Update to v1.1:
Add: Detailed discussion of the use of the Trotter product formula in the
Dyson expansion.
Minor fixes
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Changelog
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Changelog
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v1.1:
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* Add: Detailed discussion of the use of the Trotter product formula in the
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Dyson expansion.
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* Minor fixes
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144
Jauslin_2022.tex
144
Jauslin_2022.tex
@ -336,7 +336,7 @@ We now define the interaction Hamiltonian which we take to be of {\it Hubbard} f
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\label{hamintx}\end{equation}
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where the $d_\alpha$ are the vectors that give the position of each atom type with respect to the centers of the lattice $\Lambda$: $d_a:=0$, $d_b:=\delta_1$.
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\subsection{Grassmann integral representation}
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\subsection{Dyson series}
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\indent
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The {\it specific free energy} on the lattice $\Lambda$ is defined by
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\begin{equation}
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@ -344,17 +344,118 @@ The {\it specific free energy} on the lattice $\Lambda$ is defined by
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\label{freeen}
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\end{equation}
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where $\beta$ is the inverse temperature.
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We define these at finite $\beta$ and $L$, but will take $\beta,L\to\infty$.
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A straightforward application of the Trotter product formula implies that (see\-~\cite[(4.1)]{Gi10})
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\bigskip
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\indent
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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 $\mathcal H_I$.
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We can split these from each other using the Trotter product formula:
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\begin{equation}
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e^{-\beta\mathcal H}=
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\lim_{p\to\infty}
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\left(
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e^{-\frac\beta p\mathcal H_0}
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e^{-\frac\beta p\mathcal H_I}
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\right)^p
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\end{equation}
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(which follows from the Baker-Campbell-Hausdorff formula).
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Taking the limit $p\to\infty$, we can expand the exponential of $\mathcal H_I$:
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\begin{equation}
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e^{-\beta\mathcal H}=
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\lim_{p\to\infty}
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\left(
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e^{-\frac\beta p\mathcal H_0}
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\left(1-\frac\beta p\mathcal H_I\right)
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\right)^p
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.
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\end{equation}
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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.
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Thus,
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\begin{equation}
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e^{-\beta\mathcal H}=
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e^{-\beta\mathcal H_0}
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+
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\lim_{p\to\infty}
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\sum_{N=1}^p
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\left(\frac{-\beta}p\right)^{N}
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\sum_{i_{N+1}=0}^p
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\sum_{\displaystyle\mathop{\scriptstyle i_1,\cdots,i_N\in\{1,\cdots,p\}}_{i_1+\cdots+i_{N+1}=p}}
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\left(\prod_{\alpha=1}^{N}e^{-\beta\frac{i_\alpha}p\mathcal H_0}\mathcal H_I\right)
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e^{-\beta\frac{i_{N+1}}p\mathcal H_I}
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.
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\end{equation}
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Defining
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\begin{equation}
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\mathcal H_I(t):=e^{t\mathcal H_0}\mathcal H_Ie^{-t\mathcal H_0}
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\end{equation}
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and, for $j=1,\cdots,N$,
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\begin{equation}
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\tau_j:=\frac1p\sum_{\alpha=j+1}^{N+1}i_\alpha
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\end{equation}
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we prove by induction that
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\begin{equation}
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\left(\prod_{\alpha=k}^{N}e^{-\beta\frac{i_\alpha}N\mathcal H_0}\mathcal H_I\right)
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e^{-\beta\frac{i_{N+1}}N\mathcal H_I}
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=
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e^{-\tau_{k-1}\mathcal H_0}
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\prod_{j=k}^{N}\mathcal H_I(\tau_j)
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.
