Genereral:Questions about Srednicki's QFT

[The_Duck]

For question one, just insert the small-epsilon approximations for the gamma function and the integrand, multiply everything out to get a bunch of terms, drop terms proportional to epsilon (since they go to zero) and then collect the remaining terms.

For question two, you're going to want to play with logs using some manipulation like ln(mu^2/D) = -ln(D/m^2) + 2*ln(mu/m)

 

[Lapidus]

Thanks for answering, The Duck.

For question one, just insert the small-epsilon approximations for the gamma function and the integrand, multiply everything out to get a bunch of terms, drop terms proportional to epsilon (since they go to zero) and then collect the remaining terms.

What you mean with insert the small-epsilon approximations for the gamma function? How and where can I insert it? Is my first step given in the last post correct?

For question two, you're going to want to play with logs using some manipulation like ln(mu^2/D) = -ln(D/m^2) + 2*ln(mu/m)

Ahhh! But why are the two ln(mu/m) not in the integrand anymore?

 

[Avodyne]

Is my first step given in the last post correct?

Yes, and that's what The Duck meant.

But why are the two ln(mu/m) not in the integrand anymore?

Because ln(mu/m) is now just a constant times D, and D has been integrated.

 

[Lapidus]

Got it! Thank you

I hate to test more of your patience, but still two minor quibbles in chapter 14.

In 14.40, the -ln m^2 from the integrand in 14.39 is lumped into the 'linear in k^2 and m^2' part in 14.40, right?

How does Pi(-m^2) vanish via 14.41 and 14.42? I assume the two terms in 14.41 neutralize each other. How can I see this?

Now to the excellent https://www.physicsforums.com/showpost.php?p=2725516&postcount=35" by RedX, where he adresses 14.8. Can it be said that 14.7 corresponds to mass renormalization and 14.8 to field renormalization? I remember reading that somewhere.

again I would be thankful for any answers

 

[The_Duck]

How does Pi(-m^2) vanish via 14.41 and 14.42? I assume the two terms in 14.41 neutralize each other. How can I see this?

When you plug in k^2 = -m^2, then D = D0, so the log vanishes (ln 1 = 0). Furthermore k^2+m^2 = 0, so the term linear in (k^2 + m^2) also vanishes.

 

[emz]

A simple, new question on Srednicki's book:
Equation 2.26 :
U($$\Lambda$$)$$^{-1}$$ $$\varphi$$(x)U($$\Lambda$$)= $$\varphi$$($$\Lambda^$$$$^{-1}$$x)

which describes how a scalar field transforms under Lorentz transformation is not derived in the book. Instead it seems to be inspired by time-translation equation (2.24).
Anyone can point me to a proof ?

 

[Avodyne]

A simple, new question on Srednicki's book:
Equation 2.26 :
U($$\Lambda$$)$$^{-1}$$ $$\varphi$$(x)U($$\Lambda$$)= $$\varphi$$($$\Lambda^$$$$^{-1}$$x)

which describes how a scalar field transforms under Lorentz transformation is not derived in the book. Instead it seems to be inspired by time-translation equation (2.24).
Anyone can point me to a proof ?

I believe this is the definition of a scalar field.

 

[RedX]

A simple, new question on Srednicki's book:
Equation 2.26 :
U($$\Lambda$$)$$^{-1}$$ $$\varphi$$(x)U($$\Lambda$$)= $$\varphi$$($$\Lambda^$$$$^{-1}$$x)

which describes how a scalar field transforms under Lorentz transformation is not derived in the book. Instead it seems to be inspired by time-translation equation (2.24).
Anyone can point me to a proof ?

That has always been a point of confusion for me. Forgetting about operators, a solution to the KG equation can be the c-number:

$$\phi(x)=ae^{ikx}$$

This shouldn't change under Lorentz transform, so I think what happens is that if you change x to x', k changes to k' such that k'x'=kx. The coefficient 'a' just stays the same.

Now take a superposition of plane waves:

$$\phi(x)=\int d^3 \tilde{k} a(k)e^{ikx}$$

How does this behave under Lorentz transform? Well isn't it the same thing:

$$\phi(x')=\int d^3 \tilde{k} a(k)e^{ik'x'}$$

?

But this isn't the same as:

$$\phi(x')=\int d^3 \tilde{k} a(k')e^{ik'x'}$$

right?

Anyways, what if you look at it differently. What if under Lorentz transform of$$\phi(x)=\int d^3 \tilde{k} a(k)e^{ikx}$$

only the x is changed?

$$\phi(x')=\int d^3 \tilde{k} a(k)e^{ikx'}= \int d^3 \tilde{k} a(k)e^{ik\Lambda x}= \int d^3 \tilde{k} a(k)e^{i(\Lambda^{-1}k)x} =\int d^3 \tilde{k'} a(\Lambda k')e^{ik'x}$$

But is this equal to:$$\phi(x)=\int d^3 \tilde{k} a(k)e^{ikx}$$

? Anyways, I don't know. I think I confused myself.

