What is the meaning of square root cut in Zee's Quantum field theory?

[blue2script]

Hello!

I am just working through Zee's Quantum field theory in a nutshell and have a little problem. He calculates the free propagator and evaluates an integral (I doubt the integral is important for my question) when he writes "and the square root cut starting at \pm i*m leads to an exponential decay [...]". I don't really know what he means. I am german so maybe that's a problem too ;). What does he mean by "square root cut"? At google I just found some scientific papers...

Hope somebody can help me! Thanks in advance!

Blue2script

 

[muppet]

Is there a discontinuity in the function? Or in a contour if it's a complex integral?

 

[D H]

For any complex number $z$ (other than zero), there are two values (two branches) that can qualify as the square root of $z$. To make a square root function, you have to have some way to choose which one is wanted. A branch cut does just that.

 

[blue2script]

Thanks for your fast answers! I am new to this forum... but very impressed! All german forums are so small and slow...

But, well, I think I need to learn how to handle these integrals. I think I got the point of the square root cut but it doesn't really help me out. So, here is the line in Zee I don't get:

"[...] $$D\left( {\vec x} \right) = - i\int {\left[ {d^3 k/\left( {2\pi } \right)^3 2\sqrt {\vec k^2 + m^2 } } \right]e^{ - i\vec k \cdot \vec x} }$$ and the square root cut starting at $$\pm im$$ leads to an exponential $$\sim e^{-im\left|\vec x\right|}$$, as we would expect."

Could you help me with that? A big thanks in advance, I am somewhat confused...

Thanks!

Blue2script

 

[sufive]

Dear Blue2script,

Have you get know how A. Zee get his results?

I just have the same question as you did

 

[blue2script]

Hey sufive,
ufff... this is more than two years away and I am not working in this stuff any more. I don't think I got the answer at the time I wrote all this. Now, after two more years of studying, I think its just a matter of complex integration. What you do is you shift m slightly in the imaginary direction (m -> m + ie, e > 0) and integrate around the poles using Cauchys theorem and the residue. Look at these references:

http://en.wikipedia.org/wiki/Residue_(complex_analysis)
http://en.wikipedia.org/wiki/Cauchy's_integral_theorem

Especially this one should help you out:

http://en.wikipedia.org/wiki/Methods_of_contour_integration

I may have some time at the week-end but I am quite busy at the moment. If you still need help in a few days, just write again here.

Cheers!
Blue2script

 

[sufive]

Dear Blue2script,

I am very glad to get your reply. I read the materials you mentioned
but find no reasonable calculating method to support A. Zee's declaration.

On the contrary, my derivations seems to contradict A. Zee's declaration.
the propagaor amplitude at space-like separations not only does not decay
exponentially, it increases exponentially, with rather weak decay factor of
1/x

Can you provide your emails to me so I can send you my derivation
manual script? it is written by latex and compiled to be pdf. I don't know
if this forum support pdf attachments

 

[course8min]

Dear sufive,

First, thanks to you and blue2script for keeping this thread alive! I googled "square root cut" and this was the first link. I too am working through Zee's book (I went through it fast and now again more slowly). Usually he either provides more clues to the calculation or skips it entirely. This one has me confused too.

blue2script, what has me confused (maybe you feel the same way, sufive) is that the integral is performed in euclidean 3-space. Cauchy's thm/residues/etc. apply to complex path integrals. How do you do complex integration over dk^3? I find nothing on this in my textbooks or on the web.

I have a hunch that when Zee says "without evaluating the integral", he means "this will take a lot longer than it's worth to explain". My guess is that the shift to complex integration occurs not by shifting m into the complex plane, but in the radial component of the euclidean integral when the k-vector is cast in spherical coordinates. To quote Zee, let me explain "roughly how [I think] things go".

