Notebook Entry
Reading notes for Modern Particle Physics, Mark Thomson (Chapter 3)
Chapter 3 Decay rates and cross sections
Recap: Fermi’s golden rule
Given a system which Hamiltonian $H = H_0 +H^\prime$, where $H^\prime \ll H_0$. The question is, what’s the average transition rate $\varGamma_{fi}$ between state f and state i, when $t\to 0$?
Due to the property of perturbation, we can simply calculate the change of probability amplitude of states under the eigen wave functions of Hamiltonian $H_0$. Assuming the wave function $ \Psi(t) = \sum_{k} c_{k}(t) \psi_{k}(0) \mathrm{e}^{-\mathrm{i} \omega_{k} t} $, we calculate the coefficients to the first order, and get
\[P_{fi}(t)=c_f(t)c_f^*(t)=|T_{fi}|^2\frac{\sin^2\frac{\omega t}{2}}{\left(\frac{\omega}{2}\right)^2}\]| where $T_{fi}=<\psi_f(0) | H^\prime | \psi_i(0)>$, and $\omega = E_f - E_i$ (set $\hbar = 1$). What we’re interested in is the transition rate $\varGamma$: |
According to the definition of Dirac-delta function, the limitation above converge to
\[\varGamma_{fi}=2\pi|T_{fi}|^2\delta(\omega)\]Proof for the delta function:
Going back to the integral form of $\text{sinc}(x)$, we got
$$
\begin{aligned} \lim_{T\to\infty}\frac{1}{T}\frac{\sin^2{\frac{\omega T}{2}}}{\left(\frac{\omega}{2}\right)^2} &= \lim_{T\to\infty}\frac{1}{T}\int_0^T\mathrm e^{-\mathrm i \omega t}\mathrm d t \int 0^T\mathrm e ^{\mathrm i \omega t^\prime}\mathrm d t^\prime
&= \lim{T\to\infty}\frac{1}{T}\int_0^T\int 0^T \mathrm e ^{\mathrm i \omega (t^\prime-t)}\mathrm d t\mathrm d t^\prime
&=\lim{T\to\infty}\frac{1}{T}\int_0^T\mathrm d t\int_{-t}^{T-t}\mathrm e^{\mathrm i \omega\tau}\mathrm d \tau
\text{(periodic function)}&=\lim_{T\to\infty}\frac{1}{T}\cdot T\int_{-T/2}^{T/2}\mathrm e^{\mathrm i \omega\tau}\mathrm d \tau
&=\lim_{T\to\infty}\int_{-T/2}^{T/2}\mathrm e^{\mathrm i \omega\tau}\mathrm d \tau
&= 2\pi \delta(\omega) \end{aligned} $$ Q.E.D.
For systems with continuous energy levels, the transition rate can be written in the integral form, $$
| \varGamma_{fi} = \int\mathrm d \varGamma_{fi} = \int2\pi | T_{fi} | ^2\delta(\omega) \frac{\mathrm d n}{\mathrm d E_f}\mathrm d E_f = 2\pi | T_{fi} | ^2 \rho(E_i) |
\vec p = \frac{2\pi}{a}\vec n, \vec n \in \mathbb Z^3
\[Since the large numeric value of a, we get "density of momentum"\]\frac{\mathrm d n}{\mathrm d p}=\frac{V}{(2\pi)^3}\cdot 4\pi p^2
\[Due to normalization factor in $|T_{fi}|^2$, $V$ will not appear in the expression of the decay rate$\varGamma_{fi}$. One step more, we can set V=1 initially. As for multiple particle scenarios, we get\]\text{(number of states for N particles)}\mathrm d n= \prod_{i=1}^{N-1}\mathrm d n_i=\prod_{i=1}^{N-1}\frac{1}{(2\pi)^3}\mathrm d^3\vec p_i
\[The last term for $p_N$ is constrained by the law of conservation of momentum. For symmetry reasons, we can also finish this term in the form of delta-function, i.e.,\]\mathrm d n =(2\pi)^3\delta(\vec P-\sum_{i=1}^N\vec p_i)\prod_{i=1}^{N}\frac{1}{(2\pi)^3}\mathrm d^3\vec p_i
\[where $\vec P$ is the total momentum of the system of particles. #### Lorentz-invariant phase space Apparently the integral factor $\mathrm d ^3\vec p$ is not Lorentz-invariant. Let's consider another way for integration, which is\]\int {\psi^\prime}^*\psi^\prime\mathrm d^3\vec x = 2E
\[Then we can define\]| M_{fi} | ^2 = | \int {\psi^\prime_f}^*H^\prime\psi^\prime_i\mathrm d^3\vec x | ^2=2 E_i\cdot2E_f \cdot | T_{fi} | ^2 |
\begin{aligned}
\varGamma_{fi}=&\int2\pi|T_{fi}|^2\delta(E-\sum_{i=1}^NE_i)\mathrm d n
&= \int(2\pi)^4|T_{fi}|^2\delta(E-\sum_{i=1}^NE_i)\delta(\vec P-\sum_{i=1}^N\vec p_i)\prod_{i=1}^{N}\frac{1}{(2\pi)^3}\mathrm d^3\vec p_i
&= \int\frac{(2\pi)^4}{2E_i}|M_{fi}|^2\delta(E-\sum_{i=1}^NE_i)\delta(\vec P-\sum_{i=1}^N\vec p_i)\prod_{i=1}^{N}\frac{1}{(2\pi)^3}\frac{\mathrm d^3\vec p_i}{2E_i}
\end{aligned}
$$ Obviously the p/E factor is Lorentz-invariant, so we can calculate this term in any frame we want.
