Another question from Srednicki's QFT book

[MathematicalPhysicist]

Sorry for my questions, (it does seem like QFT triggers quite a lot of questions :-D).

Anyway, on page 103 (it has a preview in google books), I am not sure how did he get equation (14.40), obviously it should follow from (14.39), but I don't understand where did -ln(m^2) disappear ?

Shouldn't we have an expression like (14.40) but the term (linear in k^2 and m^2) be instead:
(linear in k^2,m^2 and ln(m^2)).

Cause as far as I can tell from (14.39) \Pi(k^2) depends also on ln(m^2), and thus the term "linear in..." should be replaced with "linear also in ln(m^2)", cause as far as I can tell ln(m^2) isn't linear function of m^2, right?

Hope someone can enlighten me.

 

[ChrisVer]

No that's not possibly the case, because even if you have the - \ln (m^{2}) it's multiplied with a D.. After integration of D dx you will get (k^{2}+m^{2})ln(m^{2})/6
so I think the ln(m^{2}) is absorbed within the κ_{A,B}

But I hope someone can be more helpful

 

[MathematicalPhysicist]

Actually the integral on Ddx yields 1/6 k^2 +m^2, it's written previously.

 

[ChrisVer]

and that's what I've written? I just didn't take in account the minus from the ln...Ah yes, you are write, just put a 6 in front of m^2 then

 

[MathematicalPhysicist]

I think the task of solving the problem of mass gap from clay institute which relates to QFT looks a lot more intimidating as I keep reading. :-D

 

[DarMM]

The mass gap problem is essentially that you have to prove Yang-Mills exists and after you have done that, you must prove it describes massive particles (rather than massless ones as you would naively think from the Lagrangian). So, yes it is very difficult!

 

[MathematicalPhysicist]

Ok, I plan to ask all of my questions from QFT books here (if the moderators want to include my other posts into this thread it's ok by me, but make it chronological orderd (or as we say use the time-ordering operator).

 

[MathematicalPhysicist]

eqs. (27.23)-(27.24) on page 166 of Srednicki's QFT.

In this case we don't have a book preview of pages 166-167 .

So I'll write the equations:

(27.23) ln |\mathcal{T}|^2_{obs} = C_1+2ln \alpha +3\alpha (ln \mu +C_2)+O(\alpha^2)

Now he says that "Differentiating wrt ln \mu then gives":

(27.24)0=\frac{d}{dln \mu} ln |\mathcal{T}|^2_{obs} = \frac{2}{\alpha} \frac{d\alpha}{dln \mu} +3\alpha +O(\alpha^2)

Now as far as I can tell when you differentiate: \frac{d}{dln \mu} (3\alpha(ln \mu +C_2))=3\alpha + 3 \frac{d\alpha}{dln \mu} (ln \mu +C_2)
so where did 3 \frac{d\alpha}{dln \mu} ln \mu disapper from eq. (27.24)?

Don't see why he didn't include this term in eq. (27.24).

Anyone?

 

[MathematicalPhysicist]

Another question from Srednicki.

Hi, so I hope there are still some folks who look at this thread of mine.

So now I am looking at Srednicki's solution to question 48.4b, here:
http://www.scribd.com/doc/87916496/Srednicki-Ms-Qft-Solutions-Rev

I don't understand how in eq (48.53) he pulled from the trace the term (1+s1s2), shouldn't it be (1-s1s2)?

I mean if you simplify the term inside the trace in the first line you should get:
Tr((1-s1s2)(...)) = (1-s1s2)Tr(...)

So how did he get the plus sign there?

Perhaps x^2=1 and not -1 as he wrote there?

Puzzled.

 

[ChrisVer]

I think there is a mistake in the signs...
\bar{x} ( -\bar{p}_{1} -m) \bar{x}

-\bar{x} \bar{p}_{1} \bar{x} - m \bar{x} \bar{x}

+\bar{x}\bar{x} \bar{p}_{1}- 2 xp_{1} + m

-\bar{p}_{1}+m

So I think the minus in front of s1s2 in the first line of 48.53 is wrong? Obviously by bared quantities I mean slash, it's just \slash{} didn't work...

 

[MathematicalPhysicist]

Another question from Srednicki's text.

On page 316, I am not sure how did he get equation (51.10) from eq. (51.9)?

