Change k_1,k_2 to k_x,k_y in bilayer graphene
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Ian Jauslin
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@ -945,7 +945,7 @@ is a symmetry.\par
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and every complex coefficient of $h_0$ and $\mathcal V$ is mapped to its complex conjugate is a symmetry.
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\bigskip
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\point{\bf Vertical reflection.} Let $R_v\mathbf k=(k_0,k_1,-k_2)$,
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\point{\bf Vertical reflection.} Let $R_v\mathbf k=(k_0,k_x,-k_y)$,
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\begin{equation}
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\left\{\begin{array}l
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\hat\xi_{\mathbf k}^\pm\longmapsto \hat\xi_{R_v\mathbf k}^\pm\\[0.2cm]
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@ -954,7 +954,7 @@ and every complex coefficient of $h_0$ and $\mathcal V$ is mapped to its complex
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is a symmetry.
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\bigskip
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\point{\bf Horizontal reflection.} Let $R_h\mathbf k=(k_0,-k_1,k_2)$,
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\point{\bf Horizontal reflection.} Let $R_h\mathbf k=(k_0,-k_x,k_y)$,
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\begin{equation}\left\{\begin{array}l
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\hat\xi_{\mathbf k}^-\longmapsto \sigma_1\hat\xi_{R_h\mathbf k}^-,\ \hat\xi_{\mathbf k}^+\longmapsto\hat\xi_{R_h\mathbf k}^+\sigma_1\\[0.2cm]
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\hat\phi_{\mathbf k}^-\longmapsto \sigma_1\hat\phi_{R_h\mathbf k}^-,\ \hat\phi_{\mathbf k}^+\longmapsto\hat\phi_{R_h\mathbf k}^+\sigma_1
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@ -962,7 +962,7 @@ is a symmetry.
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is a symmetry.\par
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\bigskip
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\point{\bf Parity.} Let $P\mathbf k=(k_0,-k_1,-k_2)$,
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\point{\bf Parity.} Let $P\mathbf k=(k_0,-k_x,-k_y)$,
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\begin{equation}\left\{\begin{array}l
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\hat\xi_{\mathbf k}^\pm\longmapsto i(\hat\xi_{P\mathbf k}^{\mp})^T\\[0.2cm]
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\hat\phi_{\mathbf k}^\pm\longmapsto i(\hat\phi_{P\mathbf k}^{\mp})^T
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@ -970,7 +970,7 @@ is a symmetry.\par
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is a symmetry.
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\bigskip
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\point{\bf Time inversion.} Let $I\mathbf k=(-k_0,k_1,k_2)$, the mapping
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\point{\bf Time inversion.} Let $I\mathbf k=(-k_0,k_x,k_y)$, the mapping
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\begin{equation}\left\{\begin{array}l
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\hat\xi_{\mathbf k}^-\longmapsto -\sigma_3\hat\xi_{I\mathbf k}^-,\ \hat\xi_{\mathbf k}^+\longmapsto\hat\xi_{I\mathbf k}^+\sigma_3\\[0.2cm]
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\hat\phi_{\mathbf k}^-\longmapsto-\sigma_3\hat\phi_{I\mathbf k}^-,\ \hat\phi_{\mathbf k}^+\longmapsto\hat\phi_{I\mathbf k}^+\sigma_3
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@ -1543,7 +1543,7 @@ good enough for performing the localization and renormalization procedure descri
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\begin{equation}
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|2^{hm_0}\partial_{k_0}^{m_0}\partial_k^{m_k}\hat g_{h}(\mathbf k)|\leqslant (\mathrm{const.})\ 2^{-h}
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\label{bilayerestgkuv}\end{equation}
