QM-Formulas and Theorems in CN

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最新更新时间戳：2021-12-24-2:10

重要的公式

1，量子力学部分

（1）H-F定理

\begin{aligned} E (λ) & = < ψ (λ) | {\hat{H}}_{λ} | ψ (λ) > \\ \frac{d E (λ)}{d λ} & = < ψ (λ) | \frac{d {\hat{H}}_{λ}}{d λ} | ψ (λ) > \end{aligned}

（2）真正交矩阵性质

\begin{matrix} R \tilde{R} = \tilde{R} R = 1 \\ d e t R = | \begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix} | = 1 \end{matrix}

$\tilde{R}$ $R$ 的转置序列, |R| = 1 or -1

（3）幺正矩阵性质

R R^{+} = R^{+} R = 1

（4）F表象

$\psi$ $\psi _{k}$ （k代表一组完备的量子数，在本节中代表离散谱），可以用来构成该态空间的一完备的基矢，称作 F表象。

$(\psi_k,\psi_j=\delta_{kj})$

$\psi$ 都可以用它们来展开：

\begin{aligned} ψ & = \sum_{k} a_{k} ψ_{k} \\ a_{k} & = (ψ_{k}, ψ) \end{aligned}

$F'$ $a'$ $a$ 的关系是

$a_{1,2,3,...列向量}'= S_{a_{1,2,3,...}k_{1,2,3...}矩阵a行k列}=a_{1,2,3,...列向量}$ $S_{ak}=(\psi'_{a},\psi_{k})$

$S^+S=SS^+=I$ $S^+_{ka}=S^*_{ak}$

（5）一维谐振子的 x , p ，H 算符等

$x_{m n}=\left(\psi_{m}, x \psi_{n}\right)=\frac{1}{\alpha}\left[\sqrt{\frac{n+1}{2}} \delta_{m, n+1}+\sqrt{\frac{n}{2}} \delta_{m, n-1}\right]$ $P_{m n}=\left(\psi_{m}, \hat{p} \psi_{n}\right)=i \hbar \alpha\left[\sqrt{\frac{n+1}{2}} \delta_{m, n+1}-\sqrt{\frac{n}{2}} \delta_{m, n-1}\right]$ ，

$H_{m n}=\left(n+\frac{1}{2}\right) \hbar \omega \delta_{m n}$ ，

$\psi_{n}(x)=A_{n} e^{-\frac{1}{2} \alpha^{2} x^{2}} H_{n}(\alpha x)$ $A_{n}=\left(\frac{\alpha}{\sqrt{\pi} 2^{n} \cdot n !}\right)^{\frac{1}{2}}$ $\alpha=\sqrt{\frac{m \omega}{\hbar}}, \omega=\sqrt{\frac{k}{m}}$ $E_{n}=\left(n+\frac{1}{2}\right) \hbar \omega, n=0,1,2, \cdots$ ，

$\begin{aligned} x \psi_{n}(x) &=\frac{1}{\alpha}\left[\sqrt{\frac{n}{2}} \psi_{n-1}(x)+\sqrt{\frac{n+1}{2}} \psi_{n+1}(x)\right] \\ x^{2} \psi_{n}(x) &=\frac{1}{2 \alpha^{2}}\left[\sqrt{n(n-1)} \psi_{n-2}(x)+(2 n+1) \psi_{n}(x)+\sqrt{(n+1)(n+2)} \psi_{n+2}(x)\right] \\ \frac{d}{d x} \psi_{n}(x) &=\alpha\left[\sqrt{\frac{n}{2}} \psi_{n-1}-\sqrt{\frac{n+1}{2}} \psi_{n+1}\right] \\ \frac{d^{2}}{d x^{2}} \psi_{n}(x) &=\frac{\alpha^{2}}{2}\left[\sqrt{n(n-1)} \psi_{n-2}-(2 n+1) \psi_{n}+\sqrt{(n+1)(n+2)} \psi_{n+2}\right] \end{aligned}$ ，

