%Paper: hep-th/9412048
%From: "Svozil Karl (TUVienna)" <E1360DAB@helios.edvz.univie.ac.at>
%Date: Tue, 06 Dec 94 11:32:51 MEZ












%\documentstyle[12pt,svozil,epsf,emlines]{article}
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   \def\bbb{\heartsuit }
   \def\000{{\bf 0}}
   \def\111{{\bf 1}}
   \def\uuu{\frak U}
   \def\mmm{\mid }
   \def\rrr{\rangle }
   \def\lll{\langle }
   \def\ppp{\psi }
   \def\qqq{\varphi }

\begin{document}
 \title{Quantum recursion theory}
\author{K. Svozil\\
 {\small Institut f\"ur Theoretische Physik}  \\
  {\small University of Technology Vienna }     \\
  {\small Wiedner Hauptstra\ss e 8-10/136}    \\
  {\small A-1040 Vienna, Austria   }            \\
  {\small e-mail: svozil@tph.tuwien.ac.at}}
\maketitle

\begin{flushright}
{\scriptsize qrt.tex}
\end{flushright}

\begin{abstract}
{\em
Incompleteness and undecidability theorems have to be revised in view of
quantum information and computation theory.}
\end{abstract}
 \clearpage

As has already been pointed out in  G\"odel's centennial paper
on the incompleteness af arithmetic
\cite{godel1},
the classical undecidability theorems of formal logic \cite{davis}
and the theory of computable functions
\cite{rogers,odi}
are based on semantical pardoxes
such as the liar
\cite{bible} or Richard's paradox.

The  method of diagonalization, which
was first applied by
Cantor for a proof of the undenumerability of real numbers
\cite{cantor}, has been applied by Turing for a proof
of the recursive undecidability of the halting problem \cite{turing}.
The halting problem is the problem of whether or not
an arbitrary algorithm terminates or produces a
particular output and terminates.

Assume that the halting problem is decidable.
Turing \cite{turing} proved that
this assumption yields a contradiction.
To construct the contradiction,
consider an arbitrary algorithm $B(x)$ whose input is a string of
symbols
$x$.
 Assume that there exists a ``halting algorithm'' ${\tt HALT}$
 which  is
 able to decide whether $B$ terminates on $x$ or not.

 Using  ${\tt HALT}(B(x))$ we shall construct another
 deterministic computing agent
$A$, which
 has as input any effective program $B$ and which proceeds as follows:
 Upon reading the program $B$ as input, $A$ makes a copy of it.
 This  can be readily achieved, since
 the program $B$ is presented to $A$  in some
 encoded form $\# (B)$, i.e., as a string of symbols. In the next
 step, the agent uses the
 code $\# (B)$ as input string for $B$ itself; i.e., $A$  forms
 $B(\#(B))$, henceforth denoted by $B(B)$. The agent now hands
 $B(B)$ over to its
 subroutine ${\tt HALT}$.
 Then, $A$ proceeds as follows:
  if ${\tt HALT}(B(B))$ decides that $B(B)$
 halts, then the agent
 $A$ does not halt;
this can for instance be realised by an infinite {\tt
 DO}-loop;
  if ${\tt HALT}(B(B))$ decides that $B(B)$
 does {\em not} halt, then
 $A$ halts.

The agent $A$ will now be confronted with the following paradoxical
task:
take the own code as input and proceed.
Assume that $A$ is restricted to classical bits of
information.
Then, whenever
 $A(A)$ halts,  ${\tt HALT}(A(A))$  forces
 $A(A)$ not to halt.
Conversely, whenever $A(A)$ does not halt, then ${\tt HALT}(A(A))$
 steers $A(A)$
 into the halting mode. In both cases one arrives at a
complete contradiction.
 Classically,
 this contradiction can only be consistently avoided by
 assuming
 the nonexistence of $A$ and, since the only nontrivial feature of $A$
 is the use of the peculiar halting algorithm
 ${\tt HALT}$, the impossibility of any such
 halting algorithm.

The task of the agent $A$ can be performed consistently if
$A$ is allowed to process quantum information.
In quantum information theory (cf.
\cite{a:8,be,deutsch-85,f-85,peres-85,b-86,m-86,deutsch:89,deutsch:92}),
the most elementary unit of information,
henceforth called {\em qbit},
may be physically represented by a coherent
superposition
of the two states which correspond to the symbols $\000$ and ${\bf 1}$.
The qbit states
are the coherent superposition
of the
classical basis states
$\{\mmm {\bf 0}\rrr , \mmm {\bf 1} \rrr \}$.
They are in the undenumerable set
\begin{equation}
\{ \vert a,b\rangle \mid \vert a,b\rangle =a\vert {\bf 0}\rangle +b\vert
{\bf 1}
\rangle
,\; \vert a\vert^2+\vert b\vert^2=1,\; a,b\in {\Bbb C} \}
\quad .
\label{e:qbit}
\end{equation}



Due to the possibility of a coherent superposition of classical bit
states, the usual {\it reductio ad absurdum} argument breaks down.
Instead, diagonalization procedures in
quantum information theory yield qbit solutions which are fixed points
of the associated unitary operators.


