Another question from Srednicki's QFT book

[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.

 

[vanhees71]

Exactly!

 

[MathematicalPhysicist]

What is the identity for four terms in the trace, i.e Tr(T^aT^bT^cT^d I need this in order to calculate the second term in eq. (83.13)?

 

[ChrisVer]

By using consecutively that:

\tau_a \tau_b= \delta_{ab} + i \epsilon_{abc} \tau_c

you can show all Tr[T T T...T ] results.
what's out the trace doesn't see those indices : a,b,c,d

 

[ChrisVer]

If you want:
\begin{align}
t_a t_b t_c t_d = & (\delta_{ab} 1+ i \epsilon_{abk} t_k ) (\delta_{cd} 1+ i \epsilon_{cdm} t_m ) \\
=& \delta_{ab} \delta_{cd} 1+ i \delta_{ab} \epsilon_{cdm}t_m + i \epsilon_{abk} \delta_{cd} t_k - \epsilon_{abk} \epsilon_{cdm} t_k t_m \\
=& \delta_{ab} \delta_{cd} 1 + i \delta_{ab} \epsilon_{cdm}t_m + i \epsilon_{abk} \delta_{cd} t_k - \epsilon_{abk} \epsilon_{cdm} \delta_{km} 1- i \epsilon_{abk} \epsilon_{cdm} \epsilon_{kmr} t_r \\
\end{align}

Now using that Tr t_{a}=0 and also that T_a = \frac{1}{2} t_a

The trace will be: Tr (T_a T_b T_c T_d ) = \frac{2}{2^4} \delta_{ab} \delta_{cd} - \epsilon_{abk} \epsilon_{cdk} \frac{2}{2^4}= \frac{1}{8} ( \delta_{ab} \delta_{cd} - \delta_{ac} \delta_{bd} + \delta_{ad} \delta_{cb} )

Maybe this derivation needs some checking, also maybe I shouldn't have written the contraction of the two epsilons, since they are antisymmetric in the first indices, but I think you contract them with something symmetric in those indices..

 

[MathematicalPhysicist]

I have another question, on page 521, he writes down the following:
"...The result is a term in the chiral lagrangian that incroporates the leading effect of the quark masses,
(83.16) \mathcal{L}_{mass} = v^3 Tr(MU+M^\daggerU^\dagger)."

"... If we expand $\mathcal{L}_{mass}$ in inverse powers of $f_\pi$ and use $M^\dagger=M$, we find:

(83.17)\mathcal{L}_{mass} = -4(v^3/f_\pi^2)Tr(MT^aT^b)\pi^a\pi^b+\ldots

My problem is that I don't see how did he arrive at equation (83.17).
We have \mathcal{L}_{mass} = v^3 Tr(M(U+U^\dagger)) = v^3 Tr(M(\exp(2i\pi^aT^a/f_\pi)+\exp(-2i\pi^b T^b / f_\pi)))

Now I should be expanding the \exp in powers of 1/f_\pi, here's what I get:

v^3Tr[M(1+2i\pi^aT^a/f_\pi-2\pi^a\pi^bT^aT^b/f_\pi^2+\ldots + 1 - 2i\pi^aT^a/f_\pi - 2\pi^a\pi^bT^aT^b/f_\pi^2+\ldots)] = v^3Tr[M(2-4\pi^a\pi^bT^aT^b/f_\pi^2)]

The matrix $M$ has the form, <br /> \left( \begin{array}{cc}<br /> m_u &amp; 0 \\<br /> 0 &amp; m_d \\ \end{array} \right)
where m_u and m_d are the masses of up quark and down quark.
So the trace of M isn't zero, so I don't see how come the Tr(2M) doesn't appear in eq. (83.17), anyone can clear it to me?
Thanks in advance.

 

[ChrisVer]

The term having only the trace of the mass matrix is just a constant number to your lagrangian... however you wanted to write the fields...

