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Diffstat (limited to 'Jauslin_2022.tex')
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1 files changed, 33 insertions, 73 deletions
diff --git a/Jauslin_2022.tex b/Jauslin_2022.tex index f200478..958cb08 100644 --- a/Jauslin_2022.tex +++ b/Jauslin_2022.tex @@ -255,8 +255,8 @@ We define the Fourier transform of the annihilation operators as \hat a_{k,\sigma}:=\frac1{\sqrt{|\Lambda|}}\sum_{x\in\Lambda}e^{ikx}a_{x,\sigma} ,\quad \hat b_{k,\sigma}:=\frac1{\sqrt{|\Lambda|}}\sum_{x\in\Lambda}e^{ikx}b_{x+\delta_1,\sigma} - . \end{equation} +where $|\Lambda|=L^2$. Note that, with this choice of normalization, $\hat a_{k,\sigma}$ and $\hat b_{k,\sigma}$ satisfy the canonical anticommutation relations: \begin{equation} \{a_{k,\sigma},a_{k',\sigma'}^\dagger\} @@ -276,7 +276,7 @@ We express $\mathcal H_0$ in terms of $\hat a$ and $\hat b$: \mathcal H_0=-\sum_{\sigma\in\{\uparrow,\downarrow\}}\sum_{ k\in\hat\Lambda}\hat A_{k,\sigma}^\dagger H_0(k)\hat A_{k,\sigma} \label{hamk} \end{equation} -where $|\Lambda|=L^2$, $\hat A_{k,\sigma}$ is a column vector whose transpose is $\hat A_{k,\sigma}^T=(\hat a_{k,\sigma},\hat{b}_{k,\sigma})$, +$\hat A_{k,\sigma}$ is a column vector whose transpose is $\hat A_{k,\sigma}^T=(\hat a_{k,\sigma},\hat{b}_{k,\sigma})$, \begin{equation} H_0( k):= \left(\begin{array}{*{2}{c}} @@ -440,7 +440,7 @@ The Gaussian Grassmann measure is specified by a {\it propagator}, which is a $2 \begin{largearray} P_{\hat g}(d\psi) := \left( \prod_{\mathbf k\in\mathcal B_{\beta,L}^*} - (\beta\det\hat g(\mathbf k))^4 + (\beta^2\det\hat g(\mathbf k))^2 \left(\prod_{\sigma\in\{\uparrow,\downarrow\}}\prod_{\alpha\in\{a,b\}}d\hat\psi_{\mathbf k,\alpha}^+d\hat\psi_{\mathbf k,\alpha}^-\right) \right) \cdot\\[0.5cm]\hfill\cdot @@ -512,7 +512,7 @@ Thus, we define the Gaussian Grassmann integration measure $P_{\leqslant M}(d\ps \end{equation} where $f_{0,\Lambda}$ is the free energy in the $U=0$ case and \begin{equation} - \mathcal V(\psi)=U\sum_{\alpha\in\{a,b\}}\frac1{|\Lambda|}\int_{0}^\beta dt \sum_{x\in \Lambda}\psi^+_{\mathbf x,\alpha,\uparrow}\psi^{-}_{\mathbf x,\alpha,\uparrow} + \mathcal V(\psi)=U\sum_{\alpha\in\{a,b\}}\int_{0}^\beta dt \sum_{x\in \Lambda}\psi^+_{\mathbf x,\alpha,\uparrow}\psi^{-}_{\mathbf x,\alpha,\uparrow} \psi^+_{\mathbf x,\alpha,\downarrow}\psi^{-}_{\mathbf x,\alpha,\downarrow} \label{V_grassmann} \end{equation} @@ -549,7 +549,7 @@ The idea is to approach the singularities $p_F^{(\omega)}$ slowly, by defining s \Phi_{h}(\mathbf k-\mathbf p_F^{(\omega)}):=(\chi_0(2^{-h}|\mathbf k-\mathbf p_F^{(\omega)}|)-\chi_0(2^{-h+1}|\mathbf k-\mathbf p_F^{(\omega)}|)) \label{fh} \end{equation} -which is a smooth function that is supported in $|\mathbf k-\mathbf p_F^{(\omega)}|\in[2^h\frac16,2^h\frac23]$, in other words, if localizes $\mathbf k$ to be at a distance from $\mathbf p_F^{(\omega)}$ that is of order $2^h$, see figure\-~\ref{fig:scale}. +which is a smooth function that is supported in $|\mathbf k-\mathbf p_F^{(\omega)}|\in[2^h\frac16,2^h\frac23]$, in other words, it localizes $\mathbf k$ to be at a distance from $\mathbf p_F^{(\omega)}$ that is of order $2^h$, see figure\-~\ref{fig:scale}. Since $|k_0|\geqslant\frac\pi\beta$, we only need to consider \begin{equation} h\geqslant -N_\beta:=\log_2\frac\pi\beta @@ -679,7 +679,6 @@ We will take the propagators to be \end{equation} \begin{equation} \int P^{[h]}(d\psi^{[h]})\ \psi_{b,\sigma}^{[h]-}(\Delta)\psi_{a,\sigma'}^{[h]+}(\Delta')=\delta_{\sigma,\sigma'}\delta_{\Delta,\Delta'} - . \end{equation} and all other propagators will be set to 0. We can now evaluate how well these propagators approximate the non-hierarchical ones. @@ -754,7 +753,7 @@ Because there are only four Grassmann fields and their conjugates per cell, $v_h In fact, by symmetry considerations, we find that $v_h$ must be of the form \begin{equation} v_h(\psi)= - \sum_{i=0}^6\alpha_i^{(h)}O_i(\psi) + \sum_{i=0}^6\ell_i^{(h)}O_i(\psi) \label{vh_rcc} \end{equation} with @@ -894,19 +893,19 @@ We expand the exponential and use\-~(\ref{vh_rcc}): \begin{largearray} \beta|\Lambda|c^{[h]} + - \sum_{i=0}^6\alpha_i^{(h-1)}O_i(\psi^{[\leqslant h-1]}(\bar\Delta)) + \sum_{i=0}^6\ell_i^{(h-1)}O_i(\psi^{[\leqslant h-1]}(\bar\Delta)) =\\\hfill= 2^{d+1}\log \int P(d\psi^{[h]}(\Delta)) \sum_{n=0}^\infty \frac1{n!