Potentially "space" is not the correct word to use in the title, please correct if wrong

I am reading "Quantum Error Correction Via codes over Gf(4)" and I came across something I can't quite get straight in my head.

Take $E$ to be the quantum error group. Elements $e \in E$ can be written $$e = \pm w_{1} \otimes \dots \otimes w_{n}, \pm i w_{1} \otimes \dots \otimes w_{n}$$

$E$ has order $2^{2n+2}$

Let $S$ be the subgroup of $E$ consisting of of errors that have no effect on the encoded state.

Let $S'$ be subgroup of $E$ consisting of the undetectable errors.

Note that $S \leqslant S'$.

We require that every element of $S'$ commutes with $S$. As such, $S$ is abelian.

Since $S$ is abelian, its elements can be simultaneously diagonalised. This induces a decomposition of $\mathbb{C^{2^{n}}}$ into orthogonal eigenspaces.

My question here is how did we find that $S$ induces a decomposition of $\mathbb{C^{2^{n}}}$?

I think I have a vague (non-rigorous) intuition for why the decomposition would occur:

When you diagonlise a matrix, you obtain the eigenvectors/eigenvalues. Eigenvectors corresponding to different eigenvalues are linearly independent and therefore orthogonal to each other.

So, the fact that $S$ can be simultaneously diagonalised induces a decomposition into orthogonal eigenspaces.

However, how do we know that it decomposes $\mathbb{C}^{2^{n}}$ specifically, is this just because the diagonal matrices are $n \times n$ binary matrices?


1 Answer 1


To fully understand this statement and the subsequent few paragraphs of the paper, you need to learn the representation theory of finite groups. However, for just the stated claim, we can get by with less. Please refer to these very short notes of which you only need page 1-2 and the appendix on the last page.

Context for the claim in the paper

$\newcommand{\ket}[1]{|#1\rangle}$ Remember that we are trying to construct a code that encoded $k$ "logical" qubits into $n$ "data" qubits. Obviously, the state of $n$ data qubits lives in the Hilbert space $\mathbb{C}^{2^n}$.

The Pauli group $E$ is made of the $n$-fold tensor products of two-dimensional Pauli operators. Hence, the elements of $E$ act naturally on $\mathbb{C}^{2^n}$. In fact, we can easily write any $p \in E$ as a $2^n \times 2^n$ matrix operator $\Gamma_p$ [1]. Remember that these operators $\Gamma_p$ are unitary.

Now, we know that the Pauli group $E$ has $4^{n+1}$ elements, but the group $S$ only has $4 \times 2^{n-k}$. All these elements can be written as $2^n \times 2^n$ matrix operators as above. These operators as well act naturally on the $\mathbb{C}^{2^n}$ Hilbert space.

We have also established that all elements of $S$ commute. The claim from the paper is that

simultaneously diagonalizing the matrix operators of $S$ will decompose $\mathbb{C}^{2^n}$ into orthogonal eigenspaces.

Eigenspaces of a single operator

First consider a single operator $p$ in $E$, with matrix representative $\Gamma_p$. Recall that any Pauli operator in $E$ only has two eigenvalues $\lambda \in \{+1, -1\}$. The eigenspace associated with the eigenvalue $\lambda$ is defined as $$ W_\lambda = \{\ket{\psi} \in \mathbb{C}^{2^n} : \Gamma_p \ket{\psi} = \lambda \ket{\psi} \}. $$ From the spectral theorem of unitary operators [2], we know that $W_{+1}$ and $W_{-1}$ are orthogonal and their direct sum $W_{+1} \oplus W_{-1} = \mathbb{C}^{2^n}$. As a sidenote, both $W_{+1}, W_{-1}$ have dimension $2^{n-1}$, i.e. they slice the Hilbert space $\mathbb{C}^{2^n}$ exactly in half.