Here's what I have done :
eqaution (51.10) is the same as the next equation:

(I am writing it down without a slash, since I am not sure how to use it here)

\frac{(-p^{\mu}\gamma_{\mu}+m)}{(p^{\mu}\gamma_{\mu}+m-i\epsilon)(-p^{\mu}\gamma_{\mu}+m)}+\int \frac{...}{p^2+s-\epsilon^2-2i\epsilon\sqrt{s}}

So as you can see it's a bit different than (51.9), the denominator in the integral, the epsilon doesn't have as a factor, the coefficient, i .
And the first term after rearranging it we should have:

\frac{ -p^{\mu}\gamma_{\mu} + m}{p^2+m^2-i\epsilon \cdot (-p^{\mu}\gamma_{\mu} +m)}

It's not the same term as in eq (51.9), perhaps he stated previously that it's meaningless factor multiplies the epsilon, I might have forgotten it.

What do you think?

P.S
There's a preview in google books:
http://books.google.co.il/books?id=5OepxIG42B4C&printsec=frontcover&hl=iw#v=onepage&q&f=false

 

[ChrisVer]

the page is unfortunately not previewed.

 

[MathematicalPhysicist]

I can see the preview in google books of pages 316-315 does exist, I think you have a problem in your computer.

Cheers!

 

[MathematicalPhysicist]

Now I am not sure as for the solution of question 51.2) in Srednicki.

It's in here:
http://www.scribd.com/doc/87916496/Srednicki-Ms-Qft-Solutions-Rev#logout

If I am not mistaken he uses the next first order approximation:

V_Y(p,p') = V_Y(0,0) + V_Y'(0,0) p\cdot p'

But I don't understand what did he plug instead of V_Y'(0,0)

Anyone, or if you have a more lengthy detailed calculation as to how did he get eq. (51.58)

 

[nrqed]

On page 316, I am not sure how did he get equation (51.10) from eq. (51.9)?

Here's what I have done :
eqaution (51.10) is the same as the next equation:

(I am writing it down without a slash, since I am not sure how to use it here)

\frac{(-p^{\mu}\gamma_{\mu}+m)}{(p^{\mu}\gamma_{\mu}+m-i\epsilon)(-p^{\mu}\gamma_{\mu}+m)}+\int \frac{...}{p^2+s-\epsilon^2-2i\epsilon\sqrt{s}}

So as you can see it's a bit different than (51.9), the denominator in the integral, the epsilon doesn't have as a factor, the coefficient, i .
And the first term after rearranging it we should have:

\frac{ -p^{\mu}\gamma_{\mu} + m}{p^2+m^2-i\epsilon \cdot (-p^{\mu}\gamma_{\mu} +m)}

It's not the same term as in eq (51.9), perhaps he stated previously that it's meaningless factor multiplies the epsilon, I might have forgotten it.

What do you think?

The key point is that if you multiply any positive quantity by epsilon, you may still write the result as epsilon since it is taken much smaller than any other quantity. \sqrt{s} is positive so we may write \sqrt{s} \epsilon \simeq \sqrt{s}. Likewise for the other term, multiplying or dividing epsilon by p^2 + m^2 does not matter.

Hope this helps.

 

[MathematicalPhysicist]

Well, if you can write \sqrt{s} \epsilon \approx \sqrt{s} then presumably you can divide by $\sqrt{s}$ both sides (besides when s=0 which we have equality), and thus you'll get: \epsilon \approx 1 but we've taken it to be approaching zero from above, so it should be \approx 0.

Best regards.

 

[nrqed]

Well, if you can write \sqrt{s} \epsilon \approx \sqrt{s} then presumably you can divide by $\sqrt{s}$ both sides (besides when s=0 which we have equality), and thus you'll get: \epsilon \approx 1 but we've taken it to be approaching zero from above, so it should be \approx 0.

Best regards.

Sorry, I had a typo in the equation. As I wrote in words, I meant that \sqrt{s} \epsilon \approx \epsilon

 

[MathematicalPhysicist]

ok, now it makes sense.
Thanks!

 

[MathematicalPhysicist]

Another question from Srednicki's.

 

[MathematicalPhysicist]

I have got another question from Srdnicki's book.
from chapter 54, pages 338-337.

He arrives at the next form of the lagrangian:

 

[nrqed]

What is your question? :)

 

[MathematicalPhysicist]

I scanned my question, hopefully you can watch the question, it's a pdf you just need to flip it 180 degrees, becasue I scanned it in the opposite direction.

 

[nrqed]

I scanned my question, hopefully you can watch the question, it's a pdf you just need to flip it 180 degrees, becasue I scanned it in the opposite direction.

