A ^ ( c 1 ψ 1 + c 2 ψ 2 ) = c 1 A ^ ψ 1 + c 2 A ^ ψ 2 \hat{A} (c_1 \psi_1 + c_2 \psi_2) = c_1 \hat{A} \psi_1 + c_2 \hat{A} \psi_2 A ^ ( c 1 ψ 1 + c 2 ψ 2 ) = c 1 A ^ ψ 1 + c 2 A ^ ψ 2
( A ^ + B ^ ) ψ = A ^ ψ + B ^ ψ \pqty{\hat{A} + \hat{B}} \psi = \hat{A} \psi + \hat{B} \psi ( A ^ + B ^ ) ψ = A ^ ψ + B ^ ψ
( A ^ B ^ ) ψ = A ^ ( B ^ ψ ) \pqty{\hat{A} \hat{B}} \psi = \hat{A} \pqty{\hat{B} \psi} ( A ^ B ^ ) ψ = A ^ ( B ^ ψ )
一般来说, 算符之积不满足交换律 , 即 A ^ B ^ ≠ B ^ A ^ \hat{A} \hat{B} \neq \hat{B} \hat{A} A ^ B ^ = B ^ A ^ 唯一不同之处 .
定义 对易式 (commutator)
\comm A ^ B ^ = A ^ B ^ − B ^ A ^ \comm{\hat{A}}{\hat{B}} = \hat{A} \hat{B} - \hat{B} \hat{A} \comm A ^ B ^ = A ^ B ^ − B ^ A ^
\comm x α p β = i ℏ δ α β \comm{x_{\alpha}}{p_{\beta}} = i \hbar \delta_{\alpha \beta} \comm x α p β = i ℏ δ α β
不难证明, 对易式满足下列代数恒等式:
\comm A ^ B ^ = − \comm B ^ A ^ \comm A ^ B ^ + C ^ = \comm A ^ B ^ + \comm A ^ C ^ \comm A ^ B ^ C ^ = B ^ \comm A ^ C ^ + \comm A ^ B ^ C ^ \comm A ^ B ^ C ^ = A ^ \comm B ^ C ^ + \comm A ^ C ^ B ^ \comm A ^ \comm B ^ C ^ + \comm B ^ \comm C ^ A ^ + \comm C ^ \comm A ^ B ^ = 0 (Jacobi 恒等式) \begin{align*} & \comm{\hat{A}}{\hat{B}} = - \comm{\hat{B}}{\hat{A}} \\ & \comm{\hat{A}}{\hat{B} + \hat{C}} = \comm{\hat{A}}{\hat{B}} + \comm{\hat{A}}{\hat{C}} \\ & \comm{\hat{A}}{\hat{B} \hat{C}} = \hat{B} \comm{\hat{A}}{\hat{C}} + \comm{\hat{A}}{\hat{B}} \hat{C} \\ & \comm{\hat{A} \hat{B}}{\hat{C}} = \hat{A} \comm{\hat{B}}{\hat{C}} + \comm{\hat{A}}{\hat{C}} \hat{B} \\ & \comm{\hat{A}}{\comm{\hat{B}}{\hat{C}}} + \comm{\hat{B}}{\comm{\hat{C}}{\hat{A}}} + \comm{\hat{C}}{\comm{\hat{A}}{\hat{B}}} = 0 & \qq{(Jacobi 恒等式)} \end{align*} \comm A ^ B ^ = − \comm B ^ A ^ \comm A ^ B ^ + C ^ = \comm A ^ B ^ + \comm A ^ C ^ \comm A ^ B ^ C ^ = B ^ \comm A ^ C ^ + \comm A ^ B ^ C ^ \comm A ^ B ^ C ^ = A ^ \comm B ^ C ^ + \comm A ^ C ^ B ^ \comm A ^ \comm B ^ C ^ + \comm B ^ \comm C ^ A ^ + \comm C ^ \comm A ^ B ^ = 0 (Jacobi 恒等式 )
