# Knill-Laflamme condition derivation in Nielsen&Chuang: issue to understand a part of the proof

I have some trouble to understand the proof in Nielsen&Chuang about Knill-Laflamme conditions.

The conditions:

Let $$C$$ be a quantum code and $$P$$ the projector onto $$C$$. Suppose $$\mathcal{E}$$ is a quantum operatiion with operation elements $$\{E_i\}$$. A necessary and sufficient condition for the existence of an error-correction operation $$\mathcal{R}$$ correcting $$\mathcal{E}$$ on $$C$$ is that:

$$P E_i^{\dagger} E_j P = \alpha_{ij} P$$

For some Hermitian matrix $$\alpha$$ of complex numbers.

To show the sufficient condition, he diagonalizes the matrix $$\alpha$$. Doing this, we end up with a simpler condition:

$$P F_k^{\dagger} F_l P = d_{kl} P$$

Where $$d_{kl}$$ is the matrix element of a diagonal matrix.

The polar decomposition tells us that there exist a unitary $$U_k$$ such that:

$$F_k P = U_k \sqrt{P F_k^{\dagger} F_k P} = \sqrt{d_{kk}} U_k \sqrt{P} = \sqrt{d_{kk}} U_k P \tag{1}$$

## My question

At this point, he says that the effect of $$F_k$$ is therefore to rotate the coding subspace into the subspace defined by the projector:

$$P_k \equiv U_k P U_k^{\dagger}$$

I am not sure to understand precisely this statement. I agree that $$P_k F_k P = F_k P$$ such that $$P_k$$ stabilizes the space $$F_k C$$. But how to be sure that $$P_k$$ is exactly a good projector. What I mean is that veryfing $$P_k F_k P = F_k P$$ is for sure not enough as $$P_k=I$$ would also do the job for instance.

• I'm not quite sure what does a good projector mean. Anyway, since the definition of $P_k$ is $U_kPU_k^\dagger=F_kPU_k^\dagger/\sqrt{d_{kk}}$. And $PF_k^\dagger F_lP=d_{kl}P$. Hence, if $l\neq k$, $P_kP_l = 0$. May 15, 2021 at 13:03
• @narip see the answer, this was what disturbed me. May 15, 2021 at 17:18

I found the answer. It is just a basic linear algebra property to show but I take the same notations as in the question for clarity.

Let $$P$$ be a projector on $$C$$

As shown below in this post, a projector on the space $$U_k C$$ can always be found as $$P_k = U_k P U_k^{\dagger}$$.

From this property, we see that as $$F_k P \propto U_k P$$, a projector onto $$Im(F_k P)$$ is $$P_k$$ which is what is claimed by the book.

## Proof

We just need to show that $$Im(P_k)=U_k C$$.

I first show $$Im(P_k) \subset U_k C$$:

Let $$|y \rangle \in Im(P_k)$$. We thus have: $$|y\rangle=U_k P U_k^{\dagger} |x\rangle$$ for some $$|x \rangle$$

And as $$P U_k^{\dagger} |x\rangle \in C$$ (because $$P$$ is a projector on $$C$$), we have $$Im(P_k) \subset U_k C$$

Now, I need to show $$U_k C \subset Im(P_k)$$:

Let $$|y \rangle \in U_k C$$:

For some $$|x \rangle$$, we have: $$|y\rangle = U_k P |x \rangle$$

Also, for some $$|z \rangle$$, we have $$|x \rangle = U_k^{\dagger} |z \rangle$$, thus in the end: $$|y\rangle = U_k P U_k^{\dagger} |z \rangle \in Im(P_k)$$

Conclusion: $$Im(P_k)=U_k C$$