Order of scalar interaction impact Feynman diagrams

Higgsy
Messages
18
Reaction score
0
On page 60 of srednicki (72 for online version) for the $$\phi^{3}$$ interaction for scalar fields he defines

$$Z_{1}(J) \propto exp\left[\frac{i}{6}Z_{g}g\int d^{4}x(\frac{1}{i}\frac{\delta}{\delta J})^{3}\right]Z_0(J)$$

Where does this come from? I.e for the quartic interaction does this just become

$$Z_{1}(J) \propto exp\left[\frac{i}{6}Z_{g}g\int d^{4}x(\frac{1}{i}\frac{\delta}{\delta J})^{4}\right]Z_0(J)$$

and for the feynman diagrams the $$\phi ^{3}$$ theory has 3-line vertices whereas the $$\phi^{4}$$ has 4-line vertices? Then how do the feynman diagrams change as we change the order of g?
 
We define the generating functional,

<br /> Z[J] = \int \mathcal{D} \phi \exp \left[i \int d^dx \left(\mathcal{L}_0(\phi) + \mathcal{L}_1(\phi) + J(x)\phi(x) \right)\right]<br />

where \mathcal{L}_0 is solvable, be which I mean I can write down the "free" generating functional

<br /> Z_0[J] = \int \mathcal{D} \phi \exp \left[i \int d^dx \left(\mathcal{L}_0(\phi) + J(x)\phi(x) \right)\right]<br />

exactly as an analytic functional of J(x). In particular, I can take functional derivatives with respect to J. Then by taking derivatives, we can evaluate the following functional integrals:
<br /> \int \mathcal{D}\phi (\phi(x))^n (\phi(y))^m \cdots \exp \left[i \int d^dx \left(\mathcal{L}_0(\phi) + J(x)\phi(x) \right)\right] = \frac{1}{i}\frac{\delta^n}{\delta \phi(x)^n}\frac{1}{i}\frac{\delta^m}{\delta \phi(y)^m} \cdots Z_0[J].<br />

This is basically already the content of your expression. We assume e^{\int d^dx\mathcal{L}_0(\phi)} is just an analytic function of \phi so that it can be defined by a polynomial power series like the above, and we can formally write
<br /> Z[J] = \exp\left( \int d^dx \mathcal{L}_1\left( \frac{1}{i}\frac{\delta}{\delta \phi} \right) \right) Z_0[J].<br />

So for ANY interaction, you just replace the interaction lagrangian with \phi(x) \rightarrow \frac{1}{i}\frac{\delta}{\delta \phi(x)} acting on the free generating function. The easiest way to deal with \phi^4 theory is with the Lagrangian
<br /> \mathcal{L}_1 = \frac{g}{24} \phi(x)^4<br />
(the factor of 24 will help you later for the same reason the factor of 6 helps you in phi^3 theory). So the generating functional is
<br /> Z[J] = \exp\left( \frac{g}{24}\int d^dx \left( \frac{1}{i}\frac{\delta}{\delta \phi} \right)^4 \right) Z_0[J].<br />
Then expanding the exponential in powers of g gives you the Feynman expansion.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

Similar threads

Back
Top