Update to v1.1:

Add: Detailed discussion of the use of the Trotter product formula in the
       Dyson expansion.

  Minor fixes

Changelog

@@ -0,0 +1,6 @@
+v1.1:
+
+  * Add: Detailed discussion of the use of the Trotter product formula in the
+         Dyson expansion.
+  
+  * Minor fixes

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}}
@@ -336,7 +336,7 @@ We now define the interaction Hamiltonian which we take to be of {\it Hubbard} f
 \label{hamintx}\end{equation}
 where the $d_\alpha$ are the vectors that give the position of each atom type with respect to the centers of the lattice $\Lambda$: $d_a:=0$, $d_b:=\delta_1$.
 
-\subsection{Grassmann integral representation}
+\subsection{Dyson series}
 \indent
 The {\it specific free energy} on the lattice $\Lambda$ is defined by
 \begin{equation}
@@ -344,17 +344,118 @@ The {\it specific free energy} on the lattice $\Lambda$ is defined by
   \label{freeen}
 \end{equation}
 where $\beta$ is the inverse temperature.
-We define these at finite $\beta$ and $L$, but will take $\beta,L\to\infty$.
+\bigskip
-A straightforward application of the Trotter product formula implies that (see\-~\cite[(4.1)]{Gi10})
+
+\indent
+The exponential of $\mathcal H$ is difficult to compute, due to the presence of both the kinetic term $\mathcal H_0$ and the interacting Hamiltonian $\mathcal H_I$.
+We can split these from each other using the Trotter product formula:
+\begin{equation}
+  e^{-\beta\mathcal H}=
+  \lim_{p\to\infty}
+  \left(
+    e^{-\frac\beta p\mathcal H_0}
+    e^{-\frac\beta p\mathcal H_I}
+  \right)^p
+\end{equation}
+(which follows from the Baker-Campbell-Hausdorff formula).
+Taking the limit $p\to\infty$, we can expand the exponential of $\mathcal H_I$:
+\begin{equation}
+  e^{-\beta\mathcal H}=
+  \lim_{p\to\infty}
+  \left(
+    e^{-\frac\beta p\mathcal H_0}
+    \left(1-\frac\beta p\mathcal H_I\right)
+  \right)^p
+  .
+\end{equation}
+We can expand the power by noting that, in each factor, either the term $1$ or $-\frac\beta p\mathcal H_I$ can be selected.
+Thus,
+\begin{equation}
+  e^{-\beta\mathcal H}=
+  e^{-\beta\mathcal H_0}
+  +
+  \lim_{p\to\infty}
+  \sum_{N=1}^p
+  \left(\frac{-\beta}p\right)^{N}
+  \sum_{i_{N+1}=0}^p
+  \sum_{\displaystyle\mathop{\scriptstyle i_1,\cdots,i_N\in\{1,\cdots,p\}}_{i_1+\cdots+i_{N+1}=p}}
+  \left(\prod_{\alpha=1}^{N}e^{-\beta\frac{i_\alpha}p\mathcal H_0}\mathcal H_I\right)
+  e^{-\beta\frac{i_{N+1}}p\mathcal H_I}
+  .
+\end{equation}
+Defining
+\begin{equation}
+  \mathcal H_I(t):=e^{t\mathcal H_0}\mathcal H_Ie^{-t\mathcal H_0}
+\end{equation}
+and, for $j=1,\cdots,N$,
+\begin{equation}
+  \tau_j:=\frac1p\sum_{\alpha=j+1}^{N+1}i_\alpha
+\end{equation}
+we prove by induction that
+\begin{equation}
+  \left(\prod_{\alpha=k}^{N}e^{-\beta\frac{i_\alpha}N\mathcal H_0}\mathcal H_I\right)
+  e^{-\beta\frac{i_{N+1}}N\mathcal H_I}
+  =
+  e^{-\tau_{k-1}\mathcal H_0}
+  \prod_{j=k}^{N}\mathcal H_I(\tau_j)
+  .
+\end{equation}
+Indeed, for $k=N$,
+\begin{equation}
+  e^{-\beta\frac{i_N}N\mathcal H_0}\mathcal H_I
+  e^{-\beta\frac{i_{N+1}}N\mathcal H_I}
+  =
+  e^{-\beta\frac{i_\alpha-i_{N+1}}N\mathcal H_0}\mathcal H_I(\tau_N)
+\end{equation}
+and
+\begin{equation}
+  \left(\prod_{\alpha=k}^{N}e^{-\beta\frac{i_\alpha}N\mathcal H_0}\mathcal H_I\right)
+  e^{-\beta\frac{i_{N+1}}N\mathcal H_I}
+  =
+  e^{-\beta\frac{i_k}N\mathcal H_0}\mathcal H_I
+  e^{-\tau_{k}\mathcal H_0}
+  \prod_{j=k+1}^{N}\mathcal H_I(\tau_j)
+  =
+  e^{-\beta(\frac{i_k}N+\tau_k)\mathcal H_0}\mathcal H_I
+  \prod_{j=k}^{N}\mathcal H_I(\tau_j)
+  .
+\end{equation}
+Therefore,
+\begin{equation}
+  e^{-\beta\mathcal H}=
+  e^{-\beta\mathcal H_0}
+  +
+  \lim_{p\to\infty}
+  e^{-\beta\mathcal H_0}
+  \sum_{N=1}^p
+  \left(\frac{-\beta}p\right)^{N}
+  \sum_{i_{N+1}=0}^p
+  \sum_{\displaystyle\mathop{\scriptstyle i_1,\cdots,i_N\in\{1,\cdots,p\}}_{i_1+\cdots+i_{N+1}=p}}
+  \prod_{j=1}^{N}\mathcal H_I(\tau_j)
+  .
+\end{equation}
+This is a Riemann sum, which converges to
+\begin{equation}
+  e^{-\beta\mathcal H}=
+  e^{-\beta\mathcal H_0}
+  +
+  e^{-\beta\mathcal H_0}
+  \sum_{N=1}^\infty
+  (-\beta)^{N}
+  \int_{\beta\geqslant t_1\geqslant\cdots\geqslant t_N\geqslant 0} dt_1\cdots dt_N
+  \prod_{j=1}^{N}\mathcal H_I(t_j)
+  .
+  \label{trotter}
+\end{equation}
+
+\subsection{Grassmann integral representation}
+\indent
+To compute the free energy\-~(\ref{freeen}), we use\-~(\ref{trotter}):
 \begin{equation}
   \mathrm{Tr}( e^{-\beta\mathcal H})
   =
   \mathrm{Tr}(e^{-\beta\mathcal H_0})
-  +\sum_{N=1}^\infty\frac{(-\beta)^N}{N!}\int_{\beta\geqslant t_1\geqslant\cdots\geqslant t_N\geqslant0}\mathrm{Tr}\left(e^{-\beta\mathcal H_0}\mathcal H_I(t_1)\cdots\mathcal H_I(t_N)\right)
+  +\sum_{N=1}^\infty(-\beta)^N\int_{\beta\geqslant t_1\geqslant\cdots\geqslant t_N\geqslant0}\mathrm{Tr}\left(e^{-\beta\mathcal H_0}\mathcal H_I(t_1)\cdots\mathcal H_I(t_N)\right)
-\end{equation}
-where
-\begin{equation}
-  \mathcal H_I(t):=e^{t\mathcal H_0}\mathcal H_Ie^{-t\mathcal H_0}
   .
 \end{equation}
 To compute this trace, we will use the {\it Wick rule}, which we will now descibe.
@@ -393,7 +494,7 @@ The Wick rule can be proved by a direct computation, and follows from the fact t
 \bigskip
 