The emergence of fixed points can be demonstrated by a simple example.
Diagonalization effectively
transforms the classical bit value ``$\000$'' into value ``$\111$'' and
``$\111$'' into ``$\000$.''
For the halting problem, identify
``$\111$'' with the halting state and ``$\000$'' with the non-halting
state. The evolution representing diagonaliation, and effectively, agent
$A$'s task, can be expressed by the
unitary operator
$
\widehat{D}
$ as follows
$
\widehat{D} \vert \000\rrr  \rightarrow  \vert \111 \rrr $, and
$\widehat{D} \vert \111\rrr  \rightarrow  \vert \000 \rrr $.
In the state basis
$\mmm {\bf 0}\rrr  \equiv
\left(
\begin{array}{c}
1 \\
0
 \end{array}
\right)
$
and
$\mmm {\bf 1}\rrr \equiv
\left(
\begin{array}{c}
0 \\
1
 \end{array}
\right)
$
($\tau_1$ stands for the Pauli spin operator),
\begin{equation}
\widehat{D}=
\tau_1 =
{\tt not}=
\left(
\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}
\right) =\vert \111\rrr \langle \000\vert
+ \vert \000 \rrr \langle \111 \vert    \quad .
\end{equation}
$
\widehat{D}
$
will be called
{\em diagonalization} operator,
despite the fact that the only nonvanishing components are
off-diagonal.
It can be realized by elementary quantum devices such as the beam
splitters and Mach-Zehnder interferometers
\cite{rzbb} depicted in Figure \ref{f:qid} with
$
\widehat{D}=
T^{bs}_{21}(0,0,0,0)=
T^{MZ}_{21}(-\,{\pi \over 2},0,0,-\,{\pi \over 2})$.
\begin{figure}
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\end{picture}
\end{center}
\caption{Elementary quantum interference device.
An elementary quantum interference device can be realized by a
4-port interferometer with two input ports ${\bf 0} ,{\bf 1} $
and two
output ports
${\bf 0} ',{\bf 1} '$.
Any twodimensional unitary transformation can be realised by the
devices.
a) shows a realization
by a single beam
splitter $S(T)$
with variable transmission $t$
and three phase shifters $P_1,P_2,P_3$;
b) shows a realization with 50:50 beam
splitters $S_1({1\over 2}) $ and $S_2 ({1\over 2})$ and four phase
shifters
$P_1,P_2,P_3,P_4$.
 \label{f:qid}}
\end{figure}

As has been pointed out earlier,
quantum information theory allows a coherent superposition
$\mmm a,b\rrr =a\, \mmm \000 \rrr +b\, \mmm \111 \rrr$
of the
classical bit states. Therefore,
$
\widehat{D}
$
has a
fixed point at
\begin{equation}
\mmm {1\over \sqrt{2}},{1\over \sqrt{2}}   \rrr ={1\over \sqrt{2}}\vert
\000 \rrr + {1\over \sqrt{2}}
\vert \111 \rrr
\quad ,
\end{equation}
which is a coherent superposition of the classical bit base and
does not give rise to inconsistencies \cite{svozil-paradox}.

Classical undecidability is recovered if one performs any irreversible
measurement
on the fixed point state. This  causes a state reduction into the
classical states $\mmm \000 \rrr$ and $\mmm \111 \rrr$.
The
probability for
the fixed point state
$\vert  {1\over \sqrt{2}},{1\over \sqrt{2}} \rangle $
to be in either
$\vert \000 \rangle  $
or
$\vert \111 \rangle $
is equal; i.e.,
\begin{equation}
\vert \langle \000 \mid  {1\over \sqrt{2}},{1\over \sqrt{2}} \rangle
\vert^2
=
\vert \langle \111 \mid  {1\over \sqrt{2}},{1\over \sqrt{2}}\rangle
\vert^2
={1  \over 2}
\quad .
\end{equation}

However, no contradiction is involved in the diagonalization argument.
Therefore,
standard proofs of the recursive unsolvability of the halting problem
do not apply to quantum recursion theory.

Another, less abstract, application for quantum information theory is
the handling of inconsistent information in databases.
Thereby,
two contradicting cbits of information
$\vert a\rangle $ and
$\vert b\rangle $ are resolved by the qbit
$\vert {1\over \sqrt{2}},{1\over \sqrt{2}} \rangle =
{1\over \sqrt{2}}(\vert a\rangle + \vert b\rangle )$.
Throughout the rest of the computation the coherence is maintained.
After the processing, the result is obtained by an irreversible
measurement. The processing of qbits requires an exponential space
overhead on classical computers in cbit base \cite{feynman}.
Thus, in order to remain tractable,
the corresponding qbits should be implemented on
truly quantum universal computers.

\begin{thebibliography}{99}

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K. G\"odel, {\sl Monatshefte f\"ur Mathematik und Physik}
{\bf 38}, 173 (1931);
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 \cite{davis}.

 \bibitem{davis}
 M. Davis, {\sl Computability \& Unsolvability}
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 \bibitem{godel-ges1}
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\bibitem{rogers}
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\bibitem{odi}
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 \bibitem{bible}
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 \bibitem{b-86}
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 \bibitem{deutsch:92}
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\bibitem{rzbb}
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\bibitem{svozil-paradox}
K. Svozil,
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 \bibitem{feynman}
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{\bf 21}, 467 (1982).

\end{thebibliography}

\end{document}

