \bar{j}_2 \cdots \bar{j}_n \bar{i}_{m+1} \cdots \bar{i}_{m+n}} E^{\dagger}_{i_{m+1} i_{m+2} \cdots i_{m+n} ,j_{n+1} j_{n+2} \cdots j_{n+m}} dU \end{aligned} \end{equation} the free indices are $\bar{i}_1, \bar{i}_2, \cdots, \bar{i}_m$ and $\bar{j}_{n+1}, \bar{j}_{n+2}, \cdots \bar{j}_{n+m}$.

The integral of the component of unitary matrices can be represented as a series over all possible permutations $\tau, \sigma$ \begin{equation} \int_{\mathcal{U}(N)} \prod_{k=1}^{n} U^{\dagger}_{\bar{i_k} i_k} U_{j_k \bar{j}_k} dU=\sum_{\tau,\sigma\in S_n} {\rm Wg}^U(\tau^{-1}\sigma)\prod_{k=1}^n \delta_{i_k,\tau(j_k)}\delta_{\bar{i}_k,\sigma(\bar{j} _k)} \end{equation} where $\rm Wg$ is the Weingarten function.

In this summation, the indices with and without bar are contracted separated, so $E$ will contract with $E^{\dagger}$ for different permutations and $X$ will contract with itself.

To further simplify this integral, you will have to explore the properties of the Weingarten function. For example when $\sigma = \tau$, the function is $1$ and all the others will be of order $\mathcal{O}(\frac{1}{d})$. If $d$ is a large number this will give you the leading order contribution.

Even the $m = 1$ case is hard. When $n > 1$, the contractions can't be expressed as trace or matrix product. There are in general $(n+m)!$ terms and I don't know how to obtain a closed form expression.

Let me demonstrate how to compute $n = 1$(which may not be trivial to OP). In that case we have $E_{i_1 j_1} X_{\bar{j}_1 \bar{i}_2 } E^{\dagger}_{i_2 j_2}$ contracted those deltas. $n+m = 2$, so there are only two possible permutations for $\sigma $ and $\tau$. Check out the Wikipedia table, when they are the same ${\rm Wg} = \frac{1}{d}$, when they are different ${\rm Wg} = \frac{-1}{d(d^2 -1)}$. Therefore the integral is \begin{equation} \begin{aligned} \frac{1}{d} \left[ E_{i_1 i_1} E^{\dagger}_{i_2 i_2 } X_{\bar{i}_1 \bar{j}_2 } + E_{i_1 i_2 }E^{\dagger}_{i_2 i_1} X_{\bar{i}_2 \bar{i}_2 } \delta_{\bar{i}_1 \bar{j}_2 } \right]\\ - \frac{1}{d(d^2 - 1 ) } \left[ E_{i_1 i_1} E^{\dagger}_{i_2 i_2 } X_{\bar{i}_2 \bar{i}_2 } \delta_{\bar{i}_1 \bar{j}_2 } + E_{i_1 i_2 }E^{\dagger}_{i_2 i_1} X_{\bar{i}_1 \bar{j}_2 } \right] \end{aligned} \end{equation}

In the matrix notation it is \begin{equation} \begin{aligned} \frac{1}{d} \left[ {\rm tr}(E) {\rm tr}(E^{\dagger}) X + {\rm tr}(E E^{\dagger}) {\rm tr}( X) I \right] - \frac{1}{d(d^2 - 1 ) } \left[ {\rm tr}(E) {\rm tr}(E^{\dagger}) {\rm tr}( X) I + {\rm tr}(E E^{\dagger}) X \right] \end{aligned} \end{equation}

  • $\begingroup$ Thanks! Is there explicit closed form for the case m=1? $\endgroup$
    – gondolf
    Oct 8, 2016 at 10:54

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