Why does the propagator converge?

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In summary, adding an oscillating exponential decreases the growth in the ultraviolet when integrated over all dkn, but for n>=2 the propagator converges. However, when considering 1 dimension, the expression $$ \int_1^\infty \frac{dx}{x^n} $$ is convergent for n>1. But for n<0, it doesn't converge.
  • #1
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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  • #3
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.

 

1. Why is it important for a propagator to converge?

A propagator represents the probability amplitude for a particle to move from one point to another in a given time. If it does not converge, the probability amplitude will diverge, making it impossible to accurately predict the behavior of a particle.

2. What factors affect the convergence of a propagator?

The convergence of a propagator can be affected by various factors, such as the potential energy, the shape of the potential, the mass of the particle, and the time step used in the calculation.

3. How do we know when a propagator has converged?

One way to determine if a propagator has converged is by looking at the change in the probability amplitude with each time step. When the change becomes smaller than a predefined threshold, the propagator can be considered to have converged.

4. What happens if a propagator does not converge?

If a propagator does not converge, it means that the probability amplitude for the particle to move from one point to another is infinite. This can lead to inaccurate predictions and make it impossible to study the behavior of the particle.

5. Can a propagator diverge after initially converging?

Yes, it is possible for a propagator to diverge after initially converging. This can happen if the particle encounters a singularity or if the calculation is performed over a long period of time, leading to accumulation of errors.

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