# Phantom_Ghost

1. 1天前
2019-01-21 01:18:09
Phantom_Ghost 更新于 Chern characteristic class

Chern character
\begin{align*}
\text{ch}(E_1\oplus E_2)&=\text{ch}(E_2\oplus E_1)=\text{ch}(E_1)+\text{ch}(E_2)\\
\text{ch}(E_1\otimes E_2)&=\text{ch}(E_1)\smile \text{ch}(E_2)\\
\text{ch}(L)&=e^{c_1(L)}=1+c_1(L)+\frac{1}{2!}(c_1(L))^2+\cdots\\
\text{ch}(L_1\otimes L_2)&=e^{c_1(L_1\otimes L_2)}=e^{c_1(L_1)+c_1(L_2)}=e^{c_1(L_1)}e^{c_1(L_1)}
\end{align*}

Proposition 4: $c_1:\text{Vect}_\mathbb{C}^1(B)\to H^2(B;\mathbb{Z})$ is a homomorphism of rings.
(Here the vector bundles equipped with operation $\oplus,\otimes$ become a ring, while the integral cohomology equipped with $+,\smile$ become a graded ring)

Proof: First show for line bundles: $c_1(L_1\otimes L_2)=c_1(L_1)+c_1(L_2)$ where $L_1\otimes L_2\to\mathbb{C}P^\infty\times\mathbb{C}P^\infty$ , $L_{1,2}\hookrightarrow\mathbb{C}P^\infty$ are the pullbacks of canonical line bundle. Since $c_1(L_i)$ is a generator of $H^2(\mathbb{C}P^\infty)$, we get $H^\bullet(\mathbb{C}P^\infty\times\mathbb{C}P^\infty)\cong\mathbb{Z}[\alpha_1,\alpha_2]$ by Künneth formula where $\alpha_i=\text{pr}_i^*(c_1(L_i))$, $\text{pr}_{1,2}:\mathbb{C}P^\infty_1\times\mathbb{C}P^\infty_2\to \mathbb{C}P^\infty_{1,2}$.
\begin{align*}
H^2(\mathbb{C}P^\infty\times\mathbb{C}P^\infty)&\cong[H^2(\mathbb{C}P^\infty)\otimes H^0(\mathbb{C}P^\infty)]\oplus[H^0(\mathbb{C}P^\infty)\otimes H^2(\mathbb{C}P^\infty)]\\
&\bigoplus_{i+j=2+1}\underset{=0\;\text{(torsion free)}}{\underbrace{\text{Tor}_\mathbb{Z}(H^i(\mathbb{C}P^\infty),H^j(\mathbb{C}P^\infty))}}
\end{align*}
$\Rightarrow$ $\mathbb{C}P^\infty\vee\mathbb{C}P^\infty\hookrightarrow\mathbb{C}P^\infty\times\mathbb{C}P^\infty$ induces isomorphism on cohomology $H^2(\mathbb{C}P^\infty\vee\mathbb{C}P^\infty)\cong H^2(\mathbb{C}P^\infty\times\mathbb{C}P^\infty)$.
Therefore to compute $c_1(L_1\otimes L_2)$ it is sufficient to compute $c_1(L_1\otimes L_2|_{\mathbb{C}P^\infty\vee\mathbb{C}P^\infty})$:
$L_2$ is trivial on the first factor of $\mathbb{C}P^\infty\vee\mathbb{C}P^\infty$ while $L_1$ is trivial on the second factor: $c_1(L_1\otimes L_2|_{\mathbb{C}P^\infty\times\{\text{pt.}\}})=c_1(L_1)=\alpha_1$ , $c_1(L_1\otimes L_2|_{\{\text{pt.}\}\times\mathbb{C}P^\infty})=c_1(L_2)=\alpha_2$ $\Rightarrow$ $c_1(L_1\otimes L_2)=c_1(L_1)+c_1(L_2)$.

For the general case $E\cong L_1\oplus\cdots\oplus L_n$, $t_i:=c_1(L_i)$, then $\text{ch}(E)=\sum_i e^{t_i}=n+(t_1+\cdots+t_n)+\cdots+\frac{1}{k!}(t_1^k+\cdots+t_n^k)+\cdots$
$t_1^k+\cdots+t_n^k=s_k(\sigma_1,\dots,\sigma_n)$ with $\sigma_i=c_i(E)$ where $s_k(\cdot)$ is Newton polynomial. Then $\text{ch}(E)=\text{dim}\,E+\sum_{k>0}s_k(c_1(E),\dots,c_n(E))$.

Proposition 5 $\text{ch}(E_1\oplus E_2)=\text{ch}(E_1)+\text{ch}(E_2)$ ; $\text{ch}(E_1\otimes E_2)=\text{ch}(E_1)\text{ch}(E_2)$.

