Solving the Callan-Symanzik Equation

[latentcorpse]

Hi,

(1) I'm trying Q3 of this paper
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2009/Paper48.pdf

and have got to the 2nd last bit where it asks me to define an appropriate running coupling i should solve for the massless case.

Using a procedure outlined in my notes which is roughly the same as that on p60 of these notes
http://www.damtp.cam.ac.uk/user/ho/Notes.pdf
I have established

$G_n( \{ x_i \} ; g(\mu) , \mu) = G_n( \{ x_i \} ; g(\mu_0) , \mu_0) e^{- \int_{\mu_0}^\mu \frac{ds}{s} \gamma(g(s))}$

I believe this is correct however, I think he wants me to solve it explicitly i.e. substitute in a value for $\gamma(g(s))$. I can't find any such value? Or how it was related to the coefficients in the original Callan-Symanzik equation?

Even if my solution is fine as it is for this part of the question, in the next bit I have to substitute in for $\gamma(g(s))$ so I clearly am expected to know its relationship with the original coefficients in the equation!

(2) Also in the same paper on 1,b) I don't really understand what's going on. I think the Taylor expansion in question is
$I=\int_{-\infty}^{+\infty} e^{-\frac{m^2}{2}x^2}e^{-\frac{x^4}{4!}}(1-\lambda+\frac{\lambda^2}{2!} - \dots)$
But I have no idea how to get any Feynman rules out of this? Or indeed how we need to use Feynman rules to determine the coefficients - from that expansion, they look to me as if they are already fixed, aren't they?

Thanks for any help!

 

[fzero]

Hi,

(1) I'm trying Q3 of this paper
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2009/Paper48.pdf

and have got to the 2nd last bit where it asks me to define an appropriate running coupling i should solve for the massless case.

Using a procedure outlined in my notes which is roughly the same as that on p60 of these notes
http://www.damtp.cam.ac.uk/user/ho/Notes.pdf
I have established

$G_n( \{ x_i \} ; g(\mu) , \mu) = G_n( \{ x_i \} ; g(\mu_0) , \mu_0) e^{- \int_{\mu_0}^\mu \frac{ds}{s} \gamma(g(s))}$

I believe this is correct however, I think he wants me to solve it explicitly i.e. substitute in a value for $\gamma(g(s))$. I can't find any such value? Or how it was related to the coefficients in the original Callan-Symanzik equation?

Even if my solution is fine as it is for this part of the question, in the next bit I have to substitute in for $\gamma(g(s))$ so I clearly am expected to know its relationship with the original coefficients in the equation!

The previous parts of the problem asked you to define $$\gamma_\phi$$ and then compute it at one-loop. Did you do those?

(2) Also in the same paper on 1,b) I don't really understand what's going on. I think the Taylor expansion in question is
$I=\int_{-\infty}^{+\infty} e^{-\frac{m^2}{2}x^2}e^{-\frac{x^4}{4!}}(1-\lambda+\frac{\lambda^2}{2!} - \dots)$
But I have no idea how to get any Feynman rules out of this? Or indeed how we need to use Feynman rules to determine the coefficients - from that expansion, they look to me as if they are already fixed, aren't they?

Thanks for any help!

That is not the Taylor expansion of the integral. It looks like you tried to say that $$e^{-\lambda \frac{x^4}{4!}} = e^{-\frac{x^4}{4!}} e^{-\lambda}$$, which is incorrect.

 

[latentcorpse]

The previous parts of the problem asked you to define $$\gamma_\phi$$ and then compute it at one-loop. Did you do those?

So $\gamma(g(s)) = \gamma_\phi$. Presumably this is because $m^2=0 \Rightarrow \gamma_{m^2}=0$

Well I defined them all and solved at one loop for $\beta_\lambda , \gamma_{m^2}$ but I couldn't get $\gamma_\phi$.

I define (from my notes) that $\gamma_\phi = \frac{\mu}{\sqrt{Z}} \frac{\partial \sqrt{Z}}{\partial \mu}$ where $Z=1+A$.

The procedure is outlined in those same set of notes but he defined his $\gamma_\phi$ differently from us so I kind of lost track of what was going on.

