## Saturday, 1 February 2014

### Permutation-invariant codes

Let $S_n$ be the group of permutations of the $n$-element set $[n]:=\{1 , 2, \dots, n\}$, and let $\pi\in S_n$. Define the unitary $U_\pi: (\mathbb{C}^2)^{\otimes n}\mapsto(\mathbb{C}^2)^{\otimes n}$ by
$U_\pi(\ket{\varphi_1}\otimes\cdots\otimes\ket{\varphi_n})=\ket{\varphi_{\pi^{-1}(1)}}\otimes\cdots\otimes\ket{\varphi_{\pi^{-1}(n)}}$
for all product states $\ket{\varphi_1}\otimes\cdots\otimes\ket{\varphi_n}$ (and extended linearly to all of $(\mathbb{C})^{\otimes{n}}$). Denote by $(i j)$ the transposition of $i,j\in[n]$ for $i\neq j$, and define the subspace
$Q=\{\ket{\Psi}\in(\mathbb{C})^2)^{\otimes{n}} \mid U_{(i j)}\ket{\Psi}=\ket{\Psi} \ \text{for all} \ i\neq j \}.$

An arbitrary state $\ket{\Psi}\in Q$ can be written generally as a linear combination
$\ket{\Psi}=\SUM{}{}\gamma_k\ket{\xi_k},$
where $\ket{\xi_k}\in\mathcal{B}$ and $\mathcal{B}$ is a complete set of basis vectors of $Q$. Since every $\ket{\xi_k}\in Q$, must also satisfy the constraint $U_{(ij)}\ket{\xi_k}=\ket{\xi_k}$ for all $(ij)\in S_n$, each state $\ket{\xi_k}$ must be comprised of states possessing a property that remains invariant under the $U_{(ij)}$ operations. In this regard, let $\bar{b}=b_1b_2\dots b_n \in\{0,1\}^n$ with $b_i\in\{0,1\}$, and let $\ket{\bar{b}}\in(\mathbb{C}^2)^{\otimes n}$ be a $n$-qubit computational basis state. More explicitly, $\ket{\bar{b}}=\bigotimes_{i=1}^{n}\ket{b_i}$, describes the $n$-qubit state where each qubit is either in the state $\ket{0}$ or $\ket{1}$. Now define the map
$\omega: \{0,1\}^n\to \{0,1,\dots,n\} \ \ \text{given as} \ \ \omega(\bar{b})=\SUM{i=1}{n}b_i,$
which simply counts the number of $1$s appearing in the bit string $\bar{b}$, and call $\omega(\bar{b})$ the \emph{weight} of $\bar{b}$.

For every $\bar{b}\in\{0,1\}^n$ and any $\pi\in S_n$, it will always be the case that $U_{\pi}\ket{\bar{b}}=\ket{\bar{b'}}$ for some $\bar{b'}\in\{0,1,\}^n$ such that $\omega(\bar{b})=\omega(\bar{b'})$, because $U_{\pi}$ will only permute the bits $b_i$ comprising the string $\bar{b}$ which preserves the  weight of the string. That is, if $\ket{\bar{b}}=\bigotimes_{i=1}^{n}\ket{b_i}$, then $U_\pi\ket{\bar{b}}=\ket{\bar{b'}}=\bigotimes_{i=1}^{n}\ket{b_{\pi^{-1}(i)}}$, where the bits of the resulting string $\bar{b'}$ are given by $b'_i=b_{\pi^{-1}(i)}$, and consequently $\omega(\bar{b})=\sum_{i=1}^{n}b_i=\sum_{i=1}^{n}b_{\pi^{-1}(i)}=\omega(\bar{b'})$.