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in which $\partial_{k_0}$ denotes the discrete derivative with respect to $k_0$ and, with a slightly abusive notation, $\partial_k$ the discrete derivative with respect to either $k_1$ or $k_2$. Indeed the derivatives over $k$ land on $ik_0\hat A^{-1}$, which does not change the previous estimate, and the derivatives over $k_0$ either land on $f_h$, $1/(ik_0)$, or $ik_0\hat A^{-1}$, which yields an extra $2^{-h}$ in the estimate.\par
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in which $\partial_{k_0}$ denotes the discrete derivative with respect to $k_0$ and, with a slightly abusive notation, $\partial_k$ the discrete derivative with respect to either $k_x$ or $k_y$. Indeed the derivatives over $k$ land on $ik_0\hat A^{-1}$, which does not change the previous estimate, and the derivatives over $k_0$ either land on $f_h$, $1/(ik_0)$, or $ik_0\hat A^{-1}$, which yields an extra $2^{-h}$ in the estimate.\par
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\medskip
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{\bf Remark}: The previous argument implicitly uses the Leibnitz rule, which must be used carefully since the derivatives are discrete. However, since the estimate is purely dimensional,
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@ -1579,7 +1579,7 @@ and for $m\leqslant7$,
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\begin{equation}
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|2^{mh}\partial_{\mathbf k}^m\hat g_{h,\omega}(\mathbf k)|\leqslant (\mathrm{const.})\ 2^{-h}
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\label{bilayerestgko}\end{equation}
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in which we again used the slightly abusive notation of writing $\partial_{\mathbf k}$ to mean any derivative with respect to $k_0$, $k_1$ or $k_2$. Equation~(\ref{bilayerestgko}) then follows from similar considerations as those in the ultraviolet regime.\par
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in which we again used the slightly abusive notation of writing $\partial_{\mathbf k}$ to mean any derivative with respect to $k_0$, $k_x$ or $k_y$. Equation~(\ref{bilayerestgko}) then follows from similar considerations as those in the ultraviolet regime.\par
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\bigskip
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\subpoint{\bf Configuration space bounds.} We estimate the real-space counterpart of $\hat g_{h,\omega}$,
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@ -1844,7 +1844,7 @@ scaling dimension {\it relevant}.
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\item We will show that in the first and third regimes $c_k=3$ and $c_g=1$, so that the scaling dimension is $3-|P_v|$. Therefore, the nodes with $|P_v|=2$ are relevant whereas all the others are irrelevant. In the second regime,
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$c_k=2$ and $c_g=1$, so that the scaling dimension is $2-|P_v|/2$. Therefore, the nodes with $|P_v|=2$ are relevant, those with $|P_v|=4$ are marginal, and all other nodes are irrelevant.
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\item The purpose of the factor $\mathfrak F_h(\underline m)$ is to take into account the dependence of the order of magnitude of the different components $k_0$, $k_1$ and $k_2$ in the different regimes. In other words, as was shown in~(\ref{bilayerestguv}), (\ref{bilayerestgo}), (\ref{bilayerestgt}) and~(\ref{bilayerestgth}), the effect of multiplying $g$ by $x_{j,i}$ depends on $i$, which is a fact the lemma must take into account.
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\item The purpose of the factor $\mathfrak F_h(\underline m)$ is to take into account the dependence of the order of magnitude of the different components $k_0$, $k_x$ and $k_y$ in the different regimes. In other words, as was shown in~(\ref{bilayerestguv}), (\ref{bilayerestgo}), (\ref{bilayerestgt}) and~(\ref{bilayerestgth}), the effect of multiplying $g$ by $x_{j,i}$ depends on $i$, which is a fact the lemma must take into account.
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\item The reason why we have stated this bound in $\mathbf x$-space is because of the estimate of $\det(G^{(h_v,T_v)})$ detailed below, which is very inefficient in $\mathbf k$-space.
