习题 64.1 - 解答
为了求解该波包的轨道角动量对磁矩的贡献，我们需要在动量空间中计算轨道磁矩算符的期望值。为避免符号混淆，下文中用 M M M Y ℓ m Y_{\ell m} Y ℓ m m m m e e e
1. 轨道磁矩算符的推导
在量子场论中，波包的磁矩定义为空间电流分布的期望值：
μ = 1 2 ∫ d 3 x x × J ( x ) \boldsymbol{\mu} = \frac{1}{2} \int d^3x \, \mathbf{x} \times \mathbf{J}(\mathbf{x}) μ = 2 1 ∫ d 3 x x × J ( x ) J conv ( p ′ , p ) = e 2 M u ˉ ( p ′ ) ( p ′ + p ) u ( p ) \mathbf{J}_{\text{conv}}(p', p) = \frac{e}{2M} \bar{u}(p') (\mathbf{p}' + \mathbf{p}) u(p) J conv ( p ′ , p ) = 2 M e u ˉ ( p ′ ) ( p ′ + p ) u ( p ) x → i ∇ p ′ \mathbf{x} \to i \nabla_{\mathbf{p}'} x → i ∇ p ′ δ \delta δ ∇ p ′ \nabla_{\mathbf{p}'} ∇ p ′ f ∗ ( p ′ ) f^*(\mathbf{p}') f ∗ ( p ′ ) μ orb = i 2 ∫ d 3 p ( 2 π ) 3 2 E f ( p ) ∇ p ′ × [ 1 2 E ′ f ∗ ( p ′ ) e 2 M u ˉ ( p ′ ) ( p ′ + p ) u ( p ) ] p ′ = p \boldsymbol{\mu}_{\text{orb}} = \frac{i}{2} \int \frac{d^3p}{(2\pi)^3 2E} f(\mathbf{p}) \nabla_{\mathbf{p}'} \times \left[ \frac{1}{2E'} f^*(\mathbf{p}') \frac{e}{2M} \bar{u}(p') (\mathbf{p}'+\mathbf{p}) u(p) \right]_{\mathbf{p}'=\mathbf{p}} μ orb = 2 i ∫ ( 2 π ) 3 2 E d 3 p f ( p ) ∇ p ′ × [ 2 E ′ 1 f ∗ ( p ′ ) 2 M e u ˉ ( p ′ ) ( p ′ + p ) u ( p ) ] p ′ = p E = p 2 + M 2 E = \sqrt{\mathbf{p}^2 + M^2} E = p 2 + M 2 ∇ p ′ \nabla_{\mathbf{p}'} ∇ p ′ f ∗ ( p ′ ) f^*(\mathbf{p}') f ∗ ( p ′ ) u ˉ ( p ) u ( p ) = 2 M \bar{u}(p)u(p) = 2M u ˉ ( p ) u ( p ) = 2 M i e 2 ∫ d 3 p ( 2 π ) 3 2 E 1 E f ( p ) ( ∇ p f ∗ ( p ) × p ) \frac{i e}{2} \int \frac{d^3p}{(2\pi)^3 2E} \frac{1}{E} f(\mathbf{p}) \left( \nabla_{\mathbf{p}} f^*(\mathbf{p}) \times \mathbf{p} \right) 2 i e ∫ ( 2 π ) 3 2 E d 3 p E 1 f ( p ) ( ∇ p f ∗ ( p ) × p ) L = − i p × ∇ p \mathbf{L} = -i \mathbf{p} \times \nabla_{\mathbf{p}} L = − i p × ∇ p ∇ p f ∗ × p = − p × ∇ p f ∗ = − i L f ∗ \nabla_{\mathbf{p}} f^* \times \mathbf{p} = - \mathbf{p} \times \nabla_{\mathbf{p}} f^* = -i \mathbf{L} f^* ∇ p f ∗ × p = − p × ∇ p f ∗ = − i L f ∗ L \mathbf{L} L ⟨ μ orb ⟩ = ∫ d 3 p ( 2 π ) 3 2 E f ∗ ( p ) ( e 2 E L ) f ( p ) \langle \boldsymbol{\mu}_{\text{orb}} \rangle = \int \frac{d^3p}{(2\pi)^3 2E} f^*(\mathbf{p}) \left( \frac{e}{2E} \mathbf{L} \right) f(\mathbf{p}) ⟨ μ orb ⟩ = ∫ ( 2 π ) 3 2 E d 3 p f ∗ ( p ) ( 2 E e L ) f ( p ) μ orb = e 2 E L \boldsymbol{\mu}_{\text{orb}} = \frac{e}{2E} \mathbf{L} μ orb = 2 E e L
2. 计算给定波包的期望值
题目给定的波包形式为：
f ( p ) = N exp ( − a 2 p 2 2 ) Y ℓ m ( p ^ ) f(\mathbf{p}) = N \exp\left(-\frac{a^2 \mathbf{p}^2}{2}\right) Y_{\ell m}(\hat{\mathbf{p}}) f ( p ) = N exp ( − 2 a 2 p 2 ) Y ℓ m ( p ^ ) N N N ∫ d 3 p ( 2 π ) 3 2 E ∣ f ( p ) ∣ 2 = 1 \int \frac{d^3p}{(2\pi)^3 2E} |f(\mathbf{p})|^2 = 1 ∫ ( 2 π ) 3 2 E d 3 p ∣ f ( p ) ∣ 2 = 1
我们需要计算的期望值为：
⟨ μ orb ⟩ = ∫ d 3 p ( 2 π ) 3 2 E f ∗ ( p ) e 2 E L f ( p ) ∫ d 3 p ( 2 π ) 3 2 E ∣ f ( p ) ∣ 2 \langle \boldsymbol{\mu}_{\text{orb}} \rangle = \frac{\int \frac{d^3p}{(2\pi)^3 2E} f^*(\mathbf{p}) \frac{e}{2E} \mathbf{L} f(\mathbf{p})}{\int \frac{d^3p}{(2\pi)^3 2E} |f(\mathbf{p})|^2} ⟨ μ orb ⟩ = ∫ ( 2 π ) 3 2 E d 3 p ∣ f ( p ) ∣ 2 ∫ ( 2 π ) 3 2 E d 3 p f ∗ ( p ) 2 E e L f ( p )
第一步：处理角向积分
角动量算符 L \mathbf{L} L Y ℓ m ( p ^ ) Y_{\ell m}(\hat{\mathbf{p}}) Y ℓ m ( p ^ ) L \mathbf{L} L ∣ ℓ , m ⟩ |\ell, m\rangle ∣ ℓ , m ⟩ ⟨ Y ℓ m ∣ L z ∣ Y ℓ m ⟩ = m \langle Y_{\ell m} | L_z | Y_{\ell m} \rangle = m ⟨ Y ℓ m ∣ L z ∣ Y ℓ m ⟩ = m ⟨ Y ℓ m ∣ L x ∣ Y ℓ m ⟩ = ⟨ Y ℓ m ∣ L y ∣ Y ℓ m ⟩ = 0 \langle Y_{\ell m} | L_x | Y_{\ell m} \rangle = \langle Y_{\ell m} | L_y | Y_{\ell m} \rangle = 0 ⟨ Y ℓ m ∣ L x ∣ Y ℓ m ⟩ = ⟨ Y ℓ m ∣ L y ∣ Y ℓ m ⟩ = 0 ∫ d Ω Y ℓ m ∗ ( p ^ ) L Y ℓ m ( p ^ ) = m z ^ \int d\Omega \, Y_{\ell m}^*(\hat{\mathbf{p}}) \mathbf{L} Y_{\ell m}(\hat{\mathbf{p}}) = m \hat{\mathbf{z}} ∫ d Ω Y ℓ m ∗ ( p ^ ) L Y ℓ m ( p ^ ) = m z ^
第二步：处理径向积分
