§P2.1 量子态、算符与薛定谔方程§P2.2 一维问题：盒中粒子与周期性边界条件（对应提示 D1）§P2.3 谐振子：升降算符与零点能（对应提示 D2）§P2.4 角动量与刚性转子（对应提示 D3）§P2.5 自旋与磁矩（对应提示 D4）§P2.6 全同粒子：交换对称性与泡利不相容原理（对应提示 D5）§P2.7 微扰论初步§P2.8 密度矩阵与量子系综自测题 本章目标
给 出第二、三部分用到的量子力学结论及其推导：量子态与算符、盒中粒子与周期性边界条件、谐振子、角动量与刚性转子、自旋与磁矩、全同粒子、微扰论（含交换作用与费米黄金规则）、密度矩阵。第2 –8章 中的【量子力学提示 D1 –D5 】分别对应 §P2.2 –§P2.6 。
§P2.1 量子态、算符与薛定谔方程 物理输入
量子力学的基本假设
(1) 系统的状态由一个复向量空间中的归一化向量 ∣ ψ ⟩ \lvert\psi\rangle ∣ ψ ⟩ ψ ( x ) \psi(x) ψ ( x ) ∣ ψ ( x ) ∣ 2 d x \lvert\psi(x)\rvert^2dx ∣ ψ ( x ) ∣ 2 d x [ x , x + d x ] [x,x+dx] [ x , x + d x ]
(2) 可观测量对应厄米算符 A ^ \hat A A ^ A ^ \hat A A ^ a n a_n a n a n a_n a n ∣ ⟨ a n ∣ ψ ⟩ ∣ 2 \lvert\langle a_n\vert\psi\rangle\rvert^2 ∣⟨ a n ∣ ψ ⟩ ∣ 2 ∣ a n ⟩ \lvert a_n\rangle ∣ a n ⟩
(3) 坐标与动量满足正则对易关系 [ x ^ , p ^ ] ≡ x ^ p ^ − p ^ x ^ = i ℏ [\hat x,\hat p]\equiv\hat x\hat p - \hat p\hat x = i\hbar [ x ^ , p ^ ] ≡ x ^ p ^ − p ^ x ^ = i ℏ p ^ = − i ℏ d / d x \hat p = -i\hbar\,d/dx p ^ = − i ℏ d / d x
(4) 状态随时间的演化由薛定谔方程 i ℏ ∂ t ∣ ψ ⟩ = H ^ ∣ ψ ⟩ i\hbar\,\partial_t\lvert\psi\rangle = \hat H\lvert\psi\rangle i ℏ ∂ t ∣ ψ ⟩ = H ^ ∣ ψ ⟩ H ^ \hat H H ^
复向量空间、厄米算符与狄拉克记号
⟨ ϕ ∣ ψ ⟩ \langle\phi\vert\psi\rangle ⟨ ϕ ∣ ψ ⟩ ⟨ ϕ ∣ ψ ⟩ ∗ = ⟨ ψ ∣ ϕ ⟩ \langle\phi\vert\psi\rangle^\ast = \langle\psi\vert\phi\rangle ⟨ ϕ ∣ ψ ⟩ ∗ = ⟨ ψ ∣ ϕ ⟩ ⟨ ϕ ∣ ψ ⟩ = ∫ ϕ ∗ ψ d x \langle\phi\vert\psi\rangle = \int\phi^\ast\psi\,dx ⟨ ϕ ∣ ψ ⟩ = ∫ ϕ ∗ ψ d x ⟨ ϕ ∣ A ^ ψ ⟩ = ⟨ A ^ ϕ ∣ ψ ⟩ \langle\phi\vert\hat A\psi\rangle = \langle\hat A\phi\vert\psi\rangle ⟨ ϕ ∣ A ^ ψ ⟩ = ⟨ A ^ ϕ ∣ ψ ⟩ { ∣ n ⟩ } \lbrace\lvert n\rangle\rbrace {∣ n ⟩} ∑ n ∣ n ⟩ ⟨ n ∣ = 1 \sum_n\lvert n\rangle\langle n\rvert = 1 ∑ n ∣ n ⟩ ⟨ n ∣ = 1 ∣ ψ ⟩ = ∑ n ∣ n ⟩ ⟨ n ∣ ψ ⟩ \lvert\psi\rangle = \sum_n\lvert n\rangle\langle n\vert\psi\rangle ∣ ψ ⟩ = ∑ n ∣ n ⟩ ⟨ n ∣ ψ ⟩ 附录 B4 。）
厄米算符的两条性质 。设 A ^ ∣ a ⟩ = a ∣ a ⟩ \hat A\lvert a\rangle = a\lvert a\rangle A ^ ∣ a ⟩ = a ∣ a ⟩ a ⟨ a ∣ a ⟩ = ⟨ a ∣ A ^ a ⟩ = ⟨ A ^ a ∣ a ⟩ = a ∗ ⟨ a ∣ a ⟩ a\langle a\vert a\rangle = \langle a\vert\hat Aa\rangle = \langle\hat Aa\vert a\rangle = a^\ast\langle a\vert a\rangle a ⟨ a ∣ a ⟩ = ⟨ a ∣ A ^ a ⟩ = ⟨ A ^ a ∣ a ⟩ = a ∗ ⟨ a ∣ a ⟩ 本征值是实数 。若 A ^ ∣ a ′ ⟩ = a ′ ∣ a ′ ⟩ \hat A\lvert a'\rangle = a'\lvert a'\rangle A ^ ∣ a ′ ⟩ = a ′ ∣ a ′ ⟩ a ′ ≠ a a'\ne a a ′ = a a ⟨ a ′ ∣ a ⟩ = ⟨ a ′ ∣ A ^ a ⟩ = ⟨ A ^ a ′ ∣ a ⟩ = a ′ ⟨ a ′ ∣ a ⟩ a\langle a'\vert a\rangle = \langle a'\vert\hat Aa\rangle = \langle\hat Aa'\vert a\rangle = a'\langle a'\vert a\rangle a ⟨ a ′ ∣ a ⟩ = ⟨ a ′ ∣ A ^ a ⟩ = ⟨ A ^ a ′ ∣ a ⟩ = a ′ ⟨ a ′ ∣ a ⟩ ( a − a ′ ) ⟨ a ′ ∣ a ⟩ = 0 (a - a')\langle a'\vert a\rangle = 0 ( a − a ′ ) ⟨ a ′ ∣ a ⟩ = 0 不同本征值的本征向量正交 。
平均值 。把 ∣ ψ ⟩ \lvert\psi\rangle ∣ ψ ⟩ A ^ \hat A A ^ ∣ ψ ⟩ = ∑ n c n ∣ a n ⟩ \lvert\psi\rangle = \sum_nc_n\lvert a_n\rangle ∣ ψ ⟩ = ∑ n c n ∣ a n ⟩ c n = ⟨ a n ∣ ψ ⟩ c_n = \langle a_n\vert\psi\rangle c n = ⟨ a n ∣ ψ ⟩
⟨ A ⟩ = ∑ n a n ∣ c n ∣ 2 = ⟨ ψ ∣ A ^ ∣ ψ ⟩ (P2.1) \langle A\rangle = \sum_na_n\lvert c_n\rvert^2 = \langle\psi\vert\hat A\vert\psi\rangle \tag{P2.1} ⟨ A ⟩ = n ∑ a n ∣ c n ∣ 2 = ⟨ ψ ∣ A ^ ∣ ψ ⟩ ( P2.1 ) （第二个等号：A ^ ∣ ψ ⟩ = ∑ n c n a n ∣ a n ⟩ \hat A\lvert\psi\rangle = \sum_nc_na_n\lvert a_n\rangle A ^ ∣ ψ ⟩ = ∑ n c n a n ∣ a n ⟩
验证对易关系 ：对任意 ψ \psi ψ [ x ^ , p ^ ] ψ = − i ℏ x ψ ′ + i ℏ ( x ψ ) ′ = i ℏ ψ [\hat x,\hat p]\psi = -i\hbar x\psi' + i\hbar(x\psi)' = i\hbar\psi [ x ^ , p ^ ] ψ = − i ℏ x ψ ′ + i ℏ ( x ψ ) ′ = i ℏ ψ { x , p } = 1 \lbrace x,p\rbrace = 1 { x , p } = 1 (P1.8) ）的量子对应：{ A , B } → [ A ^ , B ^ ] / ( i ℏ ) \lbrace A,B\rbrace\to[\hat A,\hat B]/(i\hbar) { A , B } → [ A ^ , B ^ ] / ( i ℏ )
定态 。若 H ^ ∣ E ⟩ = E ∣ E ⟩ \hat H\lvert E\rangle = E\lvert E\rangle H ^ ∣ E ⟩ = E ∣ E ⟩ ∣ ψ ( t ) ⟩ = e − i E t / ℏ ∣ E ⟩ \lvert\psi(t)\rangle = e^{-iEt/\hbar}\lvert E\rangle ∣ ψ ( t )⟩ = e − i Et /ℏ ∣ E ⟩ 定态 ：一切可观测量的分布都不随时间变化。统计力学中的"微观态"就是指这些定态（§2.5 ）。在坐标表象中，单粒子的定态方程为