 \indent
-Fermionic creation and annihilation operators do not anticommute: $\{a_i,a_i\dagger\}=1$.
+Fermionic creation and annihilation operators do not anticommute: $\{a_i,a_i^\dagger\}=1$.
 However, the time-ordering operator effectively makes them anticommute.
 We can make this more precise be re-expressing the problem in terms of {\it Grassmann variables}.
 \bigskip
@@ -440,7 +541,7 @@ 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 +613,7 @@ 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 +650,7 @@ 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
@@ -615,10 +716,12 @@ These boxes are obtained by doubling the size of the elementary cell of the hexa
 We also have a time dimension, which we split into boxes of size $2^{|h|}$.
 We thus define the set of boxes on scale $h\in\{-N_\beta,\cdots,0\}$ by
 \begin{equation}
-  \mathcal Q_h:=\left\{
+  \begin{largearray}
+    \mathcal Q_h:=\Big\{\left\{
       [i2^{|h|},(i+1)2^{|h|})\times(\Lambda\cap\{2^{|h|}(n_1+x_1)l_1+2^{|h|}(n_2+x_2)l_2,\ x_1,x_2\in[0,1)\})
-  \right\}_{i,n_1,n_2\in\mathbb Z}
+    \right\},\\\hfill,i,n_1,n_2\in\mathbb Z\Big\}
     \label{boxes}
+  \end{largearray}
 \end{equation}
 ($\mathcal Q_h$ is a set of sets).
 For every $(t,x)\in[0,\beta)\times\Lambda$ and $h\in\{-N_\beta,\cdots,0\}$, there exists a unique box $\Delta^{(h)}(t,x)\in\mathcal Q_m$ such that $(t,x)\in\Delta^{(m)}(t,x)$.
@@ -679,7 +782,6 @@ 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.
@@ -719,7 +821,7 @@ Note that\-~(\ref{psi_hierarchical}) is then
   \psi_{\alpha,\sigma}^\pm(t,x)\equiv\psi_{\alpha,\sigma}^{[\le0]\pm}(\Delta^{(1)}(t,x))
   .
 \end{equation}
-We then define, for $h\in\{-N_\beta,\cdots0\}$,
+We then define, for $h\in\{-N_\beta,\cdots,0\}$,
 \begin{equation}
   e^{\beta|\Lambda| c^{[h]}-\mathcal V^{[h-1]}(\psi^{\leqslant h-1]})}
   :=\int P^{[h]}(d\psi^{[h]})\ e^{-\mathcal V^{[h]}(\psi^{[\leqslant h]})}
@@ -754,7 +856,7 @@ 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
@@ -868,7 +970,7 @@ We have thus introduced a strategy to compute $\mathcal V^{[h]}$ inductively: st
   e^{\beta|\Lambda|c^{[h]}-\mathcal V^{[h-1]}(\psi^{[\leqslant h-1]})}=\int P(d\psi^{[h]})\ e^{-\mathcal V^{[h]}(\psi^{[h]}+2^{-\gamma}\psi^{[\leqslant h-1]})}
 \end{equation}
 where $\gamma\equiv1$ is the scaling dimension of $\psi$ in\-~(\ref{scaling_psi}).
-Now, by\-~(\ref{box_dcmp}), is
+Now, by\-~(\ref{box_dcmp}), this is
 \begin{equation}
   e^{\beta|\Lambda|c^{[h]}+\sum_{\bar\Delta\in\mathcal Q_{h-1}}v_{h-1}(\psi^{[\leqslant h-1]}(\bar\Delta))}=
   \prod_{\Delta\in\mathcal Q_h}\int P(d\psi^{[h]}(\Delta))\ e^{v_h(\psi^{[h]}(\Delta)+2^{-\gamma}\psi^{[\leqslant h-1]}(\bar\Delta))}
@@ -894,19 +996,19 @@ 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 computation is thus reduced to computing the map $\ell^{(h)}\mapsto\ell^{(h-1)}$ 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 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 +1016,7 @@ 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 +1066,13 @@ 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)}=
+  \ell_i^{(h-1)}=
-  2^{d+1-2l\gamma}\alpha_i^{(h)}
+  2^{d+1-2l\gamma}\ell_i^{(h)}
   +
   \cdots
 \end{equation}
@@ -1004,7 +1106,7 @@ 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 +1115,20 @@ 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}
@@ -1496,7 +1598,7 @@ Let us first prove a technical lemma.
 \bigskip
 
