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Fermi Golden Rule - Beta Decay

  1. Jan 28, 2015 #1
    We consider the following beta decay:

    [tex] ^A_ZX \rightarrow ^A_{Z+1} Y + e^{-} + \nu_e [/tex]

    The Fermi golden rule is given by:

    [tex] \Gamma = \frac{2\pi}{\hbar} |A_{fi}|^2 \frac{dN}{dE_f} [/tex]

    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 :

    [tex]\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 [/tex]

    [tex] d\Gamma = \frac{\pi}{\hbar}G_p^2 M_{nucl}^2 p~\ dp [/tex]


    I'm not sure how the text derived this.

    ifzme9.png
     
  2. jcsd
  3. Jan 29, 2015 #2
    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.
     
  4. Jan 29, 2015 #3
    Brilliant, thank you! It's people like you that make physics forums a great resource for everyone.
     
  5. Jan 29, 2015 #4
    Just a quick question: why is ##dE_v = 1## ?
     
  6. Jan 29, 2015 #5
    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.
     
  7. Jan 29, 2015 #6
    Then won't the result become: ## \int E_v^2 dE_v = \frac{1}{3} E_v^3##?
     
  8. Jan 29, 2015 #7
    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
     
  9. Jan 29, 2015 #8
    Got it, thanks!
     
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