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\end{equation}
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Indeed, for $k=N$,
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\begin{equation}
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e^{-\beta\frac{i_N}N\mathcal H_0}\mathcal H_I
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e^{-\beta\frac{i_{N+1}}N\mathcal H_I}
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=
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e^{-\beta\frac{i_\alpha-i_{N+1}}N\mathcal H_0}\mathcal H_I(\tau_N)
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\end{equation}
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and
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\begin{equation}
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\left(\prod_{\alpha=k}^{N}e^{-\beta\frac{i_\alpha}N\mathcal H_0}\mathcal H_I\right)
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e^{-\beta\frac{i_{N+1}}N\mathcal H_I}
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=
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e^{-\beta\frac{i_k}N\mathcal H_0}\mathcal H_I
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e^{-\tau_{k}\mathcal H_0}
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\prod_{j=k+1}^{N}\mathcal H_I(\tau_j)
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=
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e^{-\beta(\frac{i_k}N+\tau_k)\mathcal H_0}\mathcal H_I
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\prod_{j=k}^{N}\mathcal H_I(\tau_j)
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.
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\end{equation}
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Therefore,
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\begin{equation}
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e^{-\beta\mathcal H}=
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e^{-\beta\mathcal H_0}
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+
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\lim_{p\to\infty}
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e^{-\beta\mathcal H_0}
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\sum_{N=1}^p
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\left(\frac{-\beta}p\right)^{N}
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\sum_{i_{N+1}=0}^p
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\sum_{\displaystyle\mathop{\scriptstyle i_1,\cdots,i_N\in\{1,\cdots,p\}}_{i_1+\cdots+i_{N+1}=p}}
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\prod_{j=1}^{N}\mathcal H_I(\tau_j)
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.
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\end{equation}
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This is a Riemann sum, which converges to
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\begin{equation}
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e^{-\beta\mathcal H}=
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e^{-\beta\mathcal H_0}
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+
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e^{-\beta\mathcal H_0}
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\sum_{N=1}^\infty
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(-\beta)^{N}
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\int_{\beta\geqslant t_1\geqslant\cdots\geqslant t_N\geqslant 0} dt_1\cdots dt_N
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\prod_{j=1}^{N}\mathcal H_I(t_j)
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.
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\label{trotter}
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\end{equation}
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\subsection{Grassmann integral representation}
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\indent
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To compute the free energy\-~(\ref{freeen}), we use\-~(\ref{trotter}):
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\begin{equation}
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\mathrm{Tr}( e^{-\beta\mathcal H})
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=
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\mathrm{Tr}(e^{-\beta\mathcal H_0})
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+\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)
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\end{equation}
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where
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\begin{equation}
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\mathcal H_I(t):=e^{t\mathcal H_0}\mathcal H_Ie^{-t\mathcal H_0}
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+\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)
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.
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\end{equation}
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To compute this trace, we will use the {\it Wick rule}, which we will now descibe.
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@ -393,7 +494,7 @@ The Wick rule can be proved by a direct computation, and follows from the fact t
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\bigskip
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\indent
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Fermionic creation and annihilation operators do not anticommute: $\{a_i,a_i\dagger\}=1$.
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Fermionic creation and annihilation operators do not anticommute: $\{a_i,a_i^\dagger\}=1$.
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However, the time-ordering operator effectively makes them anticommute.
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We can make this more precise be re-expressing the problem in terms of {\it Grassmann variables}.
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\bigskip
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@ -615,10 +716,12 @@ These boxes are obtained by doubling the size of the elementary cell of the hexa
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We also have a time dimension, which we split into boxes of size $2^{|h|}$.
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We thus define the set of boxes on scale $h\in\{-N_\beta,\cdots,0\}$ by
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\begin{equation}
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\mathcal Q_h:=\left\{
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[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)\})
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\right\}_{i,n_1,n_2\in\mathbb Z}
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\label{boxes}
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\begin{largearray}
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\mathcal Q_h:=\Big\{\left\{
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[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)\})
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\right\},\\\hfill,i,n_1,n_2\in\mathbb Z\Big\}
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\label{boxes}
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\end{largearray}
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\end{equation}
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($\mathcal Q_h$ is a set of sets).
