Intro: 实对称矩阵的正交对角化
厄米矩阵的幺正对角化
接下来在复变量的范围内考虑, 也就是矩阵元都在复数的范围内讨论.
厄米矩阵
定义
厄米矩阵定义为满足
\begin{align} U^{\dagger} \equiv U \end{align}的矩阵.
\(2\times 2\)
比如对于一个 \(2\times 2\) 的厄米矩阵应满足
\begin{align} \left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right) = \left(\begin{array}{cc} a^{*} & c^{*} \\ b^{*} & d^{*} \\ \end{array}\right) \end{align}得
\begin{align} a^{*} &= a \\ d^{*} &= d \\ b^{*} &= c \\ c^{*} &= b \end{align}也就是说 \(a, d\) 都是实数, \(b, c\) 互为复共轭. 所以 \(2\times 2\) 的厄米矩阵最一般的 形式可以写为
\begin{align} \left(\begin{array}{cc} u & r^{e^{-\mathrm{i}\phi}} \\ r^{e^{\mathrm{i}\phi}} & v \\ \end{array}\right) \end{align}其中 \(u, v, r, \phi\) 都是实数. 可见, \(2\times 2\) 的厄米矩阵中四个独立的实变量. 本来 \(2\times 2\) 的矩阵有八个变量嘛, 对角元是实数, 去掉两个独立实变量, 非对角元的 互为复共轭, 再去年两个, 还剩下四个.
\(n\times n\)
更一般地, 可以证明, 对于 \(n\times n\) 的厄米矩阵, 对角元都是实的, 非对角元是一组 互为复共轭的复数, 有 \(n^2\) 个独立实变量, 正好相当于一个实矩阵的独立变量个数.
性质
- 厄米矩阵的本征值都是实数.
- 厄米矩阵对应不同特征值的特征向量相互正交.
幺正矩阵
定义
幺正矩阵定义为
\begin{align} U U^{\dagger} \equiv 1 \end{align}其中 \(1\) 表示单位矩阵. 等价地
\begin{align} U^{\dagger} \equiv U^{-1} \end{align}\(2\times 2\)
\(2\times 2\) 的幺正矩阵应满足
\begin{align} \left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right) \left(\begin{array}{cc} a^{*} & c^{*} \\ b^{*} & d^{*} \\ \end{array}\right) = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}\right) \end{align}也就是
\begin{align} |a|^2 + |b|^2 =& 1 \\ |c|^2 + |d|^2 =& 1 \\ a^{*}c + b^{*}d =& 0 \\ ac^{*} + bd^{*} =& 0 \end{align}同样, 由 \(U^{\dagger}U =1\) 可得
\begin{align} |a|^2 + |c|^2 =& 1 \\ |b|^2 + |d|^2 =& 1 \end{align}所以它和行列都正交
\begin{align} |a|^2 + |b|^2 =& 1 \\ |c|^2 + |d|^2 =& 1 \\ |a|^2 + |c|^2 =& 1 \\ |b|^2 + |d|^2 =& 1 \\ \end{align}只剩下四个独立的实变量.
它最一般的形式可以表示为
\begin{align} \left(\begin{array}{cc} e^{\mathrm{i}\alpha}\cos\theta & -e^{\mathrm{i}\alpha}e^{\mathrm{i}\phi}\sin\theta \\ e^{\mathrm{i}\beta}e^{-\mathrm{i}\phi}\sin\theta & e^{\mathrm{i}\beta}\cos\theta \\ \end{array}\right) \end{align}总之, 它有四个独立的实变量.
问题
厄米矩阵一定可以通过幺正变换对角化吗?
