Forest FermiGoldenRule Notes

Fermi's Golden Rule

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") $M_{i,f}$

Transition Amplitude

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

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)

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

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

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

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, $m_2=m_3 = 0$ :

$p_2^\mu = (E_2,\vec{p}_2) = (|\vec{p}_2|,\vec{p}_2) = (p_2,\vec{p}_2)$

$p_3^\mu = (E_3,\vec{p}_3) = (|\vec{p}_3|,\vec{p}_3)= (p_3,\vec{p}_3)$

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

$\int d \Gamma = \int \frac{|M|^2}{16 (2 \pi)^6 \hbar m_{\pi} }\left [ \left ( \frac{d^3\vec{p}_2}{E_2} \right )\left ( \frac{d^3\vec{p}_3}{E_3} \right ) \right ] (2 \pi)^4 \delta(m_{\pi} -p_2-p_3) \delta(0-\vec{p}_2 - \vec{p}_3)$

$= \int \frac{|M|^2}{(8 \pi)^2 \hbar m_{\pi} }\left [ \left ( \frac{d^3\vec{p}_2}{E_2} \right )\left ( \frac{d^3\vec{p}_3}{E_3} \right ) \right ] \delta(m_{\pi} -p_2-p_3) \delta(0-\vec{p}_2 - \vec{p}_3)$

$= \int \frac{|M|^2}{(8 \pi)^2 \hbar m_{\pi}p_2 p_3 }\left [ \left ( d^3\vec{p}_2 \right )\left ( d^3\vec{p}_3 \right ) \right ] \delta(m_{\pi} -p_2-p_3) \delta(0-\vec{p}_2 - \vec{p}_3)$

Integrating over$d^3\vec{p}_3 \Rightarrow \vec{p}_2 = - \vec{p}_3 , | p_2| = |p_3|$ :

$\int d \Gamma = \int \frac{|M|^2}{(8 \pi)^2 \hbar m_{\pi} p_2^2}\left [ \left ( d^3\vec{p}_2 \right ) \right ] \delta(m_{\pi} -2p_2)$

$= \int \frac{|M|^2}{(8 \pi)^2 \hbar m_{\pi} p_2^2}\left [ \left ( |\vec{p}_2|^2 d|\vec{p}_2| \sin(\theta) d \theta d \phi \right ) \right ] \delta(m_{\pi} -2p_2)$

If $|M|^2 = f( \vec{p}_2) \ne f( \vec{p}_2,\vec{p}_3)$( the transition does not depend on the momentum vector directions)

then

$\int d \Gamma = \frac{|M|^2}{(8 \pi)^2 \hbar m_{\pi}}\int \frac{1}{ p_2^2}\left [ \left ( |\vec{p}_2|^2 d|\vec{p}_2| \sin(\theta) d \theta d \phi \right ) \right ] \delta(m_{\pi} -2p_2)$

$= \frac{1}{(8 \pi)^2 \hbar m_{\pi}}\int |M|^2\delta(m_{\pi} -2p_2) d|\vec{p}_2| \sin(\theta) d \theta d \phi$

$= \frac{1}{(8 \pi)^2 \hbar m_{\pi}} (4 \pi) \int |M|^2 \delta(m_{\pi} -2p_2) d|\vec{p}_2|$

$= \frac{1}{(16 \pi) \hbar m_{\pi}} \int |M|^2\delta(m_{\pi} -2p_2) d|\vec{p}_2|$

$= \frac{|M|^2}{(16 \pi) \hbar m_{\pi}} \frac{1}{2} \;\; : \delta (kx) = \frac{\delta(x)}{k} \Rightarrow \delta (m_{\pi} -2p_2)=\frac{\delta (\frac{m_{\pi}}{2} -p_2)}{2}$

$\Gamma= \frac{|M|^2}{(32 \pi) \hbar m_{\pi}}$ with the additional conditions that $p_2 = \frac{m_{\pi}}{2}$and $\vec{p}_2 = - \vec{p}_3$ which must be applied when evaluating$|M|^2$

Units check

$\hbar m_{\pi} = \left ( 6.582 \times 10^{-10} MeV \cdot s\right ) (150 MeV) = 9.97 \times 10^{-8} MeV^2 \cdot s$

$\left [ |M|^2\right ] = MeV^2$= energy eigenvalues squared

$\left [ \Gamma \right ] = \frac{MeV^2 }{MeV^2 \cdot s} = \frac{1}{s} =$ transition probability per unit time

Caveat

Sometime $|M|$ will depend on the momentum vector directions in which case the integral must be done after evaluating the matrix element amplitude. An example of this is when the transition is spin dependent $(\vec{p} \cdot \vec{s})$ as in the hyperfine interaction or polarization based transitions.