− ℏ 2 2 m ∇ 2 ψ + V ( r ) ψ = E ψ (P2.2) -\frac{\hbar^2}{2m}\nabla^2\psi + V(\mathbf r)\psi = E\psi \tag{P2.2} − 2 m ℏ 2 ∇ 2 ψ + V ( r ) ψ = E ψ ( P2.2 ) 平均值的演化 。由薛定谔方程及其共轭 − i ℏ ∂ t ⟨ ψ ∣ = ⟨ ψ ∣ H ^ -i\hbar\,\partial_t\langle\psi\rvert = \langle\psi\rvert\hat H − i ℏ ∂ t ⟨ ψ ∣ = ⟨ ψ ∣ H ^ A ^ \hat A A ^
d ⟨ A ⟩ d t = 1 i ℏ ⟨ ψ ∣ [ A ^ , H ^ ] ∣ ψ ⟩ (P2.3) \frac{d\langle A\rangle}{dt} = \frac{1}{i\hbar}\langle\psi\vert[\hat A,\hat H]\vert\psi\rangle \tag{P2.3} d t d ⟨ A ⟩ = i ℏ 1 ⟨ ψ ∣ [ A ^ , H ^ ] ∣ ψ ⟩ ( P2.3 ) 与 (P1.8) 对照：经典的 { A , H } \lbrace A,\mathcal H\rbrace { A , H } [ A ^ , H ^ ] / ( i ℏ ) [\hat A,\hat H]/(i\hbar) [ A ^ , H ^ ] / ( i ℏ )
§P2.2 一维问题：盒中粒子与周期性边界条件（对应提示 D1） 盒中粒子 。§2.5 的提示 D1 已求出无限深势阱中的解 ψ n = A sin ( n π x / L ) \psi_n = A\sin(n\pi x/L) ψ n = A sin ( nπ x / L ) ε n = π 2 ℏ 2 n 2 / ( 2 m L 2 ) \varepsilon_n = \pi^2\hbar^2n^2/(2mL^2) ε n = π 2 ℏ 2 n 2 / ( 2 m L 2 ) ∫ 0 L sin 2 ( n π x / L ) d x = L / 2 \int_0^L\sin^2(n\pi x/L)\,dx = L/2 ∫ 0 L sin 2 ( nπ x / L ) d x = L /2 A = 2 / L A = \sqrt{2/L} A = 2/ L π 2 ℏ 2 / ( 2 m L 2 ) > 0 \pi^2\hbar^2/(2mL^2)>0 π 2 ℏ 2 / ( 2 m L 2 ) > 0 §P2.3 中谐振子的零点能也是如此）。三维立方盒的能级 (2.17) 往往简并：例如 ( n x , n y , n z ) = ( 2 , 1 , 1 ) (n_x,n_y,n_z) = (2,1,1) ( n x , n y , n z ) = ( 2 , 1 , 1 ) ( 1 , 2 , 1 ) (1,2,1) ( 1 , 2 , 1 ) ( 1 , 1 , 2 ) (1,1,2) ( 1 , 1 , 2 )
周期性边界条件 。研究宏观样品的体性质时，常要求 ψ ( x + L ) = ψ ( x ) \psi(x+L) = \psi(x) ψ ( x + L ) = ψ ( x )
ψ k ( x ) = 1 L e i k x , k = 2 π n L ( n = 0 , ± 1 , ± 2 , … ) , p ^ ψ k = ℏ k ψ k , ε k = ℏ 2 k 2 2 m (P2.4) \psi_k(x) = \frac{1}{\sqrt L}e^{ikx},\qquad k = \frac{2\pi n}{L}\ (n = 0,\pm1,\pm2,\dots),\qquad \hat p\,\psi_k = \hbar k\,\psi_k,\qquad \varepsilon_k = \frac{\hbar^2k^2}{2m} \tag{P2.4} ψ k ( x ) = L 1 e ik x , k = L 2 πn ( n = 0 , ± 1 , ± 2 , … ) , p ^ ψ k = ℏ k ψ k , ε k = 2 m ℏ 2 k 2 ( P2.4 ) 它们同时是动量的本征态，代表沿 ± x \pm x ± x k k k 2 π / L 2\pi/L 2 π / L k k k ( 2 π ) 3 / V (2\pi)^3/V ( 2 π ) 3 / V p = ℏ k p = \hbar k p = ℏ k h 3 / V h^3/V h 3 / V (2.20) 的来源。驻波边界条件（k = n π / L k = n\pi/L k = nπ / L n > 0 n>0 n > 0 §2.5 ）；后者处理传播的波（电子、声子、光子）更方便，第三部分都将使用它。
§P2.3 谐振子：升降算符与零点能（对应提示 D2） H ^ = p ^ 2 / 2 m + m ω 2 x ^ 2 / 2 \hat H = \hat p^2/2m + m\omega^2\hat x^2/2 H ^ = p ^ 2 /2 m + m ω 2 x ^ 2 /2
a ^ = m ω 2 ℏ ( x ^ + i p ^ m ω ) , a ^ † = m ω 2 ℏ ( x ^ − i p ^ m ω ) (P2.5) \hat a = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat x + \frac{i\hat p}{m\omega}\right),\qquad \hat a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat x - \frac{i\hat p}{m\omega}\right) \tag{P2.5} a ^ = 2ℏ mω ( x ^ + mω i p ^ ) , a ^ † = 2ℏ mω ( x ^ − mω i p ^ ) ( P2.5 ) 两个代数关系 。由 [ x ^ , p ^ ] = i ℏ [\hat x,\hat p] = i\hbar [ x ^ , p ^ ] = i ℏ
[ a ^ , a ^ † ] = m ω 2 ℏ ( − i m ω [ x ^ , p ^ ] + i m ω [ p ^ , x ^ ] ) = m ω 2 ℏ ⋅ 2 ℏ m ω = 1 [\hat a,\hat a^\dagger] = \frac{m\omega}{2\hbar}\left(-\frac{i}{m\omega}[\hat x,\hat p] + \frac{i}{m\omega}[\hat p,\hat x]\right) = \frac{m\omega}{2\hbar}\cdot\frac{2\hbar}{m\omega} = 1 [ a ^ , a ^ † ] = 2ℏ mω ( − mω i [ x ^ , p ^ ] + mω i [ p ^ , x ^ ] ) = 2ℏ mω ⋅ mω 2ℏ = 1 a ^ † a ^ = m ω 2 ℏ ( x ^ 2 + p ^ 2 m 2 ω 2 + i m ω [ x ^ , p ^ ] ) = H ^ ℏ ω − 1 2 \hat a^\dagger\hat a = \frac{m\omega}{2\hbar}\left(\hat x^2 + \frac{\hat p^2}{m^2\omega^2} + \frac{i}{m\omega}[\hat x,\hat p]\right) = \frac{\hat H}{\hbar\omega} - \frac12 a ^ † a ^ = 2ℏ mω ( x ^ 2 + m 2 ω 2 p ^ 2 + mω i [ x ^ , p ^ ] ) = ℏ ω H ^ − 2 1
所以