\vb ∗ l ^ = \vb ∗ r \cp \vb ∗ p ^ \hat{\vb*{l}} = \vb*{r} \cp \hat{\vb*{p}} \vb ∗ l ^ = \vb ∗ r \cp \vb ∗ p ^
\comm l ^ α x β = ε α β γ i ℏ x γ \comm{\hat{l}_{\alpha}}{x_{\beta}} = \varepsilon_{\alpha \beta \gamma} i \hbar x_{\gamma} \comm l ^ α x β = ε α β γ i ℏ x γ
式中 ε α β γ \varepsilon_{\alpha \beta \gamma} ε α β γ
{ ε α β γ = − ε β α γ = − ε α γ β ε 123 = 1 \begin{cases} \varepsilon_{\alpha\beta\gamma} = -\varepsilon_{\beta\alpha\gamma} = -\varepsilon_{\alpha\gamma\beta} \\ \varepsilon_{123} = 1 \end{cases} { ε α β γ = − ε β α γ = − ε α γ β ε 123 = 1
类似可以证明,
\comm l ^ α p ^ β = ε α β γ i ℏ p γ \comm{\hat{l}_{\alpha}}{\hat{p}_{\beta}} = \varepsilon_{\alpha\beta\gamma} i \hbar p_{\gamma} \comm l ^ α p ^ β = ε α β γ i ℏ p γ
还可以证明
\comm l ^ α l ^ β = ε α β γ i ℏ l ^ γ \comm{\hat{l}_{\alpha}}{\hat{l}_{\beta}} = \varepsilon_{\alpha\beta\gamma} i \hbar \hat{l}_{\gamma} \comm l ^ α l ^ β = ε α β γ i ℏ l ^ γ
这就是 角动量各分量的对易式 , 是很重要的, 必须牢记.
\vb ∗ l ^ \cp \vb ∗ l ^ = i ℏ \vb ∗ l ^ \hat{\vb*{l}} \cp \hat{\vb*{l}} = i \hbar \hat{\vb*{l}} \vb ∗ l ^ \cp \vb ∗ l ^ = i ℏ \vb ∗ l ^
\comm \vb ∗ l ^ 2 l ^ α = 0 , α = x , y , z \comm{\hat{\vb*{l}}^2}{\hat{l}_{\alpha}} = 0 \qc \alpha = x, y, z \comm \vb ∗ l ^ 2 l ^ α = 0 , α = x , y , z
在球坐标系中, 利用坐标变换关系, 即
{ x = r sin θ cos φ , y = r sin θ sin φ , z = r cos θ , { r = x 2 + y 2 + z 2 θ = arctan ( x 2 + y 2 / z ) φ = arctan ( y / x ) \begin{cases} x = r \sin\theta \cos\varphi, \\ y = r \sin\theta \sin\varphi, \\ z = r \cos\theta, \end{cases} \quad \begin{cases} r = \sqrt{x^2 + y^2 + z^2} \\ \theta = \arctan(\sqrt{x^2 + y^2} / z) \\ \varphi = \arctan(y / x) \end{cases} ⎩ ⎨ ⎧ x = r sin θ cos φ , y = r sin θ sin φ , z = r cos θ , ⎩ ⎨ ⎧ r = x 2 + y 2 + z 2 θ = arctan ( x 2 + y 2 / z ) φ = arctan ( y / x )
可以把 \vb ∗ l ^ \hat{\vb*{l}} \vb ∗ l ^