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\end{itemize}
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@ -2058,7 +2058,7 @@ one would find a {\it logarithmic} divergence for $\int d\mathbf x|K_{2,(\alpha,
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the dominant terms in $\hat g_{h}(\mathbf k)$ are odd in $k_0$, so they cancel when considering
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$$\sum_{k_0\in\frac{2\pi}\beta(\mathbb Z+\frac12)}\hat g_{h}(\mathbf k).$$
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From this idea, we compute an improved bound for $|g_{h}(\mathbf x)|$ with $x_0=0$:
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$$|g_{h}(0,x_1,x_2)|\leqslant \sum_{k_1,k_2}\left|\sum_{k_0}\hat g_{h}(\mathbf k)\right|\leqslant (\mathrm{const.})\ 2^{-h}.$$
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$$|g_{h}(0,x,y)|\leqslant \sum_{k_x,k_y}\left|\sum_{k_0}\hat g_{h}(\mathbf k)\right|\leqslant (\mathrm{const.})\ 2^{-h}.$$
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All in all, we find
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\begin{equation}
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\int d\mathbf x\ |\mathbf x^mK^{(h)}_{2,(\alpha,\alpha')}(\mathbf x)|\leqslant\mathfrak C_4|U|,\quad
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@ -2288,7 +2288,7 @@ for $h'>h$, which we do not have (and if we tried to prove it by induction, we w
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\begin{equation}
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\mathcal L:\bar A_{h,\omega}(\mathbf x)\longmapsto\delta(\mathbf x)\int d\mathbf y\ \bar A_{h,\omega}(\mathbf y)-\partial_\mathbf x\delta(\mathbf x)\cdot\int d\mathbf y\ \mathbf y \bar A_{h,\omega}(\mathbf y)
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\label{bilayerWreldefo}\end{equation}
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where $\delta(\mathbf x):=\delta(x_0)\delta_{x_1,0}\delta_{x_2,0}$ and in the second term, as usual, the derivative with respect to $x_1$ and $x_2$ is discrete; as well as
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where $\delta(x_0,x_1,x_2):=\delta(x_0)\delta_{x_1,0}\delta_{x_2,0}$ and in the second term, as usual, the derivative with respect to $x_1$ and $x_2$ is discrete; as well as
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the corresponding {\it irrelevator}:
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\begin{equation}
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\mathcal R:=\mathds1-\mathcal L.
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@ -3895,10 +3895,10 @@ from which the invariance of $h_0$ follows immediately. The invariance of $\math
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\indent We recall the definitions of the symmetry transformations from section~\ref{bilayersymsec}:
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\begin{equation}\begin{array}c
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T\mathbf k:=(k_0,e^{-i\frac{2\pi}3\sigma_2}k),\quad
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R_v\mathbf k:=(k_0,k_1,-k_2),\quad
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R_h\mathbf k:=(k_0,-k_1,k_2),\\[0.2cm]
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P\mathbf k:=(k_0,-k_1,-k_2),\quad
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I\mathbf k:=(-k_0,k_1,k_2).
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R_v\mathbf k:=(k_0,k_x,-k_y),\quad
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R_h\mathbf k:=(k_0,-k_x,k_y),\\[0.2cm]
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P\mathbf k:=(k_0,-k_x,-k_y),\quad
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I\mathbf k:=(-k_0,k_x,k_y).
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\end{array}\label{bilayersymdefs}\end{equation}
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Furthermore, given a $4\times4$ matrix $\mathbf M$ whose components are indexed by $\{a,\tilde b,\tilde a,b\}$, we denote the sub-matrix with components in $\{a,\tilde b\}^2$ by $\mathbf M^{\xi\xi}$, that with $\{\tilde a,b\}^2$ by $\mathbf M^{\phi\phi}$, with $\{a,\tilde b\}\times\{\tilde a,b\}$ by $\mathbf M^{\xi\phi}$ and with $\{\tilde a,b\}\times\{a,\tilde b\}$ by $\mathbf M^{\phi\xi}$. In addition, given a complex matrix $M$, we denote its component-wise complex conjugate by $M^*$ (which is not to be confused with its adjoint $M^\dagger$).\par
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\bigskip
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@ -3916,8 +3916,8 @@ for $\omega\in\{-,+\}$, then $\exists(\mu,\zeta,\nu,\varpi)\in\mathbb{R}^4$ such
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\begin{equation}\begin{array}c
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M(\mathbf p_{F}^\omega)=\mu\sigma_1,\quad
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\partial_{k_0} M(\mathbf p_{F}^\omega)=i\zeta\mathds1,\\[0.5cm]
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\partial_{k_1} M(\mathbf p_{F}^\omega)=\nu\sigma_2,\quad
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\partial_{k_2} M(\mathbf p_{F}^\omega)=\omega \varpi\sigma_1.
