3
$\begingroup$

I've started reading the book "Quantum Computation and Quantum Information" by Michael A. Nielsen and Issac L. Chuang, specifically chapter 10 (about quantum error correction), and I'm finding myself a bit confused as to the notations used.

In page 436, theorem 10.1, (10.16), I'd like to ask what is the "type" of each "object"? Is everything a $2^n \times 2^n$ matrix? Or is $E_i$ a vector, and ${E_i}^\dagger E_j$ the outer-product of two vectors?

An example for such $\epsilon$ with $\{E_i\}$ and $\alpha_{ij}$ would be appreciated.

The theorem: Let $C$ be a quantum code, and let $P$ be the projector onto $C$. Suppose $\epsilon$ is a quantum operation with operation elements $\{E_i\}$. A necessary and sufficient condition for the existence of an error-correction operation $R$ correcting $\epsilon$ on $C$ is that $ P E_i^\dagger E_j P = \alpha_{ij} P $ for some Hermitian matrix $\alpha$ of complex numbers. We call the operation elements $\{E_i\}$ for the noise $\epsilon$ errors, and if such an $R$ exists we say that $\{E_i\}$ constitutes a correctable set of errors.

Thank you very much :)

$\endgroup$

1 Answer 1

3
$\begingroup$

Generally speaking, if a quantum channel $\Phi$ sends operators in $\mathcal X$ into operators in $\mathcal Y$, and its Kraus representation reads $\Phi(X)=\sum_a A_a X A_a^\dagger$, then the Kraus operators are linear operators of the form $A_a\in\mathrm{Lin}(\mathcal X,\mathcal Y)$, that is, $A_a:\mathcal X\to\mathcal Y$. It follows that $A_a^\dagger\in\mathrm{Lin}(\mathcal Y,\mathcal X)$.

In your case, it seems you have a channel with $\mathcal X=\mathcal Y\simeq\mathbb C^N$ with $N=2^n$. Thus your projection operator is $P\in\mathrm{Lin}(\mathbb C^N)$, and $PE_i^\dagger E_j P$ is just a standard matrix multiplication, which results in some operator in $\mathrm{Lin}(\mathbb C^N)$.

$\endgroup$
1
  • $\begingroup$ Can the projection operator $P$ be seen as the identity matrix in the code space $C$? $\endgroup$
    – narip
    Nov 22, 2021 at 11:39

Your Answer

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