\section{(2 March 2009)}
\subsection*{Transport coefficients, cont.}
\underline{Result}: \( 1^{\text{st}} \)-order response to a 
perturbation \( \mathcal{L}_{\text{per}} = \mathcal{L} + 
\vec{\lambda} \cdot \vec{\delta} \), where \( \vec{\delta} = (\delta_{1} , 
\ldots , \delta_{n}) \) are commuting derivations and \( \vec{\lambda} = 
(\lambda_{1}, \ldots , \lambda_{n}) \) are complex constants.

Then
\[ 
\langle \delta_{\nu}H \rangle_{\text{per}} = \sum_{\mu} 
\sigma^{\delta}_{\nu,\mu} \lambda_{\mu}¥ + \bigoh(\lambda^{2}) \quad 
(0^{\text{th}}\text{-order response} = 0)
\]
defines the first order response (transport) coefficients
\begin{align*}
    \sigma^{\delta}_{\nu \mu} & = \text{C-} \!\! \lim_{t \rightarrow 
    \infty} \tau \left( \rho \int_{\underset{ s_{0} + s_{1} \, 
    = t}{s_{0} , s_{1} \, \geq \, 0}} \underbrace{\mathrm{e}^{-\mi s_{0} 
    \mathcal{L}}} \delta_{\mu} \mathrm{e}^{-\mi s_{1} 
    \mathcal{L}} \delta_{\nu} H \, \mathrm{d}s_{0}\mathrm{d}s_{1} \right)  \\
     & \hspace{12.3em} |  \\
     & \hspace{3em} \rule[0em]{0em}{1em}^{(\text{put this onto } \rho \text{, but } \rho 
     \text{ is invariant so cross this term off})}  \\
     & = \text{C-} \!\! \lim_{t \rightarrow \infty} \tau 
     \left( \int_{\underset{ s_{0} + s_{1} \, 
    = t}{s_{0} , s_{1} \, \geq \, 0}} \underbrace{\left( \mathrm{e}^{-\mi s_{0} 
    \mathcal{L}} (\rho) \right)} \delta_{\mu} \mathrm{e}^{-\mi s_{1} 
    \mathcal{L}} \delta_{\nu} H \, \mathrm{d}s_{0}\mathrm{d}s_{1} 
    \right)  \\
     & \hspace{12.3em} |  \\
     & \hspace{12.2em} \rule[0em]{0em}{1em}^{\rho}  \\
     & = \text{C-} \!\! \lim_{t \rightarrow \infty} \tau 
     \left( \delta_{\mu} \int_{\underset{ s_{0} + s_{1} \, 
    = t}{s_{0} , s_{1} \, \geq \, 0}} \mathrm{e}^{-\mi s_{1} 
    \mathcal{L}} \delta_{\nu} H \, \mathrm{d}s_{0}\mathrm{d}s_{1} 
    \right)  \\
     & = 0 - \text{C-} \!\! \lim_{t \rightarrow \infty} \tau \left( 
     \left( \delta_{\mu} (\rho) \right)  \int_{0}^{t} \mathrm{e}^{-\mi s_{1} 
    \mathcal{L}} \delta_{\nu} H \, \mathrm{d}s_{1} 
    \right)  \\
     & \hspace{1.5em} |  \\
     & \hspace{0em} \rule[0em]{0em}{1em}^{\text{because } 
     \tau \text{ is invariant under derivation } (\tau 
     \circ \delta = 0), \text{ then } 
     \tau \left( \delta_{\mu} \left( \rho \int \text{etc. } 
     \right) \right) = 0}  \\
     & = - \tau \left( \delta_{\mu} (\rho) \mathcal{D} 
     (\mathcal{L}) \delta_{\nu} (H) \right) \,\, \text{(which may be 
     infinite)} \\
     & \hspace{6.5em} |  \\
     & \hspace{0em} \rule[0em]{0em}{1em}^{ \lim_{t \rightarrow 
     \infty} \int_{0}^{t} \mathrm{e}^{-\mi s \omega} \mathrm{d}s 
     \overset{\text{dist'n}}{=} \mathrm{PV}\frac{1}{\mi \omega} + 
     \frac{\pi}{2} \delta (\omega) =: \mathcal{D} (\omega)}
\end{align*}
Attention if \( \delta_{\nu} (H) \) is not \( \perp \) to \( \ker 
\mathcal{L} \).

Assumption \( T \searrow 0 \) and \( \underset{\text{Fermi energy} 
\hspace{4em}}{E_{F} \in \mathrm{Gap}(H)} \)

Mathematical assumption is \( \rho = P_{F} = \) the spectral 
projection of \( H \) onto states \( \leq E_{F} \).  This avoids the 
singularity.

Suppose we have \( \{\psi_{i}\}_{i} \) an eigenbasis of \( H \) (in 
the generalized sense).  