Proof: Splitting principle works also for $\mathbb{Q}$. This time we can pull back $E_1$ to a sum of $L_1\oplus\cdots L_n$ over $F(E_1)$. Then pull back $E_2$ over $F(E_1)$ so as a result $F(E_1,E_2)\to B$, where we pull back both $E_1$ and $E_2$ to the line budle.
$\text{ch}(E_1\oplus E_2)=\text{ch}(\bigoplus_{i=1,2,j>0}L_{i,j})e^{c_1(L_{i,j})}=\text{ch}(E_1)+\text{ch}(E_2)$
$\text{ch}(E_1\otimes E_2)=\text{ch}(\bigoplus_{i,j>0}L_{1,i}\otimes L_{2,j})=\sum_{i,j}\text{ch}(L_{1,i}\otimes L_{2,j})=\text{ch}(E_1)\text{ch}(E_2)$

Fact: $K(X)\xrightarrow{\text{ch}}H^\bullet(X;\mathbb{Q})$ is homomorphism
\begin{array}[c]{ccc}
K(X)&\xrightarrow{\text{ch}}&H^\bullet(X;\mathbb{Q})\\
\downarrow&&\downarrow\\
K(\{\text{pt.}\})&\xrightarrow{\text{ch}}&H^\bullet(\{\text{pt.}\};\mathbb{Q})
\end{array}
This yields the functoriality and there is a natural transormation between the functors $K(X)\xrightarrow{\text{ch}}H^\bullet(X;\mathbb{Q})$ and $\widetilde{K}(X)\xrightarrow{\text{ch}}\widetilde{H}^\bullet(X;\mathbb{Q})$.

Proposition 6: $\widetilde{K}(S^{2n})\xrightarrow{\text{ch}}\widetilde{H}^{2n}(S^{2n};\mathbb{Q})$ injective, $\text{im}(\text{ch})\subset\widetilde{H}^{2n}(S^{2n;\mathbb{Z}})\subset\widetilde{H}^{2n}(S^{2n};\mathbb{Q})$.

Proof: Since
\begin{array}[c]{ccc}
\widetilde{K}(X)&\xrightarrow{\sim}&\widetilde{K}(S^2X)\\
\scriptstyle{\text{ch}}\downarrow&&\downarrow\scriptstyle{\text{ch}}\\
\widetilde{H}^{2n}(X;\mathbb{Q})&\xrightarrow{\sim}&\widetilde{H}^{2n}(S^2X;\mathbb{Q})
\end{array}
where $S^2X$ is the double suspension of $X$. The upper row in the diagram is due to Bott periodicity while the lower row is due to Künneth formula.
$\text{ch}(H-1)=\text{ch}(H)-\text{ch}(1)=1+c_1(H)-1=c_1(H)$ ($H$ denotes the canonical line bundle over $S^2\approx \mathbb{C}P^1$).
Perform induction on $n$:
$X=S^0$, it holds.
$n\to n+1$ follows by the commutative diagram
\begin{array}[c]{ccc}
\widetilde{K}(X)&\xrightarrow{\sim}&\widetilde{K}(S^2X)&\to& \widetilde{K}(S^4X)&\to& \cdots\\
\scriptstyle{\text{ch}}\downarrow&&\downarrow\scriptstyle{\text{ch}}&&\downarrow\\
\widetilde{H}^{2n}(X;\mathbb{Q})&\xrightarrow{\sim}&\widetilde{H}^{2n}(S^2X;\mathbb{Q})&\to&\widetilde{H}^{2n+4}(S^4X;\mathbb{Q})&\to& \cdots
\end{array}

Corollary 7: A calss in $H^{2n}(S^{2n};\mathbb{Z})$ is a Chern characteristic $c_n(E)$ $\Rightarrow$ It is divisble by $(n-1)!$

Proof: $E\to S^{2n}$ we get $c_i(E)=0$ for $n>i>0$.
$\text{ch}(E)=\text{dim}\,E+\frac{1}{n!}s_n(\sigma_1,\dots,\sigma_n)=\text{dim}\,E+\frac{n}{(n+1)!}\sigma_n$ since $s_n(\sigma_1,\dots,\sigma_n)=\sigma_1 s_{n-1}-\sigma_2 s_{n-2}+\cdots+(-1)^{n-2}\sigma_{n-1}s_1+(-1)^{n-1}n\sigma_n$.

2. 2019-01-20 18:51:41
Phantom_Ghost 更新于 Chern characteristic class

Theorem 1: There is a unique sequence of maps $c_i$ assigning to vector bundle $E\to B$ a class $c_i(E)\in H^{2i}(B;\mathbb{Z})$, which depends on isomorphism class of $E$ and the following properties:

$i)$ $c_{i}(f^*E)=f^*c_i(E)$ ($f^*E$ is a pullback bundle)

$ii)$ $c(E_1\oplus E_2)=c_1(E_1)\smile c(E_2)$ where $c=1+\sum_{i>0}c_i$ or equivalently $c_k(E_1\oplus E_2)=\sum_{i+j=k}c_i(E_1)\smile c_j(E_2)$ where $c_0=1$.

$iii)$ $c_i(E)=0$ $i>\text{rank}\,E$

$iv)$ For the canonical line bundle $E\to\mathbb{C}P^\infty$, $c_1(E)$ is a fixed generator pf $H^2(\mathbb{C}P^\infty;\mathbb{Z})$

The proof of the theorem is based on Leray-Hirsch theorem:
Theorem 2 (Leray-Hirsch): Let $\pi:E\to B$ be a fiber bundle, $H^\bullet(E;R)$ is a module over $H^\bullet(B;R)$ ($R$ is commutative ring); $\alpha\in H^\bullet(B;R)$ and $\beta\in H^\bullet(E;R)$, $\alpha\cdot\beta=\pi^*\alpha\smile\beta$; $H^\bullet(E;R)$ is free over $H^\bullet(B;R)$. Evidently, $\forall x\in B$, $F_x\overset{i_x}{\hookrightarrow}E$ induces a surjection on $H^\bullet(E;R)\to H^\bullet(B;R)$ and $H^\bullet(F;R)$ is a free $R$-modlue of finite rank. Moreover we can obtain a basis by choosing elements that map to a basis in $H^\bullet(F;R)$ under $i^*$.

