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 $C_{ab}=C_{ba}^*$


1 Answer 1


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

  • $\begingroup$ 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? $\endgroup$ Commented Jul 2 at 11:09
  • $\begingroup$ @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 $C_{ab}^*\langle \psi | \psi \rangle$. $\endgroup$
    – qubitzer
    Commented Jul 2 at 11:21
  • $\begingroup$ 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. $\endgroup$ Commented Jul 2 at 11:27
  • $\begingroup$ @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.) $\endgroup$ Commented Jul 2 at 11:31
  • 1
    $\begingroup$ Typo in my comment, indeed. $\endgroup$ Commented Jul 2 at 12:01

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