From c30cc44c40037322e64d9f076f8230f81d1a087c Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= Date: Thu, 19 Mar 2020 12:37:27 +0100 Subject: [PATCH] some work here --- thesis/chapters/conclusion.tex | 90 ++++++++++++++++++++++++++++++++++ thesis/chapters/stabilizer.tex | 2 + 2 files changed, 92 insertions(+) diff --git a/thesis/chapters/conclusion.tex b/thesis/chapters/conclusion.tex index 4a78ecc..f174cc4 100644 --- a/thesis/chapters/conclusion.tex +++ b/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 $h_j = \bigotimes\limits_{i=1}^{n} h_{j,i}$, +$A := \left(\bigotimes\limits_{li} I\right)$ 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. + + diff --git a/thesis/chapters/stabilizer.tex b/thesis/chapters/stabilizer.tex index 56aefd7..abb28db 100644 --- a/thesis/chapters/stabilizer.tex +++ b/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\}