some work here

thesis/chapters/conclusion.tex

@@ -1 +1,91 @@
+% vim: ft=tex
 \section{Conclusion and Outlook}
+
+As seen in \ref{ref:performance} simulation using stabilizers is exponentially
+faster than simulating using dense state vectors. Using a graphical
+representation for the stabilizers is on average more efficiently than using
+a stabilizer tableaux. In particular one can simulate more qbits while only
+applying Clifford gates.
+
+This is considerably useful when working on quantum error correcting strategies
+as they often include many qbits; the smallest quantum error correcting
+stabilizer code requires $5$ qbits to encode one logical qbit
+\cite{nielsen_chuang_2010}.  Several layers of data encoding increase the
+number of required qbits exponentially.
+
+Simulating in the stabilizer formalism is rather uninteresting from a physical
+point of view as basically no physically interesting simulations can be
+performed: As shown in \ref{ref:meas_stab} probability amplitudes have to be
+$0, \frac{1}{2}, 1$; this leaves very few points in time that could be
+simulated by applying a transfer matrix. Algorithms like the quantum fourier
+transform also require non-Clifford gates for qbit counts $n \neq 2, 4$.
+
+The basic idea of not simulating a state but (after imposing some conditions on
+the Hilbert space) other objects that describe the state is extremely
+interesting for physics as often the exponentially large or infinitely large
+Hilbert spaces cannot be mapped to a classical (super) computer.  One key idea
+to take from the stabilizer formalism is to simulate the Hamiltonian instead of
+the state:
+
+\begin{equation} H := -\sum\limits_{S^{(i)}} S^{(i)} \end{equation}
+
+The stabilizer state $\ket{\psi}$ as defined in \ref{ref:stab_states} is the
+ground state of this Hamiltonian. 
+
+While trying to extend the stabilizer formalism one inevitably hits the
+question:\\ \textit{Why is there a constraint on the $R_\phi$ angle? Why is it
+$\frac{\pi}{2}$?}\\ The answer to this question is hidden in the Clifford
+group. Recalling Definition \ref{def:clifford_group} the Clifford group is not
+defined to be generated by $H, S, CZ$ but by its property of normalizing the
+multilocal Pauli group. Storing and manipulating the multilocal Pauli group is
+only so efficient (or possible) because it is the tensor product of Pauli
+matrices. A general unitary on $n$ qbits would be a $2^{n} \times 2^{n}$ matrix
+which requires more space than a dense state vector. The Clifford group is
+a group preserving this tensor product property.
+
+When the constraint that $\ket{\varphi}$ is stabilized by the multilocal Pauli
+group but using $n$ arbitrary commuting hermitians $\langle h_1, ..., h_n
+\rangle$ that are the tensor product of $2\times 2$ hermitians one quickly
+realizes that one could apply any single-qbit gate to the $h_i$ and preserve
+the tensor product property. Applying the $CX$ gate however will break this
+property in general.
+
+Writing $h_j = \bigotimes\limits_{i=1}^{n} h_{j,i}$, 
+$A := \left(\bigotimes\limits_{l<j} I\right)$ 
+and $B := \left(\bigotimes\limits_{l>i} I\right)$ this can be seen easily
+by transforming a general $h_k$ with $CX_{i,j}$, $i = j+1$:
+
+
+\begin{equation}
+\begin{aligned}
+    CX_{i,j} h_k CX_{i,j}^\dagger &= \left( A\otimes |1\rangle\langle 1| \otimes X \otimes B  
+                                    + A \otimes |0\rangle\langle 0| \otimes I \otimes B\right)\\
+                                  &h_k\\
+                                  &\left(A \otimes |1\rangle\langle 1| \otimes X \otimes B 
+                                    + A \otimes |0\rangle\langle 0| \otimes I \otimes B\right) \\ 
+                                  &= h_{k,A} \otimes h_{k,j,11} |1\rangle\langle 1| \otimes Xh_{k,i}X \otimes h_{k,B}\\ 
+                                  &+  h_{k,A} \otimes h_{k,j,00}|0\rangle\langle 0| \otimes Ih_{k,i}I \otimes h_{k,B}\\ 
+                                  &+ h_{k,A} \otimes h_{k,j,01}|0\rangle\langle 1| \otimes Ih_{k,i}X\otimes h_{k,B}\\ 
+                                  &+ h_{k,A} \otimes h_{k,j,10}|1\rangle\langle 0| \otimes Xh_{k,i}I\otimes h_{k,B}\\
+\end{aligned}
+\end{equation}
+
+Searching for hermitians $h_1, h_2$ that fulfill 
+
+\begin{equation}
+CX_{1,2} (h_1 \otimes h_2) CX_{1,2} =  h_1' \otimes h_2'
+\end{equation}
+
+and 
+
+\begin{equation}
+CX_{2,1} (h_1 \otimes h_2) CX_{2,1} =  h_1'' \otimes h_2''
+\end{equation}
+
+might be a good step to find new classes of states that can be simulated efficiently
+using this method. This property has to be fulfilled by all elements of a group generated
+by such hermitian matrices.
+How computations and measurements would work using this method
+is not clear at the moment as many basic properties of the stabilizers are lost.
+
+

thesis/chapters/stabilizer.tex

@@ -104,6 +104,7 @@ used as the required properties of a set of stabilizers that can be studied on
 its generators.
 
 \subsubsection{Stabilizer States}
+\label{ref:stab_states}
 
 One important basic property of quantum mechanics is that hermitian operators
 have real eigenvalues and eigenspaces which are associated with these
@@ -170,6 +171,7 @@ hold true for an arbitrary $U$ but there exists a group for which $S'$ will be
 a set of stabilizers.
 
 \begin{definition}
+    \label{def:clifford_group}
     For $n$ qbits
     \begin{equation}
         C_n := \left\{U \in U(2^n) \middle| \forall p \in P_n: UpU^\dagger \in P_n \right\}