南京航空航天大学 2004 量子真题答案

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1. 一

1.

　　解：
(1) $∵ \hat{p} ψ = p [e^{\frac{i}{ℏ} (P X - E t)} - e^{- \frac{i}{ℏ} (P X + E t)}] \neq p ψ$
$∴ ψ$ 不是动量算符 $\hat{p}$ 的本征函数。

(2) $∵ {\hat{p}}^{2} ψ = p^{2} [e^{\frac{i}{ℏ} (P X - E t)} + e^{- \frac{i}{ℏ} (P X + E t)}] = p^{2} ψ$
$∴ ψ$ 是动量平方算符 ${\hat{p}}^{2}$ 的本征值函数。

2.

　　解:
$(\hat{p} + \hat{x}) ψ = λ ψ$
即： $(- i ℏ \frac{\partial}{\partial x} + x) ψ = λ ψ \frac{ℏ}{i} \frac{d ψ}{d x} = (λ - x) ψ$
$∴ \frac{ℏ}{i} \frac{d ψ}{ψ} = (λ - x) d x$
附注: 本征函数 $ψ = A e^{\frac{1}{ℏ} (λ x - \frac{x^{2}}{2})}$
本征值 $λ$ 为所有实数。

3.

　　解：

\begin{aligned} (1) & \overset{―}{V} (r) & = - \frac{e^{2}}{4 π ϵ_{0}} \int_{- \infty}^{+ \infty} ψ^{*} \frac{1}{r} ψ d τ = - \frac{e^{2}}{4 π ϵ_{0}} \int_{0}^{\infty} \frac{1}{r} | ψ |^{2} d τ \\ (2) & = (- \frac{e^{2}}{4 π ϵ_{0}}) (\frac{1}{π a_{0}^{3}}) \int_{0}^{π} \sin θ d θ \int_{0}^{2 π} d ϕ \int_{0}^{\infty} \frac{1}{r} r^{2} e^{- {\frac{2}{a_{0}}}^{2}} d r \\ (3) & = - \frac{1}{4 π ϵ_{0}} \frac{e^{2}}{a_{0}} \end{aligned}

4.

　　解：
(1) $∵ B = A^{†} A$
$∴ B^{2} = A^{†} A A^{†} A = A^{†} A (1 - A A^{†}) = A^{†} A - A^{†} A^{2} A^{†} = A^{†} A = B$

(2) $∵ B^{2} = B$
$∴ B v = λ v, B^{2} v = λ^{2} v λ^{2} = λ = 0, 1$
$而 B 的本征值为 0, 1 ，在 B 表示中 B 为对角矩阵。 B = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$
设 $A = (\begin{matrix} a & b \\ c & d \end{matrix}) 则 A^{†} = (\begin{matrix} a^{*} & c^{*} \\ b^{*} & d^{*} \end{matrix})$ 则有 $A^{†} A^{*} = (\begin{matrix} a^{*} a + c^{*} c & a^{*} b + c^{*} d \\ b^{*} a + d^{*} c & b^{*} b + d^{*} d \end{matrix}) = B = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) (1)$

　　 $A^{*} A^{†} = (\begin{matrix} a^{*} a + b b^{*} & c^{*} a + d^{*} b \\ a^{*} c + b^{*} d & c^{*} c + d^{*} d \end{matrix}) = 1 - B = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) (2)$

　　 $A^{2} = (\begin{matrix} a^{2} + b c & a b + b d \\ a c + d c & d^{2} + b c \end{matrix}) = 0 = (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}) (3)$ 由 (1), (2), (3) 得解: $A = (\begin{matrix} 0 & e^{i φ} \\ 0 & 0 \end{matrix}) 或 A = (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix})$

5、

　　解：一维无限深势阱中粒子的一维定态波函数： $ψ_{n} (x) = \sqrt{\frac{2}{a}} \sin \frac{n π x}{a},$ 能量 $E_{n} = n^{2} \frac{π^{2} ℏ^{2}}{2 m a^{2}}, n = 1, 2, 3, \dots$ 将 $ψ (x)$ 按 $ψ_{n}$ 展开:

\begin{aligned} (4) & ψ (x) & = \frac{4}{\sqrt{a}} \sin \frac{π x}{a} \cos^{2} \frac{π x}{a} = \frac{4}{\sqrt{a}} \sin \frac{π x}{a} \cdot \frac{1}{2} (1 + \cos \frac{2 π x}{a}) \\ (5) & = \frac{4}{\sqrt{a}} [\frac{1}{2} \sin \frac{π x}{a} + \frac{1}{2} \sin \frac{π x}{a} \cdot \cos \frac{2 π x}{a}] \\ (6) & = \frac{2}{\sqrt{a}} [\sin \frac{π x}{a} + \frac{1}{2} [\sin \frac{3 π x}{a} + \sin (- \frac{π x}{a})]] \\ (7) & = \sqrt{\frac{2}{a}} (\frac{1}{\sqrt{2}} \sin \frac{3 π x}{a} + \frac{1}{\sqrt{2}} \sin \frac{π x}{a}) \end{aligned}

