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'?


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: $$ \text{Tr}_B\left(U|\psi\rangle\langle \psi|\otimes|e_0\rangle\langle e_0|U^\dagger\right)=\sum_kA_k|\psi\rangle\langle\psi|A_k^\dagger. $$


I will give here another way to prove explicitly the equivalence of the two 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 $A_{j\ell}^{a*}\equiv U^*_{ja\ell0}$.


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.