# Peskin p.91-93

## Homework Statement

I’m trying to learn Feynman diagrams from Peskin and Schroeder. I’m stuck right now on page 91-93, especially about constant factors and symmetry factors. Equation 4.44 and the two constants 3 and 12 in it make perfect sense to me. The trouble starts with 4.45. I understand it is one possible contraction from 10368 that all give the same expression.

Why the 1/3! in 4.45?

To get similar result for 4.45 as for 4.44, we need to replace the 1/3! With 10368 and add other expressions each with an overall constants. Correct?

On page 93 they say (first paragraph, second to last line) adding at each vertex this integral times minus I times game gives the overall constant. How on earth does that give 10368?

What constant is why by a symmetry constant 8 to large?

Peskin p.91-93

see above

nrqed
Homework Helper
Gold Member

## Homework Statement

I’m trying to learn Feynman diagrams from Peskin and Schroeder. I’m stuck right now on page 91-93, especially about constant factors and symmetry factors. Equation 4.44 and the two constants 3 and 12 in it make perfect sense to me. The trouble starts with 4.45. I understand it is one possible contraction from 10368 that all give the same expression.

Why the 1/3! in 4.45?
this is the 1/n! coming from the fact that you are expanding an exponential!
To get similar result for 4.45 as for 4.44, we need to replace the 1/3! With 10368 and add other expressions each with an overall constants. Correct?

On page 93 they say (first paragraph, second to last line) adding at each vertex this integral times minus I times game gives the overall constant.

I don't see this in my edition...and beside, the sentence does not make sense...
How on earth does that give 10368?

What constant is why by a symmetry constant 8 to large?

They are saying the following: the product of all the combinatoric factors almost exactly cancels out the 1/3! (from the exponential expansion) and the powers of 1/4! (from the Feynman rule for the vertex. So let's say you would simply forget all those factors altogether and use an overall factor of one exactly. You would almost get it right except that your factor would be 8 times too large. If you use one and divide by the symmetry factor you will get the correct answer.

Hope this clarifies things.

Patrick

Hi nrqed,

the sentence does not make sense...
Ooops, I should not drink beer while formulating posts at PF. Anyway, they say in the first paragraph in the second to last line: "it is therefore conventional to associate the expression 'integral times -i times gamma' with each vertex".

Now again my question: what factor do I get by applying this scheme? 10368 times 8??? How?

Also again, to get a similar result for 4.45 as for 4.44, do we need to replace the 1/3! with 10368 and add other expressions each with an overall constants just as we did in 4.44?

you say
the product of all the combinatoric factors almost exactly cancels out the 1/3! (from the exponential expansion) and the powers of 1/4!

I don't see that. 10368 cancels out 1/3! times (1/4!)^3??

thank you

nrqed
Homework Helper
Gold Member
Hi nrqed,

Ooops, I should not drink beer while formulating posts at PF. Anyway, they say in the first paragraph in the second to last line: "it is therefore conventional to associate the expression 'integral times -i times gamma' with each vertex".

Now again my question: what factor do I get by applying this scheme? 10368 times 8??? How?

Also again, to get a similar result for 4.45 as for 4.44, do we need to replace the 1/3! with 10368 and add other expressions each with an overall constants just as we did in 4.44?

you say

I don't see that. 10368 cancels out 1/3! times (1/4!)^3??

thank you

I said almost cancels out!

If you do the full calculation, you have a result of the following form:

1/3! (for the expansion of the exponential) times (1/4!)^3 for the vertices times a messy combinatorial calculation which comes out to be equal to 10368.

So you get a huge number divided by a huge number and the result is 10368/(3! (4!)^3) =1/8.

They are saying that the huge number coming from the combinatorial is almost equal to the huge number coming from the factorials due to the vertex rule and the exponential expansion.

So they say: could we simply forget about all the combinatorial stuff, forget about the factorial coming from the exponential expnasion and simply use a vertex rule lambda instead of lambda/4! ? In that case, you would just have an overall factor of ONE.

But, obviously, you can't do that because the combinatorial stuff does not exactly cancel out the product of factorials.

So should we go back to doing the full calculation? No. A shortcut is the following:

Forget about the 1/n! coming from the exponential expansion. Do not include a factor of 1/4! in the vertex rule. Do not do all the combinatorial stuff. BUT you must divide by the symmetry factor. This will give the correct result in the end.

thanks nrqed! Clear now.

I will now try to accustom myself with symmetry factors, so maybe I have to come back and ask some more.

but for now, many thanks

nrqed
Homework Helper
Gold Member
thanks nrqed! Clear now.

I will now try to accustom myself with symmetry factors, so maybe I have to come back and ask some more.

but for now, many thanks

You are very welcome