l ^ x = i ℏ ( sin φ \pdv θ + cot θ cos φ \pdv φ ) l ^ y = i ℏ ( − cos φ \pdv θ + cot θ sin φ \pdv φ ) l ^ z = − i ℏ \pdv φ \vb ∗ l ^ 2 = − ℏ 2 \bqty 1 sin θ \pdv θ sin θ \pdv θ + 1 sin [ 2 ] θ \pdv [ 2 ] φ \begin{align*} \hat{l}_x & = i \hbar \pqty{\sin{\varphi} \pdv{\theta} + \cot{\theta} \cos{\varphi} \pdv{\varphi}} \\ \hat{l}_y & = i \hbar \pqty{- \cos{\varphi} \pdv{\theta} + \cot{\theta} \sin{\varphi} \pdv{\varphi}} \\ \hat{l}_z & = - i \hbar \pdv{\varphi} \\ \hat{\vb*{l}}^2 & = - \hbar^2 \bqty{\frac{1}{\sin{\theta}} \pdv{\theta} \sin{\theta} \pdv{\theta} + \frac{1}{\sin[2]{\theta}} \pdv[2]{\varphi}} \end{align*} l ^ x l ^ y l ^ z \vb ∗ l ^ 2 = i ℏ ( sin φ \pdv θ + cot θ cos φ \pdv φ ) = i ℏ ( − cos φ \pdv θ + cot θ sin φ \pdv φ ) = − i ℏ \pdv φ = − ℏ 2 \bqty sin θ 1 \pdv θ sin θ \pdv θ + sin [ 2 ] θ 1 \pdv [ 2 ] φ
设算符 A ^ \hat{A} A ^
A ^ A ^ − 1 = A ^ − 1 A ^ = I , \comm A ^ A ^ − 1 = 0 \hat{A} \hat{A}^{-1} = \hat{A}^{-1} \hat{A} = I \qc \comm{\hat{A}}{\hat{A}^{-1}} = 0 A ^ A ^ − 1 = A ^ − 1 A ^ = I , \comm A ^ A ^ − 1 = 0
( A ^ B ^ ) − 1 = B ^ − 1 A ^ − 1 \pqty{\hat{A} \hat{B}}^{-1} = \hat{B}^{-1} \hat{A}^{-1} ( A ^ B ^ ) − 1 = B ^ − 1 A ^ − 1
定义一个量子体系的任意两个波函数 (态) ψ \psi ψ φ \varphi φ
( ψ , φ ) = ∫ \dd τ ψ ∗ φ (\psi, \varphi) = \int \dd{\tau} \psi^* \varphi ( ψ , φ ) = ∫ \dd τ ψ ∗ φ
可以证明
{ ( ψ , ψ ) ⩾ 0 ( ψ , φ ) ∗ = ( φ , ψ ) ( ψ , c 1 φ 1 + c 2 φ 2 ) = c 1 ( ψ , φ 1 ) + c 2 ( ψ , φ 2 ) ( c 1 ψ 1 + c 2 ψ 2 , φ ) = c 1 ∗ ( ψ 1 , φ ) + c 2 ∗ ( ψ 2 , φ ) \begin{cases} (\psi, \psi) \geqslant 0 \\ (\psi, \varphi)^* = (\varphi, \psi) \\ (\psi, c_1 \varphi_1 + c_2 \varphi_2) = c_1 (\psi, \varphi_1) + c_2 (\psi, \varphi_2) \\ (c_1 \psi_1 + c_2 \psi_2, \varphi) = c_1^* (\psi_1, \varphi) + c_2^* (\psi_2, \varphi) \end{cases} ⎩ ⎨ ⎧ ( ψ , ψ ) ⩾ 0 ( ψ , φ ) ∗ = ( φ , ψ ) ( ψ , c 1 φ 1 + c 2 φ 2 ) = c 1 ( ψ , φ 1 ) + c 2 ( ψ , φ 2 ) ( c 1 ψ 1 + c 2 ψ 2 , φ ) = c 1 ∗ ( ψ 1 , φ ) + c 2 ∗ ( ψ 2 , φ )
( ψ , A ^ ~ φ ) = ( φ ∗ , A ^ ψ ∗ ) \pqty{\psi, \tilde{\hat{A}} \varphi} = \pqty{\varphi^*, \hat{A} \psi^*} ( ψ , A ^ ~ φ ) = ( φ ∗ , A ^ ψ ∗ )
