Forest FermiGoldenRule Notes

Fermi's Golden rule is used to calculate the probability (per unit time) of a quantum mechanical transition between two quantum states. Although Fermi first coined the term "Golden Rule", Dirac developed most of the machinery.

The first part of the Golden rule is the transition matrix element (or "Amplitude") $M_{i,f}$

$| M_{i,f}| ^2 \equiv \int \psi_i^{*}(\vec{r}) H_{int} \psi_f(\vec{r}) dr^3$

where

$\psi_i$ = initial quantum state of the system which is an eigenstate of the time independent ("steady state") Hamiltonian ($H_0$)

$\psi_f$ = final quantum state of system after a transition

$H_{int}$ = the part of the total Hamiltonian ($H_{tot}$) which describes the interaction responsible for the transition.

$H_0$ = Unperturbed ("steady state") Hamiltonian

$H_{tot} = H_0 + H_{int}$ = total Hamiltonian describing the quantum mechanical system

$dr^2$ integration over all space

The off diagonal elements of the $M_{i,f}$ matrix tell you the transition probablility.

Single Particle decay

Consider the case when a single particle decays into multiple fragments (several other particles)

$d \Gamma = \hbar W = | M |^2 \frac{S}{2m_1} \left [ \left ( \frac{d^3\vec{p}_2}{(2 \pi)^3 2E_2} \right )\left ( \frac{d^3\vec{p}_3}{(2 \pi)^3 2E_3} \right )\left ( \frac{d^3\vec{p}_4}{(2 \pi)^3 2E_4} \right ) \cdots \left ( \frac{d^3\vec{p}_N}{(2 \pi)^3 2E_N} \right ) \right ] \times (2 \pi)^4 \delta^4(p_1^\mu-p_2^\mu-p_3^\mu- \cdots p_n^\mu)$

where

$W$ = probability per second that the particle will decay

$S$ = a symmetry factor of $\left( \frac{1}{j!}\right )$ for every group of $j$ identical particles in the final state

$p_i^\mu = (E_i, \vec{p}_i)$= 4-momentum of the $i^{th}$ particle. ;$p_1 = (E_1,0)$

$\delta^4(p_1-p_2-p_3- \cdots p_n)$ = conservation of 4-momentum

Note

$\int \frac{d^3 \vec{p}_i}{2E_i} = \int d^4p_i \delta(p_i^2-m_i^2)$ = invariant under Lorentz transformations

Example: Pi-zero ($\pi^0$) decay

$\pi^0 \rightarrow \gamma + \gamma$

we are interested in calculating

$d \Gamma = \hbar W = | M |^2 \frac{S}{2m_1} \left [ \left ( \frac{d^3\vec{p}_2}{(2 \pi)^3 2E_2} \right )\left ( \frac{d^3\vec{p}_3}{(2 \pi)^3 2E_3} \right ) \right ] \delta^4(p_1^\mu-p_2^\mu-p_3^\mu)$

Consider the decay of a neutral pion ($\pi^0$) into two photons ($\gamma$).

The two gammas are identical particles so

$S = \frac{1}{2!} = \frac{1}{1 \cdot 2} = \frac{1}{2}$

Since the pion is initially at rest (or we can go to its rest fram and then Lorentz boost to back to the lab frame)

$p_1^\mu = (E_1,p_1) = (E_1,0) = (m_{\pi^0},0)$

Because photons have no mass, $m_2=m_3 = 0$ :

$p_2^\mu = (E_2,p_2) = (p_2,p_2)$

$p_3^\mu = (E_3,p_3) = (p_3,p_3)$

$\delta^4(p_1^\mu-p_2^\mu-p_3^\mu) = \delta(m_{\pi} -p_2-p_3) \delta(0-\vec{p}_2 - \vec{p}_3)$