（6）一维谐振子本征值问题的代数解法以及粒子数表象(N的矩阵形式)

\begin{aligned} \hat{a} = \frac{α}{\sqrt{2}} (\hat{x} + \frac{i}{m ω} \hat{p}), {\hat{a}}^{+} = \frac{α}{\sqrt{2}} (\hat{x} - \frac{i}{m ω} \hat{p}), [a, a^{+}] = 1 \\ \hat{x} = \frac{1}{\sqrt{2}} ({\hat{a}}^{+} + {\hat{a}}^{-}), \hat{p} = \frac{i}{\sqrt{2}} ({\hat{a}}^{+} - {\hat{a}}^{-}), \\ \hat{H} = (a^{+} a^{-} + \frac{1}{2}) = (\hat{N} + \frac{1}{2}), \hat{N} = a^{+} a^{-} = N^{+} \\ \hat{N} a^{+} | n ⟩ = (n + 1) a^{+} | n ⟩, \hat{N} a^{-} | n ⟩ = (n - 1) a^{-} | n ⟩ \\ \hat{N} | 0 ⟩ = 0 | 0 ⟩, \hat{N} a^{+} | 0 ⟩ = 1 \cdot a^{+} | 0 ⟩ \\ \hat{N} a^{+ 2} | 0 ⟩ = \hat{N} \cdot a^{+} | 1 ⟩ = (1 + 1) a^{+} | 1 ⟩ = 2 a^{+} | 1 ⟩ \\ \hat{N} a^{+ 3} | 0 ⟩ = \hat{N} a^{+ 2} | 1 ⟩ = \hat{N} \sqrt{2} a^{+} | 2 ⟩ = 3 \sqrt{2} a^{+} | 2 ⟩ \\ \hat{N} | n ⟩ = n | n ⟩, [\hat{N}, a^{-}] = - a^{-}, [\hat{N}, a^{+}] = a^{+} \\ | n ⟩ = \frac{1}{\sqrt{n!}} {(a^{+})}^{n} | 0 ⟩ \\ a^{+} | n ⟩ = \sqrt{n + 1} | n + 1 ⟩, a^{-} | n ⟩ = \sqrt{n} | n - 1 ⟩ \\ x_{n n} = \frac{1}{\sqrt{2}} \sqrt{\frac{ℏ}{ω m}} (\sqrt{n + 1} δ_{n^{'} n + 1} + \sqrt{n} δ_{n^{'} n - 1}) \\ P_{n^{'} n} = \frac{i}{\sqrt{2}} \sqrt{ω m ℏ} (\sqrt{n + 1} δ_{n^{'} n + 1} - \sqrt{n} δ_{n^{'} - 1}) \\ ψ_{n} (x) = ⟨ x ∣ n ⟩ = \frac{1}{\sqrt{n!}} ⟨ x | a^{+ n} | 0 ⟩ \end{aligned}

(7) 角动量相关

\begin{aligned} [j_{x}, j_{y}] = i ℏ j_{z}, [j_{y}, j_{z}] = i ℏ j_{x}, [j_{z}, j_{x}] = i ℏ j_{y} \\ j_{\pm} = j_{x} \pm i j_{y}, [j_{z}, j_{\pm}] = \pm ℏ j_{\pm}, [j_{+}, j_{-}] = 2 ℏ j_{z} \\ {[j_{+}, j_{-}]}_{+} = j_{+} j_{-} + j_{-} j_{+} = 2 (j^{2} - j_{z}^{2}), j_{\pm} j_{\mp} = j^{2} - j_{z}^{2} \pm ℏ j_{z} \end{aligned}