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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 $(t,x)\in\Delta^{(m)}(t,x)$.
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@ -718,7 +821,7 @@ Note that\-~(\ref{psi_hierarchical}) is then
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\psi_{\alpha,\sigma}^\pm(t,x)\equiv\psi_{\alpha,\sigma}^{[\le0]\pm}(\Delta^{(1)}(t,x))
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.
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\end{equation}
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We then define, for $h\in\{-N_\beta,\cdots0\}$,
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We then define, for $h\in\{-N_\beta,\cdots,0\}$,
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\begin{equation}
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e^{\beta|\Lambda| c^{[h]}-\mathcal V^{[h-1]}(\psi^{\leqslant h-1]})}
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:=\int P^{[h]}(d\psi^{[h]})\ e^{-\mathcal V^{[h]}(\psi^{[\leqslant h]})}
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@ -867,7 +970,7 @@ We have thus introduced a strategy to compute $\mathcal V^{[h]}$ inductively: st
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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]})}
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\end{equation}
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where $\gamma\equiv1$ is the scaling dimension of $\psi$ in\-~(\ref{scaling_psi}).
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Now, by\-~(\ref{box_dcmp}), is
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Now, by\-~(\ref{box_dcmp}), this is
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\begin{equation}
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e^{\beta|\Lambda|c^{[h]}+\sum_{\bar\Delta\in\mathcal Q_{h-1}}v_{h-1}(\psi^{[\leqslant h-1]}(\bar\Delta))}=
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\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))}
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@ -1495,7 +1598,7 @@ Let us first prove a technical lemma.
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\bigskip
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\indent\underline{Proof}:
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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 $n\geqslant 1$.
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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 $n\geqslant 1$.
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By\-~(\ref{fock1}),
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\begin{equation}
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e^{-t\sum_{i=1}^N\lambda_ia_i^\dagger a_i}a_j^\dagger
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@ -1716,14 +1819,15 @@ We now compute the two-point correlation function of $\left<\cdot\right>$.
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\frac1{-ik_0+\lambda_k}
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.
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\end{equation}
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Thus, if the Fourier transform can be inverted, then we have
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Thus, wherever the Fourier transform can be inverted, we have
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\begin{equation}
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s_{i,j}(t-t')
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=
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\frac1\beta\sum_{k_0\in\frac{2\pi}\beta(\mathbb Z+\frac12)}
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e^{-ik_0(t-t')}(-ik_0\mathds 1+\mu)^{-1}_{i,j}
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.
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\end{equation}
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and this is the case where $s_{i,j}(t-t')$ is continuous.
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The Fourier transform can be inverted where $s_{i,j}(t-t')$ is continuous.
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\bigskip
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\point
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@ -2275,7 +2379,7 @@ Finally, let us prove the addition property of Gaussiann Grassmann integrals.
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\bigskip
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\indent\underline{Proof}:
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It is sufficient to prove the lemma when $f$ is a monomial of the form
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Using the linearity of the integration, it suffices to prove the lemma when $f$ is a monomial of the form
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\begin{equation}
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f(\psi)=
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\prod_{i=1}^{n}
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@ -2290,7 +2394,7 @@ Finally, let us prove the addition property of Gaussiann Grassmann integrals.
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\int P_{\nu_1}(d\varphi_1)\int P_{\nu_2}(d\varphi_2)\ f(\varphi_1)
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\end{equation}
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where $\nu_1$ and $\nu_2$ can be computed from the change of variables, but this is not necessary here.
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Since $f$ is a monomial, it can be computed using the Wick rule\-~\ref{lemma:wick_grassmann}, and thus, changing variables back to $\psi$,
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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 $\psi$,
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\begin{equation}
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\begin{largearray}
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\int P_{\mu_1}(d\psi_1)\int P_{\mu_2}(d\psi_2)\ f(\psi)
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