Consider a separable Hilbert space $\mathcal{H}$, we can first define $T(\mathcal{H})$ the trace class of $\mathcal{H}$, then $D(\mathcal{H})$ to denote the set of positive operator with trace less than 1.

The set of positive maps consists of all elements that maps positive operator to positive operator. The set of completely positive maps contains all elements $a$ that $id_k\otimes a$ is still positive for any integer $k$.

We are interested in $C$, a subset of the completely positive maps contains all $c$ such that for all $b\in D(\mathcal{H})$, $$\mathrm{tr}[c(b)]\leq \mathrm{tr}[b].$$

In finite dimensional case, we always have the so-called Kraus decomposition that any $c$ can be written as $$c(b)=\sum E_i bE_i^{*},$$ with $E_i^{*}E_i\leq I$.

Moreover, we have $c^*$, the dual map of $c$, satisfied $$c^*(b)=\sum E_i^{*} bE_i.$$

The dual map is defined as $$\langle d,c(b)\rangle = \langle c^*(d),b \rangle$$ for all $b,d$.

In the separable Hilbert space, do we have similar version of Kruas decomposition, maybe with infinite sum? Do we have similar version of dual map?


1 Answer 1


This is true; though I don't know a canonical reference (perhaps someone else can provide this).

But let me sketch a proof. In infinite dimensions, we need to be a little careful about topologies. Let $\newcommand{\mc}{\mathcal}\mc T(H)$ be the trace-class operators on $H$ with norm $\newcommand{\tr}{\operatorname{tr}}$$\|x\|_1 = \tr(|x|)$, and $\mc B(H)$ the algebra of all bounded operators with norm $\|\cdot\|$. As usual, $\mc T(H)$ is the predual of $\mc B(H)$ with duality $\langle x, t \rangle = \tr(xt)$.

Let $c:\mc T(H) \rightarrow \mc T(H)$ be a linear map which is completely positive. But what exactly do we mean by "completely positive" here? For $M_n(\mc T(H))$ is not a C$^*$-algebra, and so doesn't immediately have a notion of positivity (but see note 2 below). For now, we proceed as follows. As $c$ is positive, it is bounded (see note 1). Thus we can form $c^*$, a map on $\mc B(H)$, and demand that this is completely bounded.

We now apply the usual Stinespring construction to obtain an auxiliary Hilbert space $H'$, a unital $*$-homomorphism $\pi:\mc B(H)\rightarrow\mc B(H')$, and a bounded linear map $V:H\rightarrow H'$ with $c^*(x) = V^*\pi(x)V$. We now make some observations:

  • You define $\mc D(H)$ as $\{x\in\mc T(H) : x\geq 0, \tr(x)\leq 1\}$ which is just the closed unit ball of $\mc T(H)$ intersected with $\mc T(H)_+$. Notice now that the requirement that $\tr(c(x)) \leq \tr(x)$ for all $x\in \mc D(H)$ can be multiplied by any positive number, and so is equivalent to $\tr(c(x)) \leq \tr(x)$ for all $x\in\mc T(H)_+$. This is equivalent to $\langle c^*(1), x \rangle \leq \langle 1, x \rangle$ for all $x\in\mc T(H)_+$. This is equivalent to just that $c^*(1) \leq 1$ in $\mc B(H)_+$.
  • Thus $c^*(1) = V^*\pi(1)V = V^*V \leq 1$.
  • Careful examination of the Stinespring construction shows that $\pi$ is normal because $c^*$ is.
  • A normal unital $*$-homomorphism $\mc B(H)\rightarrow\mc B(H')$ has a special form. $H' = H \otimes K$ for some Hilbert space $K$ and $\pi(x) = x\otimes 1$.
  • In fact, in this case, we can streamline the Stinespring construction (see for example page 9 of arXiv:1107.5244 math.OA).

So we have $V:H\rightarrow H\otimes K$ with $V^*V\leq 1$ and $c^*(x) = V^*(x\otimes 1)V$. Let $(e_i)$ be an orthonormal basis of $K$, and check that there are $E_i\in\mc B(H)$ with $$ V(\xi) = \sum_i E_i(\xi)\otimes e_i \qquad (\xi\in H). $$ That $V^*V\leq 1$ means exactly that $\sum_i E_i^*E_i\leq 1$. Then $$ c^*(x) = \sum_i E_i^* x E_i. $$ It is now easy to check that $c(t) = \sum_i E_itE_i^*$ to $t\in\mc T(H)$. Finally, if $H$ is separable, then $K$ will be separable and so the family $(E_i)$ will be a sequence (or finite).

Note 1: If $c$ is positive but not bounded, then firstly observe that every $t\in\mc T(H)$ can be written as the linear combination of 4 positive trace-class operators, so $c$ is not bounded on the positives. Thus we can find $(x_n)\subseteq \mc T(H)_+$ with $\tr(x_n)\leq n^{-2}$ but $\tr(c(x_n))\geq n$. Then $x=\sum_n x_n$ converges absolutely in $\mc T(H)_+$, and for any $N$, $\sum_{n\leq N}x_n \leq x$ so $\sum_{n\leq N} c(x_n) \leq c(x)$ so $\tr(c(x)) \geq \sum_{n\geq N} \tr(c(x_n)) \geq N$, a contradiction.

Note 2: We identify $M_n(\mc B(H))$ with $\mc B(H\otimes \mathbb C^n)$ which gives the canonical $C^*$-algebra structure on $M_n(\mc B(H))$ and hence the positive cone. The predual is $\mc T(H\otimes \mathbb C^n)$ which is of course just $M_n(\mc T(H))$. Furthermore, duality behaves as you expect, namely if $x=(x_{i,j})\in M_n(\mc B(H))$ and $t=(t_{i,j}) \in M_n(\mc T(H))$ then $$ \tr(xt) = \sum_{i,j} \tr(x_{i,j} t_{j,i}). $$ Then $c^*$ being completely positive is equivalent to $c$ being completely positive, where $M_n(\mc T(H))$ is given this positive cone, i.e. as operators on $H\otimes\mathbb C^n$. (Be aware that it is also common to use $\langle x,t\rangle = \sum_{i,j} \langle x_{i,j} , t_{i,j} \rangle = \sum_{i,j} \tr(x_{i,j}t_{i,j})$ which is of course different.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.