bachelor_thesis/thesis/chapters/conclusion.tex

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\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. One important property is that the
stabilization: The simulated state is the $+1$ eigenstate of the stabilizers.
This is another property that will have to be fulfilled by the hermitians as it
is a key property used in \ref{ref:dynamics_stabilizer}. To ensure that the
state is well defined one will have to demand that the eigenvalues fulfill
$\lambda_1 = 1$ and $\lambda_2 < 1$.

One should also note that the sabilizer states do not form a Hilbert (sub)space.
Linear combinations of stabilizer states are (in general) no stabilizer states.
The superposition principle is quite essential to many quantum algorithms and
quantum physics which limits the use of the stabilizer formalism drastically.