Solving the Callan-Symanzik Equation

[fzero]

Awesome!

One small thing that's still confusing me regarding QFT is the following:

Consider equation 5.46 in these notes:
damtp.cam.ac.uk/user/tong/qft/qft.pdf

(i) why does only one term have the $i \epsilon$ in it? I thought this was because the $(p-p')^2-\mu^2$ term can never vanish but this doesn't make sense since the $\phi$ particle will have momentum $p-p'$ and since momentum squares to mass squared $\mu^2$, surely we will actually have $(p-p')^2-\mu^2=0$ always?

He explains this on p. 65.

(ii)why is one term plus and the other minus? he talks earlier about how we pick up extra minus signs for fermionic diagrams from statistics but doesn't explicitly tell us how to recognise when they are needed (well he does in the calculation on p120 but quite a common exam question is to write down the amplitude from feynman rules - you aren't going to have time to reproduce p120 every time you need to check a minus sign!)
looking at diagrams 25 and 26 it seems to be that whenever two fermionic legs cross we get a minus sign (this is why one of the terms in figure 25 gets a minus sign but none in figure 26 do since the external legs are bosonic there) but then in (5.46) which corresponds to figure 27, why does that s channel diagram get a minus sign? No external legs get crossed there? There must be an easy way of recognising when we need the minus sign?

I don't know of any easier way apart from working through all of the contractions at least once and then trying to memorize the results.

(iii) Is it possible to show that your definition $$<br /> \gamma_\phi = - \mu \frac{d(\log \phi)}{d\mu}.<br />$$ is equivalent to the one in my notes $\gamma_\phi = \frac{\mu}{\sqrt{Z}} \frac{d \sqrt{Z}}{d \mu} \frac{d }{d \sqrt{Z}}$?

Thanks.

$$\gamma_\phi$$ is not a differential operator, so

$\gamma_\phi = \frac{\mu}{\sqrt{Z}} \frac{d \sqrt{Z}}{d \mu} \frac{d }{d \sqrt{Z}}$

makes no sense. In post #14 I showed that

$$\gamma_\phi = \frac{\mu}{\sqrt{Z}} \frac{d\sqrt{Z}}{d\mu} .$$

follows from the definition.

 

[latentcorpse]

He explains this on p. 65.
I don't know of any easier way apart from working through all of the contractions at least once and then trying to memorize the results.
$$\gamma_\phi$$ is not a differential operator, so

$\gamma_\phi = \frac{\mu}{\sqrt{Z}} \frac{d \sqrt{Z}}{d \mu} \frac{d }{d \sqrt{Z}}$

makes no sense. In post #14 I showed that

$$\gamma_\phi = \frac{\mu}{\sqrt{Z}} \frac{d\sqrt{Z}}{d\mu} .$$

follows from the definition.

Take for example (3.58). I'm struggling to show that those denominators never vanish. We want to take $(p_1-p_1')=(E_{p_1}-E_{p_1'}, \vec{p_1}-\vec{p_1'})$ and square it
$(p_1-p_1')^2=(E_{p_1}-E_{p_1'})^2-( \vec{p_1}-\vec{p_1'})^2$ and show this is bigger than M

But this just turns out to be a horrible mess...Also, in those QFT notes, on page 90 he says that the spinor field also satisfies the Klein Gordon equation (as shown in 4.54). Doesn't this mean that it describes fermions in the dirac equation as well as boson in the KG equation? How can it describe both fermions and bosons?

 

[fzero]

Take for example (3.58). I'm struggling to show that those denominators never vanish. We want to take $(p_1-p_1')=(E_{p_1}-E_{p_1'}, \vec{p_1}-\vec{p_1'})$ and square it
$(p_1-p_1')^2=(E_{p_1}-E_{p_1'})^2-( \vec{p_1}-\vec{p_1'})^2$ and show this is bigger than M

But this just turns out to be a horrible mess...

It's enough to show that $$(p_1-p_1')^2<0$$, which is simple to do in center of mass frame.

Also, in those QFT notes, on page 90 he says that the spinor field also satisfies the Klein Gordon equation (as shown in 4.54). Doesn't this mean that it describes fermions in the dirac equation as well as boson in the KG equation? How can it describe both fermions and bosons?

The KG equation is just the mass shell condition $$p^2=m^2$$. It must be satisfied by any field that is a solution to the appropriate equation of motion. Spin 0 particles don't satisfy an additional equation of motion, so KG is all there is.