Proof of thm1(existence): $P(\pi):P(E)\to B$ is universal bundle, remove zero section from $\mathbb{C}P^{n-1}$ out of $\mathbb{C}^\times$. Wee want to find $x_i\in H^{2i}(P(E);\mathbb{Z})$ s.t. they restrict to generators of $H^{2i}(\mathbb{C}P^{n-1};\mathbb{Z})$. From universal bundle, there is $g:E\to\mathbb{C}P^\infty$ with linear injections as fibers. Projectivizing $g$ by deleting zero section and then factoring out scalar multiplication induces $P(g):P(E)\to\mathbb{C}P^\infty$
\begin{array}[c]{cc}
P(E) & \\
\scriptstyle{i}\uparrow&\searrow{}^{P(g)}\\
\;\;\;\;\;\;\mathbb{C}P^{n-1}\hookrightarrow& \mathbb{C}P^\infty\\
\end{array}

\begin{array}[c]{cc}
& H^{2i}(P(E);\mathbb{Z})\\
{}^{{}^{P(g)^*}\nearrow}&\downarrow\scriptstyle{i^*}\\
\;\;\;\;\;\;H^{2i}(\mathbb{C}P^\infty;\mathbb{Z})\rightarrow& H^{2i}(\mathbb{C}P^{n-1};\mathbb{Z})
\end{array}
Let $\alpha$ be a generator of $H^2(\mathbb{C}P^\infty;R)$, then $x^i=P(g)^*\alpha^i$, $1,x,x^2,\dots, x^{n-1}$ form a basis of $H^\bullet(P(E);R)$ as a $H^\bullet(E;R)$-module. Their images under $i^*$ map form a basis for the cohomology of fiber $H^\bullet(\mathbb{C}P^{n-1})$. Therefore $x^n+c_1(E)x^{n-1}+\cdots+c_n(E)\cdot 1=0$.
This is how we obtain the Chern class.

Proof of uniqueness follows from the proposition below.

Proposition 3 (Splitting principle): For each vector bundle $E\to B$, there is a space $F(E)$ with a map $F(E)\xrightarrow{p}B$ s.t. the pullback bundle $p^*E\to F(E)$ splits as a direct sum of line bundles and $p^*:H^\bullet(B;\mathbb{Z})\to H^\bullet(P(E);\mathbb{Z})$ is injective.
(This is a similar phenomenon compare with Bott periodicity for K-theory)

Proof of prop.: We have the map $P(E)\xrightarrow{P(\pi)}B$ and pull back $E$ along this map. There is a subbundle in bundle on $P(E)$: $L:=\{(\ell,v)\in P(E)\times|v\in\ell\}$ , $P(\pi)^*E=L\oplus L^\perp$. $H^\bullet(P(E);\mathbb{Z})$ is a free module with basis $1,x,x^2,\dots,x^{n-1}$. Repeat the same construction for $L^\perp$ and we are done.

Remark
1) $E=B\times\mathbb{C}$, it is a pullback of a bundle over one point, since $H^{2i}(\{\text{pt.}\};\mathbb{Z})=0$ for $i>0$, hence $c_i(E)=0$ for $i>0$.

2) $E=E_1\oplus\mathbb{n}$, $c(E)=c(E_1\oplus\mathbb{n})=c(E_1)\smile \underset{=1}{\underbrace{c(\mathbb{n})}}=c(E_1)$.

3) $E\cong L_1\oplus L_2\oplus\cdots\oplus L_n$, uniqueness comes from splitting principle and 4)

4) There is no bundle over $E\to\mathbb{C}P^\infty$ whose sum with the canonical line bundle over $\mathbb{C}P^\infty$ is trivial. i.e assume $E\oplus L\cong\mathbb{n}$, then $1=c(\mathbb{n})=c(E\oplus L)=c(E)\smile c(L)=(1+c_1(E)+\cdots)(1+c_1(L))$ by 1); hence $c(E)=(1+c_1(L))^{-1}=1-c_1(L)+c_1(L)^2+\cdots$ contradicts to 3).

3. 2019-01-20 18:45:33
Phantom_Ghost 发表了帖子 Chern characteristic class

A short note for the content from the seminar of ''topological K-theory'' during WS18/19.

Reference: Allen Hatcher, Vector Bundles and K-Theory.

4. 上周
2019-01-14 22:16:06
Phantom_Ghost 更新于 非欧几何是向量空间不

这也要封？我认为只要是没有逾越理性讨论这点原则都不可以实施封禁处罚，别人能不能理解那是他自己的事，你解释得好不好也是你自己的事情，自由理性讨论是每个用户在学术论坛上应享有的基本权利。如果觉得对方不能理解而懒得进一步交流你完全可以不回复不再参与讨论。

5. 2019-01-10 03:50:36

我想看看天文专业的人是怎么理解卡西尼裂缝之类的东西

另外也不用给自己增添负担强行活跃，顺其自然来兴致就写点内容好了

6. 6周前
2018-12-06 07:44:47

-besetzt-

7. 2018-12-06 07:42:37

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8. 2018-12-06 07:39:28

This is a note written for a seminar of Lie algebra. Please feel free to read and leave comments and suggestions. If you find any typos and errors please inform me.

So far it has not yet been finished and will be updated in the succeeding period of time.

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9. 2月前
2018-11-05 03:30:04

@DTSIo 如果接下来想了解 de Rham 定理, 不妨直接进入层论, 了解一下层的上同调, 层的分解, 以及抽象的 de Rham 定理.