( A ^ B ^ ) ~ = B ^ ~ A ^ ~ \widetilde{\pqty{\hat{A} \hat{B}}} = \tilde{\hat{B}} \tilde{\hat{A}} ( A ^ B ^ ) = B ^ ~ A ^ ~
A ^ ∗ ψ = ( A ^ ψ ∗ ) ∗ \hat{A}^* \psi = \pqty{\hat{A} \psi^*}^* A ^ ∗ ψ = ( A ^ ψ ∗ ) ∗
算符 A ^ ∗ \hat{A}^* A ^ ∗
( ψ , A ^ + φ ) = ( A ^ ψ , φ ) \pqty{\psi, \hat{A}^+ \varphi} = \pqty{\hat{A} \psi, \varphi} ( ψ , A ^ + φ ) = ( A ^ ψ , φ )
A ^ + = A ^ ~ ∗ \hat{A}^+ = \tilde{\hat{A}}^* A ^ + = A ^ ~ ∗
( A ^ B ^ C ^ ⋯ ) + = ⋯ C ^ + B ^ + A ^ + \pqty{\hat{A} \hat{B} \hat{C} \cdots}^+ = \cdots \hat{C}^+ \hat{B}^+ \hat{A}^+ ( A ^ B ^ C ^ ⋯ ) + = ⋯ C ^ + B ^ + A ^ +
( ψ , A ^ φ ) = ( A ^ ψ , φ ) , 或 A ^ + = A ^ \pqty{\psi, \hat{A} \varphi} = \pqty{\hat{A} \psi, \varphi}, \qq{或} \hat{A}^+ = \hat{A} ( ψ , A ^ φ ) = ( A ^ ψ , φ ) , 或 A ^ + = A ^
( A ^ B ^ ) + = B ^ + A ^ + = B ^ A ^ \pqty{\hat{A} \hat{B}}^+ = \hat{B}^+ \hat{A}^+ = \hat{B} \hat{A} ( A ^ B ^ ) + = B ^ + A ^ + = B ^ A ^
只当 \comm A ^ B ^ = 0 \comm{\hat{A}}{\hat{B}} = 0 \comm A ^ B ^ = 0 A ^ B ^ \hat{A} \hat{B} A ^ B ^ A ^ B ^ \hat{A} \hat{B} A ^ B ^
定理
体系的任何状态下, 其厄米算符的平均值必为实数.
逆定理
在任何状态下平均值均为实的算符必为厄米算符.
实验上可观测量 , 当然要求在任何态下平均值都是实数, 因此 相应的算符必须是厄米算符 .
推论
设 A ^ \hat{A} A ^ ψ \psi ψ
A 2 ‾ = ( ψ , A ^ 2 ψ ) = ( A ^ ψ , A ^ ψ ) ⩾ 0 \overline{A^2} = \pqty{\psi, \hat{A}^2 \psi} = \pqty{\hat{A} \psi, \hat{A} \psi} \geqslant 0 A 2 = ( ψ , A ^ 2 ψ ) = ( A ^ ψ , A ^ ψ ) ⩾ 0
涨落定义为
Δ A 2 ‾ = ( A ^ − A ‾ ) 2 ‾ = ∫ ψ ∗ ( A ^ − A ‾ ) 2 ψ \dd τ \overline{\Delta{A}^2} = \overline{\pqty{\hat{A} - \overline{A}}^2} = \int \psi^* \pqty{\hat{A} - \overline{A}}^2 \psi \dd{\tau} Δ A 2 = ( A ^ − A ) 2 = ∫ ψ ∗ ( A ^ − A ) 2 ψ \dd τ
如果体系处于一种特殊的状态, 测量 A A A Δ A 2 ‾ = 0 \overline{\Delta{A}^2} = 0 Δ A 2 = 0 A A A 本征态 .
A ^ ψ n = A n ψ n \hat{A} \psi_n = A_n \psi_n A ^ ψ n = A n ψ n
A n A_n A n A ^ \hat{A} A ^ ψ n \psi_n ψ n 本征态 . 量子力学中的一个基本假定是: 测量力学量 A A A A ^ \hat{A} A ^
定理 2
厄米算符的属于不同本征值的本征函数, 彼此正交.
