# How is $(\langle \psi| E_{a}^\dagger E_{b} | \psi \rangle)^\dagger = C_{ba}^*\langle \psi| \psi \rangle$

I am reading through Daniel Gottesmans surviving as a quantum computer in a classical world. On page 36, he presents the following theorem:

Theorem 2.7 (QECC Conditions). $$(Q, \mathcal{E})$$ is a $$Q E C C$$ iff $$\forall|\psi\rangle,|\phi\rangle \in Q, \forall E_a, E_b \in \mathcal{E}$$, $$\left\langle\psi\left|E_a^{\dagger} E_b\right| \phi\right\rangle=C_{a b}\langle\psi \mid \phi\rangle .$$

It then goes onto state that by taking the adjoint of both sides of the equation, you get $$C_{ab}^\dagger=C_{ba}^*$$, showing that $$C$$ is self-adjoint, after you set $$|\psi \rangle =|\phi \rangle$$.

All I am getting is

$$(C_{a b}\langle\psi| \psi\rangle) ^\dagger = \langle \psi| E_{a}^\dagger E_{b} | \psi \rangle)^\dagger = \langle \psi|( E_{a}^\dagger E_{b})^\dagger | \psi \rangle = \langle \psi| E_{b}^\dagger E_{a} | \psi \rangle = C_{ba}\langle \psi| \psi \rangle$$

I am totally lost on how he is taking the adjoint of both sides and getting that result. Maybe I'm just being stupid, or it's been a while since I've worked with this formalism, but I am unsure of how the adjoint is being applied in this instance.

Here is a similar question, but I think it's more focused on what $$C$$ actually is, as opposed to this result.

In the QECC condition $\langle\psi|E_a^\dagger E_b|\phi\rangle=C_{ab}\langle\psi|\phi\rangle$, what is $C_{ab}$?

Edit: It's a typo in the book. Should read $$C_{ab}=C_{ba}^*$$

Your equation states $$C_{ab}^*=C_{ba}$$. This is precisely what you want: $$C$$ is self-adjoint.

• But is there a way to actually show this via taking the adjoint? This is why I feel like I have done something wrong. $C_{ab}$ is a complex number, so $C_{ab}^\dagger=C_{ab}^*$. But how has he ended up with $C_{ba}^*$ in this case? Commented Jul 2 at 11:09
• @GaussStrife You are almost there. At the beginning of your own derivation, you only need to distribute the $\dagger$ in $(C_{ab} \langle \psi | \psi \rangle)^\dagger$ and you get $C_{ab}^*\langle \psi | \psi \rangle$. Commented Jul 2 at 11:21
• But I already showed that in the comment above yours, unless I am misunderstanding you. What I want to know is how, using the notation of $(\langle i| E_{a}^\dagger E_{b} | j \rangle)^\dagger = \langle i|( E_{a}^\dagger E_{b})^\dagger |j \rangle$, where in our case $i, j = |\psi\rangle$, how ends up with $C_{ab}^\dagger=C_{ab}^*$ I haven't worked in this notation for a while and I've managed to convice myself that $\langle i| E_{a}^\dagger E_{b} | j \rangle)^\dagger = \langle i|( E_{a}^\dagger E_{b})^\dagger |j \rangle$ is the wrong way to take the adjoint on the inner product function. Commented Jul 2 at 11:27
• @GaussStrife Without doubt a typo. The goal is to show $C$ is self-adjoint (as also said right after in the lecture notes!), which does not amount to $C_{ab}^\dagger = C_{ba}$. (This equation is simply wrong, you can easily cook up counterexamples.) Commented Jul 2 at 11:31
• Typo in my comment, indeed. Commented Jul 2 at 12:01