Symmetry factors in Feyman Diagrams

[vertices]

Sorry I am spamming the forum, but I have yet another question on Feynman diagrams -

Please see attached picture.

Apparently the symmetry factor for this FD is 1 - I am trying to understand why.

My notes explain that "the symmetry factor is 1 because:

φ(x₁) contracts to φ(y₁) in 4 different ways.
φ(x₂) contracts to φ(y₁) in 3 different ways.
φ(x₃) contracts to φ(y₁) in 2 different ways.

Similarly for the φ(x₄), φ(x₅) and φ(x₆) with φ(y₂).

I don't understand this because if you write out the wick's theorem calculation, we have

$$\frac{k^2}{(4!)^2} \int d^4y_1 d^4y_2 <0|T[\phi (x_1) \phi (x_2) \phi (x_3)\phi (x_4)\phi (x_5)\phi (x_6) \phi^4 (y_1) \phi^4 (y_2)]|0>$$

Now, the braket is equal to all distinct pairs of contractions, so all of x₁, x₂, x₃ contract with φ(y₁) in 4 ways...

I know I can't see something quite trivial, so would be grateful if anyone could enlighten me on this...

 

[LAHLH]

Are you sure the symmetry factor of this diagram is 1? Looks very like figure 9.10 (P63 Srednicki) and that has symmetry factor 2^3 (since you can duplicate the swapping of two external vertices on the left with each other, by exchanging functional derivatives at the left vertex so this gives a factor of 2. You gain another factor of 2 by duplicating swapping the two external props on the right. Finally swapping the ends of the internal vertex in the middle can be duplicated by swapping the two vertices, and the set of external props on the left for those on right)

But I see his is $$\phi^3$$ and yours is $$\phi^4$$, so the symmetry factor will be different, but If its S=8 in the $$\phi^3$$, I would have expected yours to more still. I would have thought, you get a factor of 3! from duplicating the swapping of external derivatives on the right amongst each other, by permuting function derivs. Sim another factor of 3! on the left. Then a factor of 2 from the internal line. So I would have guessed S=3!*3!*2

I am new to all this myself though so I am probably wrong...

 

[vertices]

Thanks LAHLH:

Yes I am sure it is 1.

But I forgot to add a factor of:

$$\frac{k^2}{(4!)^2}$$ in front of the integral. So the "combinatorial weight is:

$$\frac{(4!)^2}{(4!)^2}=1$$

...

Would it be right to say there are 4! equivalent Feyman diagrams? If so why 4! - how can I see this is?

 

[torquil]

The reason Peskin/Schroeder defines the phi^4 model with a potential lambda/4!*phi^4 is so that the 4! is canceled by that thing you are talking about (the different ways of attaching a line to the interaction vertex).

See p.93 in Peskin/Schroeder.

EDIT: The symmetry factor caused by interchanging interaction vertices (y1 and y2) will cancel the 2! term in the Taylor expansion. And the 4! factor coming from the choice of which of the four phi(y1) to contract with, will cancel the 1/4! in the definition of the phi^4 potential in the Lagrangian.

Torquil

 

[vertices]

Thanks Torquil...

What I am really trying to get is why there are 4! equivalent FD (I'm not sure if this statement is correct though), I think it should be 4⁴, not 4! - if you write out the Green's function for the interaction, you'll see that each φ(x₁) (i, 1...6) contracts with each φ(y_i) (i, 1 and 2) the same number of times, so why are there fewer contractions with φ(x₂) and yet fewer contractions with φ(x₃)?

 

[LAHLH]

Just wondering is this treatment different from Srednicki's treatment of symmetry factors? Normally there from what I gather at least, S=3!3!*2, assuming your diagram is what I think it is (namely a E=6, V=2, P=7 term diagram. Thus you'd naivley expect (2P)!/E!=14!/6!=121 080 960 terms in the expansion corresponding to this, but with S=3!3!*2=72, the actual number after overcounting is taken into account is 1 681 680 terms)

Maybe what I'm talking about is completley different to what P&S are doing and mean by such things however? I know Srednicki uses the path integral approach and the expansion I'm talking about seemed to rely heavily on the path integral formalism. So I could be going on about something very different. I am very new to this stuff too, so forgive me if my calculations are somewhere incorrect.

 

[torquil]

Thanks Torquil...

What I am really trying to get is why there are 4! equivalent FD (I'm not sure if this statement is correct though), I think it should be 4⁴, not 4! - if you write out the Green's function for the interaction, you'll see that each φ(x₁) (i, 1...6) contracts with each φ(y_i) (i, 1 and 2) the same number of times, so why are there fewer contractions with φ(x₂) and yet fewer contractions with φ(x₃)?

The existence of each vertex contributes a factor 4! to the total number of mathematical terms that the diagram represents. The number of terms is the same as the number of ways to connect phi(x1), phi(x2), phi(x3) to phi(y1)^4. The number of ways is 4*3*2 = 4!. Not 4^4, because you cannot e.g. contract all the phi(x1), phi(x2), phi(x3) with the same factor phi(y1) in phi(y1)^4. Only pairs of field operators may be contracted.

Some authors might be a bit "careless", and define the potential term with the coupling constant 1/4!*lambda in front, but only use lambda on the vertices, and at the same time define the symmetry factors differently...

Torquil

 

[vertices]

The existence of each vertex contributes a factor 4! to the total number of mathematical terms that the diagram represents. The number of terms is the same as the number of ways to connect phi(x1), phi(x2), phi(x3) to phi(y1)^4. The number of ways is 4*3*2 = 4!. Not 4^4, because you cannot e.g. contract all the phi(x1), phi(x2), phi(x3) with the same factor phi(y1) in phi(y1)^4. Only pairs of field operators may be contracted.

Some authors might be a bit "careless", and define the potential term with the coupling constant 1/4!*lambda in front, but only use lambda on the vertices, and at the same time define the symmetry factors differently...

Torquil

Thanks Torquil - I was being really stupid actually - what you've said now makes perfect sense; it was just very confusing..