} - \left(\sum_{i=0}^6\alpha_i^{(h)}O_i\left(\psi^{[h]}(\Delta)+2^{-\gamma}\psi^{[\leqslant h-1]}(\bar\Delta)\right)\right)^n + \left(\sum_{i=0}^6\ell_i^{(h)}O_i\left(\psi^{[h]}(\Delta)+2^{-\gamma}\psi^{[\leqslant h-1]}(\bar\Delta)\right)\right)^n . \end{largearray} \label{betadef} \end{equation} -The computation is thus reduced to computing the map $\alpha^{(h)}\mapsto\alpha^{(h-1)}$ using\-~(\ref{betadef}). -The coefficients $\alpha_i^{(h)}$ are called {\it running coupling constants}, and the map $\alpha^{(h)}\mapsto\alpha^{(h-1)}$ is called the {\it beta function} of the model. +The computation is thus reduced to computing the map $\ell^{(h)}\mapsto\ell^{(h-1)}$ using\-~(\ref{betadef}). +The coefficients $\ell_i^{(h)}$ are called {\it running coupling constants}, and the map $\ell^{(h)}\mapsto\ell^{(h-1)}$ is called the {\it beta function} of the model. The running coupling constants play a very important role, as they specify the effective potential on scale $h$, and thereby the physical properties of the system at distances $\sim2^{-h}$. \bigskip @@ -914,7 +913,7 @@ The running coupling constants play a very important role, as they specify the e Having defined the hierarchical model as we have, the infinite sum in\-~(\ref{betadef}) is actually finite ($n\leqslant 4$), so to compute the beta function, it suffices to compute Gaussian Grassmann integrals of a finite number of Grassmann monomials. A convenient way to carry out this computation is to represent each term graphically, using {\it Feynman diagrams}. First, let us expand the power $n$ and graphically represent the terms that must be integrated. -For each $n$, we have $n$ possible choices of $\alpha_iO_i$. +For each $n$, we have $n$ possible choices of $\ell_iO_i$. Now, $O_i$ can be quadratic in $\psi$ ($O_0$), quartic ($O_1$, $O_2$, $O_3$, $O_4$), sextic ($O_5$) or octic ($O_6$). We will represent $O_i$ by a vertex with the label $i$, from which two, four, six or eight edges emanate, depending on the degree of $O_i$. Each edge corresponds to a factor $\psi^{[h]}+2^{-\gamma}\psi^{[\leqslant h-1]}$. @@ -964,13 +963,13 @@ In other words, no integrating is taking place. Let us denote the number of external edges by $2l$, which can either be 2, 4, 6 or 8. The contribution of this graph is (keeping track of the $2^{d+1}$ factor in\-~(\ref{betadef})) \begin{equation} - 2^{d+1-2l\gamma}\alpha_i^{(h)} + 2^{d+1-2l\gamma}\ell_i^{(h)} . \end{equation} -Furthermore, this graph will contribute to the running coupling constant $\alpha_i$, and so, on scale $h-1$, we will have +Furthermore, this graph will contribute to the running coupling constant $\ell_i$, and so, on scale $h-1$, we will have \begin{equation} - \alpha_i^{(h-1)}= - 2^{d+1-2l\gamma}\alpha_i^{(h)} + \ell_i^{(h-1)}= + 2^{d+1-2l\gamma}\ell_i^{(h)} + \cdots \end{equation} @@ -1004,7 +1003,7 @@ For a more general treatment of power counting in Fermionic models with point-si \indent In the case of graphene, we have one relevant coupling: $O_0$, which is quadratic in the Grassmann fields. This is the only relevant coupling, and all others stay small. -However, since the relevant coupling is quadratic, it merely shifts the non-interacting system (whose Hamiltonian is quadratic in the Grassmann fields) to another system with a quadratic (that is non-interacting) Hamiltonian. +However, since the relevant coupling is quadratic, it merely shifts the non-interacting system (whose Hamiltonian is quadratic in the Grassmann fields) to another system with a quadratic (that is, non-interacting) Hamiltonian. Thus the relevant coupling does {\it not} imply that the interactions are preponderant, but rather that the interaction terms shifts the system from one non-interacting system to another. Since