Can a Kraus representation act as the identity on any operator?

Question

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.

Answer 1

1. (Orthonormal operatorial bases and their completeness) 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.

2. (Choi representation of 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 $K(\Phi)=|m\rangle\!\langle m|$, or in components, $K(\Phi)_{ij,k\ell}=\delta_{k\ell}\delta_{ij}/d$, with $|m\rangle$ the maximally entangled state. The Kraus decomposition is then obtained as the spectral decomposition of the operator mapping $j\ell$ to $ik$. We thus want the spectral decomposition of the Choi operator $$J(\Phi)\equiv (\Phi\otimes I) (d\lvert m\rangle\!\langle m\rvert)=\frac1 d I\otimes I\equiv I/d.$$

3. (Eigendecomposition of Choi representation) 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 $(A_a)_{ij}\equiv (\bs v_a)_{ij}$, or equivalently, $\operatorname{vec}(A_a)=\bs v_a$, we get the Kraus decomposition $$\Phi(X) = \operatorname{unvec}\Big(\underbrace{\frac{1}{d}\sum_a (A_a \otimes \bar A_a)}_{\equiv K(\Phi)}\operatorname{vec}(X)\Big) = \frac{1}{d}\sum_a A_a X A_a^\dagger. $$ 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: $\langle A_a, A_b\rangle\equiv\operatorname{tr}(A_a^\dagger A_b)=\delta_{ab}$. Note also that here $\{A_a\}_a$ are not the Kraus operators of the channel: in the standard notation, the Kraus operators are the renormalised operators $\{\frac{1}{\sqrt d}A_a\}_a$.

(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.

Answer 2

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.

Answer 3

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.

Answer 4

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.