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Last time:Lec_02
希望作用量中总只出现一阶项, 并考虑了一般相空间中的坐标.
我们考虑 2 n 2n 2 n
ξ I e.g. ξ I = ( x i , p i ) \xi^I\quad\text{e.g.}\quad\xi^{I}=(x^i,p_i) ξ I e.g. ξ I = ( x i , p i ) Q: 为什么要写成 1st-order 的形式?A: 1st-order 的计算比较方便, 例如离散方法中的插值更容易实现, 数值上也更稳定. 这是最直观的理由.
Q: 我们既然允许不依赖任何坐标系, 为什么不能允许任意阶导数出现?A: 这里想传达的是: 到底哪些东西可以任意, 哪些必须做出一个选择.
一般的作用量
S = ∫ path [ γ I ( ξ , t ) d ξ I − H ( ξ , t ) d t ] S=\int_{\text{path}}\left[{\large\gamma}_{I}(\xi,t)\mathrm{d}\xi^I-H(\xi,t)\mathrm{d}t\right] S = ∫ path [ γ I ( ξ , t ) d ξ I − H ( ξ , t ) d t ] 考虑 Berry Connection
γ I d ξ I = p i d x i + q A i ( x , t ) d x i + ℏ a i ( p ) d p i ⏟ Berry connection {\large\gamma}_I\mathrm{d}\xi^I
=
p_i\mathrm{d}x^i
+qA_i(\mathbf{x},t)\mathrm{d}x^i
+\underbrace{\hbar a^i(\mathbf{p})\mathrm{d}p_i}_{\text{Berry connection}} γ I d ξ I = p i d x i + q A i ( x , t ) d x i + Berry connection ℏ a i ( p ) d p i Gauge of the Action
这个作用量里面哪些是物理的, 哪些是非物理的? 非物理的部分对应规范的选取. 若考虑
S → S + χ ∣ endpoints S\to S+\chi\Big|_{\text{endpoints}} S → S + χ endpoints 也就是说, 两个 action 只差一个 boundary term, 因而给出相同的物理. 这就是 action 中的 gauge. 为了看到这一点, 考虑变换
γ I ( ξ , t ) → γ I ( ξ , t ) + ∂ I χ ( ξ , t ) H ( ξ , t ) → H ( ξ , t ) − ∂ t χ ( ξ , t ) \begin{align}&{\large\gamma}_I(\xi,t)\to{\large\gamma}_I(\xi,t)+\partial_I\chi(\xi,t)\\& H(\xi,t)\to H(\xi,t)-\partial_t\chi(\xi,t)\end{align} γ I ( ξ , t ) → γ I ( ξ , t ) + ∂ I χ ( ξ , t ) H ( ξ , t ) → H ( ξ , t ) − ∂ t χ ( ξ , t ) ^de958e
则给出相同的 action.
例：电磁场
做规范变换
A i ( x , t ) → A i ( x , t ) + ∂ i φ ( x , t ) , A 0 ( x , t ) → A 0 ( x , t ) − ∂ t φ ( x , t ) A_i(\mathbf{x},t)\to A_i(\mathbf{x},t)+\partial_i\varphi(\mathbf{x},t),\qquad A_0(\mathbf{x},t)\to A_0(\mathbf{x},t)-\partial_t\varphi(\mathbf{x},t) A i ( x , t ) → A i ( x , t ) + ∂ i φ ( x , t ) , A 0 ( x , t ) → A 0 ( x , t ) − ∂ t φ ( x , t ) 这正对应上面
γ I → γ I + ∂ I χ , H → H − ∂ t χ \gamma_I\to\gamma_I+\partial_I\chi,\qquad H\to H-\partial_t\chi γ I → γ I + ∂ I χ , H → H − ∂ t χ 的结构. 同样, Berry connection 也有 gauge.