可否推荐下哪本书从这个角度讲的？

10. 3月前
2018-10-21 00:21:13
Phantom_Ghost 发表了帖子 量子输运理论：散射矩阵理论

- 2014 -

粒子数密度算符
一次量子化
$\rho(x)=\psi^\dagger(x)\psi(x)=\int\psi^\dagger(x')\delta(x-x')\psi(x')dx'$
二次量子化形式为
\begin{align*}
\hat{\rho}(x)&=\int\hat{\psi}^\dagger(x')\delta(x-x')\hat\psi(x')dx'\\
&=\sum_{\alpha\beta} \rho_{\alpha\beta} b_\alpha^\dagger b_\beta\\
\rho_{\alpha\beta}(x)&=\int\phi^*_\alpha(x')\delta(x-x')\phi_\beta(x') dx'\\
&=\phi^*_\alpha(x)\phi_\beta(x)
\end{align*}
总粒子数算符为
$\hat{N}=\int\hat\rho(x) dx=\sum_\alpha b_\alpha^\dagger b_\alpha$
流密度算符
在经典力学中 $\mathbf{j}=\rho e\mathbf{v}$，在量子力学中为
$\hat{j}=\frac{e}{2}\int\hat{\psi}^\dagger(x')\left[v(x)\delta(x-x')+\delta(x-x')v(x')\right]\hat\psi(x') dx'$
正则量子化中速度算符为 $\hat{v}=-\frac{i\hbar}{m}\nabla$。于是流密度算符表达为
$\hat{j}=\frac{\hbar e}{2mi}\left[\hat{\psi}^\dagger(\nabla\hat\psi)-(\nabla\hat{\psi}^\dagger)\hat\psi\right]$
量子统计里面有密度矩阵算符$\hat{\rho}=\frac{1}{Z}e^{-\beta(\hat{H}-\mu\hat{N})}$，配分函数 $Z=\text{Tr}\;e^{-\beta(\hat{H}-\mu\hat{N})}$。于是某算符$\hat{A}$的统计平均值为$\langle\hat{A}\rangle=\text{Tr}[\hat\rho\hat{A}]$，于是有
\begin{align*}
\langle b_\alpha^\dagger b_\alpha\rangle&=\text{Tr}\left[\frac{1}{Z}e^{-\beta(\hat{H}-\mu\hat{N})}b_\alpha^\dagger b_\alpha\right]\\
\hat{H}&=\sum_\alpha \varepsilon_\alpha b_\alpha^\dagger b_\alpha\;,\;\hat{N}=\sum_\alpha b_\alpha^\dagger b_\alpha
\end{align*}
可得到Fermi分布
\begin{align*}
f_\alpha&=\langle b_\alpha^\dagger b_\alpha\rangle\\
&=\frac{1}{e^{\beta(\varepsilon_\alpha-\mu)}+1}
\end{align*}
在线性相应理论里面，有著名的Kubo公式。假设受微扰系统为 $H=H_0+H_I$，零温密度矩阵算符为$\hat\rho=\sum_i |i\rangle\langle i|$，密度矩阵的运动方程为（注意不同于一般算符的Heisenberg方程 $i\hbar\dot{A}=[A,H]$）
$i\hbar\frac{d\hat{\rho}}{dt}=[H,\hat\rho]$
假设微扰在$t=0$时刻引入的 $H_I(t)=H_I(t)\theta(t)$，有限温度当$t<0$，密度矩阵为$\hat{\rho}_0=\frac{1}{Z}e^{-\beta(H_0-\mu N)}$ 不含时。当$t\geq 0$，$\hat\rho(t)$可使用微扰法求解。设$\hat\rho(t)$展开为
$\hat\rho(t)=\hat{\rho}_0(t)+\hat{\rho}_1(t)+\hat{\rho}_2(t)+...$
运动方程也展开为
$i\frac{d}{dt}\left\{\hat{\rho}_0(t)+\hat{\rho}_1(t)+\hat{\rho}_2(t)+...\right\}=[H_0+H_I,\hat{\rho}_0(t)+\hat{\rho}_1(t)+\hat{\rho}_2(t)+...]$
比较两边，零阶项为 $i\frac{d\rho_0(t)}{dt}=[H_0,\rho_0(t)]=0$
一阶项为 $i\frac{d\rho_1(t)}{dt}=[H_I,\rho_0(t)]+[H_0,\rho_1(t)]$
一般解为：
\begin{align*}
\hat\rho_0(t)&=\hat\rho_0=\frac{1}{Z}e^{-\beta(H_0-\mu H)}\\
\hat\rho_1(t)&=-i\int_0^t dt'\;e^{-iH_0(t-t')}[H_I(t'),\rho_0]e^{iH_0(t-t')}
\end{align*}
因此一级近似解为 $\hat\rho(t)=\hat\rho_0(t)+\hat\rho_1(t)$
由于算符$A$的统计平均定义为 $\langle A\rangle=\text{Tr}[\hat\rho A]$ ，因此有
$\langle A\rangle_t=\text{Tr}[\hat{\rho}_0 A(t)]+\text{Tr}[\hat{\rho}_1(t) A(t)]\Rightarrow \langle A\rangle_t=\langle A\rangle_0+\text{Tr}[\hat{\rho}_1(t) A(t)]$
于是对于$\hat{\rho}_1(t)$有
\begin{align*}
\text{Tr}[\hat{\rho}_1(t) A(t)]&=-i\int_0^t dt'\;\text{Tr}\left\{e^{-iH_0(t-t')}[H_I(t'),\hat{\rho}_0]e^{iH_0(t-t')} A(t)\right\}\\
&=-i\int_0^t dt'\;\text{Tr}\left\{\hat{\rho}_0[A^{H_0}(t),H_I^{H_0}(t')]\right\}\\
&=-i\int_0^t dt'\langle[A^{H_0}(t),H_I^{H_0}(t')]\rangle
\\\\
A^{H_0}(t)&=e^{iH_0t}A(t)e^{-iH_0 t}\\
H_I^{H_0}(t')&=e^{iH_0t'}H_I(t')e^{-iH_0 t'}
\end{align*}
最后便得到Kubo公式
$\langle A\rangle_t=\langle A\rangle_0-i\int_0^t dt'\langle[A^{H_0}(t),H_I^{H_0}(t')]\rangle$
流密度响应
流密度对外磁场扰动的响应，哈密顿量为
$\hat{H}=\int \hat{\psi}^\dagger(r,t)\frac{1}{2m}\left[P(r)+\frac{q}{c}A(r,t)\right]^2\hat{\psi}(r,t) dr$
在哈密顿量里保持$A$的线性项并忽略$A^2$，有
\begin{align*}
\hat{H}&=\hat{H}_0+\hat{H}_I\\
\hat{H}_0&=\frac{1}{2m}\int \hat{\psi}^\dagger(r,t) P^2(r) \hat{\psi}(r,t) dr\\
\hat{H}_I&=\frac{q}{c}\int A(r,t)\cdot J(r,t) dr\\
J(r,t)&=-\frac{i\hbar}{2m}\left[\hat{\psi}^\dagger(r,t)(\nabla\hat{\psi}(r,t) )-(\nabla \hat{\psi}^\dagger(r,t))\hat{\psi}(r,t)\right]
\end{align*}
根据Kubo定理，流$J(r)$平均为：
$\langle J(r,t)\rangle_t=\langle J(r)\rangle_0-i\int_0^t dt'\int\langle[J(r,t),J(r',t')]\rangle\cdot A(r',t') dr'$
$Q(r,t;r',t')=-i\langle[J(r,t),J(r',t')]\rangle$ 为流-流关联函数。

