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\begin{document}
% title matter {{{1
\title[Notes on The Abelian HSP]{Notes on \\ The Hidden Subgroup Problem for Abelian Groups}
\author{Matthew Moore}
\date{\today}
\maketitle
%----------------------------------------------------------------------------}}}1

\noindent
\begin{center} \begin{tabular}{rl}
  \multicolumn{2}{l}{\textbf{The Hidden Subgroup Problem}} \\ \midrule
  \texttt{Input:}   & group $\alg{G}$, function $f:G\to X$ (as a blackbox)  \\
  \texttt{Promise:} & $f$ hides a subgroup $\alg{H}\leq \alg{G}$ \\
  \texttt{Task:}    & determine $\alg{H}$
\end{tabular} \end{center}

\bigskip

By the Fundamental Theorem of Finitely Generated Abelian Groups, if $\alg{G}$ is
finitely generated and Abelian then there are $m_i\in \ZZ$ such that
\[
  \alg{G}
  \cong \prod \ZZ_{m_i}.
\]
Represent $g\in G$ as a vector $g = (g_i)$ in $\prod \ZZ_{m_i}$. Define a
bilinear map $\mu: G\times G \to \CC^*$ by
\[
  \mu(g,h)
  = \prod \omega_{m_i}^{g_ih_i}
\]
where $\omega_k = e^{2\pi i/k}$ is the $k$-th root of unity. The function $\mu$
yields a notion of orthogonality: for $\alg{H}\leq \alg{G}$,
\begin{gather*}
  \alg{H}^{\perp} 
    = \big\{ g\in G\mid \mu(g,h) = 1 \big\}, \\
    |\alg{H}^{\perp}| = [ \alg{G} : \alg{H} ] = |\alg{G}|/|\alg{H}|,
    \qquad\qquad
    (\alg{H}^{\perp})^{\perp} = \alg{H}.
\end{gather*}

\begin{lem} \label{lem:sum_inner_prod}  % {{{
$\ds{
  \sum_{h\in H} \mu(g,h)
  = \begin{cases}
    |\alg{H}| & \text{if } g\in H^{\perp}, \\
    0         & \text{otherwise}.
  \end{cases}
}$
\end{lem}
\begin{proof}
For $\lambda\in \CC^*$ define
\[
  H_{\lambda}
  = \big\{ h\in H \mid \mu(g,h) = \lambda \big\}
  \qquad\text{and}\qquad
  L = \big\{ \lambda \mid H_\lambda \neq \emptyset \big\}.
\]
Observe that if $h\in H_\lambda$ then $H_\lambda = h + H_1$. It follows that all
non-empty $H_\lambda$ are cosets and hence of the same size, say $P$.

The set $L$ is closed under multiplication and is hence a subgroup of $\CC^*$.
Since $L$ is finite it is cyclic (elements are roots of unity --- the generator
is the root of unity or order equal to the least common multiple). Therefore
\[
  \sum_{h\in H} \mu(g,h)
  = \sum_{\lambda\in L} P \lambda
  = P \sum_{k=1}^{|L|} \nu^k
\]
where $\nu$ is the generator of $L$. If $|L| > 1$ then this sum is $0$. If $|L| =
1$ then $L = \{1\}$. The lemma follows.
\end{proof} % }}}

For $A\subseteq G$, define the quantum state
\[
  \ket{A}
  = \frac{1}{\sqrt{|A|}} \sum_{a\in A} \ket{a}.
\]
Define the operators
\begin{align*}
  \cl{F}_{\alg{G}}
    &= \frac{1}{\sqrt{|\alg{G}|}} \sum_{g,h\in G} \mu(g,h) \ket{g}\bra{h}
    &\quad& \text{(the \emph{quantum Fourier transform} for $\alg{G}$),} \\
  \tau_t
    &= \sum_{g\in G} \ket{t+g}\bra{g}
    && \text{(the \emph{translation operator for $t\in G$}),} \\
  \phi_t
    &=\sum_{g\in G} \mu(t,g)\ket{g}\bra{g}
    && \text{(the \emph{phase-change operator for $t\in G$}).}
\end{align*}

