Why does the propagator converge?

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SUMMARY

The discussion centers on the convergence properties of the propagator $$\frac{1}{k^2+m^2}$$ in quantum field theory. It is established that this propagator diverges in the ultraviolet for dimensions n≥2, but becomes convergent when an oscillating exponential factor is introduced, specifically in the expression $$\Delta(x)=\int d^nk \frac{e^{ikx}}{k^2+m^2}$$ for x≠0. The conversation also highlights the contrasting convergence behaviors of integrals involving cosine functions, particularly $$\int_{-\infty}^{\infty} \cos(x^2) dx$$ and $$\int_{-\infty}^{\infty} \cos(x) dx$$, with the former converging due to diminishing oscillation amplitudes at infinity.

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  • Basic principles of renormalization in quantum field theory.
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  • Study the properties of Fourier transforms in quantum mechanics.
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This discussion is beneficial for theoretical physicists, particularly those specializing in quantum field theory, as well as students and researchers interested in the mathematical foundations of propagators and convergence in integrals.

geoduck
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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 $$\int_{-\infty}^{\infty} \cos(x) dx $$?
 
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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.
 
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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 λ[itex]ψ^{4}[/itex], 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.

 

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