H ^ = ℏ ω ( N ^ + 1 2 ) , N ^ ≡ a ^ † a ^ , [ a ^ , a ^ † ] = 1 (P2.6) \hat H = \hbar\omega\left(\hat N + \frac12\right),\qquad \hat N\equiv\hat a^\dagger\hat a,\qquad [\hat a,\hat a^\dagger] = 1 \tag{P2.6} H ^ = ℏ ω ( N ^ + 2 1 ) , N ^ ≡ a ^ † a ^ , [ a ^ , a ^ † ] = 1 ( P2.6 ) N ^ \hat N N ^ [ N ^ , a ^ ] = a ^ † [ a ^ , a ^ ] + [ a ^ † , a ^ ] a ^ = − a ^ [\hat N,\hat a] = \hat a^\dagger[\hat a,\hat a] + [\hat a^\dagger,\hat a]\hat a = -\hat a [ N ^ , a ^ ] = a ^ † [ a ^ , a ^ ] + [ a ^ † , a ^ ] a ^ = − a ^ [ N ^ , a ^ † ] = a ^ † [\hat N,\hat a^\dagger] = \hat a^\dagger [ N ^ , a ^ † ] = a ^ † N ^ ∣ ν ⟩ = ν ∣ ν ⟩ \hat N\lvert\nu\rangle = \nu\lvert\nu\rangle N ^ ∣ ν ⟩ = ν ∣ ν ⟩ N ^ ( a ^ ∣ ν ⟩ ) = ( ν − 1 ) a ^ ∣ ν ⟩ \hat N(\hat a\lvert\nu\rangle) = (\nu-1)\hat a\lvert\nu\rangle N ^ ( a ^ ∣ ν ⟩) = ( ν − 1 ) a ^ ∣ ν ⟩ N ^ ( a ^ † ∣ ν ⟩ ) = ( ν + 1 ) a ^ † ∣ ν ⟩ \hat N(\hat a^\dagger\lvert\nu\rangle) = (\nu+1)\hat a^\dagger\lvert\nu\rangle N ^ ( a ^ † ∣ ν ⟩) = ( ν + 1 ) a ^ † ∣ ν ⟩ a ^ \hat a a ^ a ^ † \hat a^\dagger a ^ † ∥ a ^ ∣ ν ⟩ ∥ 2 = ⟨ ν ∣ a ^ † a ^ ∣ ν ⟩ = ν \lVert\hat a\lvert\nu\rangle\rVert^2 = \langle\nu\vert\hat a^\dagger\hat a\vert\nu\rangle = \nu ∥ a ^ ∣ ν ⟩ ∥ 2 = ⟨ ν ∣ a ^ † a ^ ∣ ν ⟩ = ν ν ≥ 0 \nu\ge0 ν ≥ 0 a ^ \hat a a ^ a ^ ∣ ν 0 ⟩ = 0 \hat a\lvert\nu_0\rangle = 0 a ^ ∣ ν 0 ⟩ = 0 ν 0 = ∥ a ^ ∣ ν 0 ⟩ ∥ 2 = 0 \nu_0 = \lVert\hat a\lvert\nu_0\rangle\rVert^2 = 0 ν 0 = ∥ a ^ ∣ ν 0 ⟩ ∥ 2 = 0 ν \nu ν n = 0 , 1 , 2 , … n = 0,1,2,\dots n = 0 , 1 , 2 , …
ε n = ( n + 1 2 ) ℏ ω , n = 0 , 1 , 2 , … (P2.7) \varepsilon_n = \left(n + \frac12\right)\hbar\omega,\qquad n = 0,1,2,\dots \tag{P2.7} ε n = ( n + 2 1 ) ℏ ω , n = 0 , 1 , 2 , … ( P2.7 ) 这就是第6章 用作输入的谐振子能级。∥ a ^ † ∣ n ⟩ ∥ 2 = ⟨ n ∣ a ^ a ^ † ∣ n ⟩ = n + 1 \lVert\hat a^\dagger\lvert n\rangle\rVert^2 = \langle n\vert\hat a\hat a^\dagger\vert n\rangle = n + 1 ∥ a ^ † ∣ n ⟩ ∥ 2 = ⟨ n ∣ a ^ a ^ † ∣ n ⟩ = n + 1 a ^ † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ \hat a^\dagger\lvert n\rangle = \sqrt{n+1}\,\lvert n+1\rangle a ^ † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ a ^ ∣ n ⟩ = n ∣ n − 1 ⟩ \hat a\lvert n\rangle = \sqrt n\,\lvert n-1\rangle a ^ ∣ n ⟩ = n ∣ n − 1 ⟩
基态与零点能 。a ^ ψ 0 = 0 \hat a\psi_0 = 0 a ^ ψ 0 = 0 ( x + ℏ m ω d d x ) ψ 0 = 0 \left(x + \frac{\hbar}{m\omega}\frac{d}{dx}\right)\psi_0 = 0 ( x + mω ℏ d x d ) ψ 0 = 0 ψ 0 ∝ e − m ω x 2 / 2 ℏ \psi_0\propto e^{-m\omega x^2/2\hbar} ψ 0 ∝ e − mω x 2 /2ℏ ∣ ψ 0 ∣ 2 ∝ e − m ω x 2 / ℏ \lvert\psi_0\rvert^2\propto e^{-m\omega x^2/\hbar} ∣ ψ 0 ∣ 2 ∝ e − mω x 2 /ℏ (2.5) ，⟨ x 2 ⟩ = ℏ / ( 2 m ω ) \langle x^2\rangle = \hbar/(2m\omega) ⟨ x 2 ⟩ = ℏ/ ( 2 mω ) 1 2 m ω 2 ⟨ x 2 ⟩ = 1 4 ℏ ω \frac12m\omega^2\langle x^2\rangle = \frac14\hbar\omega 2 1 m ω 2 ⟨ x 2 ⟩ = 4 1 ℏ ω 1 4 ℏ ω \frac14\hbar\omega 4 1 ℏ ω 1 2 ℏ ω \frac12\hbar\omega 2 1 ℏ ω Δ x \Delta x Δ x ℏ / Δ x \hbar/\Delta x ℏ/Δ x
与经典计数的对应 ：能量不超过 E E E E / ℏ ω E/\hbar\omega E /ℏ ω 2 π E / ω 2\pi E/\omega 2 π E / ω h h h 第2章 自测题 2），验证了 (2.20) 。
§P2.4 角动量与刚性转子（对应提示 D3） 对易关系 。轨道角动量 L ^ = r ^ × p ^ \hat{\mathbf L} = \hat{\mathbf r}\times\hat{\mathbf p} L ^ = r ^ × p ^ L ^ x = y ^ p ^ z − z ^ p ^ y \hat L_x = \hat y\hat p_z - \hat z\hat p_y L ^ x = y ^ p ^ z − z ^ p ^ y [ x ^ a , p ^ a ] = i ℏ [\hat x_a,\hat p_a] = i\hbar [ x ^ a , p ^ a ] = i ℏ
[ L ^ x , L ^ y ] = [ y ^ p ^ z , z ^ p ^ x ] + [ z ^ p ^ y , x ^ p ^ z ] = y ^ p ^ x [ p ^ z , z ^ ] + x ^ p ^ y [ z ^ , p ^ z ] = i ℏ ( x ^ p ^ y − y ^ p ^ x ) = i ℏ L ^ z [\hat L_x,\hat L_y] = [\hat y\hat p_z,\hat z\hat p_x] + [\hat z\hat p_y,\hat x\hat p_z] = \hat y\hat p_x[\hat p_z,\hat z] + \hat x\hat p_y[\hat z,\hat p_z] = i\hbar(\hat x\hat p_y - \hat y\hat p_x) = i\hbar\hat L_z [ L ^ x , L ^ y ] = [ y ^ p ^ z , z ^ p ^ x ] + [ z ^ p ^ y , x ^ p ^ z ] = y ^ p ^ x [ p ^ z , z ^ ] + x ^ p ^ y [ z ^ , p ^ z ] = i ℏ ( x ^ p ^ y − y ^ p ^ x ) = i ℏ L ^ z 轮换即得另外两式。由此 [ L ^ x 2 , L ^ z ] = L ^ x [ L ^ x , L ^ z ] + [ L ^ x , L ^ z ] L ^ x = − i ℏ ( L ^ x L ^ y + L ^ y L ^ x ) [\hat L_x^2,\hat L_z] = \hat L_x[\hat L_x,\hat L_z] + [\hat L_x,\hat L_z]\hat L_x = -i\hbar(\hat L_x\hat L_y + \hat L_y\hat L_x) [ L ^ x 2 , L ^ z ] = L ^ x [ L ^ x , L ^ z ] + [ L ^ x , L ^ z ] L ^ x = − i ℏ ( L ^ x L ^ y + L ^ y L ^ x ) [ L ^ y 2 , L ^ z ] = + i ℏ ( L ^ y L ^ x + L ^ x L ^ y ) [\hat L_y^2,\hat L_z] = +i\hbar(\hat L_y\hat L_x + \hat L_x\hat L_y) [ L ^ y 2 , L ^ z ] = + i ℏ ( L ^ y L ^ x + L ^ x L ^ y ) [ L ^ 2 , L ^ z ] = 0 [\hat L^2,\hat L_z] = 0 [ L ^ 2 , L ^ z ] = 0 L ^ 2 \hat L^2 L ^ 2 L ^ z \hat L_z L ^ z