粒子数密度响应
粒子数密度对外电场扰动响应，其哈密顿量为
\begin{align*}
\hat{H}&=\int\hat{\psi}^\dagger(r,t)\left(\frac{1}{2m}P^2+V_\text{ext}(r,t)\right)\hat{\psi}(r,t) dr\\
&=\hat{H}_0+\hat{H}_I\\
\hat{H}_0&=\frac{1}{2m}\int\hat{\psi}^\dagger(r,t) P^2\hat{\psi}(r,t) dr\\
\hat{H}_I&=\int\hat{\psi}^\dagger(r,t)V_\text{ext}(r,t)\hat{\psi}(r,t) dr=\int\hat{\rho}(r,t)V_\text{ext}(r,t) dr\\
\hat{\rho}(r,t)&=\hat{\psi}^\dagger(r,t)\hat{\psi}(r,t)
\end{align*}
由Kubo定理，粒子数密度平均为
$\langle\hat{\rho}(r,t)\rangle_t=\langle\hat{\rho}(t)\rangle_0-i\int_0^t dt'\int\langle[\hat{\rho}(r,t),\hat{\rho}(r',t')]\rangle V_\text{ext}(r',t') dr'$
定义 $\Pi(r,t;r',t')=-i\langle[\hat{\rho}(r,t),\hat{\rho}(r',t')]\rangle$ 为密度-密度关联函数，或者密度响应函数（Lindahard函数）。在单粒子近似下，场算符可写为
$\hat\psi(r,t)=\sum_i\varphi_i(r)\hat{a}_i(t)\;,\;\hat{a}_i(t)=\hat{a}_i e^{-i\varepsilon_i t}\;(\hbar=1)$
代入密度响应函数中得到
\begin{align*}
\Pi(r,t;r',t')&=-i\langle[\hat{\rho}(r,t),\hat{\rho}(r',t')]\rangle\\
&=-i\langle[\hat{\psi}^\dagger(r,t)\hat\psi(r,t),\hat{\psi}^\dagger(r',t')\hat\psi(r',t')]\rangle\\
&=-i\sum_{ijkl}\varphi^*_i(r)\varphi_j(r)\varphi^*_k(r')\varphi_l(r')e^{i\varepsilon_i t-i\varepsilon_j t+i\varepsilon_k t'-i\varepsilon_l t'}\\
&(\langle a_i^\dagger a_j a_k^\dagger a_l\rangle-\langle a_k^\dagger a_l a_i^\dagger a_j\rangle)
\end{align*}
由于 $a_j a_k^\dagger+a_k^\dagger a_j=\delta_{kj}$，及 $\langle a_k^\dagger a_j\rangle=f_k\delta_{kj}$，其中$f_k$为Fermi分布。便得到
$\Pi(r,t;r',t')=-i\sum_{kl}\varphi^*_l(r)\varphi_k(r)\varphi^*_k(r')\varphi_l(r')e^{i\varepsilon_l(t-t')-i\varepsilon_k(t-t')}(f_l-f_k)$
若满足时间平移对称则可作Fourier变换到频率空间
\begin{align*}
\Pi(r,r';\omega)&=\int \Pi(r,t;r',t')e^{i\omega(t-t')}d(t-t')\\
&=\sum_{kl}\varphi^*_l(r)\varphi_k(r)\varphi^*_k(r')\varphi_l(r')\frac{f_k-f_l}{\omega+(\varepsilon_l-\varepsilon_k)+i0^+}\\
-i\lim_{\eta\to 0^+}&\int_{\infty}^0 e^{i((\varepsilon_l-\varepsilon_k)x}e^{i\omega x}e^{-\eta x}dx=-\frac{1}{\omega+(\varepsilon_l-\varepsilon_k)+i0^+}\\
&(0<t'<t\;,\;t\to\infty)
\end{align*}

