# Breit-Wigner formula

1. Jul 2, 2005

### kuengb

Hello
The Breit-Wigner formula says that the cross section for a resonance is proportional to
$$\frac{1}{(2S_a + 1)(2 S_b + 1)},$$
where Sa and Sb are the spins of the incident particles. What's the reason for this? I don't understand it because for a general cross section there appears just the inverse of that because that's what enters the density-of-states.

Thanks,
Bruno

2. Jul 3, 2005

### CarlB

I'll take a stab. Think of it in terms of lifetimes. A high cross section means a short life time because that means that the reaction has a high probability of happening. But to get a sharp (and therefore tall) resonance, you want a very long life time. So some of the terms are inverted. Resonances are sort of reversed from usual cross sections like that.

You should be aware that I answer questions like this in order to make myself think, not necessarily because I am a world expert.

Carl

3. Jul 3, 2005

### marlon

A

Could you please tell me how you get this formula ?

Look at : http://www.astro.uwo.ca/~jlandstr/p467/lec9-nucl_reacts/
to see the actual Breit Wigner formula. How do you convert this formula into the one you gave ?

And what do you mean by 'that is what enters the DOS ?'

regards
marlon

4. Jul 3, 2005

### kuengb

In Fermi's golden rule for the reaction rate, there is the DOS as a factor, and this is proportional to the number of spin states.

I got it from Perkin's book "I. to High Energy Physics". I think I'll quote the questionable paragraph:
"So far, we have not considered particle spin. The appropriate spin multiplicity factors were given in the previous section [what I mentioned above, golden rule]. Putting all these things together, we finally get the complete Breit-Wigner formula,
$$\sigma = \frac{\lambda^2 (2J+1)}{\pi (2S_a+1)(2S_b+1)} \frac{\Gamma^2 / 4}{(E-E_R)^2 + \Gamma^2/4}$$
where Sa and Sb are the spins of the incident and target particles and J is the spin of the resonant states."

That's the same as in your reference except for this prefactor.

Now, I understand the factor (2J+1) because that comes from the fact that the cross section for a given partial wave is proportional to this number. But I don't get why the other spin stuff appears in the denominator. If there are more spin states, the reaction rate is higher and therefore the cross section is larger...or isn't it?

Last edited by a moderator: May 2, 2017
5. Jul 3, 2005

### Meir Achuz

The initial spin factor arises because the cross-section is determined by the outgoing flux divided by the incident flux. The factor (2Sa+1)(2Sb+1) comes from the number of spin states in the incident beam, and therefore is in the denominator.

6. Jul 3, 2005

### kuengb

I see. Thank you!

7. Oct 23, 2007

### nucular1

statistical factor in B-W formula

Hi,
The statistical factor in the Breit-Wigner cross section formula arises from the coupling of angular momenta. For a given reaction A + a -> B + b, or A(a,b)B in nuclear physics parlance, the entrance channel A + a has an associated channel spin S, which is the sum of the individual particle spin quantum numbers. If the incident particle 'a' has spin s =1/2, for example, and the target particle 'A' is some nucleus in its ground state with spin I = 5/2, then the channel spin S is the vector sum of s + I, which by addition rules for angular momentum can add to either S = 3 (5/2 + 1/2) or 2 (5/2 - 1/2). The total angular momentum J is the sum of s, I, and the orbital angular momentum between them L: J = (s + I) + L (all are in units of h_bar). If L = 1, for example, we can now have J = 2, 3, or 4.

For a compound nucleus mechanism, which is what the Breit-Wigner formula describes, the intermediate state is the compound nucleus (CN) which is just A + a -> CN -> B + b. For example, in the reaction 20F(p,n)20Ne, the compound nucleus is 21Ne. The reaction can proceed through excited states in 21Ne of a given energy and total spin J. (I am leaving out parity, which also must be conserved, brevity) The spin of that particular resonance is the one that goes into the B-W formula.

Now each value of the channel spin S, can have 2S + 1 spatial orientations from the magnetic quantum number m_S. Recall m_S = S, S - 1, ...., -S. So including all possible configurations of both particles in the entrance channel, there are (2s + 1)(2I + 1) total states for a channel spin S = s + I. If you follow this logic through, the cross section for one particular state must be divided by this factor. The cross section is also proportional to the spin of the CN, giving the factor (2J + 1) in the numerator, so the total statistical factor for the cross section is therefore:

(2J + 1)
---------------- = g(S).
(2s + 1)(2I + 1)

Hence, the expression given in the Breit-Wigner formula. For more details see Blatt & Weisskopf, Krane or your favorite Nuclear Physics text. Note that for the special case of spinless particles in the entrance channel, this factor is simply (2L + 1) as given in the limit for the geometrical cross section.

I hope this helps, rather than adding to your confusion. Isn't physics PHUN?!? :)