本征值 。令 L ^ ± = L ^ x ± i L ^ y \hat L_\pm = \hat L_x\pm i\hat L_y L ^ ± = L ^ x ± i L ^ y [ L ^ z , L ^ ± ] = ± ℏ L ^ ± [\hat L_z,\hat L_\pm] = \pm\hbar\hat L_\pm [ L ^ z , L ^ ± ] = ± ℏ L ^ ±
L ^ ∓ L ^ ± = L ^ 2 − L ^ z 2 ∓ ℏ L ^ z (P2.8) \hat L_\mp\hat L_\pm = \hat L^2 - \hat L_z^2\mp\hbar\hat L_z \tag{P2.8} L ^ ∓ L ^ ± = L ^ 2 − L ^ z 2 ∓ ℏ L ^ z ( P2.8 ) 设 L ^ 2 ∣ λ , m ⟩ = λ ℏ 2 ∣ λ , m ⟩ \hat L^2\lvert\lambda,m\rangle = \lambda\hbar^2\lvert\lambda,m\rangle L ^ 2 ∣ λ , m ⟩ = λ ℏ 2 ∣ λ , m ⟩ L ^ z ∣ λ , m ⟩ = m ℏ ∣ λ , m ⟩ \hat L_z\lvert\lambda,m\rangle = m\hbar\lvert\lambda,m\rangle L ^ z ∣ λ , m ⟩ = m ℏ ∣ λ , m ⟩ §P2.3 同理，L ^ ± \hat L_\pm L ^ ± m m m λ \lambda λ ⟨ L ^ x 2 + L ^ y 2 ⟩ ≥ 0 \langle\hat L_x^2 + \hat L_y^2\rangle\ge0 ⟨ L ^ x 2 + L ^ y 2 ⟩ ≥ 0 m 2 ≤ λ m^2\le\lambda m 2 ≤ λ m max = l m_{\max} = l m m a x = l L ^ + ∣ λ , l ⟩ = 0 \hat L_+\lvert\lambda,l\rangle = 0 L ^ + ∣ λ , l ⟩ = 0 (P2.8) ，0 = L ^ − L ^ + ∣ λ , l ⟩ = ( λ − l 2 − l ) ℏ 2 ∣ λ , l ⟩ 0 = \hat L_-\hat L_+\lvert\lambda,l\rangle = (\lambda - l^2 - l)\hbar^2\lvert\lambda,l\rangle 0 = L ^ − L ^ + ∣ λ , l ⟩ = ( λ − l 2 − l ) ℏ 2 ∣ λ , l ⟩ λ = l ( l + 1 ) \lambda = l(l+1) λ = l ( l + 1 ) m min m_{\min} m m i n λ = m min 2 − m min \lambda = m_{\min}^2 - m_{\min} λ = m m i n 2 − m m i n m min = − l m_{\min} = -l m m i n = − l − l -l − l l l l 2 l 2l 2 l
L ^ 2 : l ( l + 1 ) ℏ 2 , L ^ z : m ℏ , m = − l , − l + 1 , … , l ( 2 l + 1 个态 ) (P2.9) \hat L^2:\ l(l+1)\hbar^2,\qquad \hat L_z:\ m\hbar,\quad m = -l,-l+1,\dots,l\qquad(2l+1\text{ 个态}) \tag{P2.9} L ^ 2 : l ( l + 1 ) ℏ 2 , L ^ z : m ℏ , m = − l , − l + 1 , … , l ( 2 l + 1 个态 ) ( P2.9 ) 由 (P2.8) 还得 ∥ L ^ ± ∣ l , m ⟩ ∥ 2 = ℏ 2 [ l ( l + 1 ) − m ( m ± 1 ) ] \lVert\hat L_\pm\lvert l,m\rangle\rVert^2 = \hbar^2[l(l+1) - m(m\pm1)] ∥ L ^ ± ∣ l , m ⟩ ∥ 2 = ℏ 2 [ l ( l + 1 ) − m ( m ± 1 )] L ^ ± ∣ l , m ⟩ = ℏ l ( l + 1 ) − m ( m ± 1 ) ∣ l , m ± 1 ⟩ \hat L_\pm\lvert l,m\rangle = \hbar\sqrt{l(l+1) - m(m\pm1)}\,\lvert l,m\pm1\rangle L ^ ± ∣ l , m ⟩ = ℏ l ( l + 1 ) − m ( m ± 1 ) ∣ l , m ± 1 ⟩ l l l L ^ z = − i ℏ ∂ / ∂ φ \hat L_z = -i\hbar\,\partial/\partial\varphi L ^ z = − i ℏ ∂ / ∂ φ e i m φ e^{im\varphi} e im φ z z z m m m §P2.5 ）。
刚性转子 。由 (P1.3) ，线形转子的经典能量为 L 2 / 2 I L^2/2I L 2 /2 I H ^ = L ^ 2 / 2 I \hat H = \hat L^2/2I H ^ = L ^ 2 /2 I (P2.9) ：
ε J = ℏ 2 2 I J ( J + 1 ) , g J = 2 J + 1 , J = 0 , 1 , 2 , … (P2.10) \varepsilon_J = \frac{\hbar^2}{2I}J(J+1),\qquad g_J = 2J+1,\qquad J = 0,1,2,\dots \tag{P2.10} ε J = 2 I ℏ 2 J ( J + 1 ) , g J = 2 J + 1 , J = 0 , 1 , 2 , … ( P2.10 ) 这就是第7章 用作输入的转动能级。
§P2.5 自旋与磁矩（对应提示 D4） 物理输入
电子、质子、中子等粒子具有内禀角动量（自旋）S ^ \hat{\mathbf S} S ^ L ^ \hat{\mathbf L} L ^ s = 1 / 2 s = 1/2 s = 1/2 S z = ± ℏ / 2 S_z = \pm\hbar/2 S z = ± ℏ/2 μ ^ = γ S ^ \hat{\boldsymbol\mu} = \gamma\hat{\mathbf S} μ ^ = γ S ^ γ \gamma γ − μ ⋅ B -\boldsymbol\mu\cdot\mathbf B − μ ⋅ B §P3.1 ）。
塞曼能级 。取 B \mathbf B B z z z H ^ = − γ B S ^ z \hat H = -\gamma B\hat S_z H ^ = − γ B S ^ z ∓ 1 2 γ ℏ B \mp\frac12\gamma\hbar B ∓ 2 1 γ ℏ B ω 0 = γ B \omega_0 = \gamma B ω 0 = γ B
Δ E = γ ℏ B = ℏ ω 0 = h ν 0 (P2.11) \Delta E = \gamma\hbar B = \hbar\omega_0 = h\nu_0 \tag{P2.11} Δ E = γ ℏ B = ℏ ω 0 = h ν 0 ( P2.11 ) 这就是 (6.8) 中用到的关系，ν 0 \nu_0 ν 0 γ p / 2 π ≈ 42.58 M H z / T \gamma_{\mathrm p}/2\pi\approx42.58\ \mathrm{MHz/T} γ p /2 π ≈ 42.58 MHz/T
