I understand that there are two ways to think about 'general quantum operators'.

Way 1

We can think of them as trace-preserving completely positive operators. These can be written in the form $$\rho'=\sum_k A_k \rho A_k^\dagger \tag{1}$$ where $A_k$ are called Kraus operators.

Way 2

As given in (An Introduction to Quantum Computing by Kaye, Laflamme and Mosca, 2010; pg59) we have a $$\rho'=\mathrm{Tr}_B\left\{ U(\rho \otimes \left| 00\ldots 0\right>\left<00\ldots 0 \right|) U^\dagger \right\} \tag{2}$$ where $U$ i s a unitary matrix and the ancilla $\left|00 \ldots 0\right>$ has at most size $N^2$.


Exercise 3.5.7 (in Kaye, Laflamme and Mosca, 2010; pg60) gets you to prove that operators defined in (2) are completely positive and trace preserving (i.e. can be written as (1)). My question is the natural inverse of this; can we show that any completely positive, trace preserving map can be written as (2)? I.e. are (1) and (2) equivalent definitions of a 'general quantum operator'?


2 Answers 2


This question is posed, and answered positively, in Nielsen & Chuang in a subsection of chapter 8 entitled "System-environment models for and operator-sum representation". In my version, it can be found on page 365.

Imagine $|\psi\rangle$ is an arbitrary pure state on the space upon which you wish to enact the operators. Let $|e_0\rangle$ be some fixed state on another quantum system (with dimension equal to at least the number of Krauss operators, and labelled 'B'). Then you can define a unitary by its action on the space of states spanned by $|\psi\rangle$: $$ U|\psi\rangle|e_0\rangle=\sum_k(A_k|\psi\rangle)|e_k\rangle, $$ where the $|e_k\rangle$ are an orthonormal basis. To check that this corresponds to a valid unitary, we just have to test it for different input states and ensure that the initial overlap is preserved: $$ \langle\psi|\phi\rangle\langle e_0|e_0\rangle=\langle\psi|\langle e_0|U^\dagger U|\phi\rangle|e_0\rangle=\langle\psi|\sum_kA_k^\dagger A_k|\phi\rangle, $$ which is true thanks to the completeness relation of the Krauss operators.

Finally, one just has to check that this unitary does indeed implement the claimed map: $$ \text{Tr}_B\left(U|\psi\rangle\langle \psi|\otimes|e_0\rangle\langle e_0|U^\dagger\right)=\sum_kA_k|\psi\rangle\langle\psi|A_k^\dagger. $$


Direct proof

We want to prove that a channel given in the form $$\Phi(\rho) = \sum_a A_a \rho A_a^\dagger\tag1$$ can equivalently be represented as a unitary evolution in an enlarged space. I'll actually consider the more general case of an arbitrary (CP) map of the form (1) (that is, I won't assume the normalisation $\sum_a A_a^\dagger A_a=I$).

A direct way to do this is to define the operator $V$ with action $$V|\psi\rangle\equiv \sum_a (A_a|\psi\rangle)\otimes |a\rangle.$$ If $\Phi$ is a channel, that is, $\sum_a A_a^\dagger A_a=I$, then $V$ is an isometry. The vectors $|a\rangle$ are here an (arbitrary) orthonormal basis in an auxiliary space. The auxiliary space needs to have a dimension equal to the number of Kraus operators for this expression to work. You could have a larger ancillary space if you're fine with $V$ being a partial isometry instead, but you don't really gain anything doing this.

With this $V$, we can see that the map can be equivalently written as $$\Phi(\rho) = \operatorname{Tr}_2[V\rho V^\dagger].\tag2$$ This is the so-called Stinespring representation of the map/channel. We can then also equivalently write (2) as $$\Phi(\rho) = \operatorname{Tr}_2[U(\rho\otimes |u\rangle\!\langle u|)U^\dagger)],$$ for any pure state $|u\rangle$ living in the auxiliary space, defining $U$ as such that $U(|i\rangle\otimes|u\rangle)=V|i\rangle$. This can always be done, and if $\Phi$ is a channel, then $V$ is an isometry, and $U$ is a unitary.

Proof with explicit componentwise expressions

From the Kraus representation, $\Phi(\rho)=\sum_a A^a \rho A^{a\dagger}$, making the indices explicit, we get $$\Phi(\rho)_{ij}=\sum_{a,k,\ell}A^a_{ik}A^{a*}_{j\ell}\rho_{k\ell}.\tag1$$

On the other hand, unravelling the second expression we have for a generic $\sigma$ (let me here use numbers instead of latin letters for the indices, for better clarity, as well as Einstein's notation for repeated indices), $$[U(\rho \otimes \sigma) U^\dagger]_{1234}=U_{1256}U^{*}_{3478}\rho_{57}\sigma_{68}.$$ Note that here the first two indices, ($1$ and $2$) correspond to the "output space" of the operator $\Phi(\rho)$, while the other two ($3$ and $4$) correspond to its "input space". Similarly, $2$ and $4$ live in the second Hilbert space, while $1$ and $3$ live in the first one.

Tracing with respect to the second Hilbert space amounts to introducing a $\delta_{24}$ factor, and we thus get

$$\left\{\mathrm{Tr}_B\left[ U(\rho \otimes \sigma) U^\dagger \right]\right\}_{13} =U_{1256}U^{*}_{3478}\rho_{57}\sigma_{68} \color{red}{\delta_{24}} =U_{1256}U^{*}_{3278}\rho_{57}\sigma_{68}.$$

If we take $\sigma$ to be a pure state, for example $\sigma=\lvert0\rangle\!\langle0\rvert$, so that $\sigma_{68}=\delta_{60}\delta_{80}$, then we have

$$\left\{\mathrm{Tr}_B\left[ U(\rho \otimes \lvert0\rangle\!\langle0\rvert) U^\dagger \right]\right\}_{13} =U_{1250}U^*_{3270}\rho_{57}.$$ Going back to using the standard notation for the indices, and making explicit the sums, we have

$$\left\{\mathrm{Tr}_B\left[ U(\rho \otimes \lvert0\rangle\!\langle0\rvert) U^\dagger \right]\right\}_{ij} =\sum_{a,k,\ell}U_{iak0}U^*_{ja\ell0}\rho_{k\ell}.\tag2$$ This expression is equivalent to (1), defining $A^a_{ik}\equiv U_{iak0}$ and $A_{j\ell}^{a*}\equiv U^*_{ja\ell0}$.


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