Proof: \(\frac{\mathrm d p^\prime}{\mathrm d p}=\gamma( 1-\beta\frac{\mathrm dE}{\mathrm d p})\)
\[\frac{E^\prime}{E}=\gamma(1-\beta\frac{p}{E})\]Using mass-energy equation: \(E^2 = p^2 +m^2\Rightarrow E\mathrm d E = p \mathrm d p\) Hence \(\frac{\mathrm d p^\prime}{E^\prime}=\frac{\mathrm d p}{E}\) Q.E.D.
Another footprint for LIPS(Lorentz-invariant phase space): \(LIPS = \int \prod_{i=1}^{N} \delta (E_i^2-\vec p_i^2-m_i^2)\mathrm d^4\vec p_i=\int \prod_{i=1}^{N}\frac{\mathrm d^3\vec p_i}{2E_i}\) Each delta meets a constrain condition.
### Particle Decays
As an example, let’s calculate the decay rate for the 2-body situation,i.e. particle interaction $a \to 1+2$. For simplicity, finish this calculation in center-of-mass frame, where total momentum equals to 0 and $E_a = m_a$ $$
\begin{aligned}
\varGamma_{fi}&= \int\frac{(2\pi)^4}{2m_a}|M_{fi}|^2\delta(E_a-E_1-E_2)\delta(\vec p_1+\vec p_2)\frac{1}{(2\pi)^6}\frac{\mathrm d^3\vec p_1}{2E_1}\frac{\mathrm d^3\vec p_2}{2E_2}
&=\frac{1}{8\pi^2m_a}\int|M_{fi}|^2\cdot\frac{1}{4E_1E_2}\cdot\delta(E_a-\sqrt{m_1^2+p_1^2}-\sqrt{m_2^2+p_1^2})p_1^2\mathrm dp_1\mathrm d^2\Omega^
&=\frac{p^}{32\pi^2m_a^2}\int|M_{fi}|^2\mathrm d^2\Omega^*
\end{aligned}
p^*=\frac{\sqrt{(m_a^2-(m_1+m_2)^2)(m_a^2-(m_1-m_2)^2)}}{2m_a}
\[### Interaction cross sections Let's introduce the definition of **cross section**. For interactions between a and b, we have\]\text{(the rate of interaction particle b)}r_b :=\frac{\mathrm d N_b}{\mathrm d t}= \text{(cross section)}\sigma\cdot\text{(flux)}\phi_a
\[where flux\]\phi_a = \frac{\mathrm d N_a}{\mathrm d t \mathrm d S}
\[For instance, let's consider this definition in the interactions between a and b. Suppose the vertical area of the flux of particle a is $A$, and the velocity of the flux is $v$, and the still particle b has the uniform distribution of $n$, hence in time $\delta t$, the number of particle b encountered with particle a is\]\delta N = Av\delta t n
\[By the definition of cross section, we get the probability of interaction\]\delta P = \frac{\delta N\cdot \sigma}{A} = nv\sigma\delta t
\[As the classical explanation of cross section, we got the cross section by how much area where the particle in can interact; in the quantum world, all the processes happen with a probability distribution, so the cross section can be an effective equivalent for this quantum process. #### Lorentz-invariant flux Let's find the relations between interaction rate and cross section. For simplicity, let's consider the interaction $a+b\to1+2$. By the definition of cross section $\sigma$, we have\]\text{(interaction rate)}r = \phi_a \sigma n_bV = n_an_b(v_a+v_b)\sigma V
\[Therefore\]\varGamma_{fi}=(v_a+v_b)\sigma
\[Hence\]| \sigma = \int\frac{(2\pi)^4}{4(v_a+v_b)E_aE_b} | M_{fi} | ^2\delta(E_a+E_b-E_1-E_2)\delta(\vec p_a+\vec p_b-\vec p_1-\vec p_2)\frac{1}{(2\pi)^6}\frac{\mathrm d^3\vec p_1}{2E_1}\frac{\mathrm d^3\vec p_2}{2E_2} |
| \sigma = \frac{p_f^}{64\pi^2(E_a^+E_b^)^2p_i^}\int | M_{fi} | ^2\mathrm d^2\Omega^* |
| \frac{\mathrm d \sigma}{\mathrm d^2 \Omega^} =\frac{p_f^}{64\pi^2(E_a^+E_b^)^2p_i^*} | M_{fi} | ^2 |
E_3^* = \gamma(E_3-\beta p_3\cos\theta)
p_3^\cos \theta^ =\gamma(p_3\cos\theta-\beta E_3)
\frac{\mathrm d \cos \theta^}{\mathrm d \cos \theta}= \frac{\cos\theta-\beta\frac{\beta E_3}{1-\beta \cos \theta}+E_3}{p_3^}\gamma
\[Align with\]\beta = \frac{E_1}{E_1+m_p}
\[We have\]| \frac{\mathrm d \sigma}{\mathrm d \Omega}= \frac{ | M_{fi} | ^2}{64\pi^2}\frac{1}{(m_p+E_1-E_1\cos\theta)^2} |
$$
I believe this derivation is much simpler and more straightforward than that in the book.