电子与原子的磁矩 。电子的磁矩与自旋反向，μ ^ = − g s μ B S ^ / ℏ \hat{\boldsymbol\mu} = -g_s\mu_{\mathrm B}\hat{\mathbf S}/\hbar μ ^ = − g s μ B S ^ /ℏ g s ≈ 2.0023 g_s\approx2.0023 g s ≈ 2.0023 μ B = e ℏ / ( 2 m e ) = 9.274 × 1 0 − 24 J / T \mu_{\mathrm B} = e\hbar/(2m_{\mathrm e}) = 9.274\times10^{-24}\ \mathrm{J/T} μ B = e ℏ/ ( 2 m e ) = 9.274 × 1 0 − 24 J/T J J J z z z 2 J + 1 2J+1 2 J + 1 − g J μ B m J -g_J\mu_{\mathrm B}m_J − g J μ B m J g J g_J g J g J μ B B m J g_J\mu_{\mathrm B}Bm_J g J μ B B m J 第6章 自测题 1 的出发点。
§P2.6 全同粒子：交换对称性与泡利不相容原理（对应提示 D5） 交换算符 。对两个全同粒子定义 P ^ 12 ψ ( x 1 , x 2 ) = ψ ( x 2 , x 1 ) \hat P_{12}\psi(x_1,x_2) = \psi(x_2,x_1) P ^ 12 ψ ( x 1 , x 2 ) = ψ ( x 2 , x 1 ) x x x [ P ^ 12 , H ^ ] = 0 [\hat P_{12},\hat H] = 0 [ P ^ 12 , H ^ ] = 0 P ^ 12 2 = 1 \hat P_{12}^2 = 1 P ^ 12 2 = 1 P ^ 12 \hat P_{12} P ^ 12 ± 1 \pm1 ± 1
物理输入
对称化假设与自旋–统计关系
全同粒子系统的状态在交换任意两个粒子时，要么完全对称（玻色子 ，自旋为整数，如光子、4 ^4 4 费米子 ，自旋为半整数，如电子、质子、中子、3 ^3 3
泡利不相容原理 。设两个无相互作用的全同粒子分别处于单粒子态 ϕ a \phi_a ϕ a ϕ b \phi_b ϕ b
ψ ± ( x 1 , x 2 ) = 1 2 [ ϕ a ( x 1 ) ϕ b ( x 2 ) ± ϕ b ( x 1 ) ϕ a ( x 2 ) ] (P2.12) \psi_\pm(x_1,x_2) = \frac{1}{\sqrt2}\left[\phi_a(x_1)\phi_b(x_2)\pm\phi_b(x_1)\phi_a(x_2)\right] \tag{P2.12} ψ ± ( x 1 , x 2 ) = 2 1 [ ϕ a ( x 1 ) ϕ b ( x 2 ) ± ϕ b ( x 1 ) ϕ a ( x 2 ) ] ( P2.12 ) （+ + + − - − a = b a = b a = b ϕ a ( x 1 ) ϕ a ( x 2 ) \phi_a(x_1)\phi_a(x_2) ϕ a ( x 1 ) ϕ a ( x 2 ) a = b a = b a = b ψ − ≡ 0 \psi_- \equiv 0 ψ − ≡ 0 两个全同费米子不能处于同一个单粒子态 。N N N ϕ a i ( x j ) \phi_{a_i}(x_j) ϕ a i ( x j ) 提示 B3 ）。
数态：与 1 / N ! 1/N! 1/ N ! 。两个粒子、两个单粒子态 a a a b b b
可区分粒子：a a aa aa a b ab ab b a ba ba b b bb bb
玻色子：a a aa aa b b bb bb ( a b + b a ) (ab+ba) ( ab + ba )
费米子：只有 ( a b − b a ) (ab-ba) ( ab − ba )
经典的"除以 2 ! 2! 2 ! 4 / 2 = 2 4/2 = 2 4/2 = 2
只有当两个粒子几乎不会处于同一个态时（单粒子态远多于粒子数），三种计数才趋于一致：这正是 §3.7 、§4.5 中 1 / N ! 1/N! 1/ N ! 第12章 将严格处理。由于全同粒子不可区分，多粒子态完全由各单粒子态上的占据数 { n j } \lbrace n_j\rbrace { n j } n j ∈ { 0 , 1 } n_j\in\lbrace0,1\rbrace n j ∈ { 0 , 1 } n j = 0 , 1 , 2 , … n_j = 0,1,2,\dots n j = 0 , 1 , 2 , … §8.6 ）。
两个自旋 1/2 的合成 。由 §P2.4 的结果（s = 1 / 2 s = 1/2 s = 1/2 S ^ + ∣ ↓ ⟩ = ℏ ∣ ↑ ⟩ \hat S_+\lvert\downarrow\rangle = \hbar\lvert\uparrow\rangle S ^ + ∣ ↓ ⟩ = ℏ ∣ ↑ ⟩ S ^ − ∣ ↑ ⟩ = ℏ ∣ ↓ ⟩ \hat S_-\lvert\uparrow\rangle = \hbar\lvert\downarrow\rangle S ^ − ∣ ↑ ⟩ = ℏ ∣ ↓ ⟩ S ^ 1 ⋅ S ^ 2 = S ^ 1 z S ^ 2 z + 1 2 ( S ^ 1 + S ^ 2 − + S ^ 1 − S ^ 2 + ) \hat{\mathbf S}_1\cdot\hat{\mathbf S}_2 = \hat S_{1z}\hat S_{2z} + \frac12(\hat S_{1+}\hat S_{2-} + \hat S_{1-}\hat S_{2+}) S ^ 1 ⋅ S ^ 2 = S ^ 1 z S ^ 2 z + 2 1 ( S ^ 1 + S ^ 2 − + S ^ 1 − S ^ 2 + )
S ^ 1 ⋅ S ^ 2 = + ℏ 2 4 （对三个对称态 ∣ ↑ ↑ ⟩ , 1 2 ( ∣ ↑ ↓ ⟩ + ∣ ↓ ↑ ⟩ ) , ∣ ↓ ↓ ⟩ ） , − 3 ℏ 2 4 （对反对称态 1 2 ( ∣ ↑ ↓ ⟩ − ∣ ↓ ↑ ⟩ ) ） (P2.13) \hat{\mathbf S}_1\cdot\hat{\mathbf S}_2 = +\frac{\hbar^2}{4}\ \text{（对三个对称态 }\lvert\uparrow\uparrow\rangle,\ \tfrac{1}{\sqrt2}(\lvert\uparrow\downarrow\rangle+\lvert\downarrow\uparrow\rangle),\ \lvert\downarrow\downarrow\rangle\text{）},\qquad -\frac{3\hbar^2}{4}\ \text{（对反对称态 }\tfrac{1}{\sqrt2}(\lvert\uparrow\downarrow\rangle-\lvert\downarrow\uparrow\rangle)\text{）} \tag{P2.13} S ^ 1 ⋅ S ^ 2 = + 4 ℏ 2 （对三个对称态 ∣ ↑↑ ⟩ , 2 1 (∣ ↑↓ ⟩ + ∣ ↓↑ ⟩) , ∣ ↓↓ ⟩ ） , − 4 3 ℏ 2 （对反对称态 2 1 (∣ ↑↓ ⟩ − ∣ ↓↑ ⟩) ） ( P2.13 ) （例如 S ^ 1 ⋅ S ^ 2 ∣ ↑ ↓ ⟩ = − ℏ 2 4 ∣ ↑ ↓ ⟩ + ℏ 2 2 ∣ ↓ ↑ ⟩ \hat{\mathbf S}_1\cdot\hat{\mathbf S}_2\lvert\uparrow\downarrow\rangle = -\frac{\hbar^2}{4}\lvert\uparrow\downarrow\rangle + \frac{\hbar^2}{2}\lvert\downarrow\uparrow\rangle S ^ 1 ⋅ S ^ 2 ∣ ↑↓ ⟩ = − 4 ℏ 2 ∣ ↑↓ ⟩ + 2 ℏ 2 ∣ ↓↑ ⟩ − ℏ 2 4 ± ℏ 2 2 -\frac{\hbar^2}{4}\pm\frac{\hbar^2}{2} − 4 ℏ 2 ± 2 ℏ 2 S ^ 2 = S ^ 1 2 + S ^ 2 2 + 2 S ^ 1 ⋅ S ^ 2 \hat S^2 = \hat S_1^2 + \hat S_2^2 + 2\hat{\mathbf S}_1\cdot\hat{\mathbf S}_2 S ^ 2 = S ^ 1 2 + S ^ 2 2 + 2 S ^ 1 ⋅ S ^ 2 S ^ i 2 = 3 4 ℏ 2 \hat S_i^2 = \frac34\hbar^2 S ^ i 2 = 4 3 ℏ 2 S 2 = 2 ℏ 2 S^2 = 2\hbar^2 S 2 = 2 ℏ 2 S = 1 S = 1 S = 1 S 2 = 0 S^2 = 0 S 2 = 0 S = 0 S = 0 S = 0 §P2.7 中交换作用的来源。