∴ 能量可能值: E_{1} = \frac{π^{2} ℏ^{2}}{2 m a^{2}} 概率, W_{1} = \frac{1}{2}

E_{3} = \frac{9 π^{2} ℏ^{2}}{2 m a^{2}} 概率, W_{3} = \frac{1}{2}

2. 二

1.

　　解：电子自旋变 $X (t)$ 满足方程 , $i ℏ \frac{d}{d t} X (t) = \hat{H} X (t)$ (1) $\hat{H} = - \hat{μ} \cdot B = μ_{0} B {\hat{σ}}_{y} = μ_{0} B (\begin{matrix} 0 & - i \\ i & 0 \end{matrix})$ $X (t) = (\begin{matrix} a (t) \\ b (t) \end{matrix})$ 则在 $S_{g}$ 表中方程(1)可表示为: $i ℏ \frac{d}{d t} (\begin{matrix} a (t) \\ b (t) \end{matrix}) = μ_{0} B (\begin{matrix} 0 & - i \\ i & 0 \end{matrix}) (\begin{matrix} a (t) \\ b (t) \end{matrix}) (2)$

　　 $∴ {\begin{cases} \frac{d a (t)}{d t} = - \frac{μ_{0} B}{ℏ} b (t) = - ω b (t) \\ \frac{d b (t)}{d t} = \frac{μ_{0} B}{ℏ} a (t) = ω a (t) \end{cases} 其中, ω = \frac{μ_{0} B}{ℏ}$

　　 $应用初始条件 X (0) = (\begin{matrix} 1 \\ 0 \end{matrix}), 有解: a (t) = \cos ω t, b (t) = \sin ω t$ $∴ t 时刻自旋变为: X (t) = (\begin{matrix} \cos ω t \\ \sin ω t \end{matrix})$

2.

\begin{aligned} (8) & {\bar{S}}_{X} (t) = X^{†} (t) {\hat{S}}_{X} X (t) = \frac{ℏ}{2} {(\begin{array}{c} \cos ω t \\ \sin ω t \end{array})}^{†} (\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}) (\begin{array}{c} \cos ω t \\ \sin ω t \end{array}) = \frac{ℏ}{2} \sin 2 ω t \\ (9) \\ (10) & {\bar{S}}_{Y} (t) = & X^{†} (t) {\hat{S}}_{Y} X (t) = \frac{ℏ}{2} {(\begin{array}{c} \cos ω t \\ \sin ω t \end{array})}^{†} (\begin{array}{c} 0 & - i \\ i & 0 \end{array}) (\begin{array}{c} \cos ω t \\ \sin ω t \end{array}) = 0 \\ (11) \\ (12) & {\bar{S}}_{Z} (t) = & X^{†} (t) {\hat{S}}_{Z} X (t) = \frac{ℏ}{2} {(\begin{array}{c} \cos ω t \\ \sin ω t \end{array})}^{†} (\begin{array}{c} 1 & 0 \\ 0 & - 1 \end{array}) (\begin{array}{c} \cos ω t \\ \sin ω t \end{array}) = \frac{ℏ}{2} \cos 2 ω t \\ (13) \\ (14) \end{aligned}

3.

　　 $将 X (t) 按 {\hat{S}}_{0} 的本征函数展开： X (t) = (\begin{matrix} \cos ω t \\ \sin ω t \end{matrix}) = \cos ω t (\begin{matrix} 1 \\ 0 \end{matrix}) + \sin ω t (\begin{matrix} 0 \\ 1 \end{matrix})$

\begin{aligned} (15) & ∴ t 时刻： & 测量电子自旋向上的概率 W_{↑} = \cos^{2} ω t \\ (16) \\ (17) & 测量电子自旋向上的概率 W_{↓} = \sin^{2} ω t \end{aligned}

4.