So \( H \psi_{i} = E_{i} \psi_{i} \) (that is, \( H \,| \, \psi_{i} \rangle = 
E_{i} \, | \, \psi_{i} \rangle \)).
\begin{align*}
    \langle \psi_{i} \, | \, \mathcal{L}(\delta_{\nu} H) \, | \, 
    \psi_{j} \rangle & = \langle \psi_{i} \, | \, (H \delta_{\nu} H - ( \delta_{\nu} H) H) \, | \, 
    \psi_{j} \rangle  \\
    | \hspace{5em} & \\
    \rule[0em]{0em}{1em}^{\mathcal{L}(\,\cdot\,) = [H, \,\cdot\,]} 
    \hspace{2em} &   \\
     & = (E_{i} - E_{j})\langle \psi_{i} \, | \, \delta_{\nu} H \, | \, 
    \psi_{j} \rangle
\end{align*}
So,
\begin{align*}
    \langle \psi_{i} \, | \, \mathcal{L}^{n}(\delta_{\nu} H) \, | \, 
    \psi_{j} \rangle & = \langle \psi_{i} \, | \, \mathcal{L} 
    (\mathcal{L}^{n - 1} (\delta_{\nu} H)) \, | \, 
    \psi_{j} \rangle  \\
     & = (E_{i} - E_{j}) \langle \psi_{i} \, | \,  
    \mathcal{L}^{n - 1} (\delta_{\nu} H) \, | \, 
    \psi_{j} \rangle
\end{align*}
Hence
\[ 
\langle \psi_{i} \, | \, \mathrm{e}^{-\mi s_{1} \mathcal{L}}
\delta_{\nu} H \, | \, \psi_{j} \rangle = \mathrm{e}^{-\mi s_{1}
(E_{i} - E_{j})} \langle \psi_{i} \, | \, \delta_{\nu} H \, | \,
\psi_{j} \rangle
\]

So we  need that \( E_{i} \neq E_{j} \) (to avoid \( \omega = 0 \) in 
the distribution \( \mathcal{D}(\omega) \)).  In that case
\[ 
\langle \psi_{i} \, | \, \lim_{t \rightarrow \infty} \int_{0}^{t}
\mathrm{e}^{-\mi s_{1} \mathcal{L}} \mathrm{d}s_{1} \delta_{\nu} H \,
| \, \psi_{j} \rangle = \frac{1}{\mi (E_{i} - E_{j})} \langle \psi_{i}
\, | \, \delta_{\nu} H \, | \, \psi_{j} \rangle
\]