I did not have any problem showing that Eq 54.24 = Eq 54.23. Did you try this?
Note that you can always rename dummy indices so you are free to exchange $\mu$ and $\nu$. Note also that you can always switch a pair of indices that is upstair and downstair when they are summed over. For example,
$A_\mu \partial^\mu \partial^\nu A_\nu = A^\nu \partial^\mu \partial_\nu A_\mu$
where I have switched order of the two partial derivatives

 

[MathematicalPhysicist]

Indeed, I get the equality as well, not sure why I didn't succeed in doing it before. Thanks!

 

[MathematicalPhysicist]

On page 347 of Srednicki.

First I'll highlight the equations that are at hand.

(56.25)\tilde{\Delta}^{\mu\nu} (k) = -\frac{\hat{t}^{\mu} \hat{t}^{\nu}}{k^2+(\hat{t}\cdot k)^2}+\frac{g^{\mu\nu} +\hat{t}^{\mu}\hat{t}^{\nu}-\hat{z}^{\mu}\hat{z}^{\nu}}}{k^2-i\epsilon

Now he aregues that from the next sub:
(56.26)\hat{z}^{\mu} \rightarrow \frac{(\hat{t}\cdot k ) \hat{t}^{\mu}}{[k^2+(\hat{t}\cdot k)^2]^{0.5}}

Then equation (56.25) becomes:
\tilde{\Delta}^{\mu\nu} (k) =\frac{1}{k^2-i\epsilon}[g^{\mu\nu}+(-\frac{k^2}{k^2+(\hat{t}k)^2} +1-\frac{(\hat{t}\cdot k )^2}{k^2+(\hat{t}\cdot k)^2})\hat{t}^{\mu}\hat{t}^{\nu}]

But I don't get this result.
BTW, I must say a few things about the expressions (though I believe that most of you have the books (electronic copy or otherwise).

(56.19)\hat{t}^{\mu} = (1,\vec{0})
where $z$ is a unit vector in the direction of \vec{k}.

Well, I get after some algebraic manipulations:
that \Delta=\frac{i\epsilon\hat{t}^{\mu}\hat{t}^{\nu}+(k^2+(\hat{t}\cdot k)^2)g^{\mu \nu} }{(k^2+(\hat{t}\cdot k )^2)(k^2-i\epsilon)

In a PDF is attached what I have done.

 

[ChrisVer]

Just make the substitution... \hat{z}^\mu \hat{z}^\nu = \hat{t}^{\mu} \hat{t}^{\nu}\frac{(\hat{t}\cdot k )^2}{k^2+(\hat{t}\cdot k)^2}
So:
\frac{g^{\mu\nu} +\hat{t}^{\mu}\hat{t}^{\nu}-\hat{z}^{\mu}\hat{z}^{\nu}}{k^2-i\epsilon} =\frac{g^{\mu\nu} +\hat{t}^{\mu}\hat{t}^{\nu} (1-\frac{(\hat{t}\cdot k )^2}{k^2+(\hat{t}\cdot k)^2})}{k^2-i\epsilon}

I'll drop the g^{ab} term since it's already in the form that you want it. So there should be a small work to be done with:

\frac{1}{k^2-i\epsilon} \Big(1-\frac{(\hat{t}\cdot k )^2}{k^2+(\hat{t}\cdot k)^2} \Big) \hat{t}^{\mu}\hat{t}^{\nu} -\frac{\hat{t}^{\mu} \hat{t}^{\nu}}{k^2+(\hat{t}\cdot k)^2}

Which gives [trivially] the one you want to have... if for one moment you forget the i\epsilon in the nominator

 

[MathematicalPhysicist]

I got after some algebra that the propagator should be:

\frac{i\epsilon t^{\mu} t^{\nu}+(k^2+(k\cdot t)^2)g^{\mu\nu}}{(k^2-i\epsilon)(k^2+(k\cdot t)^2)}

Can anyone confirm or disconfirm my result?

 

[ChrisVer]

Since the prop is
\tilde{\Delta}^{\mu\nu} (k) =\frac{1}{k^2-i\epsilon}[g^{\mu\nu}+(-\frac{k^2}{k^2+(\hat{t}k)^2} +1-\frac{(\hat{t}\cdot k )^2}{k^2+(\hat{t}\cdot k)^2})\hat{t}^{\mu}\hat{t}^{\nu}]

Do you want to start manipulating it?
For example [just by looking at your input] I say that k \cdot \hat{t} is giving k_0 alone... So

\tilde{\Delta}^{\mu\nu} (k) =\frac{1}{k^2-i\epsilon}[g^{\mu\nu}+(-\frac{k^2}{k^2+k_0^2} +1-\frac{k_0^2}{k^2+k_0^2})\hat{t}^{\mu}\hat{t}^{\nu}]

or

\tilde{\Delta}^{\mu\nu} (k) =\frac{1}{k^2-i\epsilon}[g^{\mu\nu}+(1-\frac{k^2+k_0^2}{k^2+k_0^2})\hat{t}^{\mu}\hat{t}^{\nu}]=\frac{g^{\mu\nu}}{k^2-i\epsilon}

Which is practically the same as yours, except for I didn't carry the i \epsilon in the nominator.