 \indent\underline{Proof}:
-  To prove\-~(\ref{fock1}), we write $e^{-t\sum_i\lambda_ia_i^\dagger a_i}=\prod_ie^{-t\lambda_ia_i^\dagger a_i}$ and expand the exponential, using the fact that $(a_i^\dagger a_i)^n=a_i^\dagger a_i$ for any $n\geqslant 1$.
+  The proof of\-~(\ref{fock1}) proceeds as follows: we write $e^{-t\sum_i\lambda_ia_i^\dagger a_i}=\prod_ie^{-t\lambda_ia_i^\dagger a_i}$ and expand the exponential, using the fact that $(a_i^\dagger a_i)^n=a_i^\dagger a_i$ for any $n\geqslant 1$.
   By\-~(\ref{fock1}),
   \begin{equation}
     e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}a_j^\dagger
@@ -1509,69 +1611,30 @@ 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)
+  \end{equation}
+  and since $(a_j^\dagger)^2=0$,
+  \begin{equation}
+    e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}a_j^\dagger
+    =
+    e^{-t\lambda_j}a_j^\dagger\prod_{i}
+    (1+(e^{-t\lambda_i}-1)a_i^\dagger a_i)
+    =
+    e^{-t\lambda_j}a_j^\dagger
+    e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}
     .
     \label{fock2}
   \end{equation}
-  Similarly,
+  Taking the $\dagger$ of\-~(\ref{fock2}), we find
   \begin{equation}
-    e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}a_j
+    a_je^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}
-    =
-    \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\lambda_j}
+    e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}
     a_j
-    \prod_{i\neq j}
-    (1+(e^{-t\lambda_i}-1)a_i^\dagger a_i)
     .
     \label{fock3}
   \end{equation}
-  Furthermore, taking the $\dagger$ of\-~(\ref{fock3}), we find
+  Combining\-~(\ref{fock2}) and\-~(\ref{fock3}), we find\-~(\ref{fock4}).
-  \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 
-    .
-    \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
-  \end{equation}
-  we have
-  \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
-    .
-    \label{fock3'}
-  \end{equation}
-  This implies the second of\-~(\ref{fock4}).
 \qed
 \bigskip
 