　　 $t = 0 时 X (0) = [\begin{matrix} \cos ω_{0} t \\ \sin ω_{0} t \end{matrix}] = [\begin{matrix} 1 \\ 0 \end{matrix}]$

　　 $设 t = τ 时, 相位反转向下 . X (τ) = [\begin{matrix} \cos ω τ \\ \sin ω τ \end{matrix}] = [\begin{matrix} 0 \\ 1 \end{matrix}]$

　　 $∴ {\begin{cases} \cos ω τ = 0 \\ \sin ω τ = 1 \end{cases} 解得： ω τ = (2 k + 1) - \frac{π}{2} (k = 0, 1, 2, \dots)$ $∴ τ = \frac{(2 k + 1) π}{2 ω} = \frac{(2 k + 1) π ℏ}{2 μ_{0} B} (k = 0, 1, 2, \dots)$

3. 三.

　　解：

(1)

　　转盘转动惯量 $I = \frac{1}{2} M R^{2}$ ，取 $ξ$ 轴为转轴。
哈密顿算符： ${\hat{H}}_{0} = \frac{1}{2 I} {\hat{L}}_{ξ}^{2} = - \frac{ℏ^{2}}{2 I} \frac{\partial^{2}}{\partial φ^{2}}$
本征方程： $- \frac{ℏ^{2}}{2 I} \frac{\partial^{2}}{\partial φ^{2}} ψ = E ψ$
对应本征函数： $ψ_{m}^{(0)} = \frac{1}{\sqrt{2 π}} e^{i m φ} m = 0, \pm 1, \pm 2, \dots$
能量本征值： $E_{m}^{(0)} = m^{2} \frac{ℏ^{2}}{2 I} = m^{2} \frac{ℏ^{2}}{M R^{2}}$
除基态( $m = 0$ )外，第 $| m |$ 激发态能级为二重简并： $ψ_{1 m}^{(0)} = \frac{1}{\sqrt{2 π}} e^{i m φ}, ψ_{2 m}^{(0)} = \frac{1}{\sqrt{2 π}} e^{- i m φ}$
或 $ψ_{1 m}^{(0)} = \frac{1}{\sqrt{π}} \cos m φ, ψ_{2 m}^{(0)} = \frac{1}{\sqrt{π}} \sin m φ$

(2)

　　受微扰后， $H^{'} = F_{0} δ (φ - φ_{0})$ 。微扰矩阵元： $H_{11}^{'} = \int_{0}^{2 π} \frac{1}{\sqrt{π}} \cos m_{0} φ F_{0} δ (φ - φ_{0}) \frac{1}{\sqrt{π}} \cos m_{0} φ d φ = \frac{F_{0}}{π} \cos^{2} (m φ_{0})$ $H_{22}^{'} = \int_{0}^{2 π} \frac{1}{\sqrt{π}} \sin m_{0} φ F_{0} δ (φ - φ_{0}) \frac{1}{\sqrt{π}} \sin m_{0} φ d φ = \frac{F_{0}}{π} \sin^{2} (m φ_{0})$ $H_{12}^{'} = H_{21}^{'} = \int_{0}^{2 π} \frac{1}{\sqrt{π}} \cos m_{0} φ F_{0} δ (φ - φ_{0}) \frac{1}{\sqrt{π}} \sin m_{0} φ d φ = \frac{F_{0}}{2 π} \sin (2 m φ_{0})$ 由久期方程： $| \begin{array}{cc} \frac{F_{0}}{π} \cos^{2} (m φ_{0}) - E_{m}^{(1)} & \frac{F_{0}}{2 π} \sin (2 m φ_{0}) \\ \frac{F_{0}}{2 π} \sin (2 m φ_{0}) & \frac{F_{0}}{π} \sin^{2} (m φ_{0}) - E_{m}^{(1)} \end{array} | = 0$ 解得: $E_{m}^{(1)} = {\begin{cases} \frac{E_{0}}{x} \\ 0 \end{cases}$
$∴ 能量二阶近似为： E_{m} = E_{m}^{(0)} + E_{m}^{(1)} = {\begin{cases} m^{2} \frac{ℏ^{2}}{M R^{2}} + \frac{F_{0}}{π}, \\ m^{2} \frac{ℏ^{2}}{M R^{2}} \end{cases}$ 零级波函数： $Ψ_{m}^{(0)} = C_{1} \frac{1}{\sqrt{π}} \cos m φ + C_{2} \frac{1}{\sqrt{π}} \sin m φ$ $由方程 (\begin{matrix} \frac{F_{0}}{π} \cos^{2} m φ_{0} - E_{m}^{(1)} & \frac{F_{0}}{π} \sin m φ_{0} \cdot \cos m φ_{0} \\ \frac{F_{0}}{π} \sin m φ_{0} \cdot \cos m φ_{0} & \frac{F_{0}}{π} \sin^{2} m φ_{0} - E_{m}^{(1)} \end{matrix}) (\begin{matrix} C_{1} \\ C_{2} \end{matrix}) = 0$ $当 E_{m}^{(1)} = 0 时,解得： C_{1} = - \sin m φ_{0}, C_{2} = \cos m φ_{0}$
$∴ {\overset{―}{Ψ}}_{m}^{(1)} = \frac{1}{\sqrt{π}} \sin [m (φ - φ_{0})]$ $∵ 当 E_{m}^{(1)} = \frac{F_{0}}{π} 时： C_{1} = \cos m φ_{0}, C_{2} = \sin m φ_{0}$
$∴ Φ_{2 m}^{(0)} = \frac{1}{\sqrt{π}} \cos [m (φ - φ_{0})]$

4. 四.