应用：氢分子的正、仲态与对称数 。H 2 \mathrm H_2 H 2 ( − 1 ) J (-1)^J ( − 1 ) J J J J 仲氢 ）只能配偶数 J J J 正氢 ）只能配奇数 J J J J J J T / 2 Θ r T/2\Theta_{\mathrm r} T /2 Θ r
q r q n ≈ 1 ⋅ T 2 Θ r + 3 ⋅ T 2 Θ r = 4 ⋅ T 2 Θ r q_{\mathrm r}q_{\mathrm n}\approx1\cdot\frac{T}{2\Theta_{\mathrm r}} + 3\cdot\frac{T}{2\Theta_{\mathrm r}} = 4\cdot\frac{T}{2\Theta_{\mathrm r}} q r q n ≈ 1 ⋅ 2 Θ r T + 3 ⋅ 2 Θ r T = 4 ⋅ 2 Θ r T 即核自旋简并度 ( 2 ⋅ 1 2 + 1 ) 2 = 4 (2\cdot\tfrac12+1)^2 = 4 ( 2 ⋅ 2 1 + 1 ) 2 = 4 T / ( σ Θ r ) T/(\sigma\Theta_{\mathrm r}) T / ( σ Θ r ) σ = 2 \sigma = 2 σ = 2 §7.3 中对称数的量子起源。低温下正、仲氢的比例偏离 3:1，相互转化又极慢，这正是 H 2 \mathrm H_2 H 2 §27.6 ）。
§P2.7 微扰论初步 定态微扰 。设 H ^ = H ^ 0 + V ^ \hat H = \hat H_0 + \hat V H ^ = H ^ 0 + V ^ V ^ \hat V V ^ H ^ 0 \hat H_0 H ^ 0 ∣ n ( 0 ) ⟩ \lvert n^{(0)}\rangle ∣ n ( 0 ) ⟩ E n ( 0 ) E_n^{(0)} E n ( 0 ) E n = E n ( 0 ) + E n ( 1 ) + E n ( 2 ) + ⋯ E_n = E_n^{(0)} + E_n^{(1)} + E_n^{(2)} + \cdots E n = E n ( 0 ) + E n ( 1 ) + E n ( 2 ) + ⋯ ∣ n ⟩ = ∣ n ( 0 ) ⟩ + ∣ n ( 1 ) ⟩ + ⋯ \lvert n\rangle = \lvert n^{(0)}\rangle + \lvert n^{(1)}\rangle + \cdots ∣ n ⟩ = ∣ n ( 0 ) ⟩ + ∣ n ( 1 ) ⟩ + ⋯ ⟨ n ( 0 ) ∣ n ( 1 ) ⟩ = 0 \langle n^{(0)}\vert n^{(1)}\rangle = 0 ⟨ n ( 0 ) ∣ n ( 1 ) ⟩ = 0 H ^ ∣ n ⟩ = E n ∣ n ⟩ \hat H\lvert n\rangle = E_n\lvert n\rangle H ^ ∣ n ⟩ = E n ∣ n ⟩
一阶：H ^ 0 ∣ n ( 1 ) ⟩ + V ^ ∣ n ( 0 ) ⟩ = E n ( 0 ) ∣ n ( 1 ) ⟩ + E n ( 1 ) ∣ n ( 0 ) ⟩ \hat H_0\lvert n^{(1)}\rangle + \hat V\lvert n^{(0)}\rangle = E_n^{(0)}\lvert n^{(1)}\rangle + E_n^{(1)}\lvert n^{(0)}\rangle H ^ 0 ∣ n ( 1 ) ⟩ + V ^ ∣ n ( 0 ) ⟩ = E n ( 0 ) ∣ n ( 1 ) ⟩ + E n ( 1 ) ∣ n ( 0 ) ⟩ ⟨ n ( 0 ) ∣ \langle n^{(0)}\rvert ⟨ n ( 0 ) ∣ E n ( 1 ) = V n n E_n^{(1)} = V_{nn} E n ( 1 ) = V nn ⟨ m ( 0 ) ∣ \langle m^{(0)}\rvert ⟨ m ( 0 ) ∣ m ≠ n m\ne n m = n ⟨ m ( 0 ) ∣ n ( 1 ) ⟩ = V m n / ( E n ( 0 ) − E m ( 0 ) ) \langle m^{(0)}\vert n^{(1)}\rangle = V_{mn}/(E_n^{(0)} - E_m^{(0)}) ⟨ m ( 0 ) ∣ n ( 1 ) ⟩ = V mn / ( E n ( 0 ) − E m ( 0 ) ) V m n ≡ ⟨ m ( 0 ) ∣ V ^ ∣ n ( 0 ) ⟩ V_{mn}\equiv\langle m^{(0)}\vert\hat V\vert n^{(0)}\rangle V mn ≡ ⟨ m ( 0 ) ∣ V ^ ∣ n ( 0 ) ⟩
二阶：左乘 ⟨ n ( 0 ) ∣ \langle n^{(0)}\rvert ⟨ n ( 0 ) ∣ E n ( 2 ) = ⟨ n ( 0 ) ∣ V ^ ∣ n ( 1 ) ⟩ E_n^{(2)} = \langle n^{(0)}\vert\hat V\vert n^{(1)}\rangle E n ( 2 ) = ⟨ n ( 0 ) ∣ V ^ ∣ n ( 1 ) ⟩
E n ≈ E n ( 0 ) + V n n + ∑ m ≠ n ∣ V m n ∣ 2 E n ( 0 ) − E m ( 0 ) (P2.14) E_n\approx E_n^{(0)} + V_{nn} + \sum_{m\ne n}\frac{\lvert V_{mn}\rvert^2}{E_n^{(0)} - E_m^{(0)}} \tag{P2.14} E n ≈ E n ( 0 ) + V nn + m = n ∑ E n ( 0 ) − E m ( 0 ) ∣ V mn ∣ 2 ( P2.14 ) 对基态，二阶修正的每一项分母都为负：二阶微扰总是降低基态能量 （§P3.3 的色散力就是一例）。
应用：交换作用的起源 。两个电子处于两个正交的空间轨道 a a a b b b V ^ \hat V V ^ §P2.6 ，空间部分取 ψ ± = 1 2 [ a ( r 1 ) b ( r 2 ) ± b ( r 1 ) a ( r 2 ) ] \psi_\pm = \frac{1}{\sqrt2}[a(\mathbf r_1)b(\mathbf r_2)\pm b(\mathbf r_1)a(\mathbf r_2)] ψ ± = 2 1 [ a ( r 1 ) b ( r 2 ) ± b ( r 1 ) a ( r 2 )] + + + − - − V ^ \hat V V ^ 1 ↔ 2 1\leftrightarrow2 1 ↔ 2 ⟨ ψ ± ∣ V ^ ∣ ψ ± ⟩ \langle\psi_\pm\vert\hat V\vert\psi_\pm\rangle ⟨ ψ ± ∣ V ^ ∣ ψ ± ⟩
E ± ( 1 ) = K ± J e x , K = ∫ ∣ a ( 1 ) ∣ 2 ∣ b ( 2 ) ∣ 2 V d 3 r 1 d 3 r 2 , J e x = ∫ a ∗ ( 1 ) b ∗ ( 2 ) V b ( 1 ) a ( 2 ) d 3 r 1 d 3 r 2 (P2.15) E_\pm^{(1)} = K\pm J_{\mathrm{ex}},\qquad K = \int\lvert a(1)\rvert^2\lvert b(2)\rvert^2V\,d^3r_1d^3r_2,\qquad J_{\mathrm{ex}} = \int a^\ast(1)b^\ast(2)\,V\,b(1)a(2)\,d^3r_1d^3r_2 \tag{P2.15} E ± ( 1 ) = K ± J ex , K = ∫ ∣ a ( 1 ) ∣ 2 ∣ b ( 2 ) ∣ 2 V d 3 r 1 d 3 r 2 , J ex = ∫ a ∗ ( 1 ) b ∗ ( 2 ) V b ( 1 ) a ( 2 ) d 3 r 1 d 3 r 2 ( P2.15 ) 所以 E 单态 − E 三重态 = 2 J e x E_{\text{单态}} - E_{\text{三重态}} = 2J_{\mathrm{ex}} E 单态 − E 三重态 = 2 J ex (P2.13) ，这两个能量可以用一个只含自旋 的等效哈密顿量统一写出：
