The Optical Theorem for Feynman Diagrams

[phylz]

In Peskin's textbook section 7.3 The Optical Theorem for Feynman Diagrams(Page233), he said it is easy to check that the corresponding t- and u-channel diagrams have no branch cut singularities for s above threshold.

But I can't figure out how to prove it. Can angone help me? Thanks!

 

[CAF123]

For the case of equal mass scattering, one may derive in the centre of mass frame the following relations for the Mandelstam invariants $s,t,u$:
$$s = 4(\mathbf p^2 + m^2) ,\,\,\,\,t=-2\mathbf p^2 (1- \cos \theta), \,\,\,\text{and} \,\,\,\,u = -2\mathbf p^2 (1+\cos \theta) $$ As the physical domain of the variables is such that $\mathbf p^2 \geq 0$ and $-1 \leq \cos \theta \leq 1$, means that $s \geq 4m^2, t \leq 0, u \leq 0$ with the equalities holding when the relative momentum vector $\mathbf p = 0$.

In particular, and coming to your question now, if $s$ is above threshold, meaning that $s$ is higher than the $4m^2$, then $u$ and $t$ are strictly less than zero. This all collectively maintains the constraint $s+t+u = 4m^2$. So, the $t$ and $u$ channel diagrams proportional to $1/t$ and $1/u$ respectively, at the amplitude level, do not admit singular structures as $t$ and $u$ may not be nullified when $s$ is above $4m^2.$

 

[phylz]

For the case of equal mass scattering, one may derive in the centre of mass frame the following relations for the Mandelstam invariants $s,t,u$:
$$s = 4(\mathbf p^2 + m^2) ,\,\,\,\,t=-2\mathbf p^2 (1- \cos \theta), \,\,\,\text{and} \,\,\,\,u = -2\mathbf p^2 (1+\cos \theta) $$ As the physical domain of the variables is such that $\mathbf p^2 \geq 0$ and $-1 \leq \cos \theta \leq 1$, means that $s \geq 4m^2, t \leq 0, u \leq 0$ with the equalities holding when the relative momentum vector $\mathbf p = 0$.

In particular, and coming to your question now, if $s$ is above threshold, meaning that $s$ is higher than the $4m^2$, then $u$ and $t$ are strictly less than zero. This all collectively maintains the constraint $s+t+u = 4m^2$. So, the $t$ and $u$ channel diagrams proportional to $1/t$ and $1/u$ respectively, at the amplitude level, do not admit singular structures as $t$ and $u$ may not be nullified when $s$ is above $4m^2.$

Thank you.
$1.$ You mean that the Mandelstam variables $t$ and $u$ correspond to the branch cut singularities of $t$- and $u$-channel diagrams respectively? Why?
$2.$ I can't understand your last sentence.
$3.$ $t=-2\mathbf p^2 (1- \cos \theta)^2, \,\,\,\text{} \,\,\,\,u = -2\mathbf p^2 (1+\cos \theta)^2$. You lost $^2$ above the $(1- \cos \theta)$ and $(1+\cos \theta)$?

 

[CAF123]

Thank you.
$1.$ You mean that the Mandelstam variables $t$ and $u$ correspond to the branch cut singularities of $t$- and $u$-channel diagrams respectively? Why?

Actually I confess that those statements were made with the tree level diagrams in mind and I implicitly assumed the propagator was massless. But the argument goes through with a massive propagator because then the t and u diagrams are proportional to 1/(t-m²) and 1/(u-m²) respectively and for s>4m² strictly t,u<0 so one never encounters a pole at tree level. This is a general feature under 'S-matrix analyticity' .

What happens at loop level?

One finds that the t,u diagrams may contribute an imaginary part because of their proportionality to $\sqrt{1-4m^2/q^2}$, where q² stands for either t or u. Given that s>4m² again we must have t,u<0 by the above. So the argument of the square root may never be negative and, as such, we are away from the branch cut and in the domain of parameter space where the diagrams are real.

I think there might be another argument without using the explicit form of the diagrams (indeed P&S make their statement before starting their computation of the s channel diagram) but I don't see it yet.

$2.$ I can't understand your last sentence.

Does the above make it better?

$3.$ $t=-2\mathbf p^2 (1- \cos \theta)^2, \,\,\,\text{} \,\,\,\,u = -2\mathbf p^2 (1+\cos \theta)^2$. You lost $^2$ above the $(1- \cos \theta)$ and $(1+\cos \theta)$?

Hmm I don't think so. What makes you think there is a square?

 

[phylz]

Actually I confess that those statements were made with the tree level diagrams in mind and I implicitly assumed the propagator was massless. But the argument goes through with a massive propagator because then the t and u diagrams are proportional to 1/(t-m²) and 1/(u-m²) respectively and for s>4m² strictly t,u<0 so one never encounters a pole at tree level. This is a general feature under 'S-matrix analyticity' .

What happens at loop level?

One finds that the t,u diagrams may contribute an imaginary part because of their proportionality to $\sqrt{1-4m^2/q^2}$, where q² stands for either t or u. Given that s>4m² again we must have t,u<0 by the above. So the argument of the square root may never be negative and, as such, we are away from the branch cut and in the domain of parameter space where the diagrams are real.

I think there might be another argument without using the explicit form of the diagrams (indeed P&S make their statement before starting their computation of the s channel diagram) but I don't see it yet.

Does the above make it better?

Hmm I don't think so. What makes you think there is a square?

Thank you.

$1.$ Now I understand that the $t-$ and $u-$ channel diagrams(tree level diagrams) have no branch cut singularities for $s>4m^2$. But at the loop level, why are the $t-$ and $u-$ diagrams proportional to $\sqrt{1-4m^2/q^2}$ ?

$2.$ Yeah, it is better.

$3.$ Yes, you are right, I have made a mistake.

 

[CAF123]

As I mentioned in my last post, I think there should be an argument based on general considerations of analyticity but I don't see it. So, to see the proportionality to the kinematic square root, we can simply compute the diagrams, after all phi^4 theory is not beyond us :P

Edit: See below

 

[CAF123]

Actually, I think I have an argument based on the optical theorem which I guess is fitting given the section in P&S in which the statement is presented. The imaginary part of the two to two scattering event is given by consideration of the sum of all diagrams with kinematically possible intermediate states between the in and out state of your process. This amounts to decorating the amplitude with final state on shell cuts. Now, the physical domain in the t channel and u channel processes for final state production of at least two states, assuming equal mass m, is $t,u \geq 4m^2$ respectively. However, if $s > 4m^2$, then we must have t,u<0. This is below any particle production threshold for the t and u processes and thus by the optical theorem these two diagrams do not contribute to the imaginary part of the amplitude. That is to say they are purely real and away from branch cut singularities in the t, u planes.

Think it makes sense?