\begin{aligned} ⟨ j m + 1 | j_{+} | j m ⟩ = \sqrt{(j - m) (j + m + 1)} \\ ⟨ j m - 1 | j_{-} | j m ⟩ = \sqrt{(j + m) (j - m + 1)} \\ ⟨ j m + 1 | j_{x} | j m ⟩ = \frac{1}{2} \sqrt{(j - m) (j + m - 1)} \\ ⟨ j m - 1 | j_{x} | j m ⟩ = \frac{1}{2} \sqrt{(j + m) (j - m + 1)} \\ ⟨ j m + 1 | j_{y} | j m ⟩ = - \frac{i}{2} \sqrt{(j - m) (j + m + 1)} \\ ⟨ j m - 1 | j_{y} | j m ⟩ = \frac{i}{2} \sqrt{(j + m) (j - m + 1)} \end{aligned}

(8) 自旋

\begin{aligned} ψ (r, S_{z}) = (\begin{array}{c} ψ (r, \frac{ℏ}{z}) \\ ψ (r, - \frac{ℏ}{z}) \end{array}) = ϕ (r) χ (S_{z}), \\ χ (S_{z}) = (\begin{matrix} a \\ b \end{matrix}), \\ | a |^{2} + | b |^{2} = x^{+} x = (a^{*} b^{*}) (\begin{matrix} a \\ b \end{matrix}) = 1 \\ α = X_{\frac{1}{2}} (S_{z}) = (\begin{matrix} 1 \\ 0 \end{matrix}), β = X_{- \frac{1}{2}} (S_{z}) = (\begin{matrix} 0 \\ 1 \end{matrix}) \\ X (S_{z}) = (\begin{matrix} a \\ b \end{matrix}) = a α + b β \\ ψ (r, S_{z}) = ψ (r, \frac{ℏ}{2}) α + ψ (r, - \frac{ℏ}{2}) β \end{aligned}

\begin{aligned} [S_{x}, S_{y}] = i ℏ S_{z}, [S_{y}, S_{z}] = i ℏ S_{x}, [S_{z}, S_{x}] = i ℏ S_{y} \\ S = \frac{ℏ}{2} σ \\ [σ_{x}, σ_{y}] = 2 i σ_{z}, [σ_{y}, σ_{z}] = 2 i σ_{x}, [σ_{z}, σ_{x}] = 2 i σ_{y} \\ σ_{x} = (\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}), σ_{y} = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array}), σ_{z} = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}) \end{aligned}

\begin{aligned} σ_{\pm} = \frac{1}{2} (σ_{x} \pm i σ_{y}) \\ σ_{+} = (\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}), σ_{y} = (\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}) \\ ξ (r) \cdot s \cdot l \\ ξ (r) = \frac{1}{2 μ^{2} c^{2}} \frac{1}{r} \frac{d V}{d r} \\ j = l + s, [j, s \cdot l] = 0 \\ l^{2} ϕ = l (l + 1) ℏ ϕ, j_{z} ϕ = (m + \frac{1}{2}) ℏ ϕ \end{aligned}

\begin{aligned} j^{2} & = l^{2} + s^{2} + 2 s \cdot l \\ = l^{2} + \frac{3}{4} ℏ^{2} + ℏ (σ_{x} l_{x} + σ_{y} l_{y} + σ_{z} l_{z}) \\ = (\begin{array}{cc} l^{2} + \frac{3}{4} ℏ^{2} + ℏ l_{z} & ℏ l_{-} \\ ℏ l_{+} & l^{2} + \frac{3}{4} ℏ^{2} - ℏ l_{z} \end{array}) \\ l_{\pm} & = l_{x} \pm i l_{y} \\ l_{\pm} Y_{l m} & = ℏ \sqrt{(l \pm m + 1) (l \mp m)} Y_{l, m \pm 1} \end{aligned}

(9) 微扰论

\begin{aligned} E^{(1)} & = ⟨ ψ^{(0)} | H^{'} | ψ^{(0)} ⟩ \\ E^{(2)} & = ⟨ ψ^{(0)} | H^{'} | ψ^{(1)} ⟩ \\ E^{(3)} & = ⟨ ψ^{(0)} | H^{'} | ψ^{(2)} ⟩ = ⟨ ψ^{(1)} | (H^{'} - E^{(1)}) | ψ^{(1)} ⟩ \end{aligned}

