# Kraus operators required for operations on $d$ dimensional Hilbert space

All quantum operations $$\mathcal{E}$$ on a system of Hilbert space dimension $$d$$ can be generated by an operator-sum representation containing at most $$d^2$$ elements, $$\mathcal{E}(\rho)=\sum_{k=1}^M E_k\rho E_k^\dagger$$ where $$1\leq M\leq d^2$$.

A possible attempt to prove this is given in Exercise 8.10, Page 373, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang, as

Let $$\{E_j\}$$ be a set of operation elements for $$\mathcal{E}$$. Define a matrix $$W_{jk}\equiv tr(E_j^\dagger E_k )$$. Show that the matrix $$W$$ is Hermitian and of rank at most $$d^2$$, and thus there is unitary matrix $$u$$ such that $$uWu^\dagger$$ is diagonal with at most $$d^2$$ non-zero entries. Use $$u$$ to define a new set of at most $$d^2$$ non-zero operation elements $$\{F_j\}$$ for $$\mathcal{E}$$.

Proof of this is given in Kraus operator rank, as

$$E_j=(I\otimes \langle e_j|)U(I\otimes |e_0\rangle)$$

Given that the quantum operation $$\mathcal{E}$$ acts on a Hilbert space of dimension $$d$$, i.e., $$\rho\in\mathbb{C}^{d\times d}$$ and $$E_k\in\mathbb{C}^{d\times d}$$

Therefore, a maximum of $$d^2$$ of the $$E_j$$ can be linearly independent.

$$W=\begin{bmatrix} tr(E_1^\dagger E_1)&tr(E_1^\dagger E_2)&\cdots&tr(E_1^\dagger E_M)\\ tr(E_1^\dagger E_1)&tr(E_1^\dagger E_2)&\cdots&tr(E_1^\dagger E_M)\\ \vdots&\vdots&\ddots&\vdots\\ tr(E_M^\dagger E_1)&tr(E_M^\dagger E_2)&\cdots&tr(E_M^\dagger E_M)\\ \end{bmatrix}$$ $$W_{jk}=tr(E_j^\dagger E_k)=tr((E_k^\dagger E_j)^\dagger)=tr(E_k^\dagger E_j)^*=W_{kj}^*$$

$$\therefore W$$ is hermitian.

Let the $$k^{th}$$ column of $$W$$ is, $$|w_k\rangle=\sum_{j=1}^M W_{jk}|j\rangle=\sum_{j=1}^M tr(E_j^\dagger E_k)|j\rangle$$

Let $$E_j$$ are arranged such that $$E_1,\cdots,E_{d^2}$$ are linearly independent, then for any $$k>d^2$$, we have $$E_k=\sum_{l=1}^{d^2} c_{kl}E_l$$ where $$c_{kl}\in\mathbb{C}$$.

\begin{align} |w_k\rangle=\sum_{j=1}^M tr(E_j^\dagger E_k)|j\rangle=\sum_{j=1}^M tr(E_j^\dagger\sum_{l=1}^{d^2} c_{kl} E_l)|j\rangle=\sum_{j=1}^M tr(\sum_{l=1}^{d^2} c_{kl} E_j^\dagger E_l)|j\rangle\\ =\sum_{j=1}^M \sum_{l=1}^{d^2} c_{kl} tr( E_j^\dagger E_l)|j\rangle=\sum_{l=1}^{d^2} c_{kl}\bigg[\sum_{j=1}^M tr( E_j^\dagger E_l)|j\rangle\bigg]=\sum_{l=1}^{d^2} c_{kl}|w_l\rangle \end{align} $$\implies$$ for every $$k>d^2$$, the $$k^{th}$$ column of $$W$$ is a linear combination of the first $$d^2$$ linearly independent columns of $$W$$.

$$\implies$$ the number of linearly independent columns of $$W$$ is therefore at most $$d^2$$

$$\implies rank(W)\leq d^2$$

And since $$W$$ is a hermitian matrix, there exists a decomposition $$uWu^\dagger$$ which is a diagonal matrix with at most $$d^2$$ nonzero diagonal entries.

This seems fine, but

How does showing "there is unitary matrix $$u$$ such that $$uWu^\dagger$$ is diagonal with at most $$d^2$$ non-zero entries" proves "all quantum operations $$\mathcal{E}$$ on a system of Hilbert space dimension $$d$$ can be generated by an operator-sum representation containing at most $$d^2$$ elements"?

• This is just the singular value decomposition of W, you can derive it from the eigenvalue decompositions of $WW^\dagger$ and $W^\dagger W$. Oct 20, 2022 at 7:06

The missing step that you ask about in your last paragraph is to use the following fact (labeled Theorem 8.2 in my edition of Nielsen and Chuang): If $$\{E_1,\ldots,E_n\}$$ and $$\{F_1,\ldots,F_n\}$$ are two sets of operators defining operations $$\mathcal{E}$$ and $$\mathcal{F}$$ respectively, then $$\mathcal{E}=\mathcal{F}$$ iff there is an $$n$$-by-$$n$$ unitary matrix $$u$$ such that $$E_i = \sum_j u_{ij} F_j$$.
Use the $$u$$ that diagonalizes $$W$$ to define new operators $$F_i = \sum_j u^*_{ij}E_j$$. Then for all but $$d^2$$ of the $$F_i$$ we have $$\operatorname{tr}(F_i^\dagger F_i)=0$$ (and hence $$F_i=0$$). By the above theorem the $$F_i$$ describe the same operation as the original $$E_i$$.