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\partial_{k_x} M(\mathbf p_{F}^\omega)=\nu\sigma_2,\quad
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\partial_{k_y} M(\mathbf p_{F}^\omega)=\omega \varpi\sigma_1.
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\end{array}\label{bilayerApfzt}\end{equation}
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\endtheo
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\bigskip
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@ -3943,21 +3943,21 @@ $$
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Therefore $(t_0,x_0,y_0,z_0)$ are independent of $\omega$, $x_0=y_0=z_0=0$ and $t_0\in i\mathbb{R}$.\par
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\bigskip
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\point We now turn our attention to $\partial_{k_1}M$:
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$$\partial_{k_1} M(\mathbf p_{F}^\omega)=:t_1\mathds1+x_1\sigma_1+y_1\sigma_2+z_1\sigma_3.$$
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\point We now turn our attention to $\partial_{k_x}M$:
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$$\partial_{k_x} M(\mathbf p_{F}^\omega)=:t_1\mathds1+x_1\sigma_1+y_1\sigma_2+z_1\sigma_3.$$
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We have
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$$
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\partial_{k_1} M(\mathbf p_{F}^\omega)=-(\partial_{k_1} M(\mathbf p_{F}^{-\omega}))^*=\partial_{k_1} M(\mathbf p_{F}^{-\omega})=-\sigma_1\partial_{k_1} M(\mathbf p_{F}^\omega)\sigma_1
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=-\sigma_3\partial_{k_1} M(\mathbf p_{F}^{\omega})\sigma_3.
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\partial_{k_x} M(\mathbf p_{F}^\omega)=-(\partial_{k_x} M(\mathbf p_{F}^{-\omega}))^*=\partial_{k_x} M(\mathbf p_{F}^{-\omega})=-\sigma_1\partial_{k_x} M(\mathbf p_{F}^\omega)\sigma_1
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=-\sigma_3\partial_{k_x} M(\mathbf p_{F}^{\omega})\sigma_3.
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$$
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Therefore $(t_1,x_1,y_1,z_1)$ are independent of $\omega$, $t_1=x_1=z_1=0$ and $y_1\in\mathbb{R}$.\par
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\bigskip
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\point Finally, we consider $\partial_{k_y}M$:
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$$\partial_{k_2} M(\mathbf p_{F}^\omega)=:t_2^{(\omega)}\mathds1+x_2^{(\omega)}\sigma_1+y_2^{(\omega)}\sigma_2+z_2^{(\omega)}\sigma_3.$$
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$$\partial_{k_y} M(\mathbf p_{F}^\omega)=:t_2^{(\omega)}\mathds1+x_2^{(\omega)}\sigma_1+y_2^{(\omega)}\sigma_2+z_2^{(\omega)}\sigma_3.$$
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We have
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$$
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\partial_{k_2} M(\mathbf p_{F}^\omega)=-(\partial_{k_2} M(\mathbf p_{F}^{-\omega}))^*=-\partial_{k_2} M(\mathbf p_{F}^{-\omega})=\sigma_1\partial_{k_2} M(\mathbf p_{F}^\omega)\sigma_1=-\sigma_3\partial_{k_2} M(\mathbf p_{F}^\omega)\sigma_3.
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\partial_{k_y} M(\mathbf p_{F}^\omega)=-(\partial_{k_y} M(\mathbf p_{F}^{-\omega}))^*=-\partial_{k_y} M(\mathbf p_{F}^{-\omega})=\sigma_1\partial_{k_y} M(\mathbf p_{F}^\omega)\sigma_1=-\sigma_3\partial_{k_y} M(\mathbf p_{F}^\omega)\sigma_3.