Büttiker提出散射矩阵理论（PRB,Vol.46,12485(1992)）及由Datta等发展（PRB,Vol.53,No.24(1996)），讨论低频、弱非线性介观体系中量子输运问题。纳米电子器件在此情况下要保持电流守恒以及规范不变性则须考虑电子间Coulomb作用，可在平均场意义下约化成有效势（需自洽求解），有效势和外电场（导致非平衡态）改变了波函数，波函数也反过来影响有效势。散射矩阵理论可以简明地给出自洽计算以讨论量子输运问题。

研究一个多端口介观导体，端口以$\alpha$标记，每个端口有多种模式。现在先对模型做一些理想处理：假设各端口间存在相干透射，则介观导体须足够小；每个端口与处于热平衡的热库连接，暂时假定端口均处于相同温度$T$，且每个端口载流子化学势各为$\mu_\alpha$，平衡时化学势为$\mu$，弱外场（各端口电压$V_\alpha$）激励下有$\mu_\alpha=\mu+e V_\alpha$；在介观导体上加静磁场，各端口间透射振幅可用散射矩阵$S$描述。$S$是幺正矩阵，加磁场时左右入射体系不对称 $S_{\alpha\beta}(-B)=S_{\beta\alpha}(B)$，$S(E_F,U)$与Fermi能以及有效势都有关系从而依赖于端口电势。散射矩阵理论在平均场意义下考虑电子间相互作用，形式上仍是单粒子理论，引入自洽平均场在于既要考虑相互作用又要确保流守恒以及规范不变性。
现在来具体计算端口$\alpha$的电流，可由电流守恒再推出流经器件的电流。设端口上电子的色散关系为 $E_{\alpha m}(k)=E_{\alpha m}(0)+\frac{\hbar^2 k_{\alpha m}^2}{2m}$，散射态为 $\chi_{\alpha m}^\pm(x,y)=e^{\pm ik_{\alpha m}x}\phi_{\alpha m}(y)$，这里由于是散射态故不需要归一化，$\phi_{\alpha m}(y)$为$y$轴方向驻波（量子限域），$e^{\pm ik_{\alpha m}x}$为$x$轴方向平面波（行波）。通常端口内电子波函数一般形式写为：
$\psi(r,t)=\sum_n [a_n(t)\chi_n^+(r) +b_n(t)\chi_n^-(r)]$
因此在端口内散射态的场算符为
\begin{align*}
\hat{\psi}(r,t)&=\sum_n\int\frac{dE_{\alpha m}}{2\pi\hbar}\sqrt{v_{\alpha m}}\left\{\chi^+_{\alpha m}(r)\hat{a}_{\alpha m}(E_{\alpha m})+\chi^-_{\alpha m}(r)\hat{b}_{\alpha m}(E_{\alpha m})\right\}e^{-i\frac{E_{\alpha m}}{\hbar}t}\\
\hat{\psi}^\dagger(r,t)&=\sum_n\int\frac{dE_{\alpha m}}{2\pi\hbar}\sqrt{v_{\alpha m}}\left\{\chi^{+ *}_{\alpha m}(r)\hat{a}^\dagger_{\alpha m}(E_{\alpha m})+\chi^{- *}_{\alpha m}(r)\hat{b}^\dagger_{\alpha m}(E_{\alpha m})\right\}e^{i\frac{E_{\alpha m}}{\hbar}t}
\end{align*}
$v_{\alpha m}=\frac{\hbar k_{\alpha m}}{m}$为速度，作为入射流归一化常数。二次量子化下电流密度算符为
$\hat{J}(r,t)=-i\frac{e\hbar}{2m}\left\{\hat{\psi}^\dagger(\nabla\hat{\psi})-(\nabla\hat{\psi}^\dagger)\hat{\psi}\right\}$
则通过$\alpha$端口电流算符对应着电流密度在$x_\alpha$上沿截面的积分 $I_\alpha(t)=\int \hat{J}(r,t)\left|\right._{x_\alpha}\;dy_{\alpha}$。注意只有散射态对电流有贡献（尽管电流 密度分布可依赖于局域态），于是可把散射态场算符代入电流表达式：
$\frac{\partial\hat{\psi}^\dagger}{\partial x}\hat{\psi}=\sum_{mn}\sqrt{v_{\alpha m} v_{\alpha n}}e^{-i(E_{\alpha n}-E_{\alpha m})t/\hbar}\left(\frac{d\chi_{\alpha m}^{+ *}}{dx}\hat{a}_{\alpha m}^\dagger+\frac{d\chi_{\alpha m}^{- *}}{dx}\hat{b}_{\alpha m}^\dagger\right)\left(\chi_{\alpha n}^+\hat{a}_{\alpha n}+\chi_{\alpha n}^-\hat{b}_{\alpha n}\right)$
乘积展开得到四个项
\begin{align*}
\frac{d\chi_{\alpha m}^{+ *}}{dx}\chi_{\alpha n}^+\hat{a}_{\alpha m}^\dagger\hat{a}_{\alpha n}\;&,\;\frac{d\chi_{\alpha m}^{+ *}}{dx}\chi_{\alpha n}^-\hat{a}_{\alpha m}^\dagger\hat{b}_{\alpha n}\\
\frac{d\chi_{\alpha m}^{- *}}{dx}\chi_{\alpha n}^+\hat{b}_{\alpha m}^\dagger\hat{a}_{\alpha n}\;&,\;\frac{d\chi_{\alpha m}^{- *}}{dx}\chi_{\alpha n}^-\hat{b}_{\alpha m}^\dagger\hat{b}_{\alpha n}
\end{align*}
同理