\begin{thm} \label{thm:operators} % {{{
\phantom{ }
\begin{enumerate}
  \item $\cl{F}_{\alg{G}}\ket{H} = \ket{H^{\perp}}$ for a subgroup
    $\alg{H}\leq \alg{G}$,
  \item $\mu(s,t) \tau_t \phi_s = \phi_s \tau_t$ for $s,t\in G$,
  \item $\cl{F}_{\alg{G}} \phi_s = \tau_{-s} \cl{F}_{\alg{G}}$ for $s\in G$,
  \item $\cl{F}_{\alg{G}} \tau_s = \phi_s \cl{F}_{\alg{G}}$ for $s\in G$.
\end{enumerate}
\end{thm}
\begin{proof}
\Proofitem{(1)} We have
\begin{align*}
  \cl{F}_{\alg{G}}\ket{H}
  &= \bigg( \frac{1}{\sqrt{|\alg{G}|}} 
     \sum_{g,h\in G} \mu(g,h)\ket{g}\bra{h} \bigg) \ket{H}
  = \frac{1}{\sqrt{|\alg{G}||\alg{H}|}} 
    \sum_{\stack{g\in G}{h\in H}} \mu(g,h)\ket{g} \\
  &= \frac{1}{\sqrt{|\alg{G}||\alg{H}|}}
     \sum_{g\in G} \bigg( \sum_{h\in H} \mu(g,h) \bigg) \ket{g}
  = \frac{1}{\sqrt{|\alg{G}||\alg{H}|}} \sum_{g\in H^\perp} |\alg{H}| \ket{g} \\
  &= \frac{1}{\sqrt{|\alg{H}^\perp|}} \sum_{g\in H^\perp} \ket{g}
  = \ket{H^\perp},
\end{align*}
where Lemma \ref{lem:sum_inner_prod} is applied at the fourth equality.


\Proofitem{(2)} We have
\begin{align*}
  \mu(s,t) \tau_t \phi_s
  &= \mu(s,t) \bigg( \sum_{g\in G} \ket{t+g} \bra{g} \bigg) 
     \bigg( \sum_{g\in G} \mu(s,g) \ket{g} \bra{g} \bigg) \\
  &= \sum_{g\in G} \mu(s,t+g) \ket{t+g} \bra{g} \\
  &= \bigg( \sum_{g\in G} \mu(s,t+g) \ket{t+g} \bra{t+g} \bigg) 
     \bigg( \sum_{g\in G} \ket{t+g} \bra{g} \bigg) \\
  &= \bigg( \sum_{g\in G} \mu(s,g) \ket{g} \bra{g} \bigg) 
     \bigg( \sum_{g\in G} \ket{t+g} \bra{g} \bigg)
  = \phi_s \tau_t.
\end{align*}


\Proofitem{(3)} We have
\begin{align*}
  \cl{F}_{\alg{G}} \phi_s
  &= \bigg( \frac{1}{\sqrt{|\m{G}|}} \sum_{g,h\in G} \mu(g,h) \ket{g} \bra{h} \bigg) 
     \bigg( \sum_{g\in G} \mu(g,s) \ket{g} \bra{g} \bigg) \\
  &= \frac{1}{\sqrt{|\m{G}|}} \sum_{g,h\in G} \mu(g+s,h) \ket{g} \bra{h}
  = \frac{1}{\sqrt{|\m{G}|}} \sum_{g,h\in G} \mu(g,h) \ket{g-s} \bra{h} \\
  &= \bigg( \sum_{g\in G} \ket{g-s} \bra{g} \bigg) 
    \bigg( \frac{1}{\sqrt{|\m{G}|}} \sum_{g,h\in G} \mu(g,h) \ket{g} \bra{h} \bigg)
  = \tau_{-s} \cl{F}_{\alg{G}}.
\end{align*}