Example: Two -Body decay (fission fragments)

Now consider a more general case in whigh the decay daughters have mass:

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

We don't know if the two daughter particles are identical so leave $S$ as a funtion.

$S = ?$

We can still do the calculation in the mother particles rest frame.

$p_1^\mu = (E_1,p_1) = (E_1,0) = (m_1,0)$

Now that the daughters have mass we need:

$p_2^\mu = (E_2,\vec{p}_2) \; E_2 = \sqrt{p_2^2 + m_2^2}$

$p_3^\mu = (E_3,\vec{p}_3) \; E_3 = \sqrt{p_3^2 + m_3^2}$

$\int d \Gamma = \int \frac{S|M|^2}{8 (2 \pi)^6 \hbar m_1 }\left [ \left ( \frac{d^3\vec{p}_2}{E_2} \right )\left ( \frac{d^3\vec{p}_3}{E_3} \right ) \right ] (2 \pi)^4 \delta(m_1 -p_2-p_3) \delta(0-\vec{p}_2 - \vec{p}_3)$

Recast the delta function:

$\delta^4(p_1^\mu-p_2^\mu-p_3^\mu) = \delta(m_1 -E_2-E_3) \delta(0-\vec{p}_2 - \vec{p}_3)$

Upon integrating over $d^3 \vec{p}_3$

$\delta(0-\vec{p}_2 - \vec{p}_3) \Rightarrow |\vec{p}_2| = | \vec{p}_3| \Rightarrow E_3 = \sqrt{p_3^2 + m_3^2} = E_3 = \sqrt{p_2^2 + m_3^2}$

$\delta^4(p_1^\mu-p_2^\mu-p_3^\mu) = \delta(m_1 -\sqrt{p_2^2 + m_2^2} - \sqrt{p_2^2 + m_3^2})$

After integrating over $d^3 \vec{p}_3$ the delta function gives you

$\int d \Gamma = \int \frac{S|M|^2}{8 (2 \pi)^2 \hbar m_1 }\left [ \left ( \frac{d^3\vec{p}_2}{E_2 \sqrt{p_2^2 + m_3^2}} \right ) \right ] \delta(m_1 -\sqrt{p_2^2 + m_2^2} - \sqrt{p_2^2 + m_3^2})$

Once again , if the transition amplitude does not depend on the vector directions of$\vec{p}_2$ and $\vec{p}_3$ then you can integrate over angles $(\theta , \phi)$ and get a $4 \pi$.

$\int d \Gamma = \frac{S}{8 \pi \hbar m_1 } \int \left [ \left ( |M|^2 |\vec{p}_2|^2 \frac{d |\vec{p}_2| }{E_2 \sqrt{p_2^2 + m_3^2}} \right ) \right ] \delta(m_1 -\sqrt{p_2^2 + m_2^2} - \sqrt{p_2^2 + m_3^2})$

Let

$E_{tot} \equiv \sqrt{p_2^2 + m_2^2} + \sqrt{p_3^2 + m_3^2} = \sqrt{p_2^2 + m_2^2} + \sqrt{p_2^2 + m_3^2}$

Then

$d E_{tot} = \frac{E_{tot} p_2 d p_2}{ \sqrt{p_2^2 + m_2^2} \sqrt{p_2^2 + m_3^2}}$

substituting

$\Gamma = \frac{S}{8 \pi \hbar m_1 } \int |M|^2 \frac{p_2}{E_{tot}} \delta(m_1 -E_{tot}) d E_{tot}$

$= \frac{S}{8 \pi \hbar m_1 } |M|^2 \frac{p_2}{E_{tot}}$

where $E_{tot} = m_1$

and

$p_2$ = momentum when $E_{tot} = m_1 = \sqrt{p_2^2 + m_2^2} + \sqrt{p_2^2 + m_3^2}$

or

$p_2 = \frac{1}{2m_1} \sqrt{m_1^4 + m_2^4 + m_3^4 - 2m_1^2m_2^2 - 2m_2^2m_3^2 - 2m_3^2m_1^2}$

$\Gamma = \frac{S}{8 \pi \hbar m_1^2 } |M|^2 \frac{1}{2m_1} \sqrt{m_1^4 + m_2^4 + m_3^4 - 2m_1^2m_2^2 - 2m_2^2m_3^2 - 2m_3^2m_1^2}$

2 Body scattering in CM frame

Lets consider the case when two particles ( $m_1$ and $m_2$) collide and are transformed into two separate particle ($m_3$ and $m_4$). An interaction happens during the collision which changes the two particles into two other particles.

Calculate the differential cross section, in the center of momentum frame, assuming that $|M|$ is the amplitude for this collision.

