Update to v1.1:HEADv1.1master

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

  Minor fixes

2 files changed, 130 insertions, 20 deletions

diff --git a/Changelog b/Changelog
new file mode 100644
index 0000000..67f0860
--- /dev/null
+++ b/Changelog
@@ -0,0 +1,6 @@
+v1.1:
+
+  * Add: Detailed discussion of the use of the Trotter product formula in the
+         Dyson expansion.
+  
+  * Minor fixes
diff --git a/Jauslin_2022.tex b/Jauslin_2022.tex
index 958cb08..c51c5a7 100644
--- a/Jauslin_2022.tex
+++ b/Jauslin_2022.tex
@@ -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 $\beta,L\to\infty$.
-A straightforward application of the Trotter product formula implies that (see\-~\cite[(4.1)]{Gi10})
+\bigskip
+
+\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}
-  \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)
+  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}
-where
+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(-\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}
 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
@@ -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\{
-    [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}
-  \label{boxes}
+  \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\},\\\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 $(t,x)\in\Delta^{(m)}(t,x)$.
@@ -718,7 +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]})}
@@ -867,7 +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))}
@@ -1495,7 +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
@@ -1716,14 +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
@@ -2275,7 +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}
@@ -2290,7 +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)