Another question from Srednicki's QFT book

[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, $$\left( \begin{array}{cc} m_u & 0 \\ 0 & m_d \\ \end{array} \right)$$
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)?

 

[ChrisVer]

because in 88.15 you can see why the solution is $Q=T_3+Y$

 

[MathematicalPhysicist]

Because it's the coefficient of $$A_\mu$$ the electromagnetic vector field.

 

[ChrisVer]

yup..

p.s. for the same line latex, you should use double # instead of double $.

 

[MathematicalPhysicist]

There's no preview this time.
On page 563 Srednicki writes that " We determine the covariant derivative of each field by requiring it to transform in the same way as the field itself; for example, $D_\mu U \rightarrow L(D_\mu U)R^\dagger$.
We thus find that $D_\mu U = \partial_\mu U - il_\mu U+iUr_\mu$"

where $l_\mu \rightarrow L l_\mu L^\dagger + i L\partial_\mu L^\dagger$
$r_\mu \rightarrow Rr_\mu R^\dagger+iR\partial_\mu R^\dagger$
$U\rightarrow LUR^\dagger$

How do I get the above equation: $D_\mu U = \partial_\mu U - il_\mu U+iUr_\mu$?

 

[ChrisVer]

Well I guess that's the only thing that could work out when you want to make that transformation. Inserting the l and r transfos, you can find:
$(D_\mu U)^\prime = L \partial_\mu U R^\dagger - i (L l_\mu L^\dagger + i L \partial_\mu L^\dagger ) L U R^\dagger +i L U R^\dagger (R r_\mu R^\dagger + i R \partial_\mu R^\dagger )$

Resuting:

$(D_\mu U)^\prime = L ( \partial_\mu U- i l_\mu U + i U r_\mu ) R^\dagger = L (D_\mu U) R^\dagger$

 

[MathematicalPhysicist]

But I get: $L\partial_\mu U R^\dagger - i(Ll_\mu L^\dagger+iL\partial_\mu L^\dagger)LUR^\dagger+iLUR^\dagger (Rr_\mu R^\dagger + iR\partial_\mu R^\dagger)= L\partial_\mu U R^\dagger - iLl_\mu UR^\dagger -iLl_\mu UR^\dagger + L\partial_\mu L^\dagger L UR^\dagger + i LUr_\mu R^\dagger - LU\partial_\mu R^\dagger$

where I have used that $R,L$ are unitary matrices. I can use another contraction, i.e $L\partial_\mu L^\dagger LUR^\dagger = - \partial_\mu L U R^\dagger$, but I don't see how to get the last line you wrote in the first left equality.

 

[ChrisVer]

The derivative is not supposed to act on the L,R , since these are the transformations on the left/right components. So something like $L \partial_\mu L^\dagger L U R^\dagger = L \partial_\mu U R^\dagger$

When you take out as common factors from the left and from the right the L and R*:

$= L ( \partial_\mu U - i l_\mu U + i U r_\mu ) R^\dagger$

 

[welcomeblack]

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?

This thread's a bit old but I'd like to answer this question for people in the future. You're doing fixed-order perturbation theory at order $\alpha$, which means you drop terms of order $\alpha^2$ You can recognize $d\alpha / dln \mu$ as the beta function, which starts at order $\alpha^2$, so it gets dropped from the final expression.

 

[MathematicalPhysicist]

There's another derivation I don't get from the book.

On pages 563-564, he writes down eq. (90.20)$eA_\mu = l_\mu^3+r_\mu^3+1/2b_\mu$
and he says that this equation follows from reconciling eqs (90.9) and 90.10) with the requirement that the electromagnetic covariant derivatives of the proton field $p=\mathcal{N}_1$ and the neutron field $n=\mathcal{N}_2$ be given by $(\partial_\mu -ieA_\mu)p$ and $\partial_\mu n$.

where eq. (90.9) $D_\mu N_L = (\partial_\mu - il_\mu)N_L$
eq. (90.10) $D_\mu N_R = (\partial_\mu -ir_\mu) N_R$
where $N_L \equiv P_L N = 1/2(1-\gamma_5)N , \ N_R \equiv P_R N = 1/2(1+\gamma_5)N$, where $N$ is the neucleon field.
I don't see how did they get eq. (90.20).

 

[ChrisVer]

Hmm I think you have to look at the Lagrangian for this case.
I am not sure if this is going to work out, but I'd recommend you look at the terms:
$\bar{N}\Big[ (i \partial + v + \frac{1}{2} \tilde{l} +\frac{1}{2} \tilde{r})_\mu \gamma^\mu - g_A (\alpha +\frac{1}{2} \tilde{l}- \frac{1}{2} \tilde{r} )_\mu \gamma^\mu \gamma^5 \Big] N$
and try to make the substitution for the new defined $l,r$... At the end you will be able to see which term looks like the needed electromagnetic field... for example the $r^a_\mu T^a$ for $a=1,2$ will somehow mix the neutrons with protons, and only the $a=3$ part will be diagonal... eg I think that's why in the $eA$ it only has the $r^3, l^3$.