 

[emz]

I believe this is the definition of a scalar field.

I thought the definition of a scalar field was the numerical value of the field at a given point to be Lorentz invariant. That is
U($$\Lambda$$) $$\varphi$$($$\Lambda$$x)=$$\varphi$$(x)

 

[The_Duck]

No. Think about a temperature field T(x) (a scalar) under rotations in Euclidean space. The rotated version of $$T(\vec{x})$$ is $$T(R^{-1}\vec{x})$$ where R is the rotation matrix that implements the rotation on 3-vectors. Contrast, say, the electric field, a vector, where if you rotate $$\vec{E}(\vec{x})$$ you get $$R \vec{E}(R^{-1}\vec{x})$$. A scalar transforms in the simplest possible way: more complicated fields have different components that, like the components of the electric field, rotate into each other under rotations.

Also, while states in QM transform with one transformation operator: $$| \psi \rangle \to U(\Lambda) | \psi \rangle$$, operators transform with two transformation operators: $$\phi(x) \to U(\Lambda)^{-1} \phi(x) U(\Lambda)$$

 

[RedX]

Okay, I got a question. Consider just the annihilation part of the field operator

$$\phi_+(x)=\int_R d^3 \tilde{k} a(k)e^{ikx}$$

where R is a region in momentum space. Srednicki takes R to be the entire R³, but here I'm taking it to be a connected subset of R³.

Now consider a Lorentz transform of this operator:

$$U^{-1}\phi_+(x)U=\int_R d^3 \tilde{k} U^{-1}a(k)Ue^{ikx} =\int_R d^3 \tilde{k} a(\Lambda^{-1}k)e^{ikx} =\int_{\Lambda^{-1}R} d^3 \tilde{k} a(k)e^{i(\Lambda k)x} =\int_{\Lambda^{-1}R} d^3 \tilde{k} a(k)e^{ik(\Lambda^{-1}x)}$$

But is $$U^{-1}\phi_+(x)U=\int_{\Lambda^{-1}R} d^3 \tilde{k} a(k)e^{ik(\Lambda^{-1}x)}$$ really equal to: $$\int_{R} d^3 \tilde{k} a(k)e^{ik(\Lambda^{-1}x)}=\phi(\Lambda^{-1}x)$$ ?

The integration volumes are different. So for this Lorentz transform to work, does it rely on the fact that the volume is over the entire R³? That's weird.

addendum: o okay I got it, but I won't erase my posts in case anyone else got confused like me. when you integrate over all of momentum space, then you don't have to specify a special momentum k or a special region of momentum k. therefore the final result can only depend on the transformation of the coordinate x as there aren't any special 4-vectors to contract with x. but if your wavefunction is over a special region or a special value of the momentum, you have to specify k and contract it with x. therefore k and x transform. so really you should label the wavefunction $$\phi(x,k)$$. in other words, the last integral I have is dependent on k through the region of the integration (and not the dummy indices). this has to transform to $$\Lambda^{-1}R$$, making the last two integrals equal:

$$\phi(\Lambda^{-1}x)\neq\int_{R} d^3 \tilde{k} a(k)e^{ik(\Lambda^{-1}x)}$$$$\phi(\Lambda^{-1}x)=\int_{R'} d^3 \tilde{k} a(k)e^{ikx'} =\int_{\Lambda^{-1}R} d^3 \tilde{k} a(k)e^{ik(\Lambda^{-1}x)}$$

 

[Lapidus]

Back to chapter 14.

Concerning the fixing of the two purely numerical constants, the two $$\kappa$$ by imposing the two conditions 14.7 and 14.8.

How do we get 14.43? (Srednicki says it is straightforward, I once again does not see it. Where does the 1/12 come from? When I differentiate 14.41 wrt k^2, the k^2 disappears..)

And what are now the two numerical constants that we wanted to fix?

thanks

 

[Lapidus]

I admit probably not the most sophisticated question ever asked on PF, but could nevertheless someone give me a hint...

thank you

 

[Avodyne]

When I differentiate 14.41 wrt k^2, the k^2 disappears..

Then you made a mistake. Remember that D depends on k^2.

 

[Lapidus]

Then you made a mistake. Remember that D depends on k^2.

I have to take a derivative of an ln term in an integral?? And Srednicki calls that straightforward? Sorry, I still can't see...

And what are now the two numerical constants??

thank you

 

[Avodyne]

First of all, if you set $k^2=-m^2$ in eq.(14.39), the result is supposed to be zero, so this gives you $-{1\over6}\kappa_A+\kappa_B$ as an integral over $x$ of a messy function of $x$. Next you need to differentiate eq.(14.39) with respect to $k^2$, and then set $k^2=-m^2$; once again the result is supposed to be zero. To take this derivative, you need to differentiate $D\ln(D/m^2)$ under the integral over $x$. This gives you $\kappa_A$ as another integral of another messy function of $x$.