Eigenspaces for commuting operators

Now, let us consider the set $S$. The elements of this set commute, so the corresponding set of operators $\{\Gamma_s\}$ can be simultaneously diagonalized. But what implication does this have for $\mathbb{C}^{2^n}$?

First, we define the simultaneous eigenspace of two operators, $\Gamma_s, \Gamma_{s'}$ for some $s, s' \in S$. For the eigenvalue $\lambda \in \{+1,-1\}$ for $\Gamma_s$ and eigenvalue $\tau \in \{+1, -1\}$ for $\Gamma_{s'}$, their simultaneous eigenspace is $$ W_{\lambda, \tau} = \{\ket{\psi} \in \mathbb{C}^{2^n} : \Gamma_s \ket{\psi} = \lambda \ket{\psi} \text{ and } \Gamma_{s'} \ket{\psi} = \tau \ket{\psi}\}. $$ We can easily generalize this definition to more than two operators. Let $\lambda_s$ be an eigenvalue of $s \in S$, then $$ W_{\vec\lambda} = \{\ket{\psi} \in \mathbb{C}^{2^n} : \Gamma_s \ket{\psi} = \lambda_s \ket{\psi} \quad \forall s \in S \}. $$

Now, the general theorem has a few more caveats. But for an abelian subgroup $S$ of the Pauli group $E$, we have the following very informally stated theorem.

All $W_{\vec\lambda}$ spaces exist, are orthogonal, and their direct sum is equal to $\mathbb{C}^{2^n}$, i.e. $$ \bigoplus_{\vec{\lambda}} W_{\vec\lambda} = \mathbb{C}^{2^n}. $$

It is easy to see why the first part of this theorem is true. All $W_{\vec\lambda}$ spaces exist because given any $\vec\lambda$, we can construct a $\ket{\psi_{\vec\lambda}}$ with the desired eigenvalues.

To see why different spaces are orthogonal, let $\ket{\psi_{\vec\lambda}} \in W_{\vec\lambda}$ and $\ket{\psi_{\vec\tau}} \in W_{\vec\tau}$ for some $\vec\lambda \ne \vec\tau$. Also find a $s \in S$, such that $$ \Gamma_s\ket{\psi_{\vec\lambda}} = \ket{\psi_{\vec\lambda}}, \\ \Gamma_s\ket{\psi_{\vec\tau}} = -\ket{\psi_{\vec\tau}}, $$ or vice versa. Then $$ \langle \psi_{\vec\tau} | \Gamma_s \ket{\psi_{\vec\lambda}} = \langle \psi_{\vec\tau} \ket{\psi_{\vec\lambda}}, $$ if we act with $\Gamma_s$ to the right, and $$ \langle \psi_{\vec\tau} | \Gamma_s \ket{\psi_{\vec\lambda}} = -\langle \psi_{\vec\tau} \ket{\psi_{\vec\lambda}}, $$ if we act with $\Gamma_s$ to the left. Equating the two yields $\langle \psi_{\vec\tau} \ket{\psi_{\vec\lambda}} = 0$.

To argue that the direct sum indeed covers the entire Hilbert space is left as an exercise to the reader.


[1] I like to keep a clear distinction between a group element $e$ and its representation as a matrix $\Gamma_p$.

[2] See appendix of linked notes.

[3] See page 2 of linked notes.

  • $\begingroup$ I have used this answer to inform my work many times since you posted it, so thank you very much. There is one thing that I have just realised doesn't make sense to me. We have defined $S$ as being an abelian subgroup of $E$ which has no effect on the encoded state meaning $s | v \rangle = | v \rangle$ $\forall s \in S$. As such, all eigenvalues of $s \in S$ are $\lambda = 1$. How then, is it possible that all $W_{\lambda}$ spaces exist and their direct sum is equal to $\mathbb{C}^{2^{n}}$ if $W_{-1}$ does NOT exist for $\lambda $ of $S$? $\endgroup$
    – am567
    Jan 24 at 12:56

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.