习题 49.1 - 解答
a) 相互作用拉格朗日量的厄米共轭项
给定的相互作用拉格朗日量为：
L 1 ⊃ 2 e E L † X ‾ P L Ψ + 2 e E R † X ‾ P R Ψ \mathcal{L}_1 \supset \sqrt{2} e E_L^\dagger \overline{X} P_L \Psi + \sqrt{2} e E_R^\dagger \overline{X} P_R \Psi L 1 ⊃ 2 e E L † X P L Ψ + 2 e E R † X P R Ψ P L , R = 1 2 ( 1 ∓ γ 5 ) P_{L,R} = \frac{1}{2}(1 \mp \gamma_5) P L , R = 2 1 ( 1 ∓ γ 5 ) X ‾ = X † γ 0 \overline{X} = X^\dagger \gamma^0 X = X † γ 0 γ 5 † = γ 5 \gamma_5^\dagger = \gamma_5 γ 5 † = γ 5 ( E L † X ‾ P L Ψ ) † = Ψ † P L † ( X ‾ ) † E L = Ψ † P L γ 0 X E L (E_L^\dagger \overline{X} P_L \Psi)^\dagger = \Psi^\dagger P_L^\dagger (\overline{X})^\dagger E_L = \Psi^\dagger P_L \gamma^0 X E_L ( E L † X P L Ψ ) † = Ψ † P L † ( X ) † E L = Ψ † P L γ 0 X E L γ 0 γ 5 = − γ 5 γ 0 \gamma^0 \gamma_5 = - \gamma_5 \gamma^0 γ 0 γ 5 = − γ 5 γ 0 P L γ 0 = γ 0 P R P_L \gamma^0 = \gamma^0 P_R P L γ 0 = γ 0 P R Ψ † P L γ 0 X E L = Ψ † γ 0 P R X E L = Ψ ‾ P R X E L \Psi^\dagger P_L \gamma^0 X E_L = \Psi^\dagger \gamma^0 P_R X E_L = \overline{\Psi} P_R X E_L Ψ † P L γ 0 X E L = Ψ † γ 0 P R X E L = Ψ P R X E L Ψ ‾ P L X E R \overline{\Psi} P_L X E_R Ψ P L X E R h.c. = 2 e E L Ψ ‾ P R X + 2 e E R Ψ ‾ P L X \boxed{ \text{h.c.} = \sqrt{2} e E_L \overline{\Psi} P_R X + \sqrt{2} e E_R \overline{\Psi} P_L X } h.c. = 2 e E L Ψ P R X + 2 e E R Ψ P L X
b) e + e − → γ ~ γ ~ e^+ e^- \rightarrow \tilde{\gamma} \tilde{\gamma} e + e − → γ ~ γ ~
该过程为 e − ( p 1 ) + e + ( p 2 ) → γ ~ ( k 1 ) + γ ~ ( k 2 ) e^-(p_1) + e^+(p_2) \rightarrow \tilde{\gamma}(k_1) + \tilde{\gamma}(k_2) e − ( p 1 ) + e + ( p 2 ) → γ ~ ( k 1 ) + γ ~ ( k 2 ) γ ~ \tilde{\gamma} γ ~ E L E_L E L E R E_R E R t t t u u u
利用马约拉纳费米子的费曼规则，我们可以将正电子和光微子的旋量分别记为 v ( p 2 ) v(p_2) v ( p 2 ) v ( k 1 , 2 ) v(k_{1,2}) v ( k 1 , 2 ) E L E_L E L