9.1 非简并微扰论

9.1.1一级近似

\begin{aligned} | ψ^{(1)} ⟩ = \sum_{n} a_{n}^{(1)} | ψ_{n}^{(0)} ⟩ \\ a_{n}^{(1)} = \frac{H_{n k}^{'}}{E_{k}^{(0)} - E_{n}^{(0)}}, (n \neq k) \\ E^{(1)} = E_{k}^{(1)} = H_{k k}^{'} = ⟨ ψ_{k}^{(0)} | H^{'} | ψ_{k}^{(0)} ⟩, (n = k) \\ H_{n k}^{'} = ⟨ ψ_{n}^{(0)} | H^{'} | ψ_{k}^{(0)} ⟩ \end{aligned}

\begin{aligned} | ψ_{k} ⟩ = | ψ_{k}^{(0)} ⟩ + | ψ_{k}^{(1)} ⟩ = | ψ_{k}^{(0)} ⟩ + \sum_{n}^{n \neq k} \frac{H_{n k}^{'}}{E_{k}^{(0)} - E_{n}^{(0)}} | ψ_{n}^{(0)} ⟩ \\ E_{k} = E_{k}^{(0)} + H_{k k}^{'} \end{aligned}

\begin{aligned} E^{(2)} = E_{k}^{(2)} = ⟨ ψ_{k}^{(0)} | H^{'} | ψ_{k}^{(1)} ⟩ = \sum_{n}^{'} \frac{{| H_{n k}^{'} |}^{2}}{E_{k}^{(0)} - E_{n}^{(0)}} \\ E_{k} = E_{k}^{(0)} + H_{k k}^{'} + \sum_{n}^{1} \frac{{| H_{n k}^{'} |}^{2}}{E_{k}^{(0)} - E_{n}^{(0)}} \\ | \frac{H_{n k}^{'}}{E_{k}^{(0)} - E_{n}^{(0)}} | ≪ 1 when n \neq k \end{aligned}

9.1.2二级近似

\begin{aligned} E^{(2)} = E_{k}^{(2)} = ⟨ ψ_{k}^{(0)} | H^{'} | ψ_{k}^{(1)} ⟩ = \sum_{n}^{'} \frac{{| H_{n k}^{'} |}^{2}}{E_{k}^{(0)} - E_{n}^{(0)}} \\ E_{k} = E_{k}^{(0)} + H_{k k}^{'} + \sum_{n}^{'} \frac{{| H_{n k}^{'} |}^{2}}{E_{k}^{(0)} - E_{n}^{(0)}} \\ | \frac{H_{n k}^{'}}{E_{k}^{(0)} - E_{n}^{(0)}} | ≪ 1 when n \neq k \end{aligned}

9.2简并态微扰论

利用体系的对称性以及其破缺去尽可能多地解除简并。

在此以一个例子来说明问题的解法：

$H\ = \ H_0 \ + \ H^\prime$ $H_0$ 有两条非简并能级E1和E2靠得很近，而其余能级则离开很远，

H_{0} | ψ_{1} ⟩ = E_{1} | ψ_{1} ⟩, H_{0} | ψ_{2} ⟩ = E_{2} | ψ_{2} ⟩

则 H 的对角化可以局限在|φ1>和|φ2>张开的二维态空间中进行，在此空间中 H 表示为：

\begin{matrix} H = (\begin{array}{ll} E_{1} & H_{12}^{'} \\ H_{21}^{'} & E_{2} \end{array}), H_{12}^{'} = ⟨ φ_{1} | H^{'} | φ_{2} ⟩ = H_{21}^{'} * \end{matrix}