Rewrite the integral over dk^3 in spherical coordinates. Then the relevant part of the denominator becomes simply sqrt(r^2 + m^2). The dot product in the exponential portion becomes exp(-ir|x|*angular terms). Presumably the angular integrations factor out? If I'm right about that, then the radial integral (excluding constant multipliers in the spirit of Zee) is exp(-ir|x|)/sqrt(r^2 + m^2).

Finally, something that looks like a candidate for residues! There are poles in the complex r-plane at +/-im. Now how exactly you do that integral is a mystery to me b/c I don't see why you'd go into the complex plane here at all. But replacing r with -im gives you that exp(-m|x|) factor.

Maybe it would make more sense to try the calculation from (22), setting x0-y0 = 0 and then using spherical coordinates directly from there? Maybe it's like replacing time with the radial coordinate in a black hole ;)

Sorry no latex. My skills in that sector are weak, and I have a day job!

 

[course8min]

One more thing: if you look at the appendix to I.IV, Zee demonstrates a similar integral (only without the square root in the denominator). The technique is indeed performed in spherical coordinates with the pure imaginary pole in the residue. The result includes a similar exp(-m|x|).

 

[sufive]

Dear blue2script and corse8min,

I seem to get a little on how to do that integration, but not complete.

Note, our goal is to make integration

2\pi \int dk dq k*k/sqrt{k*k+m2} e^{ik.x.q} .... (23)
q:=cos\theta
~1/x*\int_0^\infty dk k/sqrt{k*k+m2} * sin(kx) ...(phys.forum.1)

This is different from the eq(8) of A.Zee textbook's section I.4. For at least,
we even do not know the integration in (23) converges or not, while the integration
in (8) converges definitely. But if we take principal values, integration (pf1)
indeed converges. Since it can be changed into

~1/2x*\int_{-\infty}^\infty dk k/sqrt{k*k+m2} * sin(kx)

~1/2x \int_{-inf}^inf dk (k-sqrt{..}+sqrt{..})/sqrt{..} * sin(kx)

~1/2x [\int_{-inf}^inf dk (k-sqrt{..})/sqrt{..} * sin(kx) ] + cos(kx)/2x|_{k=-inf}^{inf}

~ take principal value, cox(kx)/2x is zero

~1/2x \int_{-inf}^inf dk (k-sqrt{..})/sqrt{..} * sin(kx) ……(phys.forum.2)

continuating to the complex k plane

~1/2x Im\int_{-inf}^inf dk (k-sqrt{..})/sqrt{..} * e^{ikx} ……(phys.forum.3)

By experience, we should construct a semi-circle (upper, so that e^{ikx} not diverge)
integration contour which includes (-\infty,\infty) and the branch cut (i\infty, im), (im, i\infty),
and walk around the point (im), then possibly use the residue theorem to get the final results.

But I do not know how to treat integrations of the following form

\int_{-inf}^inf dk e^{ikx}/f(\sqrt{k*k+m2})

If you two can provide helps, I will complete the integrations.

 

[course8min]

Dear sufive and blue2script,

This integral had me stumped too, though I guessed (correctly I think) that you need to do a series expansion.

I think that you evaluate it as a Laurent Series around z0 = +im. Since it's analytic in some annulus around this point with m > R1 = epsilon > 0 and R2 < 2m - 2*epsilon, the function f(z) =

e^{ikx}/f(\sqrt{k*k+m2}

can be expanded in terms of (z - im)^n, where n ranges over all integers. If you can do this expansion, you're done, since only the n = -1 term survives the integration. The non-negative powers integrate to 0 by analyticity, and the negative powers less than -1 wind up including the derivative of a constant using Cauchy's .

I have not done the expansion. There are lots of ways to play with power series to expand each piece in terms of rational functions in (z - im), at which point you simply need to collect the n = -1 terms, et voila, you have your residue. Some of the Laurent expansions involve clever tricks and have some sort of recursive series definition of coefficients. But a) I'm not working on a doctoral thesis; and b) I'm prone to addition errors, so it's all yours - but I'd love to see your result! :)

Cheers