\Proofitem{(4)} We have
\begin{align*}
  \cl{F}_{\alg{G}} \tau_s
  &= \bigg( \frac{1}{\sqrt{|\m{G}|}} \sum_{g,h\in G} \mu(g,h) \ket{g} \bra{h} \bigg) 
     \bigg( \sum_{g\in G} \ket{g+s} \ket{g} \bigg) \\
  &= \frac{1}{\sqrt{|\m{G}|}} \sum_{g,h\in G} \mu(g,h+s) \ket{g} \bra{h}
  = \frac{1}{\sqrt{|\m{G}|}} \sum_{g,h\in G} \mu(g,h) \mu(g,s) \ket{g} \bra{h} \\
  &= \bigg( \sum_{g\in G} \mu(g,s) \ket{g} \bra{g} \bigg)
    \bigg( \frac{1}{\sqrt{|\m{G}|}} \sum_{g,h\in G} \mu(g,h) \ket{g} \bra{h} \bigg)
  = \phi_s \cl{F}_{\alg{G}}.
  \qedhere
\end{align*}
\end{proof} % }}}

Define the operator
\[
  \cl{HSP}_{\alg{G},f} 
  = (\cl{F}_{\alg{G}} \otimes I) \circ \widehat{f} \circ (\cl{F}_{\alg{G}}^{-1} \otimes I),
\]
where $\widehat{f}$ is the unitary form of the blackbox function $f$, defined by
\[
  \widehat{f}\ket{x,y}
  = \ket{x,f(x)+y}.
\]

\begin{thm} \label{thm:hsp_abelian} % {{{
Let $\cl{M}_1$ be the measurement operator for the first register and define a
distribution
\[
  (z)
  = \cl{M}_1 \circ \cl{HSP}_{\alg{G},f} \ket{0,0},
\]
where the first register contains group elements (with $0\in G$ as the identity)
and the second register contains elements of $X$ (regarded as a subset of
$\{0,\dots,|G|-1\}$).

\medskip

\noindent
Then $(z)$ is a uniform distribution over $\alg{H}^{\perp}$, where $\alg{H}$ is the
hidden subgroup.
\end{thm}
\begin{proof}
Fix a transversal $T$ of $\alg{H}$ in $\alg{G}$ and note that $|T| =
|\alg{G}|/|\alg{H}| = |\alg{H}^\perp|$. We have
\begin{align*}
  &\cl{HSP}_{\alg{G},f} \ket{0,0}
    = (\cl{F}_{\alg{G}} \otimes I) \circ \widehat{f} \circ (\cl{F}_{\alg{G}}^{-1} \otimes I) 
      \ket{0,0}
    = (\cl{F}_{\alg{G}} \otimes I) \circ \widehat{f}\ 
      \ket{G}\otimes \ket{0} \\
  &\qquad
    = (\cl{F}_{\alg{G}} \otimes I)
      \frac{1}{\sqrt{|\alg{G}|}} \sum_{g\in G} \ket{g, f(g)}
    = (\cl{F}_{\alg{G}} \otimes I)
      \frac{1}{\sqrt{|T|}} \sum_{t\in T} \frac{1}{\sqrt{|\alg{H}|}} 
      \sum_{h\in H} \ket{t+h,f(t)} \\
  &\qquad
    = (\cl{F}_{\alg{G}} \otimes I) 
      \frac{1}{\sqrt{|\alg{H}^\perp|}} \sum_{t\in T} \tau_t \ket{H}\otimes \ket{f(t)}
    = \frac{1}{\sqrt{|\alg{H}^\perp|}} \sum_{t\in T} \cl{F}_{\alg{G}} \tau_t 
      \ket{H} \otimes \ket{f(t)} \\
  &\qquad
    = \frac{1}{\sqrt{|\alg{H}^\perp|}} \sum_{t\in T} \phi_t \cl{F}_{\alg{G}} 
      \ket{H} \otimes \ket{f(t)}
    = \frac{1}{\sqrt{|\alg{H}^\perp|}} \sum_{t\in T} \phi_t 
      \ket{H^\perp} \otimes \ket{f(t)}
\end{align*}
(Theorem \ref{thm:operators} is used in equalities 2, 7,and 8). The operator
$\phi_t$ is a phase shift and therefore has no effect on the probability of
measurement. It follows that the first register is a uniform distribution on
$\alg{H}^\perp$, as claimed.
\end{proof} % }}}
\end{document}