假设H的本征态为

| ψ ⟩ = c_{1} | φ_{1} ⟩ + c_{2} | φ_{2} ⟩

则H的本征方程可以化为

\begin{matrix} H | ψ ⟩ = E | ψ ⟩ \Rightarrow (\begin{array}{cc} E - E_{1} & - H_{12}^{'} \\ - H_{12}^{' *} & E - E_{2} \end{array}) (\begin{array}{l} C_{1} \\ C_{2} \end{array}) = 0 \end{matrix}

方程存在非平庸解的情况为

\begin{matrix} | \begin{array}{cc} E - E_{1} & - H_{12}^{1} \\ - H_{12}^{' *} & E - E_{2} \end{array} | = 0 \end{matrix}

行列式的解为

E_{\pm} = \frac{1}{2} [(E_{1} + E_{2}) \pm \sqrt{{(E_{1} - E_{2})}^{2} + 4 {| H_{12}^{'} |}^{2}}]

令

\begin{aligned} E_{c} & = \frac{1}{2} (E_{1} + E_{2}) \\ d & = \frac{1}{2} (E_{2} - E_{1}) (set E_{2} > E_{1}) \\ \Rightarrow E_{1} & = E_{c} - d, E_{2} = E_{c} + d \\ E_{\pm} & = E_{c} \pm \sqrt{d^{2} + {| H_{12}^{'} |}^{2}} = E_{c} \pm | H_{12}^{'} | \sqrt{1 + R^{2}} \\ R & = \frac{d}{| H_{12}^{'} |} \end{aligned}

1/R是表征微扰的重要性的一个重要参数，1/R>>1就是代表强耦合，1/R<<1代表弱耦合，改造为

\tan θ = \frac{1}{R}, H_{12}^{'} = | H_{12}^{'} | e^{- i γ}

$H_{12}^{\prime}$ 为实数，则 γ=0（斥力），或者π（引力）

\begin{aligned} \frac{C_{1}}{C_{2}} & = \frac{H_{12}^{'}}{E_{-} - E_{1}} = \frac{| H_{12}^{'} | e^{- i γ}}{- \sqrt{d^{2} + {| H_{12}^{'} |}^{2}} + d} = - \frac{e^{- i γ}}{\sqrt{R^{2} + 1} - R} \\ = - (\sqrt{R^{2} + 1} + R) e^{- i γ} = - \frac{\cos (\frac{θ}{2})}{\sin (\frac{θ}{2})} e^{- i γ} \end{aligned}

相应的本征态可以表示为

\begin{aligned} | ψ_{-} ⟩ & = \cos (\frac{θ}{2}) | φ_{1} ⟩ - \sin (\frac{θ}{2}) e^{i γ} | φ_{2} ⟩ \\ = (\begin{array}{c} \cos \frac{θ}{2} \\ - \sin \frac{θ}{2} \cdot e^{i γ} \end{array}) \\ | ψ_{+} ⟩ & = \sin (\frac{θ}{2}) | φ_{1} ⟩ + \cos (\frac{θ}{2}) e^{i γ} | φ_{2} ⟩ \\ = (\begin{array}{c} \sin \frac{θ}{2} \\ \cos \frac{θ}{2} \cdot e^{i γ} \end{array}) \end{aligned}

讨论：

$\theta=\frac \pi 2$ ,

| ψ_{\mp} ⟩ = \frac{1}{\sqrt{2}} (| φ_{1} ⟩ \pm | φ_{2} ⟩)

(b)设R>>1（弱耦合）,即

\begin{aligned} | ψ_{-} ⟩ & \approx | φ_{1} ⟩ + \frac{1}{2 R} | φ_{2} ⟩, E_{-} \approx E_{c} - R | H_{12}^{'} | \\ | ψ_{+} ⟩ & \approx \frac{1}{2 R} | φ_{1} ⟩ - | φ_{2} ⟩, E_{+} \approx E_{c} + R | H_{12}^{'} | \end{aligned}

Koit Newton