Another question from Srednicki's QFT book

[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.

 

[MathematicalPhysicist]

On page 580, eq. (92.22) of Srednicki's:
https://books.google.co.il/books?id=5OepxIG42B4C&printsec=frontcover#v=onepage&q&f=false

I don't understand why a limit of $r\rightarrow \infty$ depends on the last rhs on $r$, after taking a limit on $A(r,\phi)$, it shouldn't depend on r after taking the limit with respect to r.

 

[ChrisVer]

Then what would happen to the derivative's dimension?

 

[MathematicalPhysicist]

I don't follow. If we have a function $f(r)$, we can't have $\lim_{r\rightarrow \infty} f(r)=g(r)$, the result of the limit should be independent on $r$.

 

[ChrisVer]

I mean you have the derivative \nabla in there and your result must have a dimension [L]^-1... However it's better to look for some better answer.

 

[MathematicalPhysicist]

Maybe this equality is valid if we take both sides of the equality as distributions which their equality is valid under taking an integral on both sides, the question is integration with respect to which parameter?

 

[vanhees71]

So here's what's in Srednicki. He's looking at a global U(1) QFT with spontaneous symmetry breaking in 1+2 dimensions. Translating into the usual west-coast convention the Lagrangian reads
$$\mathcal{L}=(\partial_{\mu} \phi)^{*} (\partial_{\mu} \phi)-\frac{\lambda}{4} (\phi^* \phi-v^2)^2.$$
Now he introduces polar coordinates $(r,\phi)$ (note the difference between the field $\varphi$ and the polar angle $\phi$; Srednicki has sometimes a bit unfortunate notation). Now he looks for vacuum solutions (i.e., stationary solutions of the classical field equations) with winding number $n$,
$$\varphi(r,\phi)=v f(r) \exp[\mathrm{i} n \phi(r)].$$
Now to avoid the confusion, we can easily write everything a bit more carefully. What he assumes is that
$$f(r) \simeq 1 \quad \text{for} \quad r \rightarrow \infty, \quad f(0)=0.$$
The latter assumption is in order to make the solution well-behaved in the coordinate singularity $r=0$.

Now he's calculating the field energy and shows that the gradient term diverges. With the ansatz above you get
$$\vec{\nabla} \varphi=v \left [f'(r) \vec{e}_r + \frac{f(r)}{r} \phi'(r) \vec{e}_{\phi} \right ]\exp[\mathrm{i} n \phi(r)].$$
Since $f(r) \simeq 1$ for $r \rightarrow \infty$ the total field energy diverges like $1/r$ for $r \rightarrow \infty$. In (92.16) is an obvious typo. The integrand goes like $1/r^2$, as written correctly in (92.15).

 

[MathematicalPhysicist]

So here's what's in Srednicki. He's looking at a global U(1) QFT with spontaneous symmetry breaking in 1+2 dimensions. Translating into the usual west-coast convention the Lagrangian reads
$$\mathcal{L}=(\partial_{\mu} \phi)^{*} (\partial_{\mu} \phi)-\frac{\lambda}{4} (\phi^* \phi-v^2)^2.$$
Now he introduces polar coordinates $(r,\phi)$ (note the difference between the field $\varphi$ and the polar angle $\phi$; Srednicki has sometimes a bit unfortunate notation). Now he looks for vacuum solutions (i.e., stationary solutions of the classical field equations) with winding number $n$,
$$\varphi(r,\phi)=v f(r) \exp[\mathrm{i} n \phi(r)].$$
Now to avoid the confusion, we can easily write everything a bit more carefully. What he assumes is that
$$f(r) \simeq 1 \quad \text{for} \quad r \rightarrow \infty, \quad f(0)=0.$$
The latter assumption is in order to make the solution well-behaved in the coordinate singularity $r=0$.

Now he's calculating the field energy and shows that the gradient term diverges. With the ansatz above you get
$$\vec{\nabla} \varphi=v \left [f'(r) \vec{e}_r + \frac{f(r)}{r} \phi'(r) \vec{e}_{\phi} \right ]\exp[\mathrm{i} n \phi(r)].$$
Since $f(r) \simeq 1$ for $r \rightarrow \infty$ the total field energy diverges like $1/r$ for $r \rightarrow \infty$. In (92.16) is an obvious typo. The integrand goes like $1/r^2$, as written correctly in (92.15).

I still don't understand eq. (92.22) I mean the $r$ appears on the rhs after taking a limit on the lhs of $r\rightarrow \infty$, why does it depend on $r$?

 

[vanhees71]

This is simply bad notation. What Srednicki want's to say is that for $r \rightarrow \infty$ the function behaves in a certain way asymptotically. As I said, Srednicki sometimes has a bit strange notation ;-).

 

[MathematicalPhysicist]

Thanks.

