# Propagator - Hankel function - infinite? - Weinberg

• tiny-tim
In summary: If you're not familiar with distributions, I think they're worth learning about. They're a generalization of functions and they turn up all over the place in physics. They have a lot to do with the Fourier transform, which is why I've been thinking about them.That's the best I can do for now - I'm not sure where to point you to learn about distributions. You could try a google search. (I see that the Physics Forum has a discussion on the subject. I haven't read it, but it might be worth a look.)In summary, the conversation discusses the evaluation of the Feynman propag
tiny-tim
Homework Helper
I'm completely confused by page 202 of Weinberg's QTF, Vol.I.

In particular:

$$\Delta_+(x) = \frac{m}{4\pi^2\sqrt(x^2)} \int_{0}^{\infty}\frac{udu}{\sqrt(u^2 + 1)}sin(mu\sqrt(x^2))\hspace{2.5cm}(5.2.8)$$

or, in terms of a standard Hankel function,

$$\Delta_+(x) = \frac{m}{4\pi^2\sqrt(x^2)} K_1(m\sqrt(x^2))\hspace{0.5cm}.\hspace{4cm}(5.2.9)$$​

I can't find "Hankel" anywhere on PF, and wikipedia is no real help.

Surely integral (5.2.8) doesn't converge (the integrand oscillates between ±1)?
In fact, I think it's -∞.
So how can it be a standard function?
[size=-1]What am I missing?[/size]

Hmm... can't help with that I'm afraid! Have you tried your library for a book of mathematical methods for physicists? Specific functions like Leguerre or Hermite polynomials that often crop up can usually be found in a book like that somewhere.

There are numerous books on the theory of Bessel functions of all possible kinds; I'd reccomend J. Watson 1928 classic "The theory of Bessel functions".

good news … and bad news!

Aha! Someone on another forum has given me a very helpful reference on Wolfram:
http://functions.wolfram.com/BesselAiryStruveFunctions/BesselK/07/01/01/0005/​

Cool! Now I see what Weinberg is getting at!

Weinberg is saying (puting $$X = m\sqrt{x^2}$$):
$$K_1(X) = \int_{0}^{\infty}\frac{u\mathrm{d}u}{\sqrt{u^2 + 1}}\sin(Xu)\, ,$$​
which, putting u = sinht, is the same as:
$$K_1(X) = \int_{0}^{\infty}\sinh t.\mathrm{d}t.\sin(X\sinh t)\,.$$​
But there's still a problem -
Wolfram defines:
$$\large K_n(X) = \int_{0}^{\infty}\sinh nt.\mathrm{d}t.\sin(X\sinh t)\,,$$​
which would be the same as Weinberg for n = 1,
EXCEPT THAT WOLFRAM STIPULATES THAT |Re(n)| < 1.

Changing the variable back from t to u, and with n real, gives:
$$\Large K_n(X) = \int_{0}^{\infty}\frac{\sinh (n.\sinh^{-1} u) \mathrm{d}u}{\sqrt{u^2 + 1}}\sin(Xu)\,,$$​
and. for large values of u, we can put, approximately:
$$\large\frac{\sinh (n.\sinh^{-1} u)}{\sqrt{u^2 + 1}}\,=\,\frac{e^{n.logu}}{\sqrt{u^2 + 1}}\,=\,\frac{u^n}{\sqrt{u^2 + 1}}\,=\,u^{n\,-\,1}\,,$$​
which tends to zero,
and so the integrand tends very comfortably to zero provided n < 1.

So the Wolfram integral for $$\large K_n(X)$$ does converge for n < 1, but it doesn't seem to for n = 1 (because the integrand oscillates between ±1), and Wolfram doesn't claim it does!

: So surely Weinberg's derivation of (5.2.8) and (5.2.9) is just wrong?

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Hi,

I've been working the last few days on this exact same problem (evaluating the Feynman propagator in closed form). I haven't closed the problem out yet, but perhaps I can pass on some things I've found out and I'll be interested to see what else you find.

I came to the same integral (5.2.8) and concluded that it is indeed divergent, because as you say the integrand oscillates indefinitely. However, within the context of distributions, it can be interpreted in a rigorous and precise fashion.

In this case, the distribution is u/sqrt(u^2 + 1), which for large u is similar to a Heaviside step function, and the integral (5.2.8) asks for the Fourier transform of this distribution. The step function has a well-defined Fourier transform (there's a delta function in there somewhere - you'll have to look it up), and this is the way forward: evaluate this divergent piece first, and then what's left is finite and integrable: it's the Hankel function referred to by Weinberg.

