Add direction for magnetic field

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Ian Jauslin 2015-06-23 17:10:49 +00:00
parent 0334c43102
commit 276335e730

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@ -66,10 +66,11 @@ V_h=& -h \, \sum_{\alpha\in\uparrow,\downarrow}d^+_\alpha\sigma^3_{\alpha,\alpha
\label{e1.1}\end{array}\end{equation} \label{e1.1}\end{array}\end{equation}
where $\lambda_0,h$ are the interaction and magnetic field strengths and where $\lambda_0,h$ are the interaction and magnetic field strengths and
\begin{enumerate}[\ \ (1)\ \ ] \begin{enumerate}[\ \ (1)\ \ ]
\item$c_\alpha^\pm(x),d^\pm_\alpha, \,\alpha=\uparrow,\downarrow$ are creation and annihilation operators corresponding respectively to electrons and the impurity \item$c_\alpha^\pm(x),d^\pm_\alpha, \,\alpha=\uparrow,\downarrow$ are creation and annihilation operators corresponding respectively to electrons and the impurity,
\item$\sigma^j,\,j=1,2,3$, are the Pauli matrices \item$\sigma^j,\,j=1,2,3$, are the Pauli matrices,
\item$x$ is on the unit lattice and $-{L}/2$, ${L}/2$ are identified (periodic boundary) \item$x$ is on the unit lattice and $-{L}/2$, ${L}/2$ are identified (periodic boundary),
\item$\Delta f(x)= f(x+1)-2f(x)+f(x-1)$ is the discrete Laplacian. \item$\Delta f(x)= f(x+1)-2f(x)+f(x-1)$ is the discrete Laplacian,
\item$\bm\omega=(\bm\omega_1,\bm\omega_2,\bm\omega_3)$ is the direction of the field, which is a norm-1 vector.
\end{enumerate} \end{enumerate}
\medskip \medskip
@ -167,7 +168,7 @@ If $\beta,L$ are finite, $\int\,\frac{dk_0 dk}{(2\pi)^2}$ in Eq.(\ref{e2.5}) has
\begin{equation}\begin{array}{r@{\ }>{\displaystyle}l} \begin{equation}\begin{array}{r@{\ }>{\displaystyle}l}
V(\psi,\varphi)=& V(\psi,\varphi)=&
-\lambda_0 \sum_{{j\in\{1,2,3\}}\atop{\alpha_1,\alpha'_1,\alpha_2,\alpha_2'}}\int dt \,(\psi^+_{\alpha_1}(0,t)\sigma^j_{\alpha_1,\alpha'_1} \psi^-_{\alpha'_1}(0,t)) (\varphi^+_{\alpha_2}(t)\sigma^j_{\alpha_2,\alpha_2'} \varphi^-_{\alpha_2'}(t))\\ -\lambda_0 \sum_{{j\in\{1,2,3\}}\atop{\alpha_1,\alpha'_1,\alpha_2,\alpha_2'}}\int dt \,(\psi^+_{\alpha_1}(0,t)\sigma^j_{\alpha_1,\alpha'_1} \psi^-_{\alpha'_1}(0,t)) (\varphi^+_{\alpha_2}(t)\sigma^j_{\alpha_2,\alpha_2'} \varphi^-_{\alpha_2'}(t))\\
&-h \int dt\sum_{\alpha\in\uparrow,\downarrow}\varphi^+_{\alpha}(t)\sigma^3_{\alpha,\alpha} \varphi^-_{\alpha}(t)\\ &-h \sum_{j\in\{1,2,3\}}\bm\omega_j \int dt\sum_{\alpha\in\uparrow,\downarrow}\varphi^+_{\alpha}(t)\sigma^j_{\alpha,\alpha} \varphi^-_{\alpha}(t)
\end{array}\label{e2.6}\end{equation} \end{array}\label{e2.6}\end{equation}
Notice that $V$ only depends on the fields located at the site $x=0$. This is important because it will allow us to reduce the problem to a 1-dimensional one [\cite{aySN}, \cite{ayhSeZ}]. Notice that $V$ only depends on the fields located at the site $x=0$. This is important because it will allow us to reduce the problem to a 1-dimensional one [\cite{aySN}, \cite{ayhSeZ}].
@ -314,7 +315,7 @@ from which we see that the hierarchical model boils down to neglecting the $m'$
\begin{equation}\begin{array}{r@{\ }>{\displaystyle}l} \begin{equation}\begin{array}{r@{\ }>{\displaystyle}l}
V(\psi,\varphi)=& V(\psi,\varphi)=&
-\lambda_0 \sum_{{j\in\{1,2,3\}}\atop{\alpha_1,\alpha'_1,\alpha_2,\alpha_2'}}\int dt \,(\psi^+_{\alpha_1}(0,t)\sigma^j_{\alpha_1,\alpha'_1} \psi^-_{\alpha'_1}(0,t)) (\varphi^+_{\alpha_2}(t)\sigma^j_{\alpha_2,\alpha_2'} \varphi^-_{\alpha_2'}(t))\\ -\lambda_0 \sum_{{j\in\{1,2,3\}}\atop{\alpha_1,\alpha'_1,\alpha_2,\alpha_2'}}\int dt \,(\psi^+_{\alpha_1}(0,t)\sigma^j_{\alpha_1,\alpha'_1} \psi^-_{\alpha'_1}(0,t)) (\varphi^+_{\alpha_2}(t)\sigma^j_{\alpha_2,\alpha_2'} \varphi^-_{\alpha_2'}(t))\\
&-h \int dt\sum_{\alpha\in\uparrow,\downarrow}\varphi^+_{\alpha}(t)\sigma^3_{\alpha,\alpha} \varphi^-_{\alpha}(t)\\ &-h \sum_{j\in\{1,2,3\}}\bm\omega_j \int dt\sum_{\alpha\in\uparrow,\downarrow}\varphi^+_{\alpha}(t)\sigma^j_{\alpha,\alpha} \varphi^-_{\alpha}(t)
\end{array}\label{e3.9}\end{equation} \end{array}\label{e3.9}\end{equation}
in which $\psi^\pm_\alpha(0,t)$ and $\varphi^\pm_\alpha(t)$ are now defined in Eq.(\ref{e3.5}). in which $\psi^\pm_\alpha(0,t)$ and $\varphi^\pm_\alpha(t)$ are now defined in Eq.(\ref{e3.5}).
\medskip \medskip