 

[MathematicalPhysicist]

My next question is about the answer to question 95.3 on page 161:
https://drive.google.com/file/d/0B0xb4crOvCgTM2x6QkhKREg0WW8/edit

How did he find the solution to $C$ from the equation: $2\kappa C^\dagger (C^2-v^2) + m^2C=0$?

I mean we don't know if $C$ is unitary or something else? all we know is that $C$ is a chiral superfield.

I am asking also, because when I plug this solution to the RHS of $\partial W/ \partial A = \kappa(C^2-v^2)$, I get: $\partial W/\partial A = -1/2 m^2/\kappa$, and not as it's stated in the solution: $\partial W/\partial A =-1/2 m^2$.

 

[ChrisVer]

\kappa^2 must be a typo... It's not even solving the algebraic equation...The first term gives something proportional to $\kappa^{-2}$ and the second term gives $\kappa^{-1}$ (so they wouldn't cancel out).

The solution is taken straightforwardly from the expression you've written. First you would try to put $C^2 =v^2$, this would cancel the term in the parenthesis... At next step, you'd want to cancel out the $m^2$ term... So as a next step you would write:
C= \sqrt{v^2 + a} and for now same for C^\dagger
Inserting above you have:
2 \kappa \sqrt{v^2 +a} ~(v^2+a-v^2) +m^2 \sqrt{v^2+a}=0
(2 \kappa a + m^2 ) \sqrt{v^2+a}=0
a= -\frac{m^2}{2\kappa}

So C= \sqrt{v^2 - \frac{m^2}{2 \kappa}}

Adding an overall - sign wouldn't change the result.

You didn't use anywhere the unitarity or not (it doesn't have to be unitary at all!)
In its components it's written: C= \phi + \psi \theta + F \theta^2.

 

[MathematicalPhysicist]

@ChrisVer , it doesn't use unitarity but you did use hermiticity of C, since $C^\dagger = C$ in what you wrote.

 

[MathematicalPhysicist]

If you continue to exercise 95.3b, on the same page in the last link above, on page 162 they write: $<X>=(4\kappa^2v^2/m^2-1)^{1/2}<A>$.

But I get something else, $<X> = \cos(\theta)<A>-\sin(\theta)<B> = \cos \theta <A>+2\sin \theta \kappa <C><A>/m = <A>/\sqrt{1+4\kappa^2<C>^2/m^2} + (2\kappa /m)<C>^2 <A>/\sqrt{1+4\kappa^2<C>^2/m^2}$

Now if I plug $<C>=(v^2-m^2/(2\kappa))^{1/2}$ above I don't get what they wrote.

Can someone help me with this seemingly dubious math?

Thanks in advance.

 

[ChrisVer]

but you did use hermiticity of C,

No I didn't use hermitianity. I just wrote down a solution... It happens to give you C^\dagger =C.

Then:
&lt;X&gt; = c \theta &lt;A&gt; - s \theta &lt;B&gt;= \Big( c \theta +\frac{2 \kappa &lt;C&gt;}{m} s \theta \Big) &lt;A&gt;

The thing in the big parenthesis is:
c \theta \Big(1+ \frac{2 \kappa }{m} &lt;C&gt; \tan \theta \Big)
c \theta \Big(1+ \frac{4 \kappa^2 }{m^2} &lt;C&gt;^2 \Big)

Using that $\cos \tan^{-1} x = (x^2+1)^{-1/2}$ you get:\frac{1+ \frac{4 \kappa^2 }{m^2} &lt;C&gt;^2}{\sqrt{\frac{4 \kappa^2}{m^2} &lt;C&gt;^2 +1}}

=\sqrt{1+\frac{4 \kappa^2}{m^2} &lt;C&gt;^2}

Then put $C^2 = v^2 -\frac{m^2}{2 \kappa}$

you obtain:

\sqrt{1 + \frac{4 \kappa^2 v^2}{m^2} - \frac{4 \kappa^2 m^2}{m^2 2 \kappa} }

\sqrt{1-2 + \frac{4 \kappa^2 v^2}{m^2}}
\sqrt{\frac{4 \kappa^2 v^2}{m^2}-1}

 

[ChrisVer]

you have written:

A \frac{1}{\sqrt{1+a^2C^2}} + A \frac{aC^2}{\sqrt{1+a^2C^2}}

For which something looks wrong - you won't get the result $\sqrt{1+ a^2C^2} A$ as an alternative form of $X$. I guess your $a \equiv (2 \kappa/m)$ should be squared in the 2nd term of your long equation.
Try this in wolfram:
cos(a)*H-sin(a)*D where a=arctan(p) and D=-p*H

 

[MathematicalPhysicist]

No I didn't use hermitianity. I just wrote down a solution... It happens to give you C^\dagger =C.

