- reset_
ij BEGIN ii
KeyArray get_byte jj +
j_update swap_
s_
ij ii 255 <
WHILE ii
i_update REPEAT reset_
ij ; : rc4_byte ii
i_update jj
j_update swap_
s_
ij ii...
- {1}{2}}(w_{
ij}(1,0)-w_{
ij}(0,0))\\k_{j}&={\frac {1}{2}}(w_{
ij}(1,1)-w_{
ij}(1,0))\\k_{
ij}&={\frac {1}{2}}(w_{
ij}(0,1)+w_{
ij}(1,0)-w_{
ij}(0,0)-w_{
ij}(1,1))...
- k {\displaystyle \varkappa _{
ij}(k)=k} . Concluding,
permutations τ i ∈
S k −
S k − 1 {\displaystyle \tau _{i}\in
S_{k}-
S_{k-1}} are all representatives...
- C=\sum {
ij}\vert ii\rangle \langle jj\vert \otimes C{
ij}} for C i j := E † ( | i i ⟩ ⟨ j j | ) = ∑ i C i U i † U j C j {\displaystyle C_{
ij}:=E^{\dagger...
- ^{k}}_{
ij}{\frac {\partial {\vec {\Psi }}}{\partial x^{k}}}+{\vec {n}},{\frac {\partial {\vec {\Psi }}}{\partial x^{l}}}\right\rangle ={\Gamma ^{k}}_{
ij}\left\langle...
-
S_{ii}-2\,sc\,
S_{
ij}+
s^{2}\,
S_{jj}\\
S'_{jj}&=
s^{2}\,
S_{ii}+2sc\,
S_{
ij}+c^{2}\,
S_{jj}\\
S'_{
ij}&=
S'_{ji}=(c^{2}-
s^{2})\,
S_{
ij}+sc\,(
S_{ii}-
S_{jj})\\
S'_{ik}&=S'_{ki}=c\...
- \{{\hat {a}}_{i},{\hat {a}}_{j}^{\dagger }\}=\delta _{
ij}}
where δ i j {\displaystyle \delta _{
ij}}
denotes the
Kronecker delta. We then have a ^ i ∙ a...
- {\displaystyle \rho =\sum _{i}w_{i}\left[\sum _{j}{\bar {c}}_{
ij}(|\alpha _{
ij}\rangle \otimes |\beta _{
ij}\rangle )\right]\left[\sum _{k}c_{ik}(\langle \alpha...
-
braid groups where R {\displaystyle R}
corresponds to
swapping two strands.
Since one can
swap three strands in two
different ways, the Yang–Baxter equation...
- {\displaystyle *[F,G]^{
IJ}=-*[G,F]^{
IJ}=-[G,*F]^{
IJ}=[*F,G]^{
IJ}.}
where in the
first step we have used the anti-symmetry of the Lie
bracket to
swap F {\displaystyle...