例: BCS Mean-Field Hamiltonian 的对角化
BCS 平均场的 Hamiltonian 为
\begin{align} H_{\mathrm{MF}} = \sum_{\vec{k}}\left\{ \left(\hat{c}^{\dagger}_{\vec{k}, \uparrow}\quad \hat{c}_{-\vec{k}, \downarrow}\right) \left(\begin{array}{cc} \epsilon_{\vec{k}} - \mu & \Delta\\ \Delta^{*} & -(\epsilon_{\vec{k}} - \mu) \end{array}\right) \left( \begin{array}{c} \hat{c}_{\vec{k}, \uparrow} \\ \hat{c}^{\dagger}_{-\vec{k}, \downarrow} \end{array} \right) + (\epsilon_{\vec{k}} - \mu) \right\} - \frac{V}{g}|\Delta|^2 \end{align}下面考虑对角化其中写成矩阵的部分
\begin{align} H_0(\vec{k}) = \left(\hat{c}^{\dagger}_{\vec{k}, \uparrow}\quad \hat{c}_{-\vec{k}, \downarrow}\right) \left(\begin{array}{cc} \epsilon_{\vec{k}} - \mu & \Delta\\ \Delta^{*} & -(\epsilon_{\vec{k}} - \mu) \end{array}\right) \left( \begin{array}{c} \hat{c}_{\vec{k}, \uparrow} \\ \hat{c}^{\dagger}_{-\vec{k}, \downarrow} \end{array} \right) \end{align}为了方便, 省略一部分下标 \(\vec{k}\) , 并记 \(\epsilon_{\vec{k}} - \mu = \xi_{\vec{k}}\) , \(k = (\vec{k}, \uparrow), -k=(-\vec{k}, \downarrow)\)
\begin{align} H_0 = \left(\hat{c}^{\dagger}_k\quad \hat{c}_{-k}\right) \left(\begin{array}{cc} \xi & \Delta\\ \Delta^{*} & -\xi \end{array}\right) \left( \begin{array}{c} \hat{c}_k \\ \hat{c}^{\dagger}_{-k} \end{array} \right) \end{align}它是厄米的, 试图通过幺正变换将其对角化
\begin{align} H_0 = \left(\hat{c}^{\dagger}_k\quad \hat{c}_{-k}\right) U U^{\dagger} \left(\begin{array}{cc} \xi & \Delta\\ \Delta^{*} & -\xi \end{array}\right) U U^{\dagger} \left( \begin{array}{c} \hat{c}_k \\ \hat{c}^{\dagger}_{-k} \end{array} \right) \end{align}两样, 比较特别, 这个厄米矩阵还是 traceless 的, 也就是对角元的和为 0 .那它的本征 值可以直接得到两个本征值为 \(\pm E\) , 其中
\begin{align} E = \sqrt{|\Delta|^2 + \xi^2} \end{align}将本征值带回来解本征矢
\begin{align} \left(\begin{array}{cc} \xi & \Delta\\ \Delta^{*} & -\xi \end{array}\right) \left(\begin{array}{c} C_{1\pm}\\ C_{2\pm} \end{array}\right) = \pm E \left(\begin{array}{c} C_{1\pm}\\ C_{2\pm} \end{array}\right) \end{align}可得
\begin{align} \frac{C_{1\pm}}{C_{2\pm}} = \frac{\Delta}{\pm E - \xi} \end{align}对于幺正矩阵来说
\begin{align} |C_{1\pm}|^2 + |C_{2\pm}|^2 = 1 \end{align}有
\begin{align} \left(\begin{array}{c} C_{1\pm}\\ C_{2\pm} \end{array}\right) = \left(\begin{array}{c} \frac{\Delta}{\sqrt{2E(E\mp \xi)}}\\ \frac{\pm E - \xi}{\sqrt{2E(E\mp \xi)}} \end{array}\right) = \left(\begin{array}{c} \frac{\Delta}{\sqrt{2E(E\mp \xi)}}\\ \frac{\pm |\Delta|}{\sqrt{2E(E\pm \xi)}} \end{array}\right) \end{align}可以看出, 很明显
\begin{align} |C_{1+}|^2 + |C_{1-}|^2 = 1\\ |C_{2+}|^2 + |C_{2-}|^2 = 1 \\ C_{1\pm} = \mp C_{2\mp}e^{\mathrm{i}\phi} \end{align}其中 \(\phi\) 是 \(\Delta\) 的相位, 即 \(\phi = \mathrm{Arg} (\Delta)\) . 所以
\begin{align} U^{\dagger} = \left(\begin{array}{cc} C_{1+} & C_{2+} \\ C_{1-} & C_{2-} \end{array}\right) = \left(\begin{array}{cc} -C_{2-} e^{\mathrm{i}\phi} & C_{2+} \\ C_{2+} e^{\mathrm{i}\phi} & C_{2-} \end{array}\right) \end{align}由它们的具体表达式可以看出, \(C_{2\pm}\) 是实的. 它们有三个独立的实变量. 而一般的 \(2\times 2\) 的幺正有四个独立的实变量. 这里少了一个, 是因为厄米矩阵的两对角元中 只有一个实变量, 也少了一个.
采用新的记号
\begin{align} u &= C_{2-} \\ v &= -C_{2+} \end{align}变为
\begin{align} U^{\dagger} = \left(\begin{array}{cc} -u e^{\mathrm{i}\phi} & -v \\ -v e^{\mathrm{i}\phi} & u \end{array}\right) \end{align}这里, 关于 \(\Delta\) 的相位 \(\phi\) 的问题, 我还没有完全弄清楚, 有些地方(Reference [3])说:
These phases may beabsorbed into the definitions of the fermion operators themselves.
