I What is the derivation of the Fermi golden rule?

[James_1978]

Dear Forum,

I have a question about the derivation of the Fermi golden rule in Kenneth Krane's Introduction to Nuclear Physics. I understand everything up to equation 9.20. However, it is unclear how he goes directly to equation 9.21. Here is equation 9.20,

$d\lambda = \frac{2\pi}{\hbar}g^{2} |M_{fi}|^{2} \frac{(4\pi)^{2}}{h^{6}c^{2}} p^{2} (Q - \sqrt(p^{2}c^{2} + m_{e}^{2}c^{4}) -m_{e}c^{2})dp$

Which is then followed by equation 9.21,

$N(p)dp = Cp^{2}q^{2}dp$

I under stand that C has the constants not related to p. My question is how did he go from $d\lambda$ to $N(p)dp$. This says that $d\lambda = N(p)dp$ which is unclear to me. Can someone explain that part to me.

Thank you.

 

[mathman]

9.20 doesn't look right. LHS is a differential. RHS has two terms: the first is not a differential, second is. Can't be.

 

[malawi_glenn]

RHS has two terms: the first is not a differential

Terms?

There is a dp though

 

[Dr.AbeNikIanEdL]

dλ=2πℏg2|Mfi|2(4π)2h6c2p2(Q−(p2c2+me2c4)−mec2)dp

I think the expression you have in brackets here should be squared (what you wrote equals $q$ by 9.22, but in 9.20 it is $q^2$).

I am not sure there is much to understand here,

$d\lambda = N(p) dp$

simply defines $N(p)$ as far as I can tell. The only part of

$N(p) = C p^2 q^2$

that might be non-trivial is that indeed all other factors (in particular $|M_{fi}|$) are independent of $p^2$. However the text states this is an assumption at this point. Could you clarify what is unclear?

 

[mathman]

Terms?

There is a dp though

RHS is the sum of two terms. The second term has dp. The first term has no differential - it needs one.

 

[Dr.AbeNikIanEdL]

I think you are misreading the brackets (one is opened before the $Q$, and the corresponding closed one is right before the $dp$...)

 

[malawi_glenn]

RHS is the sum of two terms. The second term has dp. The first term has no differential - it needs one.

No RHS has three terms.
Note that the OP does not know LaTeX and has this written sqrt wrong

should read
$d\lambda = \dfrac{2\pi}{\hbar} g^2 |M_{fi}|^2 \dfrac{(4\pi)^2}{h^6c^2}p^2 (Q - \sqrt{p^2c^2+m_e^2c^4} -m_ec^2) dp$

 

[mathman]

No RHS has three terms.
Note that the OP does not know LaTeX and has this written sqrt wrong
View attachment 324103
should read
$d\lambda = \dfrac{2\pi}{\hbar} g^2 |M_{fi}|^2 \dfrac{(4\pi)^2}{h^6c^2}p^2 (Q - \sqrt{p^2c^2+m_e^2c^4} -m_ec^2) dp$

Original question error - $g^2$ replaced by $q^2$ for 9.21.