Therefore, let
$\ket{\xi_k}=\frac{1}{\sqrt{\binom{n}{k}}}\SUM{ \ \ \bar{b} \ : \ \omega(\bar{b})=k}{}\ket{\bar{b}}$
be an equally weighted superposition of the $\binom{n}{k}$ computational basis states $\ket{\bar{b}}$ with weight $\omega(\bar{b})=k$. Then it follows that $U_{(ij)}\ket{\xi_k}=\ket{\xi_k}$ for all $(ij)\in S_n$, since each $U_{(ij)}$ yields a bijection between the set of states $\ket{\bar{b}}$ having the same weight. Hence, the set of basis vectors of $Q$ is given by $\mathcal{B}=\{\ket{\xi_k} \mid k=0,1,\dots, n\}$ consisting of $n+1$ basis vectors.

A projector $P$ onto the subspace $Q$ can be expressed in terms of the basis vectors in $\mathcal{B}$ as
$P=\SUM{k=0}{n}\ket{\xi_k}\bra{\xi_k}.$

In terms of the operators $U_\pi$, the projector can be equivalently expressed as an equally weighted superposition over all permutations as
$P=\frac{1}{n!}\SUM{\pi\in S_n}{}U_\pi$

Consider the set of errors
$\mathcal{E}=\left\{\SUM{\pi\in S_n}{}a_\pi U_\pi \mid (a_\pi)_{\pi\in S_n} \subset \mathbb{C} \right\}.$
For the group symmetric $S_n$, recall that any permutation $\pi\in S_n$ can be expressed as a product of transpositions only. Actually, $S_n$ can  be generated by the $n-1$ transpositions of the form $(i,i+1)$ for $i\in\{1, 2, \dots, n-1\}$. This algebraic structure extends to the operators $U_\pi$, That is, for any $\pi\in S_n$,
$U_\pi=\PROD{(ab)\in S_n}{}U_{(ab)},$
where $(ab)=(i,i+1)$ for some $i\in\{1, 2, \dots, n-1\}$. Note that in this product, any particular $U_{(ab)}$ can appear multiple times. Therefore, for any $\pi\in S_n$ and any $\ket{\psi}\in Q$
$U_\pi\ket{\psi}=\PROD{(ab)\in S_n}{}U_{(ab)}\ket{\psi}=\ket{\psi},$
because $U_{(ab)}\ket{\psi}=\ket{\psi}$ for any $(ab)\in S_n$.

In regards to to the correctability of the error set $\mathcal{E}$, consider any $\pi,\pi\in S_n$ and any basis vectors $\ket{\xi_k},\ket{\xi_{k'}}\in\mathcal{B}$. As a consequence of what was just shown $U_{\pi}\ket{\xi_k}=\ket{\xi_k}$ and $U_{\pi'}\ket{\xi_{k'}}=\ket{\xi_{k'}}$. Then,
$\bra{\xi_{k'}}U_{\pi'}^{\dagger}U_\pi\ket{\xi_k}=\ip{\xi_{k'}}{\xi_k}=\delta_{k'k}.$
Since the error correction conditions for a code state that
$\bra{\xi_{k'}}U_{\pi'}^{\dagger}U_\pi\ket{\xi_k}=c_{\pi'\pi}\ip{\xi_{k'}}{\xi_k},$
where the $c_{\pi'\pi}$ are constants that do not depend on the states $\ket{\xi_k}$ and $\ket{\xi_k'}$, then $c_{\pi'\pi}=1$ for all $\pi,\pi'\in S_n$.
More generally, for any $\ket{\psi},\ket{\psi'}\in Q$ this implies
$\bra{\psi'}U_{\pi'}^{\dagger}U_\pi\ket{\psi}=\ip{\psi'}{\psi}$

Therefore, each $U_\pi$ for $\pi\in S_n$ is a correctable error. Then since any linear combination of correctable errors can also be corrected, any error of the form
$E=\SUM{\pi\in S_n}{}a_\pi U_\pi$
can be corrected. Thus, the set $\mathcal{E}$ is correctable.

An operator basis $\{F_j\}_j$ of $\mathcal{E}$ which diagonalizes the matrix $C=(c_{ab})$ is given by $F_\pi =U_\pi$, where $\pi\in S_n$. Since the order of $S_n$ is $n!$ (meaning there are $n!$ different permutation operators), there are $n!$ many different operators $U_\pi$ in this basis. Then upon being diagonalized, the coefficients of the matrix $C$ are given by $c_{\pi'\pi}=1$ if $\pi'= \pi$ and $c_{\pi'\pi}=0$ if $\pi'\neq \pi$.