Then:
&lt;X&gt; = c \theta &lt;A&gt; - s \theta &lt;B&gt;= \Big( c \theta +\frac{2 \kappa &lt;C&gt;}{m} s \theta \Big) &lt;A&gt;

The thing in the big parenthesis is:
c \theta \Big(1+ \frac{2 \kappa }{m} &lt;C&gt; \tan \theta \Big)
c \theta \Big(1+ \frac{4 \kappa^2 }{m^2} &lt;C&gt;^2 \Big)

Using that $\cos \tan^{-1} x = (x^2+1)^{-1/2}$ you get:\frac{1+ \frac{4 \kappa^2 }{m^2} &lt;C&gt;^2}{\sqrt{\frac{4 \kappa^2}{m^2} &lt;C&gt;^2 +1}}

=\sqrt{1+\frac{4 \kappa^2}{m^2} &lt;C&gt;^2}

Then put $C^2 = v^2 -\frac{m^2}{2 \kappa}$

you obtain:

\sqrt{1 + \frac{4 \kappa^2 v^2}{m^2} - \frac{4 \kappa^2 m^2}{m^2 2 \kappa} }

\sqrt{1-2 + \frac{4 \kappa^2 v^2}{m^2}}
\sqrt{\frac{4 \kappa^2 v^2}{m^2}-1}

Well, for the solution for $C$ you wrote the ansatz $C=\sqrt{v^2+a}$, so you suppose that $v$ and $a$ are real.

As for your second solution to b), you get that $\sqrt{1 + \frac{4 \kappa^2 v^2}{m^2} - \frac{4 \kappa^2 m^2}{m^2 2 \kappa} } \rightarrow \sqrt{1-2\kappa+ \frac{4 \kappa^2 v^2}{m^2}}$.

 

[ChrisVer]

Well, for the solution for

CC you wrote the ansatz C=v2+a−−−−−√C=\sqrt{v^2+a}, so you suppose that vv and aa are real.

Well you can try putting an imaginary part in you $<C>$. However this can get messed up for no reason.
Try replacing $<C>$ with $a+ib$, and you will have:

(a-ib) (a^2-b^2+2iab- v^2)= -\frac{m^2}{2\kappa} (a+ib)

a^3 -ab^2 + 2ia^2b - av^2 - ia^2b +i b^3 +2ab^2 +ibv^2 =-\frac{m^2}{2\kappa} (a+ib)

a^3 -ab^2 - av^2 +2ab^2 +ibv^2 + ia^2b +i b^3 =-\frac{m^2}{2\kappa} (a+ib)

(a^3 -ab^2 - av^2 +2ab^2) +i (bv^2 + a^2b + b^3) =-\frac{m^2}{2\kappa} (a+ib)

equating real/im parts:

a^3 -ab^2 - av^2 +2ab^2 = - \frac{m^2}{2 \kappa}a
bv^2 + a^2b + b^3 = - \frac{m^2}{2 \kappa}b

Or

a\Big(a^2 -b^2 - v^2 +2b^2+\frac{m^2}{2 \kappa}\Big)=0
b\Big(v^2 + a^2 + b^2 + \frac{m^2}{2 \kappa}\Big)=0

The second equation can only be true if the $b=0$,because the thing in the parenthesis is positive.
The $a$ is then solved by $a= \pm \sqrt{v^2 - (m^2/2\kappa)}$ (the solution I gave above) or $a=0$ (that would be a trivial solution), so you get 3 solutions for a 3rd order equation (also a check).

However you should be more careful with what you write. When you take the scalars $F_i^*=-\frac{\partial W(\Phi_1,...,\Phi_n)}{\partial \Phi_i}$, you are not supposed to get superchiral fields as a result. Instead you take the derivative of the superpotential wrt to a chiral superfield and then replace with the scalar components of the superfields [if any is left]. That's also why you can write $C=a+ib$ in this case (since $C$ now is just the $c$ in the multiplet $C=\begin{pmatrix}c \\ \psi_c \\ F_c \end{pmatrix}$. In some textbooks for Supersymmetry they would aslo write:
$F_i^*= -\frac{\partial W(\Phi_1,...,\Phi_n)}{\partial \phi_i}$ which does the same thing although it can be a little confusing.

 

[ChrisVer]

As for your second solution to b), you get that 1+4κ2v2m2−4κ2m2m22κ−−−−−−−−−−−−−√→1−2κ+4κ2v2m2−−−−−−−−−−−√\sqrt{1 + \frac{4 \kappa^2 v^2}{m^2} - \frac{4 \kappa^2 m^2}{m^2 2 \kappa} } \rightarrow \sqrt{1-2\kappa+ \frac{4 \kappa^2 v^2}{m^2}} .

Oops yes you are right (I was influenced by the C² solution that was written, although a typo in my mind it was still going as κ^-2). That's the right solution: $\sqrt{1-2\kappa+,..}$, as long as what is given is correct. I think the author of the solutions carried his wrong result for $C$ together with him, and that's why I got the same result as he did in the end.