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In the textbook “Quantum Computation and Quantum Information” by Nielsen and Chuang, it is stated that there exists a set of unitaries $U_i$ and a probability distribution $p_i$ for any matrix A,

$$\sum_i p_i U_i A U_i^\dagger =tr(A) I/d,$$

where $d$ is the dimension of the Hilbert space. (This is on page 517; Exercise 11.19; equation (11.85)) The left-hand side is a Kraus representation given A.

But is this possible for a general non-diagonalizable (i.e. non-normal) matrix A? For a normal matrix A, I found it is indeed the case.

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    $\begingroup$ The wording of the question is not crystal clear, but it should be understood that the unitary matrices and probabilities may not depend on $A$, they may only depend on the dimension $d$. Once you know, for a given choice of these unitary matrices and probabilities, that the equation is true for all normal matrices $A$, it follows that it is true for all matrices $A$ by linearity. For instance, every matrix $A$ can be written as $A = H + i K$ for $H$ and $K$ Hermitian. $\endgroup$ – John Watrous Jul 14 at 23:39
  • $\begingroup$ Of course every matrix can be written as a sum of the Hermitian and the anti-Hermitian part. However, for each Hermitian (and hence normal) matrix, we need to choose a specific set of unitary matrices $U_i$ to make it proportional to the identity matrix. In case of the matrix $A$ has both $H$ and $K$, I think we need to take $U_i$ which makes the lefthand side for the both $H$ and $K$ are proportional to the identity simultaneously not individually. Is this always possible? $\endgroup$ – Amplituhedron Jul 15 at 2:14
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    $\begingroup$ By the way, when I say that the "wording of the question is not crystal clear," I am referring to the exercise itself, not the question posted here. $\endgroup$ – John Watrous Jul 15 at 11:08
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    $\begingroup$ The unitary matrices and probabilities can be taken to be independent of $A$. The exercise is essentially asking you to prove that the completely depolarizing channel is a mixed-unitary channel, which it certainly is. $\endgroup$ – John Watrous Jul 15 at 11:11
  • $\begingroup$ Thanks, now I understood. $\endgroup$ – Amplituhedron Jul 17 at 8:09
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(General result) The main thing to keep in mind is that this is a result about a type of channel, not about specific states. Suppose $\operatorname{tr}(U_i U_j^\dagger)=\delta_{ij}$ for some set of matrices $U_i$. This is equivalent to $\sum_{k\ell}(U_i)_{k\ell} (U_j^*)_{k\ell}=\delta_{ij}$. If $U_i$ form a basis (i.e. there are $n^2$ of them), then we must also have $\sum_i (U_i)_{k\ell} (U_i^*)_{mn}=\delta_{km}\delta_{\ell n}$.

For such choice of matrices we have, for any matrix $\rho$, $$\sum_i U_i \rho U_i^\dagger = \sum_{ijk \ell m} \lvert j\rangle\!\langle k\rvert\,\, (U_i)_{j\ell}(U_i^*)_{km} \rho_{\ell m} = \sum_{jk\ell m} \lvert j\rangle\!\langle k\rvert\,\, \delta_{jk} \delta_{\ell m}\rho_{\ell m} \\= \sum_{j\ell} \lvert j\rangle\!\langle j\rvert \,\, \rho_{\ell\ell} = \operatorname{tr}(\rho) I. $$

Notice how the identity does not depend on what $\rho$ is. It can be an arbitrary operator. You can test it yourself with a nondiagonalisable matrix such as $\rho=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. It is a statement about the mapping $\rho\mapsto \sum_i U_i \rho U_i^\dagger$, not about $\rho$.

Notice also that I did not use any assumption on the $U_i$. They need not be unitaries (indeed, they cannot be unitaries in my choice of normalisation). To get the same factor on the RHS, you need only modify the normalisation of the matrices to have $\operatorname{tr}(U_i U_j^\dagger)=\delta_{ij}/d$, and the rest follows.