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$$
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Therefore $t_2^{(\omega)}=y_2^{(\omega)}=z_2^{(\omega)}=0$, $x_2^{(\omega)}=-x_2^{(-\omega)}\in\mathbb{R}$.\penalty10000\hfill\penalty10000$\square$\par
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@ -3967,8 +3967,8 @@ Given a $4\times4$ complex matrix $\mathbf M(\mathbf k)$ parametrized by $\mathb
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\begin{equation}\begin{array}c
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\mathbf M^{ff'}(\mathbf p_{F}^\omega)=\mu^{ff'}\sigma_1,\quad
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\partial_{k_0}\mathbf M^{ff'}(\mathbf p_{F}^\omega)=i\zeta^{ff'}\mathds1,\\[0.5cm]
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\partial_{k_1}\mathbf M^{ff'}(\mathbf p_{F}^\omega)=\nu^{ff'}\sigma_2,\quad
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\partial_{k_2}\mathbf M^{ff'}(\mathbf p_{F}^\omega)=\omega\varpi^{ff'}\sigma_1
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\partial_{k_x}\mathbf M^{ff'}(\mathbf p_{F}^\omega)=\nu^{ff'}\sigma_2,\quad
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\partial_{k_y}\mathbf M^{ff'}(\mathbf p_{F}^\omega)=\omega\varpi^{ff'}\sigma_1
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\end{array}\label{bilayerMsymhyprot}\end{equation}
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with $(\mu^{ff'}, \zeta^{ff'}, \nu^{ff'}, \varpi^{ff'})\in\mathbb R^{4}$ independent of $\omega$,
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and $\forall\mathbf k\in\mathcal B_\infty$
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@ -4016,9 +4016,9 @@ Evaluating this formula at ${\bf k}=\mathbf p_F^\omega$, recalling that $\mathbf
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$$\partial_{k_i}\mathbf M^{\phi\phi}(\mathbf p_{F}^\omega)=\sum_{j=1}^2T_{i,j}\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_j}\mathbf M^{\phi\phi}(\mathbf p_{F}^\omega)\mathcal T_{\mathbf p_{F}^\omega}$$
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with
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$$T=\frac{1}{2}\left(\begin{array}{*{2}{c}}-1&-\sqrt3\\[0.2cm]\sqrt3&-1\end{array}\right).$$
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Furthermore, recalling that $\partial_{k_1}\mathbf M^{\phi\phi}=\nu^{\phi\phi}\sigma_2$ and $\partial_{k_2}\mathbf M^{\phi\phi}=\omega\varpi^{\phi\phi}\sigma_1$,
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$$\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_1}\mathbf M^{\phi\phi}\mathcal T_{\mathbf p_{F}^\omega}=\nu^{\phi\phi}\Big(-\frac12\sigma_2-\omega\frac{\sqrt3}{2}\sigma_1\Big), \quad
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\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_2}\mathbf M^{\phi\phi}\mathcal T_{\mathbf p_{F}^\omega}=\omega\varpi^{\phi\phi}\Big(-\frac12\sigma_1+\omega\frac{\sqrt3}{2}\sigma_2\Big),$$
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Furthermore, recalling that $\partial_{k_x}\mathbf M^{\phi\phi}=\nu^{\phi\phi}\sigma_2$ and $\partial_{k_y}\mathbf M^{\phi\phi}=\omega\varpi^{\phi\phi}\sigma_1$,
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$$\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_x}\mathbf M^{\phi\phi}\mathcal T_{\mathbf p_{F}^\omega}=\nu^{\phi\phi}\Big(-\frac12\sigma_2-\omega\frac{\sqrt3}{2}\sigma_1\Big), \quad
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\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_y}\mathbf M^{\phi\phi}\mathcal T_{\mathbf p_{F}^\omega}=\omega\varpi^{\phi\phi}\Big(-\frac12\sigma_1+\omega\frac{\sqrt3}{2}\sigma_2\Big),$$
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which implies
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$$\left(\begin{array}{c}\nu^{\phi\phi}\sigma_2\\[0.2cm] \omega \varpi^{\phi\phi}\sigma_1\end{array}\right)=\frac{1}{4}\left(\begin{array}{*{2}{c}} \nu^{\phi\phi}-3 \varpi^{\phi\phi}&\omega\sqrt3(\nu^{\phi\phi}+\varpi^{\phi\phi})\\[0.2cm]-\sqrt3(\nu^{\phi\phi}+\varpi^{\phi\phi})&\omega(\varpi^{\phi\phi}-3\nu^{\phi\phi})\end{array}\right)\left(\begin{array}{c}\sigma_2\\[0.2cm]\sigma_1\end{array}\right)$$
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so $\nu^{\phi\phi}=-\varpi^{\phi\phi}$.