H ^ 等效 = 常数 − 2 J e x ℏ 2 S ^ 1 ⋅ S ^ 2 (P2.16) \hat H_{\text{等效}} = \text{常数} - \frac{2J_{\mathrm{ex}}}{\hbar^2}\,\hat{\mathbf S}_1\cdot\hat{\mathbf S}_2 \tag{P2.16} H ^ 等效 = 常数 − ℏ 2 2 J ex S ^ 1 ⋅ S ^ 2 ( P2.16 ) （验证：三重态 S ^ 1 ⋅ S ^ 2 = ℏ 2 / 4 \hat{\mathbf S}_1\cdot\hat{\mathbf S}_2 = \hbar^2/4 S ^ 1 ⋅ S ^ 2 = ℏ 2 /4 − 3 ℏ 2 / 4 -3\hbar^2/4 − 3 ℏ 2 /4 2 J e x 2J_{\mathrm{ex}} 2 J ex J e x > 0 J_{\mathrm{ex}}>0 J ex > 0 静电排斥加上泡利原理 ，大小可达 eV 量级，远大于磁偶极相互作用（§P3.2 ）。它是铁磁性的根源，也是第18章 伊辛模型的物理基础。
含时微扰与费米黄金规则 。设 t = 0 t = 0 t = 0 H ^ 0 \hat H_0 H ^ 0 ∣ i ⟩ \lvert i\rangle ∣ i ⟩ V ^ ( t ) \hat V(t) V ^ ( t ) ∣ ψ ( t ) ⟩ = ∑ n c n ( t ) e − i E n t / ℏ ∣ n ⟩ \lvert\psi(t)\rangle = \sum_nc_n(t)e^{-iE_nt/\hbar}\lvert n\rangle ∣ ψ ( t )⟩ = ∑ n c n ( t ) e − i E n t /ℏ ∣ n ⟩ ⟨ f ∣ \langle f\rvert ⟨ f ∣ i ℏ c ˙ f = ∑ n V f n ( t ) e i ω f n t c n i\hbar\,\dot c_f = \sum_nV_{fn}(t)e^{i\omega_{fn}t}c_n i ℏ c ˙ f = ∑ n V f n ( t ) e i ω f n t c n ω f n = ( E f − E n ) / ℏ \omega_{fn} = (E_f - E_n)/\hbar ω f n = ( E f − E n ) /ℏ c n = δ n i c_n = \delta_{ni} c n = δ ni
c f ( t ) ≈ 1 i ℏ ∫ 0 t V f i ( t ′ ) e i ω f i t ′ d t ′ c_f(t)\approx\frac{1}{i\hbar}\int_0^tV_{fi}(t')\,e^{i\omega_{fi}t'}\,dt' c f ( t ) ≈ i ℏ 1 ∫ 0 t V f i ( t ′ ) e i ω f i t ′ d t ′ 对 t > 0 t>0 t > 0 V ^ \hat V V ^ c f = V f i i ℏ ⋅ e i ω f i t − 1 i ω f i c_f = \frac{V_{fi}}{i\hbar}\cdot\frac{e^{i\omega_{fi}t} - 1}{i\omega_{fi}} c f = i ℏ V f i ⋅ i ω f i e i ω f i t − 1 ∣ c f ∣ 2 = ∣ V f i ∣ 2 ℏ 2 ⋅ 4 sin 2 ( ω f i t / 2 ) ω f i 2 \lvert c_f\rvert^2 = \frac{\lvert V_{fi}\rvert^2}{\hbar^2}\cdot\frac{4\sin^2(\omega_{fi}t/2)}{\omega_{fi}^2} ∣ c f ∣ 2 = ℏ 2 ∣ V f i ∣ 2 ⋅ ω f i 2 4 s i n 2 ( ω f i t /2 ) ω f i \omega_{fi} ω f i 4 sin 2 ( ω t / 2 ) / ω 2 4\sin^2(\omega t/2)/\omega^2 4 sin 2 ( ω t /2 ) / ω 2 ω = 0 \omega = 0 ω = 0 t 2 t^2 t 2 2 π / t 2\pi/t 2 π / t u = ω t / 2 u = \omega t/2 u = ω t /2 ∫ − ∞ ∞ ( sin 2 u / u 2 ) d u = π \int_{-\infty}^\infty(\sin^2u/u^2)\,du = \pi ∫ − ∞ ∞ ( sin 2 u / u 2 ) d u = π 2 π t 2\pi t 2 π t t t t 2 π t δ ( ω ) 2\pi t\,\delta(\omega) 2 π t δ ( ω )
δ 函数
δ ( x ) \delta(x) δ ( x ) ∫ f ( x ) δ ( x − x 0 ) d x = f ( x 0 ) \int f(x)\delta(x - x_0)\,dx = f(x_0) ∫ f ( x ) δ ( x − x 0 ) d x = f ( x 0 ) δ ( a x ) = δ ( x ) / ∣ a ∣ \delta(ax) = \delta(x)/\lvert a\rvert δ ( a x ) = δ ( x ) / ∣ a ∣ §A14.5 。）
跃迁概率随时间线性增长，单位时间的跃迁概率（跃迁速率）为常数。用 δ ( ω ) = ℏ δ ( E ) \delta(\omega) = \hbar\,\delta(E) δ ( ω ) = ℏ δ ( E )
W i → f = 2 π ℏ ∣ V f i ∣ 2 δ ( E f − E i ) (P2.17) W_{i\to f} = \frac{2\pi}{\hbar}\lvert V_{fi}\rvert^2\,\delta(E_f - E_i) \tag{P2.17} W i → f = ℏ 2 π ∣ V f i ∣ 2 δ ( E f − E i ) ( P2.17 ) 对能量连续分布、态密度为 ρ ( E f ) \rho(E_f) ρ ( E f ) δ \delta δ W = 2 π ℏ ∣ V f i ∣ 2 ρ ( E f ) W = \frac{2\pi}{\hbar}\lvert V_{fi}\rvert^2\rho(E_f) W = ℏ 2 π ∣ V f i ∣ 2 ρ ( E f ) 费米黄金规则 。若微扰随时间按 e ∓ i ω t e^{\mp i\omega t} e ∓ iω t δ ( E f − E i ∓ ℏ ω ) \delta(E_f - E_i\mp\hbar\omega) δ ( E f − E i ∓ ℏ ω ) ℏ ω \hbar\omega ℏ ω ∣ V f i ∣ = ∣ V i f ∣ \lvert V_{fi}\rvert = \lvert V_{if}\rvert ∣ V f i ∣ = ∣ V i f ∣ i → f i\to f i → f f → i f\to i f → i
§P2.8 密度矩阵与量子系综 统计力学处理的系统往往不处于确定的量子态，而是以概率 p k p_k p k ∣ ψ k ⟩ \lvert\psi_k\rangle ∣ ψ k ⟩ 密度矩阵
