# Fermi Golden Rule - Beta Decay

• unscientific
In summary: So it's basically just applying the conservation of energy in the form of a delta function. Thanks for clearing that up for me!Got it, thanks! So it's basically just applying the conservation of energy in the form of a delta function. Thanks for clearing that up for me!
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.

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.

unscientific
Einj said:
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.

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

Einj
Einj said:
Integrating out the energy of the neutrino you are left with:
$$dN=\frac{4}{(2\pi)^4}E_\nu^2p_e^2dp_e=\frac{4}{(2\pi)^4}(Q-E_e)^2p_e^2dp_e.$$

Which is the result you are looking for. Note that I used ##\hbar=1##.
I hope this is what you were looking for.

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

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.

Einj said:
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.

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

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

unscientific
Einj said:
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

Got it, thanks!

## What is Fermi Golden Rule?

Fermi Golden Rule is a principle in quantum mechanics that describes the rate at which a quantum system transitions from one state to another. It is often used to calculate the probability of nuclear beta decay.

## How does Fermi Golden Rule relate to beta decay?

Fermi Golden Rule is used to calculate the transition probability for beta decay, which is a type of radioactive decay where a nucleus emits a beta particle (electron) and transforms into a different element.

## What factors affect the rate of beta decay according to Fermi Golden Rule?

The rate of beta decay is affected by the strength of the weak nuclear force, the energy of the emitted beta particle, and the overlap of the initial and final states of the nucleus.

## What is the significance of the beta decay constant in Fermi Golden Rule?

The beta decay constant, also known as the decay rate, is a fundamental parameter in the calculation of the transition probability for beta decay. It is related to the half-life of the decaying nucleus and provides information about the stability of a radioactive element.

## How is Fermi Golden Rule used in practical applications?

Fermi Golden Rule is used in various fields, including nuclear physics, astrophysics, and medical imaging. It allows scientists to predict the behavior of radioactive materials and understand the underlying mechanisms of beta decay. It is also used in the development of new technologies, such as nuclear power and radiation therapy.

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