The general expression for the cross section via Fermi's golden rule is given as

$d^2 \sigma = \left( \frac{\hbar }{8 \pi}\right )^2 \frac{S |M|^2}{\sqrt{ \left [ \left ( p_1 \right )_{\mu} \left ( p_2\right )^{\mu})\right ]^2 - (m_1 m_2)^2}} \left [ \left ( \frac{d^3p_3}{E_3}\right) \left ( \frac{d^3p_4}{E_4}\right) \right ] \delta^4(p_1^{\mu}+p_2^{\mu}-p_3^{\mu}-p_4^{\mu})$

In the CM frame

$\vec{p}_1 = -\vec{p}_2$

$\left ( p_1 \right )_{\mu} \left ( p_2\right )^{\mu}$

$= (E_1,-\vec{p}_1)\cdot (E_2,\vec{p}_2)$

$= E_1E_2 -\vec{p_1} \cdot \vec{p}_2$

$= E_1 E_2 + p_1^2$

$\left [ \left ( p_1 \right )_{\mu} \left ( p_2\right )^{\mu})\right ]^2= \left [ E_1E_2 + p_1^2 \right ] ^2$

$= (E_1^2E_2^2 + 2 E_1 E_2p_1^2 + p_1^4$

$=\left [ \sqrt{p_1^2 + m_1^2}\sqrt{p_1^2 + m_2^2} \right ]^2 + p_1^4 + 2 E_1E_2p_1^2$

$= p_1^4 + (m_1 + m_2)p_1^2 + (m_1m_2)^2 + p_1^4 + 2E_1E_2p_1^2$

$= (p_1^2 + m_1^2 + p_1^2 + m_2^2)p_1^2 + (m_1 m_2)^2 + 2E_1E_2p_1^2$

$= (E_1^2 + E_2^2)p_1^2 + 2E_1E_2 + (m_1 m_2)^2$

$= (E_1 + E_2)^2p_1^2 +(m_1 m_2)^2$

$\left [ \left ( p_1 \right )_{\mu} \left ( p_2\right )^{\mu})\right ]^2= (E_1 + E_2)^2p_1^2 +(m_1 m_2)^2$

or

$\sqrt{ \left [ \left ( p_1 \right )_{\mu} \left ( p_2\right )^{\mu})\right ]^2 - (m_1 m_2)^2}= (E_1 + E_2)p_1$

substituting the above and

$\delta^4(p_1^{\mu}+p_2^{\mu}-p_3^{\mu}-p_4^{\mu})=\delta(E_1+E_2-E_3-E_4) \delta^3(\vec{p}_1 + \vec{p}_2 - \vec{p}_3 - \vec{p}_4)$

$= \delta(E_1+E_2-E_3-E_4) \delta^3( -\vec{p}_3 - \vec{p}_4)$ : $\vec{p}_1 =- \vec{p}_2$

into the cross ection equation we have

$d \sigma = \left( \frac{\hbar }{8 \pi}\right )^2 \frac{S |M|^2}{(E_1 + E_2)p_1} \left [ \left ( \frac{d^3p_3}{E_3}\right) \left ( \frac{d^3p_4}{E_4}\right) \right ] \delta(E_1+E_2-E_3-E_4) \delta^3( -\vec{p}_3 - \vec{p}_4)$

integrating over $d^3 \vec{p}_4$ we have

$\int \int d^2\sigma = \int \int \left( \frac{\hbar }{8 \pi}\right )^2 \frac{S |M|^2}{(E_1 + E_2)p_1} \left [ \left ( \frac{d^3p_3}{E_3}\right) \left ( \frac{d^3p_4}{E_4}\right) \right ] \delta(E_1+E_2-E_3-E_4) \delta^3( -\vec{p}_3 - \vec{p}_4)$

$=\int \int \left( \frac{\hbar }{8 \pi}\right )^2 \frac{S |M|^2}{(E_1 + E_2)p_1} \left [ \left ( \frac{d^3p_3}{E_3}\right) \left ( \frac{d^3p_4}{\sqrt{p_4^2 + m_4^2}}\right) \right ] \delta(E_1+E_2-E_3-\sqrt{p_4^2 + m_4^2}) \delta^3( -\vec{p}_3 - \vec{p}_4)$

$=\int \left( \frac{\hbar }{8 \pi}\right )^2 \frac{S |M|^2}{(E_1 + E_2)p_1} \left [ \left ( \frac{d^3p_3}{E_3}\right) \left ( \frac{1}{\sqrt{p_3^2 + m_4^2}}\right) \right ] \delta(E_1+E_2-E_3-\sqrt{p_3^2 + m_4^2})$

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