ρ ^ = ∑ k p k ∣ ψ k ⟩ ⟨ ψ k ∣ , ⟨ A ⟩ = ∑ k p k ⟨ ψ k ∣ A ^ ∣ ψ k ⟩ = T r ( ρ ^ A ^ ) (P2.18) \hat\rho = \sum_kp_k\lvert\psi_k\rangle\langle\psi_k\rvert,\qquad \langle A\rangle = \sum_kp_k\langle\psi_k\vert\hat A\vert\psi_k\rangle = \mathrm{Tr}(\hat\rho\hat A) \tag{P2.18} ρ ^ = k ∑ p k ∣ ψ k ⟩ ⟨ ψ k ∣ , ⟨ A ⟩ = k ∑ p k ⟨ ψ k ∣ A ^ ∣ ψ k ⟩ = Tr ( ρ ^ A ^ ) ( P2.18 ) （最后一步：T r ( ρ ^ A ^ ) = ∑ n ⟨ n ∣ ρ ^ A ^ ∣ n ⟩ = ∑ k p k ⟨ ψ k ∣ A ^ ( ∑ n ∣ n ⟩ ⟨ n ∣ ) ∣ ψ k ⟩ \mathrm{Tr}(\hat\rho\hat A) = \sum_n\langle n\vert\hat\rho\hat A\vert n\rangle = \sum_kp_k\langle\psi_k\rvert\hat A\left(\sum_n\lvert n\rangle\langle n\rvert\right)\lvert\psi_k\rangle Tr ( ρ ^ A ^ ) = ∑ n ⟨ n ∣ ρ ^ A ^ ∣ n ⟩ = ∑ k p k ⟨ ψ k ∣ A ^ ( ∑ n ∣ n ⟩ ⟨ n ∣ ) ∣ ψ k ⟩ ρ ^ \hat\rho ρ ^ T r ρ ^ = ∑ k p k = 1 \mathrm{Tr}\,\hat\rho = \sum_kp_k = 1 Tr ρ ^ = ∑ k p k = 1 ρ ^ 2 = ρ ^ \hat\rho^2 = \hat\rho ρ ^ 2 = ρ ^ 纯态 ），否则称为混合态 。
混合态与叠加态不同 。例如自旋 1/2：ρ ^ = 1 2 ( ∣ ↑ ⟩ ⟨ ↑ ∣ + ∣ ↓ ⟩ ⟨ ↓ ∣ ) \hat\rho = \frac12(\lvert\uparrow\rangle\langle\uparrow\rvert + \lvert\downarrow\rangle\langle\downarrow\rvert) ρ ^ = 2 1 (∣ ↑ ⟩ ⟨ ↑ ∣ + ∣ ↓ ⟩ ⟨ ↓ ∣) ⟨ S ^ ⟩ = 0 \langle\hat S\rangle = 0 ⟨ S ^ ⟩ = 0 1 2 ( ∣ ↑ ⟩ + ∣ ↓ ⟩ ) \frac{1}{\sqrt2}(\lvert\uparrow\rangle + \lvert\downarrow\rangle) 2 1 (∣ ↑ ⟩ + ∣ ↓ ⟩) x x x ⟨ S ^ x ⟩ = ℏ / 2 \langle\hat S_x\rangle = \hbar/2 ⟨ S ^ x ⟩ = ℏ/2
演化与平衡 。由薛定谔方程，每个 ∣ ψ k ⟩ \lvert\psi_k\rangle ∣ ψ k ⟩ i ℏ ∂ t ∣ ψ k ⟩ = H ^ ∣ ψ k ⟩ i\hbar\,\partial_t\lvert\psi_k\rangle = \hat H\lvert\psi_k\rangle i ℏ ∂ t ∣ ψ k ⟩ = H ^ ∣ ψ k ⟩
i ℏ ∂ ρ ^ ∂ t = [ H ^ , ρ ^ ] (P2.19) i\hbar\frac{\partial\hat\rho}{\partial t} = [\hat H,\hat\rho] \tag{P2.19} i ℏ ∂ t ∂ ρ ^ = [ H ^ , ρ ^ ] ( P2.19 ) 这是刘维尔方程（§P1.2 ：∂ ρ / ∂ t = − { ρ , H } \partial\rho/\partial t = -\lbrace\rho,\mathcal H\rbrace ∂ ρ / ∂ t = − { ρ , H } ∂ t ρ ^ = 0 \partial_t\hat\rho = 0 ∂ t ρ ^ = 0 [ H ^ , ρ ^ ] = 0 [\hat H,\hat\rho] = 0 [ H ^ , ρ ^ ] = 0 ρ ^ \hat\rho ρ ^ H ^ \hat H H ^
ρ ^ 微正则 = 1 Ω ∑ E ≤ E i ≤ E + δ E ∣ i ⟩ ⟨ i ∣ , ρ ^ 正则 = e − β H ^ Z , Z = T r e − β H ^ , ρ ^ 巨正则 = e − β ( H ^ − μ N ^ ) Ξ (P2.20) \hat\rho_{\text{微正则}} = \frac1\Omega\sum_{E\le E_i\le E+\delta E}\lvert i\rangle\langle i\rvert,\qquad \hat\rho_{\text{正则}} = \frac{e^{-\beta\hat H}}{Z},\ Z = \mathrm{Tr}\,e^{-\beta\hat H},\qquad \hat\rho_{\text{巨正则}} = \frac{e^{-\beta(\hat H - \mu\hat N)}}{\Xi} \tag{P2.20} ρ ^ 微正则 = Ω 1 E ≤ E i ≤ E + δ E ∑ ∣ i ⟩ ⟨ i ∣ , ρ ^ 正则 = Z e − β H ^ , Z = Tr e − β H ^ , ρ ^ 巨正则 = Ξ e − β ( H ^ − μ N ^ ) ( P2.20 ) 在能量本征态基中，e − β H ^ e^{-\beta\hat H} e − β H ^ T r e − β H ^ = ∑ i e − β E i \mathrm{Tr}\,e^{-\beta\hat H} = \sum_ie^{-\beta E_i} Tr e − β H ^ = ∑ i e − β E i (4.3) ；冯·诺伊曼熵 S = − k B T r ( ρ ^ ln ρ ^ ) S = -k_{\mathrm B}\mathrm{Tr}(\hat\rho\ln\hat\rho) S = − k B Tr ( ρ ^ ln ρ ^ ) (4.6) 。所以第二部分中"对微观态求和"的做法，就是密度矩阵表述在能量本征态基中的写法。
例：热平衡的自旋 1/2 。H ^ = − 1 2 γ ℏ B σ ^ z \hat H = -\frac12\gamma\hbar B\hat\sigma_z H ^ = − 2 1 γ ℏ B σ ^ z σ ^ z = d i a g ( 1 , − 1 ) \hat\sigma_z = \mathrm{diag}(1,-1) σ ^ z = diag ( 1 , − 1 ) x = β γ ℏ B / 2 x = \beta\gamma\hbar B/2 x = β γ ℏ B /2 σ ^ z 2 = 1 \hat\sigma_z^2 = 1 σ ^ z 2 = 1 e x σ ^ z = cosh x + σ ^ z sinh x e^{x\hat\sigma_z} = \cosh x + \hat\sigma_z\sinh x e x σ ^ z = cosh x + σ ^ z sinh x
ρ ^ = e x σ ^ z T r e x σ ^ z = 1 2 ( 1 + σ ^ z tanh x ) , ⟨ σ ^ z ⟩ = T r ( ρ ^ σ ^ z ) = tanh x \hat\rho = \frac{e^{x\hat\sigma_z}}{\mathrm{Tr}\,e^{x\hat\sigma_z}} = \frac12\left(1 + \hat\sigma_z\tanh x\right),\qquad \langle\hat\sigma_z\rangle = \mathrm{Tr}(\hat\rho\hat\sigma_z) = \tanh x ρ ^ = Tr e x σ ^ z e x σ ^ z = 2 1 ( 1 + σ ^ z tanh x ) , ⟨ σ ^ z ⟩ = Tr ( ρ ^ σ ^ z ) = tanh x 与 (6.6) 、(6.8) 的极化度完全一致。
自测题
三维立方盒中，求能量为 π 2 ℏ 2 2 m L 2 × 14 \frac{\pi^2\hbar^2}{2mL^2}\times14 2 m L 2 π 2 ℏ 2 × 14 ［答：6，即 ( 1 , 2 , 3 ) (1,2,3) ( 1 , 2 , 3 )
利用 x ^ = ℏ / 2 m ω ( a ^ + a ^ † ) \hat x = \sqrt{\hbar/2m\omega}\,(\hat a + \hat a^\dagger) x ^ = ℏ/2 mω ( a ^ + a ^ † ) ⟨ n ∣ x ^ 2 ∣ n ⟩ = ℏ 2 m ω ( 2 n + 1 ) \langle n\vert\hat x^2\vert n\rangle = \frac{\hbar}{2m\omega}(2n+1) ⟨ n ∣ x ^ 2 ∣ n ⟩ = 2 mω ℏ ( 2 n + 1 )
对 D 2 \mathrm D_2 D 2 J J J ［答：核自旋对称的 6 个态配偶数 J J J J J J
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