Why are Kraus operators $E_k=\langle e_k|U|e_0\rangle$ not just numbers?

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?

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.

TL/DR

$$E_{k} = \langle e_{k} |U|e_{0}\rangle$$ 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.

• 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 $Ops(\mathcal{H})$? Commented Aug 22, 2022 at 8:39
• $Ops(\mathcal{H})$, 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.
– JSdJ
Commented Aug 22, 2022 at 9:33