Show that there exists a set of unitary matrices $\{U_i\}$, and probability $\{p_i\}$, such that for any $n \times n$ matrix $A$ \begin{equation} \tag{1} \sum_{i} p_i U_i A U^{\dagger}_i = \text{tr}(A) \frac{I}{n} \end{equation}


For $n=2$, it is easy to show \begin{equation} \frac{1}{4} ( \sigma^x A \sigma^x + \sigma^y A \sigma^y + \sigma^z A \sigma^z + I A I ) = \text{tr}(A) I / 2 \end{equation} where $\sigma^{x,y,z}$ are Pauli sigma matrices. The idea comes from kraus operator sum representation.

We can then generalize to dimension $n = 2^m$, where $U_i$ can be taken as the tensor products of these basis, but not arbitrary dimension.

In indices, Eq.(1) is equivalent to \begin{equation} \sum_i p_i (U_i)_{ab} (U_i^*)_{dc} = \delta_{bc} \delta_{ad} / n \end{equation} This looks like the identity from the finite dimensional irreducible unitary representation of finite group, see Peter-Weyl theorem. But again this only works when group $G$ has irreducible representation at dimension $n$, and all $p_i$ in this case are equal.

I feel that "right proof" should not utilize these additional structures.

  • $\begingroup$ It may help to observe that the Choi matrix of your operator is given by $$ C_{\Phi} = \frac 1n I_n \otimes I_n $$ since the desired decomposition is a Kraus decomposition, the matrices $U_i$ must satisfy $$ \sum_{i}p_i \operatorname{vec}(U_i)\operatorname{vec}(U_i)^{\dagger} = C_{\Phi} = \frac 1n I_n \otimes I_n $$ $\endgroup$ – Omnomnomnom Mar 20 at 19:00
  • $\begingroup$ An educated guess: I think that $U_i = W^i$ should work, where $W$ is the $n \times n$ DFT matrix, with all $p_i = \frac 1n$ $\endgroup$ – Omnomnomnom Mar 20 at 19:07
  • $\begingroup$ Never mind, my guess fails. The $U_i$ must span the set of all $n \times n$ matrices, which my guess fails to do. Notably, the $U_i$ cannot be simultaneously diagonalizable. $\endgroup$ – Omnomnomnom Mar 20 at 19:10
  • $\begingroup$ Perhaps it helps though to know that you're looking for a set of unitary matrices $U_i$ whose span is $\Bbb C^{n \times n}$ $\endgroup$ – Omnomnomnom Mar 20 at 19:31
  • $\begingroup$ @Omnomnomnom, yes problem solved if you find a basis for matrix. $\endgroup$ – anecdote Mar 21 at 20:56

An attempted proof of existence that doesn't actually construct the spanning set and distribution.

First, we note that the set of unitary matrices spans $\Bbb C^{n \times n}$; we could prove this nicely using polar decomposition. From there, we note that there must exist a basis of $\Bbb C^{n \times n}$ $\{U_1,U_2,\dots,U_{n^2}\}$ consisting of unitary matrices.

It follows that the vectors $\operatorname{vec}(U_1),\dots,\operatorname{vec}(U_{n^2})$ span $\Bbb C^{n^2}$.

The argument below is incorrect

(Thus, there necessarily exist (positive) $p_k$ such that $$ \frac 1n I_{n^2} = \sum_{i} p_i \operatorname{vec}(U_i)\operatorname{vec}(U_i)^\dagger $$ We correspondingly find that these $U_i$ satisfy $\sum_{i} p_i U_iA U_i^\dagger = \frac 1n \operatorname{tr}(A) I$, as desired.)

Some clarification:

First of all, the linear span bit. Let $\langle \cdot, \cdot \rangle$ denote the Frobenius (Hilbert-Schmidt) inner product. Suppose that $A$ lies in the orthogonal complement of the span of the unitary matrices. Let $A = UP$ be a polar decomposition. Then we have $$ 0 = \langle U, A \rangle = \operatorname{trace}(U^\dagger A) = \operatorname{trace}(U^\dagger UP) = \operatorname{trace}(P) $$ but $P$ is positive semidefinite, so $\operatorname{trace}(P) = 0$ implies that $P = 0$. Thus, $A$ must be zero.

So, the span of the unitary matrices is all $\Bbb C^{n \times n}$.

Another obsrevation:

Let $\mathcal C_U$ denote the convex cone generated by the set $\{uu^* : u = \operatorname{vec}(U) \text{ for some unitary } U \}$. Showing that $\sum_{i} p_i \operatorname{vec}(U_i)\operatorname{vec}(U_i) = I$ can be achieved with non-negative coefficients $p_i$ means that we're trying to show that $I \in \mathcal C_U$.

One orthogonal basis for $\Bbb C^{n \times n}$ consisting of unitary matrices is as follows: let $$ X = \pmatrix{0&&&&1\\1&0\\&1&0\\&&\ddots\\&&&1}, Z = \pmatrix{1\\ & \omega \\ && \ddots \\ &&& \omega^{n-1}} $$ Then the matrices $\{Z^j X^k : 0 \leq j,k \leq n-1\}$ form our orthogonal basis.

  • $\begingroup$ I am a bit slow here... Can you show how to get the linear span from the polar decomposition? $\endgroup$ – anecdote Mar 22 at 3:17
  • $\begingroup$ About the resolution identity: let me write $\text{vec}(U_i)$ as $v_i$. Now $\{ v_i \}$ is a set of basis, if $I = n \sum_i p_i v_i v_i^{\dagger}$, then $G_{ij} = n G_{ik} p_k G_{kj}$ where $G_{ij}$ is the Gram matrix. Then how do we show that $p_k $ exist? I thought $v_i$ here are not necessarily orthogonal. $\endgroup$ – anecdote Mar 22 at 3:21
  • $\begingroup$ I clarified the polar decomposition argument. It seems clear that I'm wrong about that second argument though. Nevertheless, I have a feeling that the fact that the unitaries "are a large enough set" will somehow be enough here. $\endgroup$ – Omnomnomnom Mar 22 at 13:46
  • $\begingroup$ Following the argument in proposition 2.2 of this paper, I can show that there exist real coefficients $a_i$ and unitary matrices $U_i$ such that $$ \sum_{i} a_i \operatorname{vec}(U_i)\operatorname{vec}(U_i)^\dagger = \frac 1n I $$ That being said, what we would need is a way to guarantee that the coefficients $a_i$ are non-negative, which I haven't been able to come up with. $\endgroup$ – Omnomnomnom Mar 22 at 14:00
  • 1
    $\begingroup$ @anecdote Let me rephrase the argument: define $W \subseteq \Bbb C^{n \times n}$ to be the subspace spanned by the unitary matrices. What I show is that if $A \in W^\perp$, then $A = 0$. It follows that $W = \Bbb C^{n \times n}$ $\endgroup$ – Omnomnomnom Mar 22 at 16:19

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