t t t e − e^- e − k 1 k_1 k 1 e + e^+ e + k 2 k_2 k 2 i M L , t = ( i 2 e ) 2 u ˉ ( k 1 ) P L u ( p 1 ) i t − M L 2 v ˉ ( p 2 ) P R v ( k 2 ) i\mathcal{M}_{L,t} = (i\sqrt{2}e)^2 \bar{u}(k_1) P_L u(p_1) \frac{i}{t - M_L^2} \bar{v}(p_2) P_R v(k_2) i M L , t = ( i 2 e ) 2 u ˉ ( k 1 ) P L u ( p 1 ) t − M L 2 i v ˉ ( p 2 ) P R v ( k 2 ) u u u e − e^- e − k 2 k_2 k 2 e + e^+ e + k 1 k_1 k 1 i M L , u = − ( i 2 e ) 2 u ˉ ( k 2 ) P L u ( p 1 ) i u − M L 2 v ˉ ( p 2 ) P R v ( k 1 ) i\mathcal{M}_{L,u} = -(i\sqrt{2}e)^2 \bar{u}(k_2) P_L u(p_1) \frac{i}{u - M_L^2} \bar{v}(p_2) P_R v(k_1) i M L , u = − ( i 2 e ) 2 u ˉ ( k 2 ) P L u ( p 1 ) u − M L 2 i v ˉ ( p 2 ) P R v ( k 1 )
为了方便后续计算，我们利用电荷共轭矩阵 C C C v v v u u u v ( p ) = C u ˉ T ( p ) v(p) = C \bar{u}^T(p) v ( p ) = C u ˉ T ( p ) v ˉ ( p ) = − u T ( p ) C − 1 \bar{v}(p) = -u^T(p) C^{-1} v ˉ ( p ) = − u T ( p ) C − 1 v ˉ ( p 2 ) P R , L v ( k ) = − u ˉ ( k ) P R , L u ( p 2 ) \bar{v}(p_2) P_{R,L} v(k) = - \bar{u}(k) P_{R,L} u(p_2) v ˉ ( p 2 ) P R , L v ( k ) = − u ˉ ( k ) P R , L u ( p 2 ) i i i E L E_L E L M L , t = 2 e 2 t − M L 2 [ u ˉ ( k 1 ) P L u ( p 1 ) ] [ u ˉ ( k 2 ) P R u ( p 2 ) ] \mathcal{M}_{L,t} = \frac{2e^2}{t - M_L^2} [\bar{u}(k_1) P_L u(p_1)][\bar{u}(k_2) P_R u(p_2)] M L , t = t − M L 2 2 e 2 [ u ˉ ( k 1 ) P L u ( p 1 )] [ u ˉ ( k 2 ) P R u ( p 2 )] M L , u = − 2 e 2 u − M L 2 [ u ˉ ( k 2 ) P L u ( p 1 ) ] [ u ˉ ( k 1 ) P R u ( p 2 ) ] \mathcal{M}_{L,u} = \frac{-2e^2}{u - M_L^2} [\bar{u}(k_2) P_L u(p_1)][\bar{u}(k_1) P_R u(p_2)] M L , u = u − M L 2 − 2 e 2 [ u ˉ ( k 2 ) P L u ( p 1 )] [ u ˉ ( k 1 ) P R u ( p 2 )]
同理，对于 E R E_R E R P L ↔ P R P_L \leftrightarrow P_R P L ↔ P R M L → M R M_L \rightarrow M_R M L → M R M = 2 e 2 [ [ u ˉ ( k 1 ) P L u ( p 1 ) ] [ u ˉ ( k 2 ) P R u ( p 2 ) ] t − M L 2 − [ u ˉ ( k 2 ) P L u ( p 1 ) ] [ u ˉ ( k 1 ) P R u ( p 2 ) ] u − M L 2 + [ u ˉ ( k 1 ) P R u ( p 1 ) ] [ u ˉ ( k 2 ) P L u ( p 2 ) ] t − M R 2 − [ u ˉ ( k 2 ) P R u ( p 1 ) ] [ u ˉ ( k 1 ) P L u ( p 2 ) ] u − M R 2 ] \boxed{ \begin{aligned} \mathcal{M} &= 2e^2 \left[ \frac{[\bar{u}(k_1) P_L u(p_1)][\bar{u}(k_2) P_R u(p_2)]}{t - M_L^2} - \frac{[\bar{u}(k_2) P_L u(p_1)][\bar{u}(k_1) P_R u(p_2)]}{u - M_L^2} \right. \\ &\quad \left. + \frac{[\bar{u}(k_1) P_R u(p_1)][\bar{u}(k_2) P_L u(p_2)]}{t - M_R^2} - \frac{[\bar{u}(k_2) P_R u(p_1)][\bar{u}(k_1) P_L u(p_2)]}{u - M_R^2} \right] \end{aligned} } M = 2 e 2 [ t − M L 2 [ u ˉ ( k 1 ) P L u ( p 1 )] [ u ˉ ( k 2 ) P R u ( p 2 )] − u − M L 2 [ u ˉ ( k 2 ) P L u ( p 1 )] [ u ˉ ( k 1 ) P R u ( p 2 )] + t − M R 2 [ u ˉ ( k 1 ) P R u ( p 1 )] [ u ˉ ( k 2 ) P L u ( p 2 )] − u − M R 2 [ u ˉ ( k 2 ) P R u ( p 1 )] [ u ˉ ( k 1 ) P L u ( p 2 )] ]
c) 忽略电子质量且 ∣ t ∣ , ∣ u ∣ ≪ M L = M R |t|, |u| \ll M_L = M_R ∣ t ∣ , ∣ u ∣ ≪ M L = M R
在 m e → 0 m_e \to 0 m e → 0 M L = M R ≡ M ≫ s M_L = M_R \equiv M \gg \sqrt{s} M L = M R ≡ M ≫ s − M 2 -M^2 − M 2 M ≈ − 2 e 2 M 2 ( T L − U L + T R − U R ) \mathcal{M} \approx -\frac{2e^2}{M^2} (T_L - U_L + T_R - U_R) M ≈ − M 2 2 e 2 ( T L − U L + T R − U R ) T L = [ u ˉ ( k 1 ) P L u ( p 1 ) ] [ u ˉ ( k 2 ) P R u ( p 2 ) ] T_L = [\bar{u}(k_1) P_L u(p_1)][\bar{u}(k_2) P_R u(p_2)] T L = [ u ˉ ( k 1 ) P L u ( p 1 )] [ u ˉ ( k 2 ) P R u ( p 2 )] m e = 0 m_e = 0 m e = 0 L L L R R R M L \mathcal{M}_L M L M R \mathcal{M}_R M R 没有干涉 。
自旋平均的模方为：
∣ M ∣ 2 ‾ = 1 4 ∑ spins ∣ M ∣ 2 = 1 4 ( ∑ ∣ M L ∣ 2 + ∑ ∣ M R ∣ 2 ) \overline{|\mathcal{M}|^2} = \frac{1}{4} \sum_{\text{spins}} |\mathcal{M}|^2 = \frac{1}{4} \left( \sum |\mathcal{M}_L|^2 + \sum |\mathcal{M}_R|^2 \right) ∣ M ∣ 2 = 4 1 ∑ spins ∣ M ∣ 2 = 4 1 ( ∑ ∣ M L ∣ 2 + ∑ ∣ M R ∣ 2 ) L ↔ R L \leftrightarrow R L ↔ R ∑ ∣ M L ∣ 2 = ∑ ∣ M R ∣ 2 \sum |\mathcal{M}_L|^2 = \sum |\mathcal{M}_R|^2 ∑ ∣ M L ∣ 2 = ∑ ∣ M R ∣ 2 ∣ M ∣ 2 ‾ = 1 2 ∑ ∣ M L ∣ 2 \overline{|\mathcal{M}|^2} = \frac{1}{2} \sum |\mathcal{M}_L|^2 ∣ M ∣ 2 = 2 1 ∑ ∣ M L ∣ 2