 

[latentcorpse]

It's enough to show that $$(p_1-p_1')^2<0$$, which is simple to do in center of mass frame.

Is it like this:
centre of mass frame implies both particles have no energy so $p_1=(0,\vec{p_1}) , \quad p_1' = ( 0 , \vec{p_1'})$

therefore $p_1-p_1' = ( 0 , p_1 - p_1') \Rightarrow (p_1 - p_1')^2 = - | \vec{p_1} - \vec{p_1'} |^2 < 0$

and so clearly there is never going to be a pole on the denominator?

The KG equation is just the mass shell condition $$p^2=m^2$$. It must be satisfied by any field that is a solution to the appropriate equation of motion. Spin 0 particles don't satisfy an additional equation of motion, so KG is all there is.

So if we were given a random field, say $\chi(x)$, that satisfies some complicated equation of motion, you're saying that it must be the case that $\chi(x)$ also satisfies the Klein-Gordon equation?

 

[fzero]

Is it like this:
centre of mass frame implies both particles have no energy so $p_1=(0,\vec{p_1}) , \quad p_1' = ( 0 , \vec{p_1'})$

therefore $p_1-p_1' = ( 0 , p_1 - p_1') \Rightarrow (p_1 - p_1')^2 = - | \vec{p_1} - \vec{p_1'} |^2 < 0$

and so clearly there is never going to be a pole on the denominator?

The energy of a massive particle is never zero!

Nevertheless, you can use conservation of energy and momentum to show that $$E_1 = E_{1'}$$, so the rest of your calculation works out.

So if we were given a random field, say $\chi(x)$, that satisfies some complicated equation of motion, you're saying that it must be the case that $\chi(x)$ also satisfies the Klein-Gordon equation?

Yex.

 

[latentcorpse]

The energy of a massive particle is never zero!

Nevertheless, you can use conservation of energy and momentum to show that $$E_1 = E_{1'}$$, so the rest of your calculation works out.



Yex.

So I have been trying another CS question. I am asked at the end of q3 in this paper:
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2006/Paper49.pdf
to explain why $\mu \frac{dF}{d \mu}=0$

For this I wrote that we can substitute $h=\mu^{\frac{1}{2}\epsilon} Z^{\frac{1}{2}}h_0$ (where a subscript 0 denotes a bare quantity) to rewrite the definition of F all in terms of bare quantities

i.e. $e^{iF(\lambda,m^2,h;\mu)}=\frac{1}{N_0} \int d[\phi] e^{iS_0[\phi]+h_0 \int d^dx \phi_0}$
$\Rightarrow {iF(\lambda,m^2,h;\mu)}=\ln{\frac{1}{N_0} \int d[\phi] e^{iS_0[\phi]+h_0 \int d^dx \phi_0}}$
Now since bare quantities are by construction independent of the RG scale $\mu$, the RHS of this equality will be $\mu$ independent and therefore by equality so will the LHS.
Therefore $\frac{dF}{d \mu}=0 \Rightarrow \mu \frac{dF}{d \mu}=0$

Does that look ok?

What about the next part though? We need to derive the CS equation. At first glance it looks like you just bash out the same chain rule procedure as we used previously but it turns out to be a bit more complicated. Previously, we defined the $\gamma_\phi$ term to go with the $\frac{\partial}{\partial \sqrt{Z}}$ term so that when we acted it on $G_n^B=Z^{n/2}G_n$ we pulled down a factor of n by the chain rule. However, this was because we were able to write the bare green's function in terms of the renormalised greens function with that factor of $Z^{n/2}$ to operate on.

But now we have the bare quantity F (which we aren't able to write in terms of any renormalised quantity) so I don't understand how we will get the h next to the $\gamma_h$ term?

Do we just define $\gamma_h$ differently from $\gamma_\phi$? I guess I could just write $\gamma_h=\frac{\mu}{h} \frac{dh}{d \mu}$. Does this work ok?

Thanks again!

 

[fzero]

Do we just define $\gamma_h$ differently from $\gamma_\phi$? I guess I could just write $\gamma_h=\frac{\mu}{h} \frac{dh}{d \mu}$. Does this work ok?

That's the definition they're using. It's very natural for anomalous dimensions to be related to logarithmic derivatives, since they naturally appear as factors like $$\mu^\gamma$$ in scaling relations.