将角向积分结果代入期望值表达式，并在球坐标下展开动量测度 d 3 p = p 2 d p d Ω d^3p = p^2 dp \, d\Omega d 3 p = p 2 d p d Ω N N N ∫ d Ω ∣ Y ℓ m ∣ 2 = 1 \int d\Omega |Y_{\ell m}|^2 = 1 ∫ d Ω∣ Y ℓ m ∣ 2 = 1 ⟨ μ orb ⟩ = e 2 m z ^ ∫ 0 ∞ p 2 d p ( 2 π ) 3 2 E 1 E exp ( − a 2 p 2 ) ∫ 0 ∞ p 2 d p ( 2 π ) 3 2 E exp ( − a 2 p 2 ) \langle \boldsymbol{\mu}_{\text{orb}} \rangle = \frac{e}{2} m \hat{\mathbf{z}} \frac{\int_0^\infty \frac{p^2 dp}{(2\pi)^3 2E} \frac{1}{E} \exp(-a^2 p^2)}{\int_0^\infty \frac{p^2 dp}{(2\pi)^3 2E} \exp(-a^2 p^2)} ⟨ μ orb ⟩ = 2 e m z ^ ∫ 0 ∞ ( 2 π ) 3 2 E p 2 d p e x p ( − a 2 p 2 ) ∫ 0 ∞ ( 2 π ) 3 2 E p 2 d p E 1 e x p ( − a 2 p 2 ) 1 2 ( 2 π ) 3 \frac{1}{2(2\pi)^3} 2 ( 2 π ) 3 1 ⟨ μ orb ⟩ = e 2 m z ^ ∫ 0 ∞ d p p 2 E 2 exp ( − a 2 p 2 ) ∫ 0 ∞ d p p 2 E exp ( − a 2 p 2 ) \langle \boldsymbol{\mu}_{\text{orb}} \rangle = \frac{e}{2} m \hat{\mathbf{z}} \frac{\int_0^\infty dp \frac{p^2}{E^2} \exp(-a^2 p^2)}{\int_0^\infty dp \frac{p^2}{E} \exp(-a^2 p^2)} ⟨ μ orb ⟩ = 2 e m z ^ ∫ 0 ∞ d p E p 2 e x p ( − a 2 p 2 ) ∫ 0 ∞ d p E 2 p 2 e x p ( − a 2 p 2 ) E = p 2 + M 2 E = \sqrt{p^2 + M^2} E = p 2 + M 2
3. 物理极限分析（非相对论极限）
作为物理检验，考虑非相对论极限（即波包在动量空间极度局域化于 p = 0 p=0 p = 0 a M ≫ 1 a M \gg 1 a M ≫ 1 p → 0 p \to 0 p → 0 E ≈ M E \approx M E ≈ M E ≈ M E \approx M E ≈ M ⟨ μ orb ⟩ ≈ e 2 M m z ^ \langle \boldsymbol{\mu}_{\text{orb}} \rangle \approx \frac{e}{2M} m \hat{\mathbf{z}} ⟨ μ orb ⟩ ≈ 2 M e m z ^
最终答案
轨道角动量对磁矩的精确贡献（期望值）为：
⟨ μ orb ⟩ = e m z ^ 2 ∫ 0 ∞ d p p 2 p 2 + M 2 exp ( − a 2 p 2 ) ∫ 0 ∞ d p p 2 p 2 + M 2 exp ( − a 2 p 2 ) \boxed{ \langle \boldsymbol{\mu}_{\text{orb}} \rangle = \frac{e m \hat{\mathbf{z}}}{2} \frac{\int_0^\infty dp \frac{p^2}{p^2 + M^2} \exp(-a^2 p^2)}{\int_0^\infty dp \frac{p^2}{\sqrt{p^2 + M^2}} \exp(-a^2 p^2)} } ⟨ μ orb ⟩ = 2 e m z ^ ∫ 0 ∞ d p p 2 + M 2 p 2 exp ( − a 2 p 2 ) ∫ 0 ∞ d p p 2 + M 2 p 2 exp ( − a 2 p 2 )