Zee's QFT book: equal time propagator

[geoduck]

The propagator in 4-dimensions for a free scalar field has the form:

Δ(x,0)=Θ(t)A(x,t)+Θ(-t)B(x,t)

where Θ is the step function (eq 23 of Zee's QFT book, 2nd edition). He then makes the claim that for spacelike x, one can set t=0, and define Θ(0)=1/2.

The going through all the math, he derives:

\Delta(x,0)|_{t=0}=-\frac{1}{8 \pi^2 r} \int^{\infty}_{-\infty} \frac{dk k}{\sqrt{k^2+m^2}}e^{ikr}

This looks to me like it fails to converge as at large k is just oscillates without decreasing amplitude or period.

But then he does this:

\Delta(x,0)|_{t=0}=-\frac{1}{8 \pi^2 r} \int^{\infty}_{-\infty} \frac{dk k}{\sqrt{k^2+m^2}}e^{ikr}=<br /> \frac{i}{8 \pi^2 r}\frac{\partial}{\partial r} \int^{\infty}_{-\infty} \frac{dk }{\sqrt{k^2+m^2}}e^{ikr}

and now the integral converges by Jordan's lemma.

Is this to be viewed as an analytic continuation? The final results seems to be correct as he gets a Bessel function. But this doesn't seem to be correct mathematically unless he's claiming a continuation.

 

[strangerep]

\Delta(x,0)|_{t=0}=-\frac{1}{8 \pi^2 r} \int^{\infty}_{-\infty} \frac{dk k}{\sqrt{k^2+m^2}}e^{ikr}

This looks to me like it fails to converge as at large k is just oscillates without decreasing amplitude or period.

If you remember that these sort of fields are usually tempered distributions, you won't go too far wrong. In this case, it might not matter if it formally diverges at infinity -- as long as when you integrate it against an arbitrary Schwartz function the result is finite.

Another example: any function that goes as $x^n$ for $|x|\to\infty$, where $n \ge 0$, is not square integrable, yet is a perfectly good tempered distribution, since the result of integrating it against a Schwarz function is finite.

But then he does this:

\Delta(x,0)|_{t=0}=-\frac{1}{8 \pi^2 r} \int^{\infty}_{-\infty} \frac{dk k}{\sqrt{k^2+m^2}}e^{ikr}=<br /> \frac{i}{8 \pi^2 r}\frac{\partial}{\partial r} \int^{\infty}_{-\infty} \frac{dk }{\sqrt{k^2+m^2}}e^{ikr}

and now the integral converges by Jordan's lemma.

Is this to be viewed as an analytic continuation? The final results seems to be correct as he gets a Bessel function. But this doesn't seem to be correct mathematically unless he's claiming a continuation.

Tempered distributions are great, because you can differentiate them (in the sense of a weak derivative), and all the nice properties of Fourier transforms still apply. The fact that a simple manipulation in terms of a derivative works here is just a manifestation of how tempered distributions are really nice to work with.

Of course, to prove the underlying foundational results, you'll have to study the Schwartz theory of distributions.

 

[geoduck]

Thanks. I have a popular "math for physicists" book that asks to calculate:

\int_0^\infty \sin(bx) dx

and they give a hint to use a convergence factor:

\int_0^\infty e^{-ax}\sin(bx)dx= \frac{b}{a^2+b^2}

so setting a=0 gives that the original integral is 1/b.

This seems to be a continuation. I'm quite uncomfortable using analytic continuation in physics, and if I ever got that sine integral then I feel I did something wrong. If you get 1+2+3+... then something is wrong and the answer is not -1/12 unless you can claim something like renormalization, but for the free-field there should be no renormalization for the propagator.

To me order matters. Had Zee shown:

<br /> \Delta(x,0)|_{t=0}=<br /> \frac{i}{8 \pi^2 r}\frac{\partial}{\partial r} \int^{\infty}_{-\infty} \frac{dk }{\sqrt{k^2+m^2}}e^{ikr}

then the result is clear.

But instead Zee shows:

<br /> \Delta(x,0)|_{t=0}=-\frac{1}{8 \pi^2 r} \int^{\infty}_{-\infty} \frac{dk k}{\sqrt{k^2+m^2}}e^{ikr}<br />

first which is not finite.

 

[dextercioby]

I think your last 2 equations are identical, because there are good reasons to put the derivative inside the integral.

 

[strangerep]

Thanks. I have a popular "math for physicists" book that asks to calculate:
\int_0^\infty \sin(bx) dx
and they give a hint to use a convergence factor:
\int_0^\infty e^{-ax}\sin(bx)dx= \frac{b}{a^2+b^2}
so setting a=0 gives that the original integral is 1/b.

This seems to be a continuation.

That's not quite the right word in this context. Rather, they're taking a limit of a sequence of functions, as $a\to 0$. Strictly speaking, one should analyze whether it's ok to swap the order of limits (i.e., the $a\to 0$ limit and the $\infty$ limit in the integral). There is a large body of math theory on certain types of convergence, etc, to help one decide when such swapping is legitimate.

One should also examine the limit and analyze whether it's in the same function space as the elements of the sequence. Sometimes it is (e.g., in a Hilbert space), and sometimes not.

I'm quite uncomfortable using analytic continuation in physics, and if I ever got that sine integral then I feel I did something wrong. If you get 1+2+3+... then something is wrong and the answer is not -1/12 unless you can claim something like renormalization, but for the free-field there should be no renormalization for the propagator.

This is why it's important to specify a rigorous meaning for things like infinite sums, and what it means for two elements in an infinite-dimensional vector space to be "near" each other. This is called "equipping the vector space with a topology".

Zee is simply wielding the impressive machinery of the Schwartz theory of distributions, as most physicists do, without going into the gory details of topologies on function spaces. I will say that when I was first studying advanced field theory, I also was rather perplexed at the apparent extreme sloppiness. But after learning some distribution theory and a little general topology, it became clear that such manipulations are quite ok.

I recently found a good, concise (30-page) introduction to Schwartz distributions in Appendix A of this book: H.M. Nussenzveig, "Causality & Dispersion Relations".

Edit: Expanding on Dex's cryptic remark about the derivative: Zee is using the Leibniz integral rule. That link also gives the conditions under which it's valid to move a derivative operator through an integral sign. Although you perceive Zee's first form as "not finite", that's only because you're not interpreting the integral correctly as a Cauchy principal value integral.