　　解： $t > 0$ 时，总能量算符 $H = H_{0} + H^{'}$ 波函数 $ψ (x, t)$ 满足方程。 $i ℏ \frac{\partial}{\partial t} ψ (x, t) = H ψ (x, t) (1)$
将 $ψ (x, t)$ 按 $H_{0}$ 定态波函数： $ψ_{n} (x) e^{- \frac{i}{ℏ} E_{n} t}$ 层开:
$ψ (x, t) = \sum_{n} c_{n} (t) ψ_{n} (x) e^{- \frac{i}{ℏ} E_{n} t} (2)$
初始条件为 $t = 0$ ， $ψ (x, 0) = ψ_{1} (x)$ ，即 $c_{n} (0) = δ_{n 1}$ 。
将(2)式化入(1)式得: $i ℏ \sum_{n} \frac{d c_{n}}{d t} ψ_{n} e^{- \frac{i}{ℏ} E_{n} t} = \sum_{n} H^{'} ψ_{n} c_{n} e^{- \frac{i}{ℏ} E_{n} t}$
以 $ψ_{n}^{*} (x)$ 左乘上式,对全空问积分,并应用正文归一条件： $\int ψ_{m}^{*} (x) ψ_{n} (x), d x = δ_{m n}$
得: $\frac{d C_{m}}{d t} e^{- \frac{i}{ℏ} E_{m} t} = \sum_{n} H_{m n}^{'} C_{n} (t) e^{- \frac{i}{ℏ} E_{n} t} (3)$
其中绝对矩阵元: $H_{m n}^{'} = ψ_{m} | H^{'} | ψ_{n} ⟩ = \int ψ_{m}^{*} (x) H^{'} ψ_{n} (x), d x$
对一级近似(3)式中取 $C_{n} (t) \approx C_{n} (0) = δ_{n 1},$ $得： i ℏ \frac{d C_{n}}{d t} e^{- \frac{i}{ℏ} E_{n} t} = H_{n 1}^{'} e^{- \frac{i}{ℏ} E_{1} t}$ $即： i ℏ \frac{d C_{n}}{d t} = H_{n 1}^{'} e^{i ω_{n 1} t}, ω_{n 1} = \frac{E_{n} - E_{1}}{ℏ}$ $解得: C_{n} (t) = \frac{1}{i ℏ} \int_{0}^{t} H_{n 1}^{'} (t^{'}) e^{i ω_{n 1} t^{'}} d t$ 其中

\begin{aligned} (18) & H_{n 1}^{^{'} (t)} & = \int ψ_{n}^{*} (x) F (x) e^{- \frac{t}{τ}} ψ_{1} (x) d x = e^{- \frac{t}{τ}} \int ψ_{n}^{*} (x) F (x) ψ_{1} (x) d x \\ (19) \\ (20) & = e^{- \frac{t}{τ}} F_{n 1}, F_{n 1} = \int ψ_{n}^{*} (x) F (x) ψ_{1} (x) d x \end{aligned}

∴ C_{n} (t) = \frac{F_{n 1}}{i ℏ} \int_{0}^{t} . e^{- \frac{t}{τ}} e^{i ω_{n} t} d t = \frac{F_{n 1}}{i ℏ} \frac{e^{i ω_{n 1}} t - \frac{t}{τ} - 1}{i ω_{n 1} - \frac{1}{τ}}

当

t ≫ τ (t \to \infty) 时, e^{i ω_{n l} t - \frac{t}{τ}} \to 1

∴ C_{n} (\infty) = \frac{F_{n l}}{E_{n} - E_{l} + i ℏ \frac{1}{τ}} .

厚子处于

ψ_{n}

的概率为：

W = {| C_{n} (\infty) |}^{2} = \frac{| F_{n 1} |^{2}}{(E_{n} - E_{1})^{2} + (\frac{ℏ^{2}}{τ^{2}})} = \frac{{| \int ψ_{n}^{*} (x) F (x) ψ_{1} (x) d x |}^{2}}{(E_{n} - E_{1})^{2} + \frac{ℏ^{2}}{τ^{2}}} .