例 1
求角动量 z z z l ^ z = − i ℏ \pdv φ \hat{l}_z = - i \hbar \pdv{\varphi} l ^ z = − i ℏ \pdv φ
l z ′ = m ℏ , m = 0 , ± 1 , ± 2 , ⋯ ψ m ( φ ) = 1 2 π e i m φ , m = 0 , ± 1 , ± 2 , ⋯ \begin{gather*} l_z' = m \hbar \qc m = 0, \pm 1, \pm 2, \cdots \\ \psi_m(\varphi) = \frac{1}{\sqrt{2 \pi}} e^{i m \varphi} \qc m = 0, \pm 1, \pm 2, \cdots \end{gather*} l z ′ = m ℏ , m = 0 , ± 1 , ± 2 , ⋯ ψ m ( φ ) = 2 π 1 e im φ , m = 0 , ± 1 , ± 2 , ⋯
例 2
平面转子的能量本征值与本征态.
ψ m ( φ ) = 1 2 π e i m φ , m = 0 , ± 1 , ± 2 , ⋯ E m = m 2 ℏ 2 / 2 I ⩾ 0 \begin{gather*} \psi_m(\varphi) = \frac{1}{\sqrt{2 \pi}} e^{i m \varphi} \qc m = 0, \pm 1, \pm 2, \cdots \\ E_m = m^2 \hbar^2 / 2 I \geqslant 0 \end{gather*} ψ m ( φ ) = 2 π 1 e im φ , m = 0 , ± 1 , ± 2 , ⋯ E m = m 2 ℏ 2 /2 I ⩾ 0
例 3
求动量的 x x x p ^ x = − i ℏ \pdv x \hat{p}_x = - i \hbar \pdv{x} p ^ x = − i ℏ \pdv x
ψ p x ′ ( x ) = 1 2 π ℏ e i p x ′ x / ℏ \psi_{p_x'}(x) = \frac{1}{\sqrt{2 \pi \hbar}} e^{i p_x' x / \hbar} ψ p x ′ ( x ) = 2 π ℏ 1 e i p x ′ x /ℏ
例 4
一维自由粒子的能量本征态.
ψ E ( x ) ∼ e ± i k x , k = 2 m E / ℏ ⩾ 0 E = ℏ 2 k 2 / 2 m ⩾ 0 \begin{gather*} \psi_E(x) \sim e^{\pm i k x} \qc k = \sqrt{2 m E} / \hbar \geqslant 0 \\ E = \hbar^2 k^2 / 2 m \geqslant 0 \end{gather*} ψ E ( x ) ∼ e ± ik x , k = 2 m E /ℏ ⩾ 0 E = ℏ 2 k 2 /2 m ⩾ 0
Δ A \vdot Δ B ⩾ 1 2 \abs \comm A B ‾ \Delta{A} \vdot \Delta{B} \geqslant \frac{1}{2} \abs{\overline{\comm{A}{B}}} Δ A \vdot Δ B ⩾ 2 1 \abs \comm A B
这就是 任意两个力学量 A A A B B B , 即不确定度关系 (uncertainty relation).