(Representations of the completely depolarising channel) Consider the linear map $\Phi(X)=\operatorname{tr}(X) I/d$. You can verify that it is a CPTP map and thus admits a Kraus decomposition.

Its natural representation reads $\Phi_{i|j}^{k|\ell}=K(\Phi)_{ij,k\ell}=\delta_{k\ell}\delta_{ij}/d=\lvert m\rangle\!\langle m\rvert$ with $|m\rangle$ a maximally entangled state. The Kraus decomposition is then obtained as the spectral decomposition of the operator mapping $j\ell$ to $ik$. Stated more precisely, we need the spectral decomposition of the Choi operator $$J(\Phi)\equiv (\Phi\otimes I)\lvert m\rangle\!\langle m\rvert=\frac1 d I\otimes I\equiv I/d.$$

The eigendecomposition of this operator is trivial: its eigenvalues are all equal to $1/d$, thus any orthonormal set of vectors will be a suitable set of eigenvectors. Write these as $\newcommand{\bs}[1]{\boldsymbol{#1}} \{\bs v_a\}_a$, so that $J(\Phi)\bs v_a=\frac1 d \bs v_a$ for all $a=1,...,d^2$. In terms of the natural representation, these satisfy $$\sum_{j\ell} K(\Phi)_{ij,k\ell}(\bs v_a)_{j\ell} = \frac1 d(\bs v_a)_{ik} \Longleftrightarrow K(\Phi) = \frac1 d \sum_a \bs v_a \otimes \bs v_a^\dagger.$$ $$K(\Phi)_{ij,k\ell}=\frac1 d\sum_a (\bs v_a)_{ik}(\bs v_a^*)_{j\ell}.$$ Defining the operators $A_a$ as $(A_a)_{ij}\equiv (\bs v_a)_{ij}$ we thus get the Kraus decomposition $\Phi(X) = \sum_a A_a X A_a^\dagger. $ Note that the orthogonality of the vectors $\bs v_a$, $\langle \bs v_a,\bs v_b\rangle=\delta_{ab}$, translates into the orthogonality of the matrices $A_a$ in the $L_2$ norm: $\operatorname{tr}(A_a A_b^\dagger)=\delta_{ab}$.

(Result from Kraus representation) This proves that, for any set of matrices $A_a$ such that $\operatorname{tr}(A_a A_b^\dagger)=\delta_{ab}$, we have for all $X$ $$\frac1 d\sum_a A_a X A_a^\dagger= \operatorname{tr}(X) \frac I d.$$ Of course, we already showed this in the first paragraph. This is just a different angle to get to the same result.

(Finding Kraus decompositions made up of unitaries) In the above, $A_a$ are not unitaries. However, the freedom in the choice of vectors $\bs v_a$, or equivalently the freedom in the choice of $A_a$, can be used to find a decomposition in terms of Kraus operators that are (proportional to) unitaries. A basis of unitaries can be constructed e.g. using clock and shift matrices. Have a look at (Durt 2010), around page 10, and these nice notes by Wheeler (pdf alert), around page 12.

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Since it hasn't been mentioned so far, and I think it's an interesting aspect: A weighted ensemble $(p_i,U_i)$ of unitaries in $U(d)$ such that $$ \sum_i p_i U_i X U_i^\dagger = \operatorname{tr}(X) \mathbb{I}/d, $$ is called a weighted unitary 1-design. If the weights can be chosen uniformly, i.e. $p_i \equiv 1/N$ where $N$ is the size of the ensemble, this reduces to the definition of a "normal" unitary 1-design.

There are many examples for unitary 1-designs:

  1. Unitary 1-designs are exactly tight operator frames with frame constant $N/d$
  2. In particular, any orthogonal operator basis of unitaries is a unitary 1-design, e.g. the Weyl operators
  3. In fact any irreducible unitary representation of a group is a unitary 1-design, e.g. the Heisenberg-Weyl (=generalised Pauli) and Clifford group.
  4. For any large enough Haar-random ensemble of unitaries there are weights such that the above equation holds with high probability.
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If it holds for hermitian matrices, it holds for all matrices due to linearity: Over $\mathbb C$, the hermitian matrices span the full matrix space.