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@ -4031,10 +4031,10 @@ $$
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Evaluating this formula and its derivative with respect to $k_0$ at ${\bf k}=\mathbf p_F^\omega$, we obtain
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$\mu^{\phi\xi}=\zeta^{\phi\xi}=0$.
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Evaluating the derivative of this formula with respect to $k_i$ at ${\bf k}=\mathbf p_F^\omega$, we obtain
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$$\partial_{k_i}\mathbf M^{\phi\xi}(\mathbf p_{F}^\omega)=\sum_{j=1}^2T_{i,j}\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_j}\mathbf M^{\phi\xi}(\mathbf p_{F}^\omega).$$
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Furthermore,
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$$\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_1}\mathbf M^{\phi\xi}=\nu^{\phi\xi}\Big(-\frac12\sigma_2+\omega\frac{\sqrt3}{2}\sigma_1\Big), \quad
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\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_2}\mathbf M^{\phi\xi}=\omega\varpi^{\phi\xi}\Big(-\frac12\sigma_1-\omega\frac{\sqrt3}{2}\sigma_2\Big),$$
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$$\partial_{k_i}\mathbf M^{\phi\xi}(\mathbf p_{F}^\omega)=\sum_{j=1}^2T_{i,j}\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_j}\mathbf M^{\phi\xi}(\mathbf p_{F}^\omega)$$
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where we used the following notation $k_1\equiv k_x$ and $k_2\equiv k_y$. Furthermore,
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$$\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_x}\mathbf M^{\phi\xi}=\nu^{\phi\xi}\Big(-\frac12\sigma_2+\omega\frac{\sqrt3}{2}\sigma_1\Big), \quad
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\mathcal T_{\mathbf p_{F}^\omega}^\dagger\partial_{k_y}\mathbf M^{\phi\xi}=\omega\varpi^{\phi\xi}\Big(-\frac12\sigma_1-\omega\frac{\sqrt3}{2}\sigma_2\Big),$$
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which implies
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$$\left(\begin{array}{c}\nu^{\phi\xi}\sigma_2\\[0.2cm] \omega \varpi^{\phi\xi}\sigma_1\end{array}\right)=\frac{1}{4}\left(\begin{array}{*{2}{c}} \nu^{\phi\xi}+3 \varpi^{\phi\xi}&-\omega\sqrt3(\nu^{\phi\xi}-\varpi^{\phi\xi})\\[0.2cm]-\sqrt3(\nu^{\phi\xi}-\varpi^{\phi\xi})&\omega(\varpi^{\phi\xi}+3\nu^{\phi\xi})\end{array}\right)\left(\begin{array}{c}\sigma_2\\[0.2cm]\sigma_1\end{array}\right)$$
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so that $\nu^{\phi\xi}_h=\varpi^{\phi\xi}_h$.
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@ -4050,7 +4050,7 @@ Therefore for $i\in\{1,2\}$,
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$$
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\partial_{k_i}\mathbf M^{\xi\xi}(\mathbf p_{F}^\omega)=\sum_{j=1}^2T_{i,j}\partial_{k_j}\mathbf M^{\xi\xi}(\mathbf p_{F}^\omega)
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$$
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so that
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where we used the following notation $k_1\equiv k_x$ and $k_2\equiv k_y$, so that
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$\partial_{k_i}\mathbf M^{\xi\xi}(\mathbf p_{F}^\omega)=0$, that is $\nu^{\xi\xi}=\varpi^{\xi\xi}=0$.\penalty10000\hfill\penalty10000$\square$\par
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\vfill
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