若两个力学量 A A A B B B Δ A \Delta{A} Δ A Δ B \Delta{B} Δ B , 即 A A A B B B \comm A B ‾ = 0 \overline{\comm{A}{B}} = 0 \comm A B = 0 的特殊态可能是例外 ), 或者说 它们不能有共同本征态 . 反之, 若两个厄米算符 A ^ \hat{A} A ^ B ^ \hat{B} B ^ Δ A = 0 \Delta{A} = 0 Δ A = 0 Δ B = 0 \Delta{B} = 0 Δ B = 0
由于角动量的三个分量不对易, 一般无共同本征态. 但由于 \comm \vb ∗ l 2 , l α = 0 ( α = x , y , z ) \comm{\vb*{l}^2, l_{\alpha}} = 0 (\alpha = x, y, z) \comm \vb ∗ l 2 , l α = 0 ( α = x , y , z ) \vb ∗ l 2 \vb*{l}^2 \vb ∗ l 2 l z l_z l z
采用球坐标, \vb ∗ l 2 \vb*{l}^2 \vb ∗ l 2
\vb ∗ l 2 = − ℏ \bqty 1 sin θ \pdv θ sin θ \pdv θ + 1 sin [ 2 ] θ \pdv [ 2 ] φ = − ℏ 2 sin θ \pdv θ sin θ \pdv θ + 1 sin [ 2 ] θ l z 2 \begin{equation*} \begin{split} \vb*{l}^2 & = - \hbar \bqty{\frac{1}{\sin{\theta}} \pdv{\theta} \sin{\theta} \pdv{\theta} + \frac{1}{\sin[2]{\theta}} \pdv[2]{\varphi}} \\ & = - \frac{\hbar^2}{\sin{\theta}} \pdv{\theta} \sin{\theta} \pdv{\theta} + \frac{1}{\sin[2]{\theta}} l_z^2 \end{split} \end{equation*} \vb ∗ l 2 = − ℏ \bqty sin θ 1 \pdv θ sin θ \pdv θ + sin [ 2 ] θ 1 \pdv [ 2 ] φ = − sin θ ℏ 2 \pdv θ sin θ \pdv θ + sin [ 2 ] θ 1 l z 2
令
Y ( θ , φ ) = Θ ( θ ) ψ m ( φ ) Y(\theta, \varphi) = \Theta(\theta) \psi_m(\varphi) Y ( θ , φ ) = Θ ( θ ) ψ m ( φ )
Θ l m ( θ ) = ( − 1 ) m ( 2 l + 1 ) 2 \vdot ( l − m ) ! ( l + m ) ! P l m ( cos θ ) m = l , l − 1 , ⋯ , − l + 1 , − l \begin{gather*} \Theta_{l m}(\theta) = (-1)^m \sqrt{\frac{(2 l + 1)}{2} \vdot \frac{(l - m)!}{(l + m)!}} P_l^m(\cos{\theta}) \\ m = l, l - 1, \cdots, - l + 1, -l \end{gather*} Θ l m ( θ ) = ( − 1 ) m 2 ( 2 l + 1 ) \vdot ( l + m )! ( l − m )! P l m ( cos θ ) m = l , l − 1 , ⋯ , − l + 1 , − l
( \vb ∗ l 2 , l z ) (\vb*{l}^2, l_z) ( \vb ∗ l 2 , l z )
Y l m ( θ , φ ) = ( − 1 ) m 2 l + 1 4 π ( l − m ) ! ( l + m ) ! P l m ( cos θ ) e i m φ Y_{l m}(\theta, \varphi) = (-1)^m \sqrt{\frac{2 l + 1}{4 \pi} \frac{(l - m)!}{(l + m)!}} P_l^m(\cos{\theta}) e^{i m \varphi} Y l m ( θ , φ ) = ( − 1 ) m 4 π 2 l + 1 ( l + m )! ( l − m )! P l m ( cos θ ) e im φ
Y l m Y_{l m} Y l m