For the subspace $Q$, the set of detectable errors is defined as
$\mathcal{E}_D:=\left\{E \mid \bra{\psi}E\ket{\phi}=c(E)\ip{\psi}{\phi}, \forall \ket{\psi},\ket{\phi}\in Q\right\}.$
Consider a single bit flip error, say, $E=X_1$ that applies the bit flip $X$ to the first qubit of the register. Moreover, consider the two basis states $\ket{\xi_0}$ and $\ket{\xi_1}$ of Q. Here, $\ket{\xi_0}=\bigotimes_{i=1}^{n}\ket{0}$ is the unique state with weight $0$, and $\ket{\xi_1}$ is the superposition of all states labelled by weight $1$ bit strings. Observe that $X_1\ket{\xi_0}=\ket{1}\bigotimes_{i=2}^{n}\ket{0}$, which is a state that appears in the superposition of states expressed in $\ket{\xi_1}$. Therefore,
$\bra{\xi_1}X_1\ket{\xi_0}=\frac{1}{\sqrt{\binom{n}{k}}}\neq 0,$
but yet $\ket{\xi_1}$ and $\ket{\xi_0}$ are orthogonal basis states so that $\ip{\xi_1}{\xi_0}=0$. Hence, the single bit flip error $E=X_1$ is an undetectable error by definition.

Let $E=X^{\otimes n}$, and consider the two basis states $\ket{\xi_0},\ket{\xi_n}\in\mathcal{B}$ of $Q$. Then
$E\ket{\xi_0}=X^{\otimes n}\TENSOR{i=1}{n}\ket{0}=\TENSOR{i=1}{n}\ket{1}=\ket{\xi_1},$
Which implies that $\bra{\xi_n}E\ket{\xi_0}=\ip{\xi_n}{\xi_n}=1$. However, $\ip{\xi_n}{\xi_0}=0$ since they are orthogonal basis states, and thus
$E=X^{\otimes n}$ is an undetectable error.

More generally, let $U\in SU(2)$, with $U\neq I$, and consider the error $E=U^{\otimes n}$. According to the Shur-Weyl duality, $[E,U_\pi]=0$ for all $U\in SU(2)$ and $\pi\in S_n$. Consider an arbitrary state $\ket{\psi}\in Q$ so that $U_\pi\ket{\psi}=\ket{\psi}$ for all $\pi\in S_n$ as argued in above. Then,
$U_\pi E\ket{\psi}=EU_\pi\ket{\psi}=E\ket{\psi},$
Where the first equality follows from the Shur-Weyl duality. This implies that $E\ket{\psi}\in Q$.  Since, $\ket{\psi}$ was assumed to be arbitrary, this is equivalent to the statement $EQ=Q$.

Consider any basis state $\ket{\xi_j}$. Then since $EQ=Q$, it must be the case that $E\ket{\xi_j}\in Q$ so that $E\ket{\xi_j}$  itself can be expressed in terms of the basis $\mathcal{B}$ of $Q$. That is,
$E\ket{\xi_j}=U^{\otimes n}\ket{\xi_j}=\SUM{k=1}{n=1}\alpha_k\ket{\xi_k}.$
Then there must exist at least one $\alpha_{k'}\neq 0$, and perhaps more. Moreover, since $U\neq I$ by assumption, $k'\neq j$. Choose one such basis state $\ket{\xi_{k'}}$ with $\alpha_{k'}\neq 0$. With this choice,
$\bra{\xi_{k'}}E\ket{\xi_j}>0.$
However, since $\ip{\xi_k'}{\xi_j}=0$ this implies that
$\bra{\xi_{k'}}E\ket{\xi_j}\neq c(E)\ip{\xi_k'}{\xi_j}.$
Therefore, the error $E=U^{\otimes n}$ is undetectable.