More precisely, write

u/ sqrt(u^2 + 1) = H(u) + ( u/sqrt(u^2 + 1) - 1 )

H(u) is the step function; it's Fourier transform has a delta function in it somewhere. (Weinberg didn't mention it because it's concentrated on the light cone. It is therefore irrelevant to his purpose of evaluating the propagator in the spacelike region.) The second bit (in parentheses) is O(1/u^2) for large u, so it's absolutely integrable and its Fourier transform is well-defined within the context of standard real analysis. You can transform it to one of the standard integral representations of Bessel functions.

And that's where I'm up to right now. I also wanted to figure out the form of the propagator in the timelike region, and then I wanted to the whole thing over again for spin-half particles (satisfying the Dirac equation) as well.

Hope that helps,

Dave

schieghoven said:
In this case, the distribution is u/sqrt(u^2 + 1), which for large u is similar to a Heaviside step function,

More precisely, write
$$\Large\frac{u}{\sqrt{(u^2 + 1)}} = H(u) + \left(\frac{u}{\sqrt{(u^2 + 1)}} - 1\right)$$​

The second bit (in parentheses) is O(1/u^2) for large u, so it's absolutely integrable and its Fourier transform is well-defined within the context of standard real analysis.

H(u) is the step function; it's Fourier transform has a delta function in it somewhere. (Weinberg didn't mention it because it's concentrated on the light cone. It is therefore irrelevant to his purpose of evaluating the propagator in the spacelike region.)

I'm not following this - Weinberg only integrates from 0 to ∞, and so isn't interested in u < 0, so you might as well use the constant function 1 as the step function H.

And doesn't the Fourier transform integrate from -∞ to ∞ anyway? So his integral
$$K_1(X) = \int_{0}^{\infty}\frac{u\mathrm{d}u}{\sqrt{u^2 + 1}}\sin(Xu)$$​
is half of the Fourier transform
$$\int_{-\infty}^{\infty}\frac{u\mathrm{d}u}{\sqrt{u^2 + 1}}\sin(Xu)\,,$$​
in which the distribution is certainly not similar to a step function?

I agree that your second bit (in the parentheses) is O(1/u^2) for large u, but that still leaves the integral
$$\int_{0}^{\infty}H(u)\mathrm{d}u\sin(Xu)\,,\,=\,\int_{0}^{\infty}\mathrm{d}u\sin(Xu)\,,$$​
which doesn't converge.

And where does the light-cone come into it? Both u and x are one-dimensional variables.

Ah, but integral
\int H(u)sin(Xu) du
is actually referring to (the imaginary part of) the Fourier transform of the step function, which is well defined:

http://mathworld.wolfram.com/FourierTransformHeavisideStepFunction.html

It's a distribution, not a function, so its Fourier transform is not necessarily defined in terms of a convergent integral.

In the context of this problem, it brings in a contribution \delta(X), where your notation X stands for m\sqrt(x^2) in your original posting. That's what I was referring to when I meant the delta function was concentrated on the light cone, because the light cone is precisely the region X=0.

Dave

## 1. What is a propagator in the context of physics?

A propagator in physics is a mathematical concept that describes the probability amplitude for a particle to move from one point in space to another. It is often used in quantum mechanics to calculate the probability of a particle being in a certain position at a certain time.

## 2. What is the Hankel function and how is it related to the propagator?

The Hankel function is a mathematical function that describes the behavior of waves in cylindrical and spherical coordinate systems. In the context of the propagator, the Hankel function is used to represent the outgoing spherical wave that emanates from a point source. It is an essential component in calculating the propagator for a spherical potential in quantum mechanics.

## 3. How does the propagator behave in the infinite limit?

In the infinite limit, the propagator approaches a constant value. This is because as the distance between two points becomes infinitely large, the probability of a particle moving from one point to another becomes independent of the initial position. This is known as the stationary phase approximation and is commonly used in calculating the propagator for large distances.

## 4. What is the significance of Weinberg in relation to the propagator and Hankel function?

Weinberg is a prominent physicist who has made significant contributions to the field of quantum field theory. He has published several papers on the use of the propagator and Hankel function in calculating scattering amplitudes and has provided valuable insights into their behavior in various physical systems.

## 5. How is the propagator and Hankel function used in practical applications?

The propagator and Hankel function are used in a wide range of practical applications, including in quantum mechanics, electromagnetism, and signal processing. They are used to calculate the behavior of particles and waves in different systems and are essential tools for understanding the fundamental laws of physics.

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