@@ -1756,14 +1819,15 @@ We now compute the two-point correlation function of $\left<\cdot\right>$.
     \frac1{-ik_0+\lambda_k}
     .
   \end{equation}
-  Thus, if the Fourier transform can be inverted, then we have
+  Thus, wherever the Fourier transform can be inverted, we have
   \begin{equation}
     s_{i,j}(t-t')
     =
     \frac1\beta\sum_{k_0\in\frac{2\pi}\beta(\mathbb Z+\frac12)}
     e^{-ik_0(t-t')}(-ik_0\mathds 1+\mu)^{-1}_{i,j}
+    .
   \end{equation}
-  and this is the case where $s_{i,j}(t-t')$ is continuous.
+  The Fourier transform can be inverted where $s_{i,j}(t-t')$ is continuous.
   \bigskip
 
   \point
@@ -2315,7 +2379,7 @@ Finally, let us prove the addition property of Gaussiann Grassmann integrals.
 \bigskip
 
 \indent\underline{Proof}:
-  It is sufficient to prove the lemma when $f$ is a monomial of the form
+  Using the linearity of the integration, it suffices to prove the lemma when $f$ is a monomial of the form
   \begin{equation}
     f(\psi)=
     \prod_{i=1}^{n}
@@ -2330,7 +2394,7 @@ Finally, let us prove the addition property of Gaussiann Grassmann integrals.
     \int P_{\nu_1}(d\varphi_1)\int P_{\nu_2}(d\varphi_2)\ f(\varphi_1)
   \end{equation}
   where $\nu_1$ and $\nu_2$ can be computed from the change of variables, but this is not necessary here.
-  Since $f$ is a monomial, it can be computed using the Wick rule\-~\ref{lemma:wick_grassmann}, and thus, changing variables back to $\psi$,
+  Since $f$ is a monomial, it can be computed using the Wick rule, see lemma\-~\ref{lemma:wick_grassmann}, and thus, changing variables back to $\psi$,
   \begin{equation}
     \begin{largearray}
       \int P_{\mu_1}(d\psi_1)\int P_{\mu_2}(d\psi_2)\ f(\psi)

README

@@ -28,6 +28,8 @@ Some extra functionality is provided in custom style files, located in the
   gnuplot
   meankondo v1.5
 
+meankondo is available from http://ian.jauslin.org/software/meankondo
+
 
 * Files:
 

figs/hierarchical_graphene.fig/graphene_vector_field.gnuplot

@@ -1,5 +1,5 @@
-set ylabel "$\\alpha_1$" norotate
+set ylabel "$\\ell_1$" norotate
-set xlabel "$\\alpha_0$"
+set xlabel "$\\ell_0$"
 
 # default output canvas size: 12.5cm x 8.75cm
 set term lua tikz size 8,6 standalone