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| -rw-r--r-- | Changelog | 6 | ||||
| -rw-r--r-- | Jauslin_2022.tex | 144 |
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) |