100 lines
5.2 KiB
TeX
100 lines
5.2 KiB
TeX
% vim: ft=tex
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\section{Conclusion and Outlook}
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As seen in \ref{ref:performance} simulation using stabilizers is exponentially
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faster than simulating using dense state vectors. Using a graphical
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representation for the stabilizers is on average more efficiently than using
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a stabilizer tableaux. In particular one can simulate more qbits while only
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applying Clifford gates.
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This is considerably useful when working on quantum error correcting strategies
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as they often include many qbits; the smallest quantum error correcting
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stabilizer code requires $5$ qbits to encode one logical qbit
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\cite{nielsen_chuang_2010}. Several layers of data encoding increase the
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number of required qbits exponentially.
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Simulating in the stabilizer formalism is rather uninteresting from a physical
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point of view as basically no physically interesting simulations can be
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performed: As shown in \ref{ref:meas_stab} probability amplitudes have to be
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$0, \frac{1}{2}, 1$; this leaves very few points in time that could be
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simulated by applying a transfer matrix. Algorithms like the quantum fourier
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transform also require non-Clifford gates for qbit counts $n \neq 2, 4$.
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The basic idea of not simulating a state but (after imposing some conditions on
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the Hilbert space) other objects that describe the state is extremely
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interesting for physics as often the exponentially large or infinitely large
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Hilbert spaces cannot be mapped to a classical (super) computer. One key idea
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to take from the stabilizer formalism is to simulate the Hamiltonian instead of
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the state:
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\begin{equation} H := -\sum\limits_{S^{(i)}} S^{(i)} \end{equation}
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The stabilizer state $\ket{\psi}$ as defined in \ref{ref:stab_states} is the
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ground state of this Hamiltonian.
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While trying to extend the stabilizer formalism one inevitably hits the
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question:\\ \textit{Why is there a constraint on the $R_\phi$ angle? Why is it
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$\frac{\pi}{2}$?}\\ The answer to this question is hidden in the Clifford
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group. Recalling Definition \ref{def:clifford_group} the Clifford group is not
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defined to be generated by $H, S, CZ$ but by its property of normalizing the
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multilocal Pauli group. Storing and manipulating the multilocal Pauli group is
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only so efficient (or possible) because it is the tensor product of Pauli
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matrices. A general unitary on $n$ qbits would be a $2^{n} \times 2^{n}$ matrix
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which requires more space than a dense state vector. The Clifford group is
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a group preserving this tensor product property.
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When the constraint that $\ket{\varphi}$ is stabilized by the multilocal Pauli
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group but using $n$ arbitrary commuting hermitians $\langle h_1, ..., h_n
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\rangle$ that are the tensor product of $2\times 2$ hermitians one quickly
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realizes that one could apply any single-qbit gate to the $h_i$ and preserve
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the tensor product property. Applying the $CX$ gate however will break this
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property in general.
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Writing $h_j = \bigotimes\limits_{i=1}^{n} h_{j,i}$,
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$A := \left(\bigotimes\limits_{l<j} I\right)$
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and $B := \left(\bigotimes\limits_{l>i} I\right)$ this can be seen easily
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by transforming a general $h_k$ with $CX_{i,j}$, $i = j+1$:
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\begin{equation}
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\begin{aligned}
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CX_{i,j} h_k CX_{i,j}^\dagger &= \left( A\otimes |1\rangle\langle 1| \otimes X \otimes B
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+ A \otimes |0\rangle\langle 0| \otimes I \otimes B\right)\\
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&h_k\\
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&\left(A \otimes |1\rangle\langle 1| \otimes X \otimes B
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+ A \otimes |0\rangle\langle 0| \otimes I \otimes B\right) \\
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&= h_{k,A} \otimes h_{k,j,11} |1\rangle\langle 1| \otimes Xh_{k,i}X \otimes h_{k,B}\\
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&+ h_{k,A} \otimes h_{k,j,00}|0\rangle\langle 0| \otimes Ih_{k,i}I \otimes h_{k,B}\\
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&+ h_{k,A} \otimes h_{k,j,01}|0\rangle\langle 1| \otimes Ih_{k,i}X\otimes h_{k,B}\\
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&+ h_{k,A} \otimes h_{k,j,10}|1\rangle\langle 0| \otimes Xh_{k,i}I\otimes h_{k,B}\\
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\end{aligned}
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\end{equation}
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Searching for hermitians $h_1, h_2$ that fulfill
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\begin{equation}
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CX_{1,2} (h_1 \otimes h_2) CX_{1,2} = h_1' \otimes h_2'
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\end{equation}
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and
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\begin{equation}
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CX_{2,1} (h_1 \otimes h_2) CX_{2,1} = h_1'' \otimes h_2''
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\end{equation}
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might be a good step to find new classes of states that can be simulated
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efficiently using this method. This property has to be fulfilled by all
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elements of a group generated by such hermitian matrices. How computations and
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measurements would work using this method is not clear at the moment as many
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basic properties of the stabilizers are lost. One important property is that the
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stabilization: The simulated state is the $+1$ eigenstate of the stabilizers.
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This is another property that will have to be fulfilled by the hermitians as it
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is a key property used in \ref{ref:dynamics_stabilizer}. To ensure that the
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state is well defined one will have to demand that the eigenvalues fulfill
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$\lambda_1 = 1$ and $\lambda_2 < 1$.
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One should also note that the sabilizer states do not form a Hilbert (sub)space.
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Linear combinations of stabilizer states are (in general) no stabilizer states.
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The superposition principle is quite essential to many quantum algorithms and
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quantum physics which limits the use of the stabilizer formalism drastically.
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