Questions concerning integrals in Schwartz's QFT text

[Lapidus]

Two (supposedly) trival questions in Schwartz's QFT notes. The notes can be found http://isites.harvard.edu/fs/docs/icb.topic521209.files/QFT-Schwartz.pdf.

1. page 155, equation 15.2, how does the integrand reduce to k dk? I would guess that there must be some logarithm, but k dk?

2. page 172, equtiation 17.11, why is it legal to shift the variable that way? Will the x integral not notice when we add the x to the equation? I mean, the one x term comletlely disappears after the shifting.

thank you!

 

[rubbergnome]

Hi :) let's see about the questions:

1. The integral reduces to that because it's in 4 dimensions, so that d⁴k = k³dk dΩ changing to (hyper-)spherical coordinates. This cancels the k² in the denominator leaving k, and you can trivially do the angular integral of course.

2. If you swap the integration order as it's implicitly done in the notes, you can shift the k variable without problems. The moral issue is "can you swap the integration order?" the integral dx is on a compact region and the integrand has no problems w.r.t. x, while the k integral has a logarithmic divergence. But in the notes a PV regulator is introduced, so that the integral is convergent and you can do the swap. You can also justify it with DimReg, which I think would be more elegant.

I hope I got those right :) great notes by the way! Is this an actual book?

 

[Lapidus]

Thanks rubbergnome, much appreciated! Everything crystal-clear now!

And yes, these notes have been turned into a book: "https://www.amazon.com/dp/1107034736/?tag=pfamazon01-20". I bought it and it is fantastic. For me it's the best QFT text out there.

 

[rubbergnome]

That will be a great reference for my paper :) you're welcome! Even though I made a typo (d³ is supposed to be k³) and it seems I can't edit my post.

 

[Lapidus]

I have yet another question concerning these same http://isites.harvard.edu/fs/docs/icb.topic521209.files/QFT-Schwartz.pdf, this time on page 192 concerning (19.34).

Simply, how do I get from the lhs to the rhs of (19.34). I know that is just a Taylor expansion in delta_2, but I can't see how this ends up in the expression on the rhs.

any answers are much appreciated!

 

[ChrisVer]

Hmmm... can't you just write:

$\frac{1}{Z} \frac{i}{q -m } \approx \frac{i}{q-m} - \frac{i \delta_2}{q-m} = \frac{i}{q-m} + i^2 \frac{i \delta_2}{q-m} = \frac{i}{q-m} + \frac{i}{q-m} ( i \delta_2 [q-m]) \frac{i}{q-m}$
?

 

[Lapidus]

Ahhhh, of course! Thanks Chris!

 

[Orodruin]

Even though I made a typo (d3 is supposed to be k3) and it seems I can't edit my post.

Fixed. :)

 

[vanhees71]

Hm, isn't this a bit complicated? I'd say you have both a "wave-function renormalizatio" and a "mass" counter term. They are not restricted by a Ward identity and thus free to choose independently. So you could simply start from (19.40) to begin with.

I also like Schartz's book. Finally there's a QFT book one can recommend as a good starting point (not full of typos and even with some conceptual bugs as Peskin and Schroeder, as far as I've seen from using Schwartz's book not for such a long time as Peskin and Schroeder).

 

[Lapidus]

A bit embarrassed to come back, but concerning my question in post 5 and ChrisVer answer in post 6, I am confused what Schwartz does in the textbook what is a bit different from what goes on in its notes

with

and

If I expand and multipy as shown in post 6 of this thread, then how do I get the m_r in the denominators? As I understand after exanding and multiplying I have everywhere m_0. After that II can set

, but that would not look like the last line given here.

Any help?

THANKS!

 

[ChrisVer]

I don't remember how to use slash notation, so I'll write just p ,denoting p-slashed
$\frac{1}{1+ \delta_2} \frac{ i }{p - m_R - \delta_m m_R}$.

$(1 - \delta_2) \frac{i}{p - m_R} \frac{1}{1 - \frac{\delta_m m_R}{p - m_R}}$

$\frac{i}{p-m_R} (1- \delta_2) ( 1 + \frac{\delta_m m_R}{p-m_R})$

$\frac{i}{p-m_R} + \frac{i}{p-m_R}\frac{\delta_m m_R}{p-m_R} - \frac{i}{p-m_R} \delta_2 - \frac{i}{p-m_R} \delta_2\frac{\delta_m m_R}{p-m_R}$

time to drop things that are not-needed...
The first term= your first term ... the rest is so:

$\frac{i}{p-m_R}\frac{\delta_m m_R}{p-m_R} - \frac{i}{p-m_R} \delta_2 - \frac{i}{p-m_R} \delta_2\frac{\delta_m m_R}{p-m_R}= \frac{i}{p-m_R} (\delta_m m_R - \delta_2 \delta_m m_R) \frac{1}{p-m_R}-A$

Where $A = \frac{i}{p-m_R} \delta_2 = \frac{i}{p-m_R} \delta_2 \frac{ p - m_R}{p-m_R}$

Inserting this $A$ above you obtain:

$\frac{i}{p-m_R} (\delta_m m_R - \delta_2 \delta_m m_R- \delta_2 (p -m_R)) \frac{1}{p-m_R}$

So far I've been dragging the $\delta_2 \delta_m$ for too long, it's time to drop it out, since it's a second order (delta squared) contribution...

$\frac{i}{p-m_R} (\delta_m m_R- \delta_2 (p -m_R)) \frac{1}{p-m_R}$

$\frac{i}{p-m_R} ( - \delta_2 p + (\delta_2 + \delta_m) m_R ) \frac{1}{p-m_R}$

$\frac{i}{p-m_R} ( i^2 \delta_2 p - i^2 (\delta_2 + \delta_m) m_R ) \frac{1}{p-m_R} = \frac{i}{p-m_R} ( i \delta_2 p - i (\delta_2 + \delta_m) m_R ) \frac{i}{p-m_R}$

or finally what you have:
$\frac{i}{p-m_R} ( i[ \delta_2 p - (\delta_2 + \delta_m) m_R ]) \frac{i}{p-m_R}$As an overall note, by what I understand that is your problem, you needed to expand in the deltas... in other words if you have:

$\frac{1}{1+a + x}$ and you want to expand wrt to $x$, the common way to do that is by taking $1+a$ as a common factor out, so you have:
$\frac{1}{1+a} \frac{1}{1+ \frac{x}{1+a}}$
and then you can expand the second fraction in terms of x...
Here you expand in deltas...

 

[Lapidus]

AWESOME!

A big thank you, Chris!

And you are totally right in determing what my problem was. Must admit I didn't know how to expand in the delta as you have shown it.