计算 ∑ ∣ M L ∣ 2 = 4 e 4 M 4 ∑ ∣ T L − U L ∣ 2 \sum |\mathcal{M}_L|^2 = \frac{4e^4}{M^4} \sum |T_L - U_L|^2 ∑ ∣ M L ∣ 2 = M 4 4 e 4 ∑ ∣ T L − U L ∣ 2 m γ ~ m_{\tilde{\gamma}} m γ ~ ∑ ∣ T L ∣ 2 = Tr [ ̸ k 1 P L ̸ p 1 P R ] Tr [ ̸ k 2 P R ̸ p 2 P L ] = 4 ( k 1 ⋅ p 1 ) ( k 2 ⋅ p 2 ) = ( m γ ~ 2 − t ) 2 \sum |T_L|^2 = \text{Tr}[\not{k}_1 P_L \not{p}_1 P_R] \text{Tr}[\not{k}_2 P_R \not{p}_2 P_L] = 4(k_1 \cdot p_1)(k_2 \cdot p_2) = (m_{\tilde{\gamma}}^2 - t)^2 ∑ ∣ T L ∣ 2 = Tr [ k 1 P L p 1 P R ] Tr [ k 2 P R p 2 P L ] = 4 ( k 1 ⋅ p 1 ) ( k 2 ⋅ p 2 ) = ( m γ ~ 2 − t ) 2 ∑ ∣ U L ∣ 2 = Tr [ ̸ k 2 P L ̸ p 1 P R ] Tr [ ̸ k 1 P R ̸ p 2 P L ] = 4 ( k 2 ⋅ p 1 ) ( k 1 ⋅ p 2 ) = ( m γ ~ 2 − u ) 2 \sum |U_L|^2 = \text{Tr}[\not{k}_2 P_L \not{p}_1 P_R] \text{Tr}[\not{k}_1 P_R \not{p}_2 P_L] = 4(k_2 \cdot p_1)(k_1 \cdot p_2) = (m_{\tilde{\gamma}}^2 - u)^2 ∑ ∣ U L ∣ 2 = Tr [ k 2 P L p 1 P R ] Tr [ k 1 P R p 2 P L ] = 4 ( k 2 ⋅ p 1 ) ( k 1 ⋅ p 2 ) = ( m γ ~ 2 − u ) 2 P L ̸ p 1 P R = ̸ p 1 P R P_L \not{p}_1 P_R = \not{p}_1 P_R P L p 1 P R = p 1 P R ∑ T L U L † = Tr [ ( ̸ k 1 + m γ ~ ) ̸ p 1 P R ( ̸ k 2 + m γ ~ ) ̸ p 2 P L ] = m γ ~ 2 Tr [ ̸ p 1 P R ̸ p 2 P L ] = 2 m γ ~ 2 ( p 1 ⋅ p 2 ) = m γ ~ 2 s \sum T_L U_L^\dagger = \text{Tr}[(\not{k}_1 + m_{\tilde{\gamma}}) \not{p}_1 P_R (\not{k}_2 + m_{\tilde{\gamma}}) \not{p}_2 P_L] = m_{\tilde{\gamma}}^2 \text{Tr}[\not{p}_1 P_R \not{p}_2 P_L] = 2m_{\tilde{\gamma}}^2 (p_1 \cdot p_2) = m_{\tilde{\gamma}}^2 s ∑ T L U L † = Tr [( k 1 + m γ ~ ) p 1 P R ( k 2 + m γ ~ ) p 2 P L ] = m γ ~ 2 Tr [ p 1 P R p 2 P L ] = 2 m γ ~ 2 ( p 1 ⋅ p 2 ) = m γ ~ 2 s ∑ ∣ T L − U L ∣ 2 = ( m γ ~ 2 − t ) 2 + ( m γ ~ 2 − u ) 2 − 2 m γ ~ 2 s \sum |T_L - U_L|^2 = (m_{\tilde{\gamma}}^2 - t)^2 + (m_{\tilde{\gamma}}^2 - u)^2 - 2m_{\tilde{\gamma}}^2 s ∑ ∣ T L − U L ∣ 2 = ( m γ ~ 2 − t ) 2 + ( m γ ~ 2 − u ) 2 − 2 m γ ~ 2 s s + t + u = 2 m γ ~ 2 s + t + u = 2m_{\tilde{\gamma}}^2 s + t + u = 2 m γ ~ 2 ∑ ∣ T L − U L ∣ 2 = t 2 + u 2 − 2 m γ ~ 4 \sum |T_L - U_L|^2 = t^2 + u^2 - 2m_{\tilde{\gamma}}^4 ∑ ∣ T L − U L ∣ 2 = t 2 + u 2 − 2 m γ ~ 4 β = 1 − 4 m γ ~ 2 / s \beta = \sqrt{1 - 4m_{\tilde{\gamma}}^2/s} β = 1 − 4 m γ ~ 2 / s t , u = m γ ~ 2 − s 2 ( 1 ∓ β cos θ ) t, u = m_{\tilde{\gamma}}^2 - \frac{s}{2}(1 \mp \beta \cos\theta) t , u = m γ ~ 2 − 2 s ( 1 ∓ β cos θ ) t 2 + u 2 − 2 m γ ~ 4 = s 2 2 β 2 ( 1 + cos 2 θ ) t^2 + u^2 - 2m_{\tilde{\gamma}}^4 = \frac{s^2}{2} \beta^2 (1 + \cos^2\theta) t 2 + u 2 − 2 m γ ~ 4 = 2 s 2 β 2 ( 1 + cos 2 θ ) ∣ M ∣ 2 ‾ = 1 2 ( 4 e 4 M 4 ) s 2 2 β 2 ( 1 + cos 2 θ ) = e 4 s 2 M 4 β 2 ( 1 + cos 2 θ ) \overline{|\mathcal{M}|^2} = \frac{1}{2} \left( \frac{4e^4}{M^4} \right) \frac{s^2}{2} \beta^2 (1 + \cos^2\theta) = \frac{e^4 s^2}{M^4} \beta^2 (1 + \cos^2\theta) ∣ M ∣ 2 = 2 1 ( M 4 4 e 4 ) 2 s 2 β 2 ( 1 + cos 2 θ ) = M 4 e 4 s 2 β 2 ( 1 + cos 2 θ )
最后，代入质心系微商截面公式 d σ d Ω = 1 64 π 2 s ∣ k ⃗ ∣ ∣ p ⃗ ∣ ∣ M ∣ 2 ‾ \frac{d\sigma}{d\Omega} = \frac{1}{64\pi^2 s} \frac{|\vec{k}|}{|\vec{p}|} \overline{|\mathcal{M}|^2} d Ω d σ = 64 π 2 s 1 ∣ p ∣ ∣ k ∣ ∣ M ∣ 2 ∣ k ⃗ ∣ ∣ p ⃗ ∣ = β \frac{|\vec{k}|}{|\vec{p}|} = \beta ∣ p ∣ ∣ k ∣ = β α = e 2 4 π \alpha = \frac{e^2}{4\pi} α = 4 π e 2 e 4 = 16 π 2 α 2 e^4 = 16\pi^2 \alpha^2 e 4 = 16 π 2 α 2 d σ d Ω = 1 64 π 2 s β [ 16 π 2 α 2 s 2 M 4 β 2 ( 1 + cos 2 θ ) ] \frac{d\sigma}{d\Omega} = \frac{1}{64\pi^2 s} \beta \left[ \frac{16\pi^2 \alpha^2 s^2}{M^4} \beta^2 (1 + \cos^2\theta) \right] d Ω d σ = 64 π 2 s 1 β [ M 4 16 π 2 α 2 s 2 β 2 ( 1 + cos 2 θ ) ] d σ d Ω = α 2 s 4 M 4 ( 1 − 4 m γ ~ 2 s ) 3 / 2 ( 1 + cos 2 θ ) \boxed{ \frac{d\sigma}{d\Omega} = \frac{\alpha^2 s}{4 M^4} \left( 1 - \frac{4m_{\tilde{\gamma}}^2}{s} \right)^{3/2} (1 + \cos^2\theta) } d Ω d σ = 4 M 4 α 2 s ( 1 − s 4 m γ ~ 2 ) 3/2 ( 1 + cos 2 θ )