一般地, 若
p i d x i ↦ − x i d p i p_i\mathrm{d}x^i\mapsto -x^i\mathrm{d}p_i p i d x i ↦ − x i d p i 则取
χ = − p i x i \chi=-p_ix^i χ = − p i x i
接下来要讨论的就是 Simple Harmonic Oscillator 和 Symmetry 与 Noether theorem
Change of coordinate
既然这个理论不依赖于具体坐标, 现在就来试着改变坐标系. 考虑
ξ I = ξ I ( ξ ~ I ~ , t ) \xi^I=\xi^I(\tilde{\xi}^{\tilde{I}},t) ξ I = ξ I ( ξ ~ I ~ , t )
那么
d ξ I = ∂ ξ I ∂ ξ ~ I ~ d ξ ~ I ~ + ∂ ξ I ∂ t d t \mathrm{d}\xi^I=\frac{\partial\xi^I}{\partial\tilde{\xi}^{\tilde{I}}}\mathrm{d}\tilde{\xi}^{\tilde{I}}+\frac{\partial\xi^I}{\partial t}\mathrm{d}t d ξ I = ∂ ξ ~ I ~ ∂ ξ I d ξ ~ I ~ + ∂ t ∂ ξ I d t 如果要求作用量描述相同的物理, 我们需要新的 γ \gamma γ
γ ~ I ~ ( ξ ~ I ~ , t ) = γ I ( ξ ( ξ ~ , t ) , t ) ∂ ξ I ∂ ξ ~ I ~ \tilde{{\large\gamma}}_{\tilde{I}}(\tilde{\xi}^{\tilde{I}},t)={\large\gamma}_I(\xi(\tilde{\xi},t),t)\frac{\partial\xi^I}{\partial\tilde{\xi}^{\tilde{I}}} γ ~ I ~ ( ξ ~ I ~ , t ) = γ I ( ξ ( ξ ~ , t ) , t ) ∂ ξ ~ I ~ ∂ ξ I 同样, Hamiltonian 也跟着变换.
H ~ ( ξ ~ , t ) = H ( ξ ( ξ ~ , t ) , t ) − ∂ t ξ I γ I ( ξ ( ξ ~ , t ) , t ) \tilde{H}(\tilde{\xi},t)=H(\xi(\tilde{\xi},t),t)-\partial_t\xi^I{\large\gamma}_I(\xi(\tilde{\xi},t),t) H ~ ( ξ ~ , t ) = H ( ξ ( ξ ~ , t ) , t ) − ∂ t ξ I γ I ( ξ ( ξ ~ , t ) , t ) 关于上下标
d ξ I = ∂ ξ I ∂ ξ ~ I ~ d ξ ~ I ~ + ∂ ξ I ∂ t d t \mathrm{d}\xi^I=\frac{\partial\xi^I}{\partial\tilde{\xi}^{\tilde{I}}}\mathrm{d}\tilde{\xi}^{\tilde{I}}+\frac{\partial\xi^I}{\partial t}\mathrm{d}t d ξ I = ∂ ξ ~ I ~ ∂ ξ I d ξ ~ I ~ + ∂ t ∂ ξ I d t 上下指标都配对消掉了, 对 γ \gamma γ
什么时候需要这样的坐标变换?
SHO 是一个例子.
先做一个无量纲化, 取
x ′ = m ω x , p ′ = p m ω x'=\sqrt{m\omega}\,x,\qquad p'=\frac{p}{\sqrt{m\omega}} x ′ = mω x , p ′ = mω p p d x − ω ( p 2 2 m ω + m ω 2 x 2 ) d t = p ′ d x ′ − ω 2 ( p ′ 2 + x ′ 2 ) d t p\mathrm{d}x-\omega(\frac{p^2}{2m\omega}+\frac{m\omega}{2}x^2)\mathrm{d}t=p'\mathrm{d}x'-\frac{\omega}{2}(p'^2+x'^2)\mathrm{d}t p d x − ω ( 2 mω p 2 + 2 mω x 2 ) d t = p ′ d x ′ − 2 ω ( p ′2 + x ′2 ) d t 这迫使我们想到采用相空间的极坐标
− ρ 2 sin 2 θ d θ + ρ sin θ cos θ d ρ − ω 2 ρ 2 d t -\rho^2\sin^2\theta \mathrm{d}\theta+\rho\sin\theta\cos\theta \mathrm{d}\rho-\frac{\omega}{2}\rho^2\mathrm{d}t − ρ 2 sin 2 θ d θ + ρ sin θ cos θ d ρ − 2 ω ρ 2 d t 现在还很丑, 需要找一个 χ \chi χ
− ρ 2 2 d θ − ω 2 ρ 2 d t + d ( ρ 2 2 cos θ sin θ ) -\frac{\rho^2}{2}\mathrm{d}\theta-\frac{\omega}{2}\rho^2\mathrm{d}t+\mathrm{d}(\frac{\rho^2}{2}\cos\theta\sin\theta) − 2 ρ 2 d θ − 2 ω ρ 2 d t + d ( 2 ρ 2 cos θ sin θ ) 作用量中不显含 θ \theta θ
δ S δ θ ∝ d ρ = 0 δ S δ ρ = 0 ⟹ d θ d t = − ω \begin{align}&\frac{\delta S}{\delta \theta}\propto \mathrm{d}\rho=0\\& \frac{\delta S}{\delta \rho}=0\implies \frac{\mathrm{d}\theta}{\mathrm{d}t}=-\omega \end{align} δ θ δ S ∝ d ρ = 0 δ ρ δ S = 0 ⟹ d t d θ = − ω 这跟 action-angle coordinates 很像.