Since \( \rho = P_{F} \) is a projection and \( \delta_{\mu} \)  is a 
derivation (using the beautiful formula \( (\ast) \) from the 
previous lecture)
\[ 
\delta_{\mu} P_{F} = P_{F} (\delta_{\mu} P_{F}) P_{F}^{\perp} + 
P_{F}^{\perp} (\delta_{\mu} P_{F}) P_{F}
\]
Now
\[ 
\langle \psi_{i} \, | \, \delta_{\nu} H \, | \,
\psi_{j} \rangle \neq 0 \Leftrightarrow ((E_{i} \leq E_{F}) \wedge 
(E_{j} > E_{F})) \vee ((E_{j} \leq E_{F}) \wedge 
(E_{i} > E_{F}))
\]
Since \( E_{F} \in (\tilde{E}_{0}, \tilde{E}_{1}) \), a gap in \( 
\sigma (H) \), we have
\[ 
|E_{i} - E_{j}| \geq |\tilde{E}_{1} - \tilde{E}_{0}|
\]
\underline{As a consequence},
\begin{align*}
     & \hspace{1.5em} \rule[0em]{0em}{1em}_{\underset{\vee}{1 = P_{F}
     + P_{F}^{\perp}}} \\
    \sigma^{\delta}_{\nu \mu} & = - \tau \left( \delta_{\mu}
    (P_{F}) \mathcal{D} (\mathcal{L}) \delta_{\nu} (H) \right) \\
     & = - \tau \left( P_{F} \delta_{\mu} (P_{F}) (\mathcal{D}
     \mathcal{L}) \delta_{\nu} (H) \right) - \tau
     \left( P_{F}^{\perp} \delta_{\mu} (P_{F}) \mathcal{D}
     (\mathcal{L}) \delta_{\nu} (H) \right) \\
%      & \hspace{7.3em}
%      \rule[0em]{0em}{1em}_{\underset{\vee}{P_{F}^{\perp}}}
%      \hspace{13.3em} \rule[0em]{0em}{1em}_{\underset{\vee}{P_{F}}} \\
     & = - \tau \left( P_{F} \delta_{\mu} (P_{F}) (\mathcal{D}
     (\mathcal{L}) \delta_{\nu} (H)) P_{F} \right) - \tau
     \left( P_{F}^{\perp} \delta_{\mu} (P_{F}) (\mathcal{D}
     (\mathcal{L}) \delta_{\nu} (H)) P_{F}^{\perp} \right) \\
     & = - \tau \left( P_{F} \delta_{\mu} (P_{F}) P_{F}^{\perp} 
     \frac{1}{\mi}
     \mathcal{L}^{-1} (\delta_{\nu} H) P_{F} \right) - \tau
     \left( P_{F}^{\perp} \delta_{\mu} (P_{F}) P_{F} \frac{1}{\mi} 
     \mathcal{L}^{-1} (\delta_{\nu} H) P_{F}^{\perp} \right) \\
%      & \hspace{8em}
%      \rule[0em]{0em}{1em}_{\underset{\vee}{P_{F}^{\perp}}}
%      \hspace{1.8em} \rule[0em]{0em}{1em}_{\underset{\vee}{P_{F}}} \\
     & \!\!\!  \overset{\text{claim}}{=} -\mi \tau\left( P_{F} \delta_{\mu} 
     (P_{F}) \delta_{\nu} (P_{F}) - \delta_{\nu} (P_{F}) \delta_{\mu} 
     (P_{F}) \right) \\
     & \overset{\text{or}}{=} -\mi \tau\left( P_{F} \delta_{\mu} 
     (P_{F}) P_{F}^{\perp} \delta_{\nu} (P_{F}) P_{F} - \delta_{\nu} (P_{F}) \delta_{\mu} 
     (P_{F}) \right)
\end{align*}
In other words the claim is
\begin{align*}
    P_{F}^{\perp} \mathcal{L}^{-1} (\delta_{\nu} H) P_{F} & = 
    -P_{F}^{\perp} (\delta_{\nu} P_{F}) P_{F} \quad \text{and}  \\
    P_{F} \mathcal{L}^{-1} (\delta_{\nu} H) P_{F}^{\perp} & = 
    P_{F} (\delta_{\nu} P_{F}) P_{F}^{\perp}
\end{align*}
\begin{proof}
    Hit both sides of the first equation with \( \mathcal{L} \) (\(
    \mathcal{L} \) leaves \( P_{F} \) invariant).
    \begin{align*}
	\text{left-hand side: } \mathcal{L} \left( P_{F}^{\perp}
	\mathcal{L}^{-1} (\delta_{\nu} H) P_{F} \right) & = 
	P_{F}^{\perp} ( \delta_{\nu} H )P_{F} \\
        \text{right-hand side: } \mathcal{L} \left( -P_{F}^{\perp} 
	(\delta_{\nu} P_{F}) P_{F} \right) & = -P_{F}^{\perp} [H, 
	\delta_{\nu} P_{F}] P_{F}  \\
    & \hspace{5em} | \\
    & \hspace{3em} \rule[0em]{0em}{1em}^{\mathcal{L}(\,\cdot\,) = [H, \,\cdot\,]}  \\
         & = -P_{F}^{\perp} \left( \delta_{\nu} \left( [H, 
	  P_{F}] \right) - [\delta_{\nu} H, P_{F}]  \right) P_{F}  \\
    & \hspace{7em} | \\
    & \hspace{5em} \rule[0em]{0em}{1em}^{ [H, P_{F}] = 0}  \\
    & = P_{F}^{\perp} \left( ( \delta_{\nu} H ) P_{F} - P_{F} 
    (\delta_{\nu} H )  \right) P_{F}  \\
    & = P_{F}^{\perp} ( \delta_{\nu} H ) P_{F} \quad \text{(QED first 
    equation)}
    \end{align*}
    The proof of the second equation is similar and picks up an 
    extra minus sign.
\end{proof}

So this is the result

\[ \sigma^{\delta}_{\nu \mu} = \mi \, \tau \! \left( P_{F} \left[ 
( \delta_{\nu}P_{F} ) ,  ( \delta_{\mu}P_{F} ) \right] \right) \]

\underline{Consequence}: \( \sigma^{\delta}_{\nu \mu} \) \emph{is a topological 
invariant.}
\begin{remark*}
    \( \sigma^{\delta}_{\nu \mu} \) is anti-symmetric.
\end{remark*}

\begin{exa*}
    QHE in \( \mathbb{R}^{2} \) (the quantum Hall effect).
    \begin{align*}
        \delta & = ( \delta_{1} . \delta_{2} ) & \delta_{\nu} & = 
	[ \hat{q}_{\nu} , \, \cdot \, ]  \\
        \lambda & = ( \lambda_{1} , \lambda_{2} ) & \lambda_{\nu} & = 
	e \, E_{\nu} \\
        ¥ & ¥ & ¥ & \rule[0em]{0em}{1em}_{\mathrm{e} \text{ is the electric charge}}  \\
        ¥ & ¥ & ¥ & \rule[0em]{0em}{1em}^{E_{\nu} \text{ is the 
	external electric field}}
    \end{align*}
\end{exa*}