 

[MathematicalPhysicist]

Yeah right, we disregard constant terms in the lagrangian.

 

[Hepth]

If you want:
\begin{align}
t_a t_b t_c t_d = & (\delta_{ab} 1+ i \epsilon_{abk} t_k ) (\delta_{cd} 1+ i \epsilon_{cdm} t_m ) \\
=& \delta_{ab} \delta_{cd} 1+ i \delta_{ab} \epsilon_{cdm}t_m + i \epsilon_{abk} \delta_{cd} t_k - \epsilon_{abk} \epsilon_{cdm} t_k t_m \\
=& \delta_{ab} \delta_{cd} 1 + i \delta_{ab} \epsilon_{cdm}t_m + i \epsilon_{abk} \delta_{cd} t_k - \epsilon_{abk} \epsilon_{cdm} \delta_{km} 1- i \epsilon_{abk} \epsilon_{cdm} \epsilon_{kmr} t_r \\
\end{align}

Now using that Tr t_{a}=0 and also that T_a = \frac{1}{2} t_a

The trace will be: Tr (T_a T_b T_c T_d ) = \frac{2}{2^4} \delta_{ab} \delta_{cd} - \epsilon_{abk} \epsilon_{cdk} \frac{2}{2^4}= \frac{1}{8} ( \delta_{ab} \delta_{cd} - \delta_{ac} \delta_{bd} + \delta_{ad} \delta_{cb} )

Maybe this derivation needs some checking, also maybe I shouldn't have written the contraction of the two epsilons, since they are antisymmetric in the first indices, but I think you contract them with something symmetric in those indices..

Just for reference, The full form I believe is :

$Tr[T[a,b,c,d]] = \frac{\delta _{ad} \delta _{bc}-\delta _{ac} \delta _{bd}+\delta _{ab} \delta _{cd}}{4 N}+\frac{1}{8} \left(-i f_{ade} d_{bce}+i d_{ade} f_{bce}+d_{ade} d_{bce}-d_{bde} d_{ace}+d_{cde} d_{abe}\right)$

For any SU(N).

For SU(2) (xPT with 2 flavors), the symmetric constants are $$ d_{abc} = 0$$ and the antisymmetric structure constants are $$ f_{abc} = \epsilon_{abc}$$

for SU(3) you can look them up on wikipedia.

 

[ChrisVer]

Just for reference, The full form I believe is :

Tr[T[a,b,c,d]]=δadδbc−δacδbd+δabδcd4N+18(−ifadedbce+idadefbce+dadedbce−dbdedace+dcdedabe)Tr[T[a,b,c,d]] = \frac{\delta _{ad} \delta _{bc}-\delta _{ac} \delta _{bd}+\delta _{ab} \delta _{cd}}{4 N}+\frac{1}{8} \left(-i f_{ade} d_{bce}+i d_{ade} f_{bce}+d_{ade} d_{bce}-d_{bde} d_{ace}+d_{cde} d_{abe}\right)

For any SU(N).

If you find the derivation, let me know...
Yes I should have said that what I wrote applies only or SU[2] but 1. that's what Srednicki's using in that context, 2. I mentioned pauli matrices . For SU(N) it becomes more complicated, and I think I have seen the SU(3). I don't remember where though, neither the form.

 

[MathematicalPhysicist]

On page 547 of Srednicki's in problem 87.1 he asks us to find the generator #Q# of the unbroken #U(1)# subgroup as a linear combination of the #T^a#s and #Y#.

Now in the solution on page 141, https://drive.google.com/file/d/0B0xb4crOvCgTM2x6QkhKREg0WW8/edit , he writes to see eq (88.15) and eq(88.16) which are previewed in google books on page 550:
http://books.google.co.il/books?id=5OepxIG42B4C&printsec=frontcover#v=onepage&q&f=false

Isn't the solution simply, #Q=T^3+Y#, if it is then why do I need to look at eq. (88.15)?