 

[MathematicalPhysicist]

Can one of the experts let me know when can we take epsilon as zero and when put it as it is. I understand that it's negligble but I don't understand when is it legal to neglect it, anyone?

 

[ChrisVer]

Recall: Why did you put the epsilon in the propagator in the first place?
The straight answer, I guess, would be that there are no problems for neglecting it when it appears in the nominator... in contrast to neglecting it in the denominator.

 

[MathematicalPhysicist]

The big question is why can we do it, why when it appears in the numerator we can neglect it and when it appears in the denominator we cannot?

 

[ChrisVer]

Because one of the reasons that you inserted that epsilon in the first place is because \frac{1}{k^2} is problematic in the limit k^2 \rightarrow 0. In order to solve this you do a regularization by working in the complex plane rather than the real axis alone. The result is that you solve the problem and then you can take epsilon to be zero...
That is not needed when epsilon is in the nominator.

 

[MathematicalPhysicist]

In your last sentence you mean denominator, right?
Since it gets zero in the numerator.

 

[ChrisVer]

whether you have k^2+i \epsilon in the numerator or k^2 it doesn't change anything...that's what I said.
Because the k^2 \rightarrow k^2 + i \epsilon is done in the denominator to move the pole a little off the real axis, on to the complex plane.

 

[MathematicalPhysicist]

I wonder how do they do it by rigorous maths, you say it's called regularization, right?

 

[ChrisVer]

you just allow a real valued quantity to become complex..I think that's called regularization of this quantity.

 

[MathematicalPhysicist]

Another question from me.

On page 396:http://books.google.co.il/books?id=5OepxIG42B4C&printsec=frontcover#v=onepage&q&f=false

How did he get in eq (65.11) to the second line, especially how did they get the second term in the second line?

 

[MathematicalPhysicist]

Never mind, I noticed that D =x(1-x)k^2 +m^2 so that fixes the term there.

 

[MathematicalPhysicist]

In the solution of problem 66.3:
https://drive.google.com/file/d/0B0xb4crOvCgTM2x6QkhKREg0WW8/edit?pli=1

I don't understand how come Z_m/Z_2 is independent of $\xi$, we get:

Z_m/Z_2 = 1-3e^2/(8\pi^2 \epsilon) / (1-\xi e^2 /(8\pi ^2 \epsilon)

Perhaps with the first term if I expand the denominator with geometric series, but otherwise it depends on $\xi$.

 

[ChrisVer]

I don't think geometric series will help, because your numerator also has \xi dependence, it's not just a 3 but \xi +3...
Maybe the fact that the same powers of \xi appear in the numerator and the denominator can help?

 

[MathematicalPhysicist]

Well, you can eliminate the numerator dependence on #\xi#, as I did with the 1 that is outside the fraction here. But I really don't know what tell here...

 

[vanhees71]

The point is indeed that you have to expand the ratios to the order taken into account in your perturbative calculation, because numerator and denominator are accurate only to this order (in fact it's the expansion in powers of $\hbar$ that is gauge invariant not necessarily in powers of the coupling constant, but in fermionic QED that's no issue). So let's see, whether this is right for your example:

According to the solutions we have
$$Z_2=1-C \xi, \quad Z_m=1-C(3+\xi) \quad \text{with} \quad C=\frac{e^2}{8 \pi^2 \epsilon}.$$
Since the counter terms are of order $e^2$ (one-loop diagrams in QED), you have to expand also the ratio to this order, i.e., to order $C$ in my short-hand notation. Indeed this is most easily done, using the geometric series:
$$\frac{Z_m}{Z_2}=[1-C(3+\xi)]\frac{1}{1-C \xi}=[1-C(3+\xi)][1+C \xi +\mathcal{O}(C^2)]=1+C \xi -C(3+\xi)+\mathcal{O}(C^2)=1-3 C + \mathcal{O}(C^2).$$
Thus, indeed the ratio is independent of $\xi$ to the given order in the loop expansion as it must be.