graphene only has one relevant coupling, and that one is quadratic, graphene is called {\it super-renormalizable}. \bigskip @@ -1013,20 +1012,20 @@ Since graphene only has one relevant coupling, and that one is quadratic, graphe As was mentioned above, the beta function can be computed {\it explicitly} for the hierarchical model, so the claims in the previous paragraph can be verified rather easily. The exact computation involves many terms, but it can be done easily using the {\tt meankondo} software package\-~\cite{mk}. The resulting beta function contains 888 terms, and will not be written out here. -A careful analysis of the beta function shows that there is an equilibrium point at $\alpha_i=0$ for $i=1,2,3,4,5,6$ and +A careful analysis of the beta function shows that there is an equilibrium point at $\ell_i=0$ for $i=1,2,3,4,5,6$ and \begin{equation} - \alpha_0\in\{0,1\} + \ell_0\in\{0,1\} . \end{equation} -The point with $\alpha_0=0$ is unstable, whereas $\alpha_0=1$ is stable. +The point with $\ell_0=0$ is unstable, whereas $\ell_0=1$ is stable. \bigskip \begin{figure} \hfil\includegraphics[width=12cm]{graphene_vector_field.pdf} \caption{ - The projection of the directional vector field of the beta function for hierarchical graphene onto the $(\alpha_0,\alpha_1)$ plane. + The projection of the directional vector field of the beta function for hierarchical graphene onto the $(\ell_0,\ell_1)$ plane. (Each arrow shows the direction of the vector field, the color corresponds to the logarithm of the amplitude, with red being larger and blue smaller.) - The stable equilibrium point at $\alpha_0=1$ and $\alpha_i=0$ is clearly visible. + The stable equilibrium point at $\ell_0=1$ and $\ell_i=0$ is clearly visible. } \label{fig:vector_field} \end{figure} @@ -1509,69 +1508,30 @@ Let us first prove a technical lemma. = e^{-t\lambda_j}a_j^\dagger\prod_{i\neq j} (1+(e^{-t\lambda_i}-1)a_i^\dagger a_i) - . - \label{fock2} \end{equation} - Similarly, + and since $(a_j^\dagger)^2=0$, \begin{equation} - e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}a_j - = - \left(\prod_{i=1}^n(1+(e^{-t\lambda_i}-1)a_i^\dagger a_i)\right)a_j - \end{equation} - and so, using $a_i^2=0$, we find - \begin{equation} - e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}a_j + e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}a_j^\dagger = - a_j - \prod_{i\neq j} + e^{-t\lambda_j}a_j^\dagger\prod_{i} (1+(e^{-t\lambda_i}-1)a_i^\dagger a_i) - . - \label{fock3} - \end{equation} - Furthermore, taking the $\dagger$ of\-~(\ref{fock3}), we find - \begin{equation} - a_j^\dagger e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i} - = - \left( - \prod_{i\neq j} - (1+(e^{-t\lambda_i}-1)a_i^\dagger a_i) - \right) - a_j^\dagger - \end{equation} - and since - \begin{equation} - (1+(e^{-t\lambda_j}-1)a_j^\dagger a_j)a_j^\dagger = e^{-t\lambda_j}a_j^\dagger - \end{equation} - we have - \begin{equation} - a_j^\dagger e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i} - = - e^{t\lambda_j}e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}a_j^\dagger + e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i} . - \label{fock2'} - \end{equation} - This implies the first of\-~(\ref{fock4}). - Taking the $\dagger$ of\-~(\ref{fock2}) yields - \begin{equation} - a_je^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i} - =e^{-t\lambda_j}\prod_{i\neq j} - (1+(e^{-t\lambda_i}-1)a_i^\dagger a_i)a_j - \end{equation} - and since - \begin{equation} - (1+(e^{-t\lambda_i}-1)a_j^\dagger a_k)a_j - =a_j + \label{fock2} \end{equation} - we have + Taking the $\dagger$ of\-~(\ref{fock2}), we find \begin{equation} a_je^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i} - =e^{-t\lambda_j}e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}a_j + = + e^{-t\lambda_j} + e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i} + a_j . - \label{fock3'} + \label{fock3} \end{equation} - This implies the second of\-~(\ref{fock4}). + Combining\-~(\ref{fock2}) and\-~(\ref{fock3}), we find\-~(\ref{fock4}). \qed \bigskip |