Fermi's Golden Rule Density of States

  • #1

Main Question or Discussion Point

In my particles course, it says we will use Fermi's golden rule to work out rates.

FGR is:
Γ=2π|Mfi

For the case of non-relativistic phase space, my notes say the density of states can be found as follows (pretty much word for word):
Apply boundary conditions
Wave-function vanishing at box boundaries ⇒ quantised particle momenta
Volume of single state in momentum space:
(2π/a)3 = (2π)3/V
Normalizing to one particle/unit volume gives:
Number of states in element d3p=dpxdpydpz
[itex]dn=\frac{d^{3}p}{\frac{(2\pi)^3}{V}} \frac{1}{V}[/itex]

Then a bit more algebra to get some result

I've normally only seen density of states come into the picture for systems with periodic boundary conditions. If we're talking about particle physics and the rate of a collision producing another particle, I don't see why the wavefunction should vanish at the boundary of some box. The only constraint is that the wavefunction should be 0 at infinity. Surely there should be a continuous spectrum of possible wavevectors for the final particle?

Thanks
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,340
6,735
It's
[tex]\gamma=2 \pi |M_{fi}|^2 \rho,[/tex]
and the phase-space density comes from the single-particle wave function for a box with periodic boundary conditions.

The real challenge is to understand, why there is not trouble with the energy-momentum-conserving [itex]\delta[/itex] distribution since you have
[tex]S_{fi}=\delta_{fi} - (2 \pi)^4 \mathrm{i} T_{fi} \delta^{(4)}(p_f-p_i).[/tex]
The answer can be found in

Peskin, Schroeder, Introduction to Quantum Field Theory.
 
  • #3
It's
[tex]\gamma=2 \pi |M_{fi}|^2 \rho,[/tex]
and the phase-space density comes from the single-particle wave function for a box with periodic boundary conditions.

The real challenge is to understand, why there is not trouble with the energy-momentum-conserving [itex]\delta[/itex] distribution since you have
[tex]S_{fi}=\delta_{fi} - (2 \pi)^4 \mathrm{i} T_{fi} \delta^{(4)}(p_f-p_i).[/tex]
The answer can be found in

Peskin, Schroeder, Introduction to Quantum Field Theory.
My question was where do periodic boundary conditions come into the problem at all?

Since I've never seen that last equation before, I feel like you might be hijacking my thread.

No offence,
I<3NickTesla
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,340
6,735
I'm not hijacking your thread but point you to an important point when it comes to calculating cross sections ;-).

This [itex]\delta[/itex]-distribution has to be regularized in some way, because you cannot square it, and you must take the square of the transition-matrix element (which is an amplitude) to get a transition probability rate which figures into the cross section, which after all is what's measured in the lab.

To this end there are two ways: One is physical, and you never leave the usual infinite space and also take the idea of asymptotic states for the incoming and outgoing particles, using the Gell-Mann-Low-description with the appropriate [itex]\epsilon[/itex] regulators. Then you have to start with square integrable wave packet instead of generalized momentum eigen states which are plane waves and not true Hilbert space vectors.

The other is mathematically simpler. You put everything in a finite cubic box in position space and impose periodic boundary conditions, i.e., your wave functions are subject to the condition
[tex]\psi(\vec{x}+L \vec{e}_j)=\psi(\vec{x}), \quad \text{for} \quad j \in \{1,2,3 \}.[/tex]
Then your momenta become the discrete set
[tex]\vec{p} \in \frac{2 \pi \hbar}{L} \mathbb{Z}^3.[/tex]
The Hilbert-space scalar product in position space is then over the finite cube, and thus you have true momentum eigenstates.

The reason for using periodic and not other boundary conditions (e.g., "rigid boundary conditions", making the wave function vanish on the edges of the cube) is that then you are sure to have well-defined self-adjoint momentum operators at all.

Finally you also restrict your time to a large intervall [itex][-T,T][/itex]. Then your S-matrix elements don't contain any [itex]\delta[/itex] distributions. With this regularization of the energy-momentum conserving [itex]\delta[/itex] distributions you evaluate the transition property per unit time and volume (making it a transition-rate density). At the very end you can then take [itex]L[/itex] and [itex]T[/itex] to infinity to get the infinite-volume and true asymptotic-state limit.

For details for the relativistic case, see my QFT manuscript (where this is treated in a very handwaving physicist's short-cut way :-)):

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

pp. 224 ff
 

Related Threads on Fermi's Golden Rule Density of States

Replies
8
Views
8K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
15K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
699
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
0
Views
2K
Replies
3
Views
843
Top