\vb ∗ l 2 Y l m = l ( l + 1 ) ℏ 2 Y l m l z Y l m = m ℏ Y l m l = 0 , 1 , 2 , ⋯ , m = l , l − 1 , ⋯ , − l + 1 , − l ∫ 0 2 π \dd φ ∫ 0 π sin θ \dd θ Y l m ∗ ( θ , φ ) Y l ′ m ′ ( θ , φ ) = δ l l ′ δ m m ′ \begin{gather*} \vb*{l}^2 Y_{l m} = l (l + 1) \hbar^2 Y_{l m} \\ l_z Y_{l m} = m \hbar Y_{l m} \\ l = 0, 1, 2, \cdots \qc m = l, l - 1, \cdots, - l + 1, -l \\ \int_0^{2 \pi} \dd{\varphi} \int_0^{\pi} \sin{\theta} \dd{\theta} Y_{l m}^*(\theta, \varphi) Y_{l' m'}(\theta, \varphi) = \delta_{l l'} \delta_{m m'} \end{gather*} \vb ∗ l 2 Y l m = l ( l + 1 ) ℏ 2 Y l m l z Y l m = m ℏ Y l m l = 0 , 1 , 2 , ⋯ , m = l , l − 1 , ⋯ , − l + 1 , − l ∫ 0 2 π \dd φ ∫ 0 π sin θ \dd θ Y l m ∗ ( θ , φ ) Y l ′ m ′ ( θ , φ ) = δ l l ′ δ m m ′
l l l 轨道角动量量子数 , m m m 磁量子数 . 对于给定 l l l \vb ∗ l 2 \vb*{l}^2 \vb ∗ l 2 m = l , l − 1 , ⋯ , − l m = l, l - 1, \cdots, -l m = l , l − 1 , ⋯ , − l ( 2 l + 1 ) (2 l + 1) ( 2 l + 1 ) Y l m Y_{l m} Y l m l ^ z \hat{l}_z l ^ z
设给定一组量子数 α \alpha α , 则我们称 ( A ^ 1 , A ^ 2 , ⋯ ) (\hat{A}_1, \hat{A}_2, \cdots) ( A ^ 1 , A ^ 2 , ⋯ ) 对易可观测量完全集 .
如体系的 Hamilton 量不显含时间 t t t \pdv ∗ H t = 0 \pdv*{H}{t} = 0 \pdv ∗ H t = 0 H H H 对易守恒量完全集 (CSCCO) 包括 H H H 好量子数 .
关于 CSCO, 再做几点说明:
CSCO 是限于 最小集合 , 即从集合中抽出任何一个可观测量后, 就不再构成体系的 CSCO. 所以要求 CSCO 中各观测量是 函数独立的 . 一个给定体系的 CSCO 中, 可观测量的数目一般等于体系自由度的数目, 但也可以大于体系自由度的数目. 一个给定体系往往可以找到多个 CSCO, 或 CSCCO. 在处理具体问题时, 应视其侧重点来进行选择. 一个 CSCCO 的成员的选择, 涉及体系的对称性. 量子体系的可观测量 (力学量) 用一个线性厄米算符来描述 , 也是 量子力学的一个基本假定, 它们的正确性应该由实验来判定 . “量子力学中力学量用相应的线性厄米算符来表达 ”, 其含义是多方面的.
在给定状态 ψ \psi ψ A A A A ‾ \overline{A} A
A ‾ = ( ψ , A ^ ψ ) / ( ψ , ψ ) \overline{A} = \pqty{\psi, \hat{A} \psi} / \pqty{\psi, \psi} A = ( ψ , A ^ ψ ) / ( ψ , ψ )
在实验上观测某力学量 A A A 可能取值 A ′ A' A ′ 就是算符 A ^ \hat{A} A ^
力学量之间的关系也通过相应的算符之间的关系反映出来. 例如, 两个力学量 A A A B B B \comm A ^ B ^ = 0 \comm{\hat{A}}{\hat{B}} = 0 \comm A ^ B ^ = 0 \comm A ^ B ^ ≠ 0 \comm{\hat{A}}{\hat{B}} \neq 0 \comm A ^ B ^ = 0 A A A B B B
特别是对于 H H H t t t 一个力学量 A A A , 可以根据 A ^ \hat{A} A ^ H ^ \hat{H} H ^
δ \delta δ
\var ( x − x 0 ) = { 0 , x ≠ x 0 ∞ , x = x 0 \var(x - x_0) = \begin{cases} 0, & x \neq x_0 \\ \infty, & x = x_0 \end{cases} \var ( x − x 0 ) = { 0 , ∞ , x = x 0 x = x 0
