- #1

tomdodd4598

- 138

- 13

When I first approached the topic, my understanding was that the differential transition rate from Fermi's golden rule is given by: $$d{ \Gamma }_{ if }=2\pi { \left| { M }_{ if } \right| }^{ 2 }{ \left( 2\pi \right) }^{ 4 }{ \delta }^{ \left( 4 \right) }\left( \sum { { k }_{ f } } -\sum { { k }_{ i } } \right) \prod { \frac { { d }^{ 3 }\vec { { k }_{ f } } }{ { \left( 2\pi \right) }^{ 3 } } }$$ However, if I use this as the basis for calculating various differential scattering cross sections ##\frac { d\sigma }{ d\Omega }##, for example scattering from a potential or 2→2 scattering, I ended up being a factor of ##2E## or ##16{ E }_{ i1 }{ E }_{ i2 }{ E }_{ f1 }{ E }_{ f2 }## out, respectively.

I recalled such factors appearing in places such as the Lorentz-invariant measure ##\frac { 1 }{ 2E } \frac { { d }^{ 3 }\vec { k } }{ { \left( 2\pi \right) }^{ 3 } }##, defining 'four-momentum states' ##\left| k \right> ={ \left( 2\pi \right) }^{ 3/2 }{ \left( 2E \right) }^{ 1/2 }\left| \vec { k } \right>##, so I thought that maybe these factors of ##2E## would appear in the calculations for ##M_{ if }## (due to the state normalisation), and would cancel with factors of ##2E## in some sort of Lorentz invariant form of the golden rule above.

I guess my question is whether this is indeed the case, and if so, how to modify the formula for ##d{ \Gamma }_{ if }## to account for the new factors in ##M_{ if }##. It doesn't seem to me that one can just stick factors of ##\frac { 1 }{ 2E }## into the phase space measure, as that would not give me the correct energies (such as for 2→2 scattering, for example), though I may be wrong. As a side query, it seems the units of the matrix element can vary depending on the process being studied - is this correct?

Thanks in advance for any help!