Hi, can you fix this equation? thank you before
In the Heisenberg model the spin flips can be implemented using masks, with ‘zeroes’ everywhere, except in the positions of the spins to be flipped. To illustrate this let us show the effect of the off-diagonal terms on one of the Néel configurations: $$\begin{eqnarray} S_{0}^{+}S_{1}^{-},+S_{0}^{-}S_{1}^{+},|0101\rangle &\equiv &(0011)\mathtt{% .XOR.}(0101)=3\mathtt{.XOR.}5=(0110)=6\equiv ,|0110\rangle , \ S_{1}^{+}S_{2}^{-}+S_{1}^{-}S_{2}^{+},,|0101\rangle &\equiv &(0110)\mathtt{% .XOR.}(0101)=6\mathtt{.XOR.}5=(0011)=3\equiv ,|0011\rangle , \ S_{2}^{+}S_{3}^{-}+S_{2}^{-}S_{3}^{+},,|0101\rangle &\equiv &(1100)\mathtt{% .XOR.}(0101)=12\mathtt{.XOR.}5=(1001)=9\equiv ,|1001\rangle , \ S_{3}^{+}S_{0}^{-},+S_{3}^{-}S_{0}^{+}|0101\rangle &\equiv &(1001)\mathtt{% .XOR.}(0101)=9\mathtt{.XOR.}5=(1100)=12\equiv ,|1100\rangle .% \end{eqnarray}$$
After applying the Hamiltonian ($$h4$$) on our basis states, the reader can verify as an exercise that we obtain $$\begin{eqnarray} {{\hat{H}}},|0101\rangle &=&-J,|0101\rangle +\frac{J}{2}\left[ ,|1100\rangle +,|1001\rangle +,|0011\rangle +,|0110\rangle \right] , \ {{\hat{H}}},|1010\rangle &=&-J,|1010\rangle +\frac{J}{2}\left[ ,|1100\rangle +,|1001\rangle +,|0011\rangle +,|0110\rangle \right] , \ {{\hat{H}}},|0011\rangle &=&\frac{J}{2}\left[ ,,|0101\rangle +,|1010\rangle \right] , \ {{\hat{H}}},|0110\rangle &=&\frac{J}{2}\left[ ,,|0101\rangle +,|1010\rangle \right] , \ {{\hat{H}}},|1001\rangle &=&\frac{J}{2}\left[ ,|0101\rangle +,|1010\rangle \right] , \ {{\hat{H}}},|1100\rangle &=&\frac{J}{2}\left[ ,|0101\rangle +,|1010\rangle \right] .% \end{eqnarray}$$
Hi, can you fix this equation? thank you before
In the Heisenberg model the spin flips can be implemented using masks, with ‘zeroes’ everywhere, except in the positions of the spins to be flipped. To illustrate this let us show the effect of the off-diagonal terms on one of the Néel configurations: $$\begin{eqnarray} S_{0}^{+}S_{1}^{-},+S_{0}^{-}S_{1}^{+},|0101\rangle &\equiv &(0011)\mathtt{% .XOR.}(0101)=3\mathtt{.XOR.}5=(0110)=6\equiv ,|0110\rangle , \ S_{1}^{+}S_{2}^{-}+S_{1}^{-}S_{2}^{+},,|0101\rangle &\equiv &(0110)\mathtt{% .XOR.}(0101)=6\mathtt{.XOR.}5=(0011)=3\equiv ,|0011\rangle , \ S_{2}^{+}S_{3}^{-}+S_{2}^{-}S_{3}^{+},,|0101\rangle &\equiv &(1100)\mathtt{% .XOR.}(0101)=12\mathtt{.XOR.}5=(1001)=9\equiv ,|1001\rangle , \ S_{3}^{+}S_{0}^{-},+S_{3}^{-}S_{0}^{+}|0101\rangle &\equiv &(1001)\mathtt{% .XOR.}(0101)=9\mathtt{.XOR.}5=(1100)=12\equiv ,|1100\rangle .% \end{eqnarray}$$
After applying the Hamiltonian ($$h4$$ ) on our basis states, the reader can verify as an exercise that we obtain $$\begin{eqnarray} {{\hat{H}}},|0101\rangle &=&-J,|0101\rangle +\frac{J}{2}\left[ ,|1100\rangle +,|1001\rangle +,|0011\rangle +,|0110\rangle \right] , \ {{\hat{H}}},|1010\rangle &=&-J,|1010\rangle +\frac{J}{2}\left[ ,|1100\rangle +,|1001\rangle +,|0011\rangle +,|0110\rangle \right] , \ {{\hat{H}}},|0011\rangle &=&\frac{J}{2}\left[ ,,|0101\rangle +,|1010\rangle \right] , \ {{\hat{H}}},|0110\rangle &=&\frac{J}{2}\left[ ,,|0101\rangle +,|1010\rangle \right] , \ {{\hat{H}}},|1001\rangle &=&\frac{J}{2}\left[ ,|0101\rangle +,|1010\rangle \right] , \ {{\hat{H}}},|1100\rangle &=&\frac{J}{2}\left[ ,|0101\rangle +,|1010\rangle \right] .% \end{eqnarray}$$