# Fermi Golden Rule - Beta Decay

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1. Jan 28, 2015

### unscientific

We consider the following beta decay:

$$^A_ZX \rightarrow ^A_{Z+1} Y + e^{-} + \nu_e$$

The Fermi golden rule is given by:

$$\Gamma = \frac{2\pi}{\hbar} |A_{fi}|^2 \frac{dN}{dE_f}$$

Reaction amplitude is given by $A_{fi} = G_F M_{nucl}$ while density of states is given by $dN = \frac{4 \pi p^2 dp}{(2\pi)^3}$ and $\frac{dp}{dE} = \frac{1}{2}$.

Putting all these together, it gives :

$$\Gamma = \frac{2\pi}{\hbar} G_p^2M_{nucl}^2 \frac{p^2}{2\pi^2} \frac{1}{2} = \frac{\pi}{2\hbar}G_p^2 M_{nucl}^2 p^2$$

$$d\Gamma = \frac{\pi}{\hbar}G_p^2 M_{nucl}^2 p~\ dp$$

I'm not sure how the text derived this.

2. Jan 29, 2015

### Einj

There are two mistakes in your calculation:

1) You are forgetting the neutrino phase space $\frac{d^3p_\nu}{(2\pi)^3}$
2) It is not true that $dp/dE=1/2$. In fact, $p=\sqrt{E^2-m^2}$ and hence $dp/dE=E/p \Rightarrow pdp=EdE$. This is often a useful relation to keep in mind.

Then you have to do the following. Consider the phase space element for both electron and neutrino:
$$dN=\frac{d^3p_e}{(2\pi)^3}\frac{d^3p_\nu}{(2\pi)^3}=\frac{4\pi p_e^2dp_e}{(2\pi)^3}\frac{4\pi p_\nu^2dp_\nu}{(2\pi)^3}=\frac{4\pi p_e^2dp_e}{(2\pi)^3}\frac{4\pi E_\nu^2dE_\nu}{(2\pi)^3},$$
where I considered the neutrino as massless ($p_\nu=E_\nu$). Integrating out the energy of the neutrino you are left with:
$$dN=\frac{4}{(2\pi)^4}E_\nu^2p_e^2dp_e=\frac{4}{(2\pi)^4}(Q-E_e)^2p_e^2dp_e.$$
Therefore:
$$d\Gamma=2\pi G_F^2\left|M_\text{nucl}\right|^2\frac{4}{(2\pi)^4}(Q-E_e)^2p_e^2dp_e=\frac{G_F^2}{2\pi^3}\left|M_\text{nucl}\right|^2(Q-E_e)^2p_e^2dp_e.$$
Which is the result you are looking for. Note that I used $\hbar=1$.
I hope this is what you were looking for.

3. Jan 29, 2015

### unscientific

Brilliant, thank you! It's people like you that make physics forums a great resource for everyone.

4. Jan 29, 2015

### unscientific

Just a quick question: why is $dE_v = 1$ ?

5. Jan 29, 2015

### Einj

No, it's not one. I simply integrated over it. If you want to be rigorous I should have written $d^2N=f(E_\nu,p_e)dp_edp_\nu$, i.e. a double differential. In the second step I integrated over the energy of the neutrino and so I arrived to a single differential.

6. Jan 29, 2015

### unscientific

Then won't the result become: $\int E_v^2 dE_v = \frac{1}{3} E_v^3$?

7. Jan 29, 2015

### Einj

Sorry, maybe I've been a little bit too implicit. Inside the matrix element you always have a delta function for conservation of energy. Therefore you have should have something like:
$$\int dE_\nu E_\nu^2\delta(E_\nu+E_e-Q)=(Q-E_e)^2.$$
Is it clear? Sorry for being obscure :P

8. Jan 29, 2015

### unscientific

Got it, thanks!