对于 polar coordinates, 也会采用 ρ e i θ \rho e^{i\theta} ρ e i θ a , a † a,a^{\dagger} a , a †
要求 EoM
δ S δ ξ I = 0 ⟹ ∂ I γ J d ξ J − ( d ξ J ∂ J γ I + d t ∂ t γ I ) − ∂ I H d t = 0 \frac{\delta S}{\delta{\xi^I}}=0\implies \partial_I{\large\gamma}_J\mathrm{d}\xi^J-(\mathrm{d}\xi^J\partial_J{\large\gamma}_I+\mathrm{d}t\partial_t{\large\gamma}_I)-\partial_IH\mathrm{d}t=0 δ ξ I δ S = 0 ⟹ ∂ I γ J d ξ J − ( d ξ J ∂ J γ I + d t ∂ t γ I ) − ∂ I H d t = 0 即
( ∂ I γ J − ∂ J γ I ) d ξ J = ( ∂ I H + ∂ t γ I ) d t (\partial_I{\large\gamma}_J-\partial_J{\large\gamma}_I)\mathrm{d}\xi^J=(\partial_I H+\partial_t{\large\gamma}_I)\mathrm{d}t ( ∂ I γ J − ∂ J γ I ) d ξ J = ( ∂ I H + ∂ t γ I ) d t 这两部分在 χ \chi χ
γ I ( ξ , t ) → γ I ( ξ , t ) + ∂ I χ ( ξ , t ) , H ( ξ , t ) → H ( ξ , t ) − ∂ t χ ( ξ , t ) {\large\gamma}_I(\xi,t)\to{\large\gamma}_I(\xi,t)+\partial_I\chi(\xi,t),\qquad
H(\xi,t)\to H(\xi,t)-\partial_t\chi(\xi,t) γ I ( ξ , t ) → γ I ( ξ , t ) + ∂ I χ ( ξ , t ) , H ( ξ , t ) → H ( ξ , t ) − ∂ t χ ( ξ , t ) 这两部分看起来就像磁场和电场的结构.
于是定义记号
ω I J ≔ 2 ∂ [ I γ J ] = ∂ I γ J − ∂ J γ I \omega_{IJ}\coloneqq 2\partial_{[I}{\large\gamma}_{J]}=\partial_I{\large\gamma}_J-\partial_J{\large\gamma}_I ω I J : = 2 ∂ [ I γ J ] = ∂ I γ J − ∂ J γ I
[ ⋯ ] [\cdots] [ ⋯ ]
在一般的形式 ξ I = ( x i , p i ) \xi^I=(x^i,p_i) ξ I = ( x i , p i )
γ = p i d x i , γ x i = p i , γ p i = 0 \gamma=p_i\,\mathrm{d}x^i,\qquad \gamma_{x^i}=p_i,\qquad \gamma_{p_i}=0 γ = p i d x i , γ x i = p i , γ p i = 0 这里 γ I \gamma_I γ I I I I γ x i , γ p i \gamma_{x^i},\gamma_{p_i} γ x i , γ p i x i , p i x^i,p_i x i , p i γ \gamma γ i i i
ω I J = ( ω x i , x j = 0 ω x i , p j = − δ i j ω p i , x j = δ j i ω p i , p j = 0 ) \omega_{IJ}
=
\left(
\begin{array}{c|c}
\omega_{x^i,\;x^j}=0 & \omega_{x^i,\;p_j}=-\delta_i^{\ j} \\[4pt]
\omega_{p_i,\;x^j}=\delta_j^{\ i} & \omega_{p_i,\;p_j}=0
\end{array}
\right) ω I J = ( ω x i , x j = 0 ω p i , x j = δ j i ω x i , p j = − δ i j ω p i , p j = 0 ) 这里矩阵的行列顺序固定为 ( x i , p i ) (x^i,p_i) ( x i , p i )
其中
ω p i , x j = ∂ p i γ x j − ∂ x j γ p i = ∂ p i p j = δ j i , \omega_{p_i,\;x^j}
=
\partial_{p_i}\gamma_{x^j}-\partial_{x^j}\gamma_{p_i}
=
\partial_{p_i}p_j
=
\delta_j^{\ i}, ω p i , x j = ∂ p i γ x j − ∂ x j γ p i = ∂ p i p j = δ j i , 所以左下角块必须写成 δ j i \delta_j^{\ i} δ j i
将上式除以 d t \mathrm{d}t d t
ω I J d ξ J d t = ∂ I H + ∂ t γ I \omega_{IJ}\frac{\mathrm{d}\xi^J}{\mathrm{d}t}=\partial_I H+\partial_t{\large\gamma}_I ω I J d t d ξ J = ∂ I H + ∂ t γ I 因此