 

[nrqed]

In the solution of problem 66.3:
https://drive.google.com/file/d/0B0xb4crOvCgTM2x6QkhKREg0WW8/edit?pli=1

I don't understand how come Z_m/Z_2 is independent of $\xi$, we get:

Z_m/Z_2 = 1-3e^2/(8\pi^2 \epsilon) / (1-\xi e^2 /(8\pi ^2 \epsilon)

Perhaps with the first term if I expand the denominator with geometric series, but otherwise it depends on $\xi$.

He is saying that to this order in e^2 , the ratio is independent of \xi. This is indeed true, if you work to order e^2 only, the \xi dependence drops out.

EDIT: I had not seen VanHees' reply. Sorry for duplicating his post.

 

[MathematicalPhysicist]

On page 520,http://books.google.co.il/books?id=5OepxIG42B4C&printsec=frontcover#v=onepage&q&f=false.

I am not sure I see how did they get eq. (83.13) from eq.(83.12) \mathcal{L}=-1/4 f_\pi Tr(\partial^\mu U^\dagger \partial_\mu U and from eq. (83.10) U(x)=\exp(2i\pi^a(x)T^a/f_\pi).

If I substitute U into (83.12), I believe I get: -1/4 f_\pi^2 Tr(4T^aT^a/f_\pi^2\partial^\mu\pi^a \partial_\mu\pi^a)

I don't see how did they do this expansion, anyone care to explain?

Thanks!

 

[vanhees71]

Be careful! The derivatives of the exponential do not commute with it, since $\partial_{\mu} \pi_a T^a$ does not commute with $\pi^a T^a$. I'd start to expand the exponential to the disired order, then take the derivatives and multiply out the result and then take the trace.

 

[MathematicalPhysicist]

So how does the term look like?

Is it?: -1/4 f_\pi^2 Tr(4 \partial^\mu \pi^a T^a \exp(-2i (T^a)^\dagger\pi^a/f_\pi)\partial_\mu \pi^a T^a \exp(2i \pi^a T^a/f_\pi)/f_\pi^2) how to expand it in powers of 1/f_\pi^2? I mean, the f_\pi^2 before the trace gets canceled with the 1/f_\pi^2 inside the trace.
Now, I need to expand both \exp with respect to 1/f_\pi^2, I am little bit confused with the index, a.

 

[ChrisVer]

I am little bit confused with the index, a.

You can avoid writting it...it's there to denote a product with the \pi and the T.

 

[MathematicalPhysicist]

Hi Chris, but am I right in the term that I wrote?

 

[vanhees71]

Ok, let's see, what we can do (from the signs I guess your book follows the east-coast metric, sigh. I'll do my best not to confuse anything): If you only want the leading order, i.e., the kinetic term to order $\mathcal{O}(1/f_{\pi}^2)$, you just need to expand the exponential to first oder:
$$U=\exp(2 \mathrm{i} \pi^a(x)T^a/f_\pi)=1+2 \mathrm{i} \pi^a T^1/f_{\pi} + \mathcal{O}(1/f_{\pi}^2).$$
Then you have
$$\partial_{\mu} U = 2 \mathrm{i} \partial_{\mu} \pi^a T^a/f_{\pi} +\mathcal{O}(1/f_{\pi}^2).$$
The second factor in your expression we get by hermitean conjugation, using the fact that the SU(N) generators $T^a$ are hermitean. However, we have to use another summation index:
$$\partial_{\mu} U^{\dagger}=-2 \mathrm{i} \partial_{\mu} \pi^b T^{b}/f_{\pi}+\mathcal{O}(1/f_{\pi}^2).$$
Now we can calculate the kinetic piece of the Lagrangian. For dimensional reasons it must read
$$\mathcal{L}_0=-\frac{f_{\pi}^2}{4} \mathrm{Tr} (\partial_{\mu} U^{\dagger} \partial^{\mu} U)=-\mathrm{Tr}(\partial_{\mu} \pi^b \partial^{\mu} \pi^a T^b T^a).$$
Now, in the usually used basis of your SU(N) generators, these are normalized such that
$$\mathrm{Tr}(T^b T^a)=\mathrm{Tr}(T^a T^b)=\frac{1}{2} \delta^{ab}.$$
This gives the correct term
$$\mathcal{L}_0=-\frac{1}{2} \partial_{\mu} \pi^a \partial^{\mu} \pi^b,$$
i.e., the usual normalization for real scalar (or better said pseudoscalar) fields. The minus sign is from the east-coast convention of the Minkowski pseudo-metric, i.e., $(\eta_{\mu \nu})=\mathrm{diag}(1,1,1,-1)$.

 

[MathematicalPhysicist]

Ok, I see thanks. and the second term follows from expanding yet more terms in the taylor expansion of U.