∫ x 0 − ε x 0 + ε \var ( x − x 0 ) \dd x = ∫ − ∞ + ∞ \var ( x − x 0 ) \dd x = 1 ( ε > 0 ) \int_{x_0 - \varepsilon}^{x_0 + \varepsilon} \var(x - x_0) \dd{x} = \int_{-\infty}^{+\infty} \var(x - x_0) \dd{x} = 1 \quad (\varepsilon > 0) ∫ x 0 − ε x 0 + ε \var ( x − x 0 ) \dd x = ∫ − ∞ + ∞ \var ( x − x 0 ) \dd x = 1 ( ε > 0 )
或等价地表示为: 对于在 x = x 0 x = x_0 x = x 0 f ( x ) f(x) f ( x )
∫ − ∞ + ∞ f ( x ) \var ( x − x 0 ) \dd x = f ( x 0 ) \int_{-\infty}^{+\infty} f(x) \var(x - x_0) \dd{x} = f(x_0) ∫ − ∞ + ∞ f ( x ) \var ( x − x 0 ) \dd x = f ( x 0 )
δ \delta δ
\var ( x − x 0 ) = 1 2 π ∫ − ∞ + ∞ \dd k e i k ( x − x 0 ) \var(x - x_0) = \frac{1}{2 \pi} \int_{-\infty}^{+\infty} \dd{k} e^{i k (x - x_0)} \var ( x − x 0 ) = 2 π 1 ∫ − ∞ + ∞ \dd k e ik ( x − x 0 )
为了保证动量算符 p ^ x = − i ℏ \pdv x \hat{p}_x = - i \hbar \pdv{x} p ^ x = − i ℏ \pdv x . 动量本征态 ψ p ( x ) ∼ e i p x / ℏ \psi_p(x) \sim e^{i p x / \hbar} ψ p ( x ) ∼ e i p x /ℏ
ψ p ( − L / 2 ) = ψ p ( L / 2 ) \psi_p(- L / 2) = \psi_p(L / 2) ψ p ( − L /2 ) = ψ p ( L /2 )
正交完备的箱归一化波函数为
ψ p ( \vb ∗ r ) = 1 L 3 / 2 e i \vb ∗ p \vdot \vb ∗ r / ℏ \psi_p(\vb*{r}) = \frac{1}{L^{3 / 2}} e^{i \vb*{p} \vdot \vb*{r} / \hbar} ψ p ( \vb ∗ r ) = L 3/2 1 e i \vb ∗ p \vdot \vb ∗ r /ℏ
式中
p x = h L n , p y = h L l , p z = h L m , n , l , m = 0 , ± 1 , ± 2 , ⋯ p_x = \frac{h}{L} n, p_y = \frac{h}{L} l, p_z = \frac{h}{L} m \qc n, l, m = 0, \pm 1, \pm 2, \cdots p x = L h n , p y = L h l , p z = L h m , n , l , m = 0 , ± 1 , ± 2 , ⋯
\label e q : 22 \var ( \vb ∗ r − \vb ∗ r ′ ) = 1 ℏ 3 ∫ − ∞ + ∞ \dd [ 3 ] p e i \vb ∗ p \vdot ( \vb ∗ r − \vb ∗ r ′ ) / ℏ \begin{equation} \label{eq:22} \var(\vb*{r} - \vb*{r}') = \frac{1}{\hbar^3} \int_{-\infty}^{+\infty} \dd[3]{p} e^{i \vb*{p} \vdot (\vb*{r} - \vb*{r}') / \hbar} \end{equation} \label e q : 22 \var ( \vb ∗ r − \vb ∗ r ′ ) = ℏ 3 1 ∫ − ∞ + ∞ \dd [ 3 ] p e i \vb ∗ p \vdot ( \vb ∗ r − \vb ∗ r ′ ) /ℏ
式 \eqref e q : 22 \eqref{eq:22} \eqref e q : 22 相空间一个体积元 ℏ 3 \hbar^3 ℏ 3 .
