Why does the propagator converge?

  • Thread starter Thread starter geoduck
  • Start date Start date
  • Tags Tags
    Propagator
geoduck
Messages
257
Reaction score
2
The propagator $$\frac{1}{k^2+m^2} $$ diverges in the ultraviolet when integrated over all dkn, with n>=2. However, when you throw in an exponential,

$$\Delta(x)=\int d^nk \frac{e^{ikx}}{k^2+m^2} $$ is convergent for x≠0

Intuitively adding the oscillating exponential decreases the ultraviolet growth by alternating it with + and - that cancel when added.

But consider 1 dimension, and the expression $$ \int_1^\infty \frac{dx}{x^n} $$. This expression is convergent for n>1. Now add an exponential: $$ \int_1^\infty dx \frac{ e^{ix}}{x^n} $$. This improves the convergence: now the expression is convergent for n>0. But for n<0, it doesn't converge.

So intuitively, $$\Delta(x)=\int d^nk \frac{e^{ikx}}{k^2+m^2} $$ should not be convergent for n>=2. But it obviously is, since the propagator is convergent in dimensions greater than 2, I believe.

What exactly is happening with these oscillations at infinity? Also I'm a bit bewildered at why $$\int_{-\infty}^{\infty} \cos(x^2) dx $$ is convergent. I can show it mathematically, but intuitively why does that expression converge and not $$\int_{-\infty}^{\infty} \cos(x) dx $$?
 
Physics news on Phys.org
geoduck said:
Also I'm a bit bewildered at why $$\int_{-\infty}^{\infty} \cos(x^2) dx $$ is convergent. I can show it mathematically, but intuitively why does that expression converge and not $$\int_{-\infty}^{\infty} \cos(x) dx $$?

Well, consider the functions ##f(M) = \int_{-M}^{M} \cos(x^2) dx## and ##g(M) = \int_{-M}^{M} \cos(x) dx##. Plot the two functions as ##M \to \infty## and it will be clear that ##f## goes to a limit, while ##g## doesn't. Intuitively, the area under each positive and negative oscillation of ##\cos(x^2)## shrinks as ##x \to \infty##, so the integral ##f## oscillates less and less. But the oscillations of ##\cos(x)## don't have smaller areas as ##x \to \infty##, so the integral ##g## oscillates with a constant amplitude and never settles down to a limit.
 
  • Like
Likes 1 person
geoduck said:
The propagator $$\frac{1}{k^2+m^2} $$ diverges in the ultraviolet when integrated over all dkn, with n>=2. However, when you throw in an exponential,

$$\Delta(x)=\int d^nk \frac{e^{ikx}}{k^2+m^2} $$ is convergent for x≠0

Intuitively adding the oscillating exponential decreases the ultraviolet growth by alternating it with + and - that cancel when added.

But consider 1 dimension, and the expression $$ \int_1^\infty \frac{dx}{x^n} $$. This expression is convergent for n>1. Now add an exponential: $$ \int_1^\infty dx \frac{ e^{ix}}{x^n} $$. This improves the convergence: now the expression is convergent for n>0. But for n<0, it doesn't converge.

So intuitively, $$\Delta(x)=\int d^nk \frac{e^{ikx}}{k^2+m^2} $$ should not be convergent for n>=2. But it obviously is, since the propagator is convergent in dimensions greater than 2, I believe.

What exactly is happening with these oscillations at infinity? Also I'm a bit bewildered at why $$\int_{-\infty}^{\infty} \cos(x^2) dx $$ is convergent. I can show it mathematically, but intuitively why does that expression converge and not $$\int_{-\infty}^{\infty} \cos(x) dx $$?

Hi Geoduck,

You are right when you say the Fourier transform to the propagator $$\frac{1}{k^2+m^2} $$ only diverges for x=0 and that's quite logical because the contribution of a position to itself must be a delta function and not only an infinitesimal contribution.

From what I have read, the divergence problem arises when introducing the self-interacting lagrangian that depends on λψ^{4}, this new term modifies the propagator causing a mass renormalization, this mass renormalization turns to be divergent for high energy contributions. The Renormalization Group tries to solve these problems. However, I must recognise this issues related with the interacting lagrangian and its fluctuations scape quite from my comprehension.

 
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!
Back
Top