d ξ I d t ∣ EoM = ( ω − 1 ) I J ( ∂ J H + ∂ t γ J ) \frac{\mathrm{d}\xi^I}{\mathrm{d}t}\Big|_{\text{EoM}}=(\omega^{-1})^{IJ}(\partial_J H+\partial_t{\large\gamma}_J) d t d ξ I EoM = ( ω − 1 ) I J ( ∂ J H + ∂ t γ J ) ^d87288
看逆矩阵
( ω − 1 ) I J = ( ( ω − 1 ) x i , x j = 0 ( ω − 1 ) x i , p j = δ j i = { x i , p j } ( ω − 1 ) p i , x j = − δ i j = { p i , x j } ( ω − 1 ) p i , p j = 0 ) (\omega^{-1})^{IJ}
=
\left(
\begin{array}{c|c}
(\omega^{-1})^{x^i,\;x^j}=0
&
(\omega^{-1})^{x^i,\;p_j}=\delta^i_{\ j}=\{x^i,p_j\}
\\[4pt]
(\omega^{-1})^{p_i,\;x^j}=-\delta_i^{\ j}=\{p_i,x^j\}
&
(\omega^{-1})^{p_i,\;p_j}=0
\end{array}
\right) ( ω − 1 ) I J = ( ( ω − 1 ) x i , x j = 0 ( ω − 1 ) p i , x j = − δ i j = { p i , x j } ( ω − 1 ) x i , p j = δ j i = { x i , p j } ( ω − 1 ) p i , p j = 0 ) Poisson bracket 自动出现了. 这无非是在一种特殊坐标系下, 取 2-form 逆的分量.
Σ 到这里我们做了什么?
到这里我们做了什么?
我们从一般的 action 出发, 讨论它的 gauge invariance, 并计算了 EoM.
将方程整理为在 χ \chi χ
从中写出最一般的 EoM, 在 usual case 中计算逆, 发现 Poisson bracket 自动出现. 这表明这个矩阵的逆更为重要, 它不依赖于具体坐标.
所以现在需要去找这个矩阵有哪些不依赖于坐标的性质.
ω ~ I ~ J ~ = ω I J ∂ ξ I ∂ ξ ~ I ~ ∂ ξ J ∂ ξ ~ J ~ \tilde{\omega}_{\tilde{I}\tilde{J}}=\omega_{IJ}\frac{\partial\xi^I}{\partial\tilde{\xi}^{\tilde{I}}}\frac{\partial\xi^J}{\partial\tilde{\xi}^{\tilde{J}}} ω ~ I ~ J ~ = ω I J ∂ ξ ~ I ~ ∂ ξ I ∂ ξ ~ J ~ ∂ ξ J 需要检查这与 ω ~ \tilde{\omega} ω ~
这是一个作业
Abstractly, a symplectic 2-form:
is antisymmetric:
ω I J = − ω J I \omega_{IJ}=-\omega_{JI} ω I J = − ω J I
transforms as
ω ~ I ~ J ~ = ω I J ∂ ξ I ∂ ξ ~ I ~ ∂ ξ J ∂ ξ ~ J ~ \tilde{\omega}_{\tilde{I}\tilde{J}}=\omega_{IJ}\frac{\partial\xi^I}{\partial\tilde{\xi}^{\tilde{I}}}\frac{\partial\xi^J}{\partial\tilde{\xi}^{\tilde{J}}} ω ~ I ~ J ~ = ω I J ∂ ξ ~ I ~ ∂ ξ I ∂ ξ ~ J ~ ∂ ξ J
is non-degenerate, i.e. det ω ≠ 0 \det\omega\neq 0 det ω = 0
is closed: ∂ [ K ω I J ] = 0 \partial_{[K}\omega_{IJ]} = 0 ∂ [ K ω I J ] = 0
The First Very Important Property of ω I J \omega_{IJ} ω I J
ω I J \omega_{IJ} ω I J
Darboux theorem
Locally in phase space, one can always find a coordinate system such that
ω I J = ( ω x i , x j = 0 ω x i , p j = − δ i j ω p i , x j = δ j i ω p i , p j = 0 ) \omega_{IJ}
=
\left(
\begin{array}{c|c}
\omega_{x^i,\;x^j}=0 & \omega_{x^i,\;p_j}=-\delta_i^{\ j} \\[4pt]
\omega_{p_i,\;x^j}=\delta_j^{\ i} & \omega_{p_i,\;p_j}=0
\end{array}
\right) ω I J = ( ω x i , x j = 0 ω p i , x j = δ j i ω x i , p j = − δ i j ω p i , p j = 0 ) ^bb0d22
这就是 “canonical” 的定义.
不同的坐标在局部没有本质区别, 但它们有时没有明确的物理意义, 所以很多时候我们不会采用这种方便的坐标.
证明留作作业
