Update to v1.1:

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

  Minor fixes
This commit is contained in:
Ian Jauslin 2023-03-21 18:42:51 -04:00
parent fddcf91b6a
commit 8d1562d39c
2 changed files with 130 additions and 20 deletions

6
Changelog Normal file
View File

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

View File

@ -336,7 +336,7 @@ We now define the interaction Hamiltonian which we take to be of {\it Hubbard} f
\label{hamintx}\end{equation} \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 $\Lambda$: $d_a:=0$, $d_b:=\delta_1$. 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 \indent
The {\it specific free energy} on the lattice $\Lambda$ is defined by The {\it specific free energy} on the lattice $\Lambda$ is defined by
\begin{equation} \begin{equation}
@ -344,17 +344,118 @@ The {\it specific free energy} on the lattice $\Lambda$ is defined by
\label{freeen} \label{freeen}
\end{equation} \end{equation}
where $\beta$ is the inverse temperature. 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} \begin{equation}
\mathrm{Tr}( e^{-\beta\mathcal H}) \mathrm{Tr}( e^{-\beta\mathcal H})
= =
\mathrm{Tr}(e^{-\beta\mathcal H_0}) \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} \end{equation}
To compute this trace, we will use the {\it Wick rule}, which we will now descibe. 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 \bigskip
\indent \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. 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}. We can make this more precise be re-expressing the problem in terms of {\it Grassmann variables}.
\bigskip \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 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 We thus define the set of boxes on scale $h\in\{-N_\beta,\cdots,0\}$ by
\begin{equation} \begin{equation}
\mathcal Q_h:=\left\{ \begin{largearray}
[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)\}) \mathcal Q_h:=\Big\{\left\{
\right\}_{i,n_1,n_2\in\mathbb Z} [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)\})
\label{boxes} \right\},\\\hfill,i,n_1,n_2\in\mathbb Z\Big\}
\label{boxes}
\end{largearray}
\end{equation} \end{equation}
($\mathcal Q_h$ is a set of sets). ($\mathcal Q_h$ 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 $(t,x)\in\Delta^{(m)}(t,x)$. 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)) \psi_{\alpha,\sigma}^\pm(t,x)\equiv\psi_{\alpha,\sigma}^{[\le0]\pm}(\Delta^{(1)}(t,x))
. .
\end{equation} \end{equation}
We then define, for $h\in\{-N_\beta,\cdots0\}$, We then define, for $h\in\{-N_\beta,\cdots,0\}$,
\begin{equation} \begin{equation}
e^{\beta|\Lambda| c^{[h]}-\mathcal V^{[h-1]}(\psi^{\leqslant h-1]})} 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]})} :=\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]})} 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} \end{equation}
where $\gamma\equiv1$ is the scaling dimension of $\psi$ in\-~(\ref{scaling_psi}). 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} \begin{equation}
e^{\beta|\Lambda|c^{[h]}+\sum_{\bar\Delta\in\mathcal Q_{h-1}}v_{h-1}(\psi^{[\leqslant h-1]}(\bar\Delta))}= 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))} \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 \bigskip
\indent\underline{Proof}: \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}), By\-~(\ref{fock1}),
\begin{equation} \begin{equation}
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}a_j^\dagger
@ -1716,14 +1819,15 @@ We now compute the two-point correlation function of $\left<\cdot\right>$.
\frac1{-ik_0+\lambda_k} \frac1{-ik_0+\lambda_k}
. .
\end{equation} \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} \begin{equation}
s_{i,j}(t-t') s_{i,j}(t-t')
= =
\frac1\beta\sum_{k_0\in\frac{2\pi}\beta(\mathbb Z+\frac12)} \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} e^{-ik_0(t-t')}(-ik_0\mathds 1+\mu)^{-1}_{i,j}
.
\end{equation} \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 \bigskip
\point \point
@ -2275,7 +2379,7 @@ Finally, let us prove the addition property of Gaussiann Grassmann integrals.
\bigskip \bigskip
\indent\underline{Proof}: \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} \begin{equation}
f(\psi)= f(\psi)=
\prod_{i=1}^{n} \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) \int P_{\nu_1}(d\varphi_1)\int P_{\nu_2}(d\varphi_2)\ f(\varphi_1)
\end{equation} \end{equation}
where $\nu_1$ and $\nu_2$ can be computed from the change of variables, but this is not necessary here. 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{equation}
\begin{largearray} \begin{largearray}
\int P_{\mu_1}(d\psi_1)\int P_{\mu_2}(d\psi_2)\ f(\psi) \int P_{\mu_1}(d\psi_1)\int P_{\mu_2}(d\psi_2)\ f(\psi)