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This problem can be approached without regards to Kraus representations (even if the motivation is to prove the convexity of entropy) or whether A is a normal matrix or not. Rather, this is a feature of the choice of $\{ U_{j} \}$. In particular, there exists a choice such that their action is to ``coarse-grain'' all the information in a state.

Here's a single qubit example to illustrate my point: consider the set $p_{j} = \frac{1}{4}, U_{j} = \sigma_{j}$ for $j \in \{ 1,2,3,4 \}$, where, $\sigma_{j}$ are the Pauli matrices (with $\sigma_{0} = \mathbb{I}$). Then, its action on a single qubit is, $$ \sum\limits_{j} p_{j} U_{j} \rho U^{\dagger}_{j} = \frac{1}{4} \left( \mathbb{I} \rho \mathbb{I} + \sigma_{x} \rho \sigma_{x} + \sigma_{y} \rho \sigma_{y} + \sigma_{z} \rho \sigma_{z} \right) = \cdots = \operatorname{Tr}\left( \rho \right) \frac{\mathbb{I}}{2},$$ where the $\cdots$ can be evaluated using the anticommutativity of the Pauli matrices (Hint: use the relation $\sigma_{j} \sigma_{k} \sigma_{j} = - \sigma_{k}$ for $j \neq k$).

Now, since any matrix $A$ can be written as $A = H + iK$ for hermitian matrices $H,K$; and any hermitian matrix $H$ can be written as $H = H_{1} - H_{2}$ for positive semidefinite matrices, you can write $A = H_{1} - H_{2} + i(K_{1} - K_{2})$. Rewriting each of the matrices as $H_{1} = \operatorname{Tr}\left( H_{1} \right) (\frac{1}{\operatorname{Tr}\left( H_{1} \right)} H_{1})$, we have that $\frac{1}{\operatorname{Tr}\left( H_{1} \right)} H_{1}$ is a density matrix and so the above result applies. Continuing this, you'll find, using the linearity of trace, that for the $2 \times 2$ case, the above unitaries give you $\mathrm{Tr}(A) \frac{\mathbb{I}}{d}$.

The generalization to $n \times n$ matrices is left as an exercise to the OP (where you need to find a set of unitaries analogous to the Pauli matrices).

Edit: One way to obtain the result in $d$ dimensions is to use the $d^2$ Heisenberg-Weyl operators (or the finite dimensional representation of the Heisenberg-Weyl algebra). If $X(i)Z(j)$ is the $(i,j)$th operator then, we have, $\frac{1}{d^{2}} \sum_{i, j=0}^{d-1} X(i) Z(j) \rho Z^{\dagger}(j) X^{\dagger}(i)=\frac{\mathbb{I}}{d}$. See, for example, Page 176 of this book.

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  • $\begingroup$ Are you assuming the state $\rho$ is written in terms of the Pauli matrices? For a higher dimension, I found it very difficult to compute the commutation relation with unitaries and Hermitian matrices given in a quantum tomography for $\rho$. (I tried with the U(n) basis.) Nevertheless in some dimensions like 2^n dimensions, we can construct such a basis with tensor products of Pauli matrices. $\endgroup$ – Amplituhedron Jul 17 at 8:16
  • $\begingroup$ One way to obtain the result in $d$ dimensions is to use the $d^2$ Heisenberg-Weyl operators (or the finite dimensional representation of the Heisenberg-Weyl algebra). If $X(i)Z(j)$ is the $(i,j)$th operator then, we have, $\frac{1}{d^{2}} \sum_{i, j=0}^{d-1} X(i) Z(j) \rho Z^{\dagger}(j) X^{\dagger}(i)=\frac{\mathbb{I}}{d}$. See, for example, Page 176 of this book. $\endgroup$ – keisuke.akira Jul 18 at 0:52

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