Graduate Amplitudes for ##\phi^4## theory

  • Thread starter Thread starter PeroK
  • Start date Start date
  • Tags Tags
    Amplitudes Theory
Click For Summary
The discussion focuses on calculating amplitudes in ##\phi^4## theory using Feynman diagrams and rules, specifically addressing the challenges encountered with various terms leading to infinities. The calculations involve the expression for the S-matrix element, where Wick's theorem is applied to obtain non-zero terms, but questions arise regarding the validity of certain terms and their contributions to divergences. The presence of integrals that yield delta functions and infinities is highlighted, suggesting the need for renormalization to handle these divergences effectively. The conversation references the LSZ theorem, which helps in extracting physical amplitudes from correlation functions, indicating that vacuum diagrams do not contribute to scattering amplitudes. Overall, the need for a careful treatment of divergences and the modification of states in the interacting theory is emphasized.
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2024 Award
Messages
29,411
Reaction score
21,112
TL;DR
I'm trying to work out theamplitudes for the various Feynman diagrams from the Dyson series expansion of the S-matrix. Sometimes I'm left with constant integrals over all space and other anomalies.
This is QFT for the gifted amateur, chapter 19, which is generating the various Feynman diagrams and rules. Some calculations are given but I encounter various problems when trying to work them all out.

The starting point is that we want to calculate:
$$\langle q| \hat S | p \rangle = (2\pi)^3 \sqrt{2E_q}\sqrt{2E_p}\langle 0|\hat a_{\vec q} \hat S a_{\vec p}^{\dagger}| 0 \rangle$$
Where:
$$\hat S = T[1 + (\frac{-i\lambda}{4!})\int d^4x \ \hat \phi(x)^4 + (\frac{-i\lambda}{4!})^2(\frac 1 {2!})\int d^4xd^4y \ \hat \phi(x)^4 \hat \phi(y)^4 + \dots]$$
First, if we take the term in ##\lambda##, we have the term:
$$\langle 0|T[\hat a_{\vec q} \hat \phi(x)^4 a_{\vec p}^{\dagger}]| 0 \rangle$$
And, using Wick's theorem I get two non-zero terms coming out of this. The first is covered in the book:
$$12\langle 0|\ [\hat a_{\vec q} \hat \phi(x)] \ [\hat \phi(x) \hat \phi(x)] \ [\hat \phi(x) a_{\vec p}^{\dagger}]| 0 \rangle$$
Where I've used ##[ \ \ ]## to indicate a contraction.

But, I was also looking at the term:
$$3\langle 0|\ [\hat a_{\vec q} a_{\vec p}^{\dagger} ] \ [\hat \phi(x) \hat \phi(x)] \ [\hat \phi(x) \hat \phi(x)]| 0 \rangle$$
Is this term valid? In any case, it leads to an infinity:
$$\delta^4(q-p) \int d^4x \bigg ( \int \frac{d^4k}{(2\pi)^4} \big ( \frac{i}{k^2 - m^2 + i\epsilon} \big ) \bigg )^2$$

A similar thing happens for the ##\lambda^2## term. I have an extra term in the integral that does not correspond to any diagram:
$$[\hat a_{\vec q} a_{\vec p}^{\dagger} ] \ [\hat \phi(x) \hat \phi(y)]^4$$
I can see why the diagram would not make sense, but I can't see why that term vanishes from the integral.

Finally, similar terms crop up in trying to calculate the integrals for other diagrams. I get the right answer except I still have an integral of the form ##\int d^4 x## in front of everything. Mathematically, it all comes back to the same issue, as above.

Any help would be very welcome.

Thanks.
 
Last edited:
  • Like
Likes etotheipi
Physics news on Phys.org
PS I think it might boil down to what to do with the Feynman propagator when ##x = y##. If this generates a delta function ##\delta^4(0)##, then that might sort things out?
 
To continue to answer my own question, I found this, which suggests we do need something of a fudge to get rid of the infinities:

https://physicspages.com/pdf/Lahiri%20QFT/Lahiri%20&%20Pal%20Problems%2005%20No%20equal%20time%20commutators.pdf
 
That's one of the most simple examples for divergent contributions to higher-order corrections depicted by socalled tadpole diagrams, i.e., a line in a Feynman diagram that connects one spacetime-point with a loop. These are divergent and have to be renormalized. In ##\phi^4## theory the most simple example is the one-loop contribution to the self-energy. It provides simply an infinite constant contribution to ##m^2## and has to be subtracted such that you get at one-loop order the physical mass of the particle.

For details, see Sect. 5.4.1 in

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf
 
  • Like
Likes dextercioby and PeroK
These "anomolous" terms all go away when you calculate amplitudes from correlation functions via the LSZ theorem. Specifically, the expression
$$\delta^4(q-p) \int d^4x \bigg ( \int \frac{d^4k}{(2\pi)^4} \big ( \frac{i}{k^2 - m^2 + i\epsilon} \big ) \bigg )^2$$
represents a vacuum diagram which factors out and does not appear in scattering amplitudes. It corresponds to this Feynman diagram:
124141ad.png

See Peskin & Schroeder Chapter 4 for a discussion of vacuum diagrams and how they factor out in calculations, and Chapter 7 for how scattering amplitudes are extracted from correlation functions. If you continue with your current approach to calculating scattering amplitudes you'll also find strange divergences with external leg corrections, i.e. loops on external legs, which as you might expect are also fixed by the correlation function approach. The way to think about these problems is that your states ##|q\rangle## and ##|p\rangle## are not "good" states in the interacting theory and need to be modified to get rid of all the odd divergences (besides the genuine divergences in the QFT).
 
  • Like
Likes vanhees71 and PeroK
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K