$\hat{\psi}^\dagger\frac{\partial\hat{\psi}}{\partial x}=\sum_{mn}\sqrt{v_{\alpha m} v_{\alpha n}}e^{-i(E_{\alpha n}-E_{\alpha m})t/\hbar}\left(\chi_{\alpha m}^{+*}\hat{a}_{\alpha m}^\dagger+\chi_{\alpha m}^{-*}\hat{b}_{\alpha m}^\dagger\right)\left(\frac{d\chi_{\alpha n}^{+}}{dx}\hat{a}_{\alpha n}+\frac{d\chi_{\alpha n}^{-}}{dx}\hat{b}_{\alpha n}\right)$
展开也得四项
\begin{align*}
\chi_{\alpha m}^{+*}\frac{d\chi_{\alpha n}^{+}}{dx}\hat{a}_{\alpha m}^\dagger\hat{a}_{\alpha n}\;&,\;\chi_{\alpha m}^{+*}\frac{d\chi_{\alpha n}^{-}}{dx}\hat{a}_{\alpha m}^\dagger\hat{b}_{\alpha n}\\
\chi_{\alpha n}^{+*}\frac{d\chi_{\alpha n}^{-}}{dx}\hat{b}_{\alpha m}^\dagger\hat{a}_{\alpha n}\;&,\;\chi_{\alpha m}^{-*}\frac{d\chi_{\alpha n}^{-}}{dx}\hat{b}_{\alpha m}^\dagger\hat{b}_{\alpha n}
\end{align*}
由于 $\hat{J}\sim\hat{\psi}^\dagger(\nabla\hat{\psi})-(\nabla\hat{\psi}^\dagger)\hat{\psi}$，由乘积项差再作横向积分可得
$I_{mn}^{\sigma_1\sigma_2}(E,E')=\frac{e\hbar}{2mi}\int dy_\alpha\left[\chi_{\alpha m}^{\sigma_1 *}(E)\frac{d\chi_{\alpha n}^{\sigma_2}(E')}{dx}-\frac{d\chi_{\alpha m}^{\sigma_1 *}(E)}{dx}\chi_{\alpha n}^{\sigma_2 *}(E')\right]$
设$E=k^2,E'=q^2$（原子单位），求出微分后积分化为 $I_{mn}^{\sigma_1\sigma_2}(E,E')=\frac{e\hbar}{2m}\delta_{mn}(\sigma_2 q+\sigma_1 k) e^{-i(\sigma_1 k-\sigma_2 q)x}$，当$E\simeq E',k\simeq q$，得到 $I_{mn}^{\sigma_1\sigma_2}(E,E')=\frac{e\hbar}{m}\delta_{mn}\delta_{\sigma_1\sigma_2}\sigma_1 k$，再利用速度定义 $v_n(E)=\frac{\hbar k}{m}$，得到 $I_{mn}^{\sigma_1\sigma_2}(E,E')=\delta_{mn}\delta_{\sigma_1\sigma_2}\sigma_1 e v_n(E)$。若假定是低频 $E'=E+\hbar\omega$，则$E\simeq E'$，上述积分可近似为
$I_{mn}^{\sigma_1\sigma_2}(E,E')=\delta_{mn}\delta_{\sigma_1\sigma_2}\sigma_1 e \sqrt{v_{\alpha m}(E)v_{\alpha n}(E')}$
注意到所有$a^\dagger b,b^\dagger a$项涉及的积分是$\sigma_1\neq\sigma_2$，故这些项为零。最后电流算符表达为
$\hat{I}_\alpha(t)=\frac{e}{(2\pi\hbar)^2}\sum_m\iint dE dE'\;e^{i(E-E')t/\hbar}\left[\hat{a}_{\alpha m}^\dagger(E)\hat{a}_{\alpha m}(E')-\hat{b}_{\alpha m}^\dagger(E)\hat{b}_{\alpha m}(E')\right]$
$\hat{a}_{\alpha m}^\dagger\hat{a}_{\alpha m}$为入射流的粒子数密度，$\hat{b}_{\alpha m}^\dagger\hat{b}_{\alpha m}$则为出射流的粒子数密度。净电流由通过端口的电子数确定。
对电流作Fourier变换
\begin{align*}
\hat{I}_\alpha(\omega)&=\int dt\;\hat{I}_\alpha(t)e^{i\omega t}\\
&=\frac{1}{2\pi}\frac{e}{\hbar^2}\int dE dE'\sum_m\left[\hat{a}_{\alpha m}^\dagger(E)\hat{a}_{\alpha m}(E')-\hat{b}_{\alpha m}^\dagger(E)\hat{b}_{\alpha m}(E')\right]\int e^{i(E-E')t/\hbar}e^{i\omega t}dt\\
\hat{I}_\alpha(\omega)&=\frac{e}{2\pi\hbar}\sum_m\iint dE dE'\;\delta(E-E'+\hbar\omega)\left[\hat{a}_{\alpha m}^\dagger(E)\hat{a}_{\alpha m}(E')-\hat{b}_{\alpha m}^\dagger(E)\hat{b}_{\alpha m}(E')\right]\\