而同老师讲义中说
对空间平移对称的情况, 可取为实数, 即定相位为零.
(当空间不均匀时, 及讨论不同超导体连接时, \(\phi\) 不可忽略).
接下来的处理, 可以理解为将相位吸收到 Fermionic 算符的定义中去, 而不放在 \(U^{\dagger}\) 中, 这样新的基矢为
\begin{align} \left(\begin{array}{c} \alpha\\ \beta' \end{array}\right) = \left(\begin{array}{cc} -u e^{\mathrm{i}\phi} & -v \\ -v e^{\mathrm{i}\phi} & u \end{array}\right) \left( \begin{array}{c} -e^{\mathrm{i}\phi}\hat{c}_k \\ \hat{c}^{\dagger}_{-k} \end{array} \right) \\ = \left( \begin{array}{cc} u & -v \\ v & u \end{array} \right) \left( \begin{array}{c} \hat{c}_k \\ \hat{c}^{\dagger}_{-k} \end{array} \right) \end{align}或者是理解这里选 \(\phi = -\pi\) , 那么 \(-e^{\mathrm{i}\phi} = 1\) , 同样也得到
\begin{align} \left(\begin{array}{c} \alpha\\ \beta' \end{array}\right) = \left( \begin{array}{cc} u & -v \\ v & u \end{array} \right) \left( \begin{array}{c} \hat{c}_k \\ \hat{c}^{\dagger}_{-k} \end{array} \right) \end{align}总之, 搞定这个相位之后有
\begin{align} U^{\dagger} = \left( \begin{array}{cc} u & -v \\ v & u \end{array} \right) \end{align}而新的基矢
\begin{align} \left(\begin{array}{c} \alpha\\ \beta' \end{array}\right) = U^{\dagger} \left( \begin{array}{c} \hat{c}_k \\ \hat{c}^{\dagger}_{-k} \end{array} \right) = \left( \begin{array}{cc} u & -v \\ v & u \end{array} \right) \left( \begin{array}{c} \hat{c}_k \\ \hat{c}^{\dagger}_{-k} \end{array} \right) \end{align}这样, 就得到, 经过变换后
\begin{align} H_0 =& \left(\hat{c}^{\dagger}_k\quad \hat{c}_{-k}\right)U \left(\begin{array}{cc} E & 0\\ 0 & -E \end{array}\right) U^{\dagger} \left( \begin{array}{c} \hat{c}_k \\ \hat{c}^{\dagger}_{-k} \end{array} \right) \\ =& \left(\alpha^{\dagger}\quad \beta'^{\dagger}\right) \left(\begin{array}{cc} E & 0\\ 0 & -E \end{array}\right) \left(\begin{array}{c} \alpha\\ \beta' \end{array}\right) \\ =& E \left(\alpha^{\dagger} \alpha - \beta'^{\dagger} \beta' \right) \end{align}可以看出, 准粒子 \(\alpha\) 的激发能是 \(E\) , 而准粒子 \(\beta'\) 的激发能是 \(-E\) . 为了使基态是准粒子的真空态, 所有准粒子的激发能应该都是非负的, 所以定义新的准粒 子
\begin{align} \beta = \beta'^{\dagger} \end{align}这时
\begin{align} H_0 = E(\alpha^{\dagger}\alpha - \beta\beta^{\dagger}) = E(\alpha^{\dagger}\alpha + \beta\beta^{\dagger}) -E \end{align}所以最终有
\begin{align} H_{MF} = \sum_{\vec{k}} \left[ E_{\vec{k}} (\alpha_{\vec{k}}^{\dagger}\alpha_{\vec{k}} + \beta_{\vec{k}}^{\dagger}\beta_{\vec{k}}) + (\epsilon_{\vec{k}} - \mu -E_{\vec{k}} ) \right] - \frac{V}{g}|\Delta|^2 \end{align}where
\begin{align} \left(\begin{array}{c} \alpha_{\vec{k}}\\ \beta_{\vec{k}}^{\dagger} \end{array}\right) = U^{\dagger} \left( \begin{array}{c} \hat{c}_k \\ \hat{c}^{\dagger}_{-k} \end{array} \right) = \left( \begin{array}{cc} u_{\vec{k}} & -v_{\vec{k}} \\ v_{\vec{k}} & u_{\vec{k}} \end{array} \right) \left( \begin{array}{c} \hat{c}_k \\ \hat{c}^{\dagger}_{-k} \end{array} \right) \end{align}