Quantum operations can be represented in an elegant form known as the operator-sum representationn, namely

$\mathcal{E}(\rho)=\sum_kE_k\rho E_k^\dagger$, where $E_k=\langle e_k|U|e_0\rangle$, the operators $\{E_k\}$ are known as operation elements for the quantum operation $\mathcal{E}$, also called the Kraus operator, but why this form $E_k=\langle e_k|U|e_0\rangle$ not a number as a matrix element?


1 Answer 1


Where did you get the expression $E_{k} = \langle e_{k} | U | e_{0} \rangle$ from? It seems to me that this comes from a Stinespring dilation, and in that case you need to be careful with the spaces all of these operators are acting upon.


$$ E_{k} = \langle e_{k} |U|e_{0}\rangle $$ is shorthand notation for $$ E_{k} = \big(I_{\mathcal{H}}\otimes \langle e_{k} |_{\mathrm{env}}\big)U\big(I_{\mathcal{H}}\otimes |e_{0}\rangle_{\mathrm{env}}\big), $$ where $\mathcal{H}$ is the original input Hilbert space and $\mathrm{env}$ indicates some environment Hilbert space. $U$ acts on the composition of both these spaces. Therefore, the dimensions only line up if you associate the identities on $\mathcal{H}$ with the basis elements $|e_{k}\rangle$ of the environment in the expression for the Kraus operators.

Let's make it a little more precise.

For the sake of readability, suppose that the operation $\mathcal{E}$ maps from a Hilbert space to itself: $\mathcal{E} : \mathcal{H} \rightarrow \mathcal{H}$. That means that we expect $E_{k}: \mathcal{H} \rightarrow \mathcal{H}$, too, which indeed can be represented by matrices, and not scalars.

You have written down a Stinespring dilation of the channel $\mathcal{E}$. Basically, the idea is that any channel (i.e. CPTP map) acting on a Hilbert space $\mathcal{H}$, is equivalent to the following:

  • Identify an environment space $\mathcal{H}_{\mathrm{env}}$ and associated it with the original space $\mathcal{H}$. Initialize this environment in some fixed (pure) state, w.l.o.g. let's pick $|0\rangle _{\mathrm{env}} \in \mathcal{H}_{\mathrm{env}}$.
  • Let the composite system $\mathcal{H} \otimes \mathcal{H}_{\mathrm{env}}$ evolve under a unitary operation $U: \mathcal{H} \otimes \mathcal{H}_{\mathrm{env}} \rightarrow \mathcal{H} \otimes \mathcal{H}_{\mathrm{env}}$.
  • Discard the environment $\mathcal{H}_{\mathrm{env}}$ by tracing it away.

The equivalence is then manifest if we write this out explicitly for a particular input state $\rho_{\mathrm{in}} \in Ops(\mathcal{H})$, a state in the space of operators on $\mathcal{H}$: $$ \begin{split} \mathcal{E}(\rho_{\mathrm{in}}) &= \mathrm{tr}_{\mathrm{env}} \big[U(\rho_{in} \otimes |0\rangle_{\mathrm{env}}\langle0|_{\mathrm{env}})U^{\dagger}\big] \\ &= \sum_{k} \big(I_{\mathcal{H}}\otimes \langle e_{k} |_{\mathrm{env}}\big)\big[U(\rho_{in} \otimes |0\rangle_{\mathrm{env}}\langle0|_{\mathrm{env}})U^{\dagger}\big]\big(I_{\mathcal{H}}\otimes |e_{k} \rangle_{\mathrm{env}}\big) \\ &= \sum_{k} E_{k} \rho_{\mathrm{in}} E_{k}^{\dagger}, \end{split} $$ if we define $$ E_{k} = \big(I_{\mathcal{H}}\otimes \langle e_{k} |_{\mathrm{env}}\big)U\big(I_{\mathcal{H}}\otimes |e_{0}\rangle_{\mathrm{env}}\big). \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) $$ Above, I have used the fact that tracing over a subsystem is the same as summing over (the sandwich product of) a basis for that space, while leaving the rest unaffected.

Taking a closer look at $E_{k}$, we see that it maps elements from $\mathcal{H}$ to $\mathcal{H}$, as we expect it to.

The expression you wrote down is not necessarily incorrect, but it can be ambiguous: it's shorthand notation for the expression $(1)$, if you realize that the $|e_{0}\rangle$ and $\langle e_{k}|$ are elements of $\mathcal{H}$ and its dual, whereas $U$ is an element of $Ops(\mathcal{H} \otimes \mathcal{H}_{\mathrm{env}})$. I have just been more explicit, by writing out every $I_\mathcal{H}$ that pops up.

  • $\begingroup$ I get the expression $E_k=\langle e_k|U|e_0\rangle$ from Nilsen's book of Equation (8.28). BTW, what's the meaning of $Ops(\mathcal{H})$? $\endgroup$
    – karry
    Commented Aug 22, 2022 at 8:39
  • $\begingroup$ $Ops(\mathcal{H})$, as I wrote down, is the space of operators acting on $\mathcal{H}$. Both the unitaries acting on $\mathcal{H}$ and the density operators from/acting on $\mathcal{H}$ are elements of this set. $\endgroup$
    – JSdJ
    Commented Aug 22, 2022 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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