&=\frac{e}{h}\sum_m\int dE\left[\hat{a}_{\alpha m}^\dagger(E)\hat{a}_{\alpha m}(E+\hbar\omega)-\hat{b}_{\alpha m}^\dagger(E)\hat{b}_{\alpha m}(E+\hbar\omega)\right]
\end{align*}
矩阵表达形式 $\hat{\boldsymbol{a}}_\alpha^\dagger=\{\hat{a}_{\alpha m}^\dagger\}$，求和变为矩阵求迹，电流公式有简洁的表达式
$\hat{I}_\alpha(\omega)=\frac{e}{h}\int dE\;\text{Tr}\left[\hat{\boldsymbol{a}}_\alpha^\dagger(E)\hat{\boldsymbol{a}}_\alpha(E+\hbar\omega)-\hat{\boldsymbol{b}}_\alpha^\dagger(E)\hat{\boldsymbol{b}}_\alpha(E+\hbar\omega)\right]$
关于低频情况，能量$E$接近Fermi能，若$k\sim q$，则要求$\hbar\omega\ll E_F$，对通常金属$\omega\sim 10^{10}Hz$也满。由于波函数一般为$\psi\sim a\chi^++b\chi^-$，其中$a$为入射波振幅，$b$为出射波振幅。根据散射矩阵的定义 $b=S a$，考虑二次量子化场算符表达下，$\hat\psi=\hat{a}\chi^++\hat{b}\chi^-$，故也有$\hat{\boldsymbol{b}}=\boldsymbol{S}\hat{\boldsymbol{a}}$，得到
$\hat{I}_\alpha=\frac{e}{h}\int dE\;\text{Tr}\left[\hat{\boldsymbol{a}}_\alpha^\dagger(E)\hat{\boldsymbol{a}}_\alpha(E)-\sum_{\beta\gamma}S_{\alpha\beta}^\dagger\hat{\boldsymbol{a}}_\beta^\dagger(E)S_{\alpha\gamma}\hat{\boldsymbol{a}}_\gamma(E)\right]$
于是直流情形（$\omega=0$）的多端口Büttiker公式便导出为
$\hat{I}_\alpha(\omega)=\frac{e}{h}\int dE\;\text{Tr}\left[\hat{\boldsymbol{a}}_\alpha^\dagger(E)\hat{\boldsymbol{a}}_\alpha(E)-\hat{\boldsymbol{b}}_\alpha^\dagger(E)\hat{\boldsymbol{b}}_\alpha(E)\right]$
此时 $\hat{b}_{\alpha m}=\sum_{\beta n}S_{\alpha m,\beta n}\hat{a}_{\beta n}\;,\;\hat{\boldsymbol{b}}_\alpha=S_{\alpha\beta}\hat{\boldsymbol{a}}_\beta$，$\boldsymbol{S}$即为散射矩阵。
注意到流过端口的电流因为电流守恒的缘故必然流过器件中心区，而上面电流时在端口内计算的，可观测量必须是统计平均值，在端口内做统计平均 $I_\alpha=\langle\hat{I}_\alpha\rangle$。假设散射与自旋无关，因此散射矩阵中不含自旋指标，所以引入自旋简并因子2得到
\begin{align*}
I_\alpha&=\frac{2e}{h}\int dE\;\text{Tr}\left[\langle\hat{\boldsymbol{a}}_\alpha^\dagger\hat{\boldsymbol{a}}_\alpha\rangle-\sum_{\beta\gamma}S^\dagger_{\alpha\beta}S_{\alpha\gamma}\langle\hat{\boldsymbol{a}}_\beta^\dagger\hat{\boldsymbol{a}}_\gamma\rangle\right]\\
&=\frac{2e}{h}\int dE\;\sum_{\beta}\text{Tr}\left[\delta_{\alpha\beta}\delta_{\alpha\gamma}-S^\dagger_{\alpha\beta}S_{\alpha\gamma}\right]\langle\hat{\boldsymbol{a}}_\beta^\dagger\hat{\boldsymbol{a}}_\gamma\rangle\\
\langle\hat{\boldsymbol{a}}_\beta^\dagger\hat{\boldsymbol{a}}_\gamma\rangle&=f_\beta\delta_{\beta\gamma}\;,\;f_\beta=\frac{1}{e^{\frac{E-\mu_\beta}{k_B T}}+1}\\
I_\alpha&=\frac{2e}{h}\int dE\;\sum_{\beta\gamma}\text{Tr}\left[\delta_{\alpha\beta}\delta_{\alpha\gamma}-S^\dagger_{\alpha\beta}S_{\alpha\gamma}\right]f_\beta\delta_{\beta\gamma}\\
&=\frac{2e}{h}\sum_\beta\int dE\;\text{Tr}\left[\delta_{\alpha\beta}-S^\dagger_{\alpha\beta}S_{\alpha\beta}\right]f_\beta
\end{align*}
定义矩阵 $A_{\alpha\beta}=\delta_{\alpha\beta}-S^\dagger_{\alpha\beta}S_{\alpha\beta}$，直流Büttiker公式简化为
$I_\alpha=\frac{2e}{h}\sum_\beta\int dE\;\text{Tr}A_{\alpha\beta}f_\beta$
物理