PH450 HW4

5.1: $5.1$ Draw the two time-ordered diagrams for the $s$-channel process shown in Figure $5.5$. By repeating the steps of Section $5.1.1$, show that the propagator has the same form as obtained for the $t$-channel process. Hint: one of the time-ordered diagrams is non-intuitive; remember that in second-order perturbation theory the intermediate state does not conserve energy.

Sol)

$$T_{fi}^{ab} = \frac{\langle c+d|V|X\rangle \langle X|V|a+b \rangle}{(E_a + E_b) - (E_X)}.$$

The non-invariant matrix element $V_{ji}$ is related to the Lorentz invariant matrix element $\mathcal{M}{ji}$ by:

$$V{ji} = \mathcal{M}{ji} \prod_k (2E_k)^{-1/2},$$

where the index $k$ runs over the particles involved at the interaction vertex. For the simplest scalar interaction $\mathcal{M}{a+b \to X} = \mathcal{M}{X \to c+d} = g$, we obtain:

$$T{fi}^{ab} = \frac{1}{2E_X} \cdot \frac{1}{(2E_a 2E_b 2E_c 2E_d)^{1/2}} \cdot \frac{g^2}{(E_a + E_b) - (E_X)}.$$

$$T_{fi}^{cd} = \frac{\langle c+d+\tilde{X}|V|0\rangle \langle 0|V|a+b+\tilde{X} \rangle}{(E_a + E_b) - (E_a + E_b + E_c + E_d + E_X)},$$

where $0$ denotes the vacuum state. The corresponding LI matrix element is:

$$\mathcal{M}_{fi}^{cd} = -\frac{1}{2E_X} \frac{g^2}{E_c + E_d + E_X}.$$

Thus:

$$T_{fi}^{cd} = -\frac{1}{2E_X} \cdot \frac{g^2}{(E_a + E_b) + E_X}.$$

$$\mathcal{M} = \frac{g^2}{2E_X} \left[ \frac{1}{E_a + E_b - E_X} - \frac{1}{E_a + E_b + E_X} \right].$$

Simplifying:

$$\mathcal{M} = \frac{g^2}{2E_X} \cdot \frac{2E_X}{(E_a + E_b)^2 - E_X^2}.$$

Therefore:

$$\mathcal{M} = \frac{g^2}{(E_a + E_b)^2 - E_X^2}.$$

Using $E_X^2 = (p_a + p_b)^2 + m_X^2$, we rewrite the amplitude as:

$$\mathcal{M} = \frac{g^2}{(E_a + E_b)^2 - (p_a + p_b)^2 - m_X^2}.$$

Finally:

$$\mathcal{M} = \frac{g^2}{q^2 - m_X^2},$$

where $q^2 = (p_a + p_b)^2$. This is exactly the same form as for the $t$-channel process, demonstrating equivalence of the propagator forms.

5.2: Draw the two lowest-order Feynman diagrams for the Compton scattering process $\gamma e^{-1}\rightarrow \gamma e^{-}$.

5.3: Draw the lowest-order t-channel and u-channel Feynman diagrams for $e^+e^-\rightarrow \gamma\gamma$ and use the Feynman rules for QED to write down the corresponding matrix elements.

$$-i \mathcal{M}t = \left[ \varepsilon\mu^(p_3) i e \gamma^\mu u(p_1) \right] \cdot \left[ \frac{-i (\gamma^\rho q_\rho + m_e)}{q^2 - m_e} \right] \cdot \left[ \bar{v}(p_2) i e \gamma^\nu \varepsilon_\nu^(p_4) \right]$$

$$-i \mathcal{M}u = \left[ \varepsilon\mu^(p_4) i e \gamma^\mu u(p_1) \right] \cdot \left[ \frac{-i (\gamma^\rho q_\rho + m_e)}{q^2 - m_e} \right] \cdot \left[ \bar{v}(p_2) i e \gamma^\nu \varepsilon_\nu^(p_3) \right]$$