I know I should start by writing $\phi_B = \sqrt{1+A} \phi$ and then I take $\mu \frac{d}{d \mu}$ of both sides giving

$\mu \frac{d \phi_B}{d \mu}=0$ on the LHS as bare quantities are indep of RG scale

So now by evaluating the RHS, we see that $\mu \frac{d}{d \mu} \sqrt{1+A(\lambda , \epsilon)} \phi$

Now I think I use chain rule write $\mu \frac{d}{d \mu} = \sqrt{Z} \gamma_{\phi}$

Therefore we get $\sqrt{Z} \gamma_{\phi} \sqrt{Z} \phi=0$ as $Z=1+A$

which implies $\sqrt{Z} \frac{\mu}{\sqrt{Z}} \frac{d \mu}{d \sqrt{Z}} \phi =0$

I appear to have gotten rid of all the derivatives though which has created a problem as now I can just cross multiply everything till I get $\gamma_\phi=0$?

That is not the Taylor expansion of the integral. It looks like you tried to say that $$e^{-\lambda \frac{x^4}{4!}} = e^{-\frac{x^4}{4!}} e^{-\lambda}$$, which is incorrect.

Ok. That was a typo. Even with $I=\int_{-\infty}^{+\infty} e^{-\frac{m^2}{2}x^2}(1-\frac{x^4}{4!}\lambda + \frac{x^8}{8!} \lambda^2 + \dots)$

I have the same questions..

Cheers.

 

[fzero]

So $\gamma(g(s)) = \gamma_\phi$. Presumably this is because $m^2=0 \Rightarrow \gamma_{m^2}=0$

Well I defined them all and solved at one loop for $\beta_\lambda , \gamma_{m^2}$ but I couldn't get $\gamma_\phi$.

I define (from my notes) that $\gamma_\phi = \frac{\mu}{\sqrt{Z}} \frac{\partial \sqrt{Z}}{\partial \mu}$ where $Z=1+A$.

The procedure is outlined in those same set of notes but he defined his $\gamma_\phi$ differently from us so I kind of lost track of what was going on.

I don't know what you mean by $\gamma_\phi$ being defined differently. The pdf notes that you've linked to use the same definition as above.

I know I should start by writing $\phi_B = \sqrt{1+A} \phi$ and then I take $\mu \frac{d}{d \mu}$ of both sides giving

$\mu \frac{d \phi_B}{d \mu}=0$ on the LHS as bare quantities are indep of RG scale

So now by evaluating the RHS, we see that $\mu \frac{d}{d \mu} \sqrt{1+A(\lambda , \epsilon)} \phi$

Now I think I use chain rule write $\mu \frac{d}{d \mu} = \sqrt{Z} \gamma_{\phi}$

Therefore we get $\sqrt{Z} \gamma_{\phi} \sqrt{Z} \phi=0$ as $Z=1+A$

which implies $\sqrt{Z} \frac{\mu}{\sqrt{Z}} \frac{d \mu}{d \sqrt{Z}} \phi =0$

I appear to have gotten rid of all the derivatives though which has created a problem as now I can just cross multiply everything till I get $\gamma_\phi=0$?

You're missing the terms where $$\mu d/d\mu$$ acts on $$\phi$$. It's more obvious to define $$\gamma_\phi = \mu d\phi/d\mu$$ and then derive the relationship with $$d\sqrt{Z}/d\mu$$.

Ok. That was a typo. Even with $I=\int_{-\infty}^{+\infty} e^{-\frac{m^2}{2}x^2}(1-\frac{x^4}{4!}\lambda + \frac{x^8}{8!} \lambda^2 + \dots)$

You can add a source term $$j x$$ to the "action" in that integral and then mimic the functional methods used in scalar field theory to derive Feynman rules. They're not necessary here, because we can actually do all the integrals, but this toy example is illustrative of the method.

 

[latentcorpse]

I don't know what you mean by $\gamma_\phi$ being defined differently. The pdf notes that you've linked to use the same definition as above.



You're missing the terms where $$\mu d/d\mu$$ acts on $$\phi$$. It's more obvious to define $$\gamma_\phi = \mu d\phi/d\mu$$ and then derive the relationship with $$d\sqrt{Z}/d\mu$$.

Adding in the terms I missed I find
$\sqrt{Z} \gamma_\phi ( sqrt{Z} ) \phi + \sqrt{Z} \sqrt{Z} \gamma_\phi \phi = 0 \Rightarrow \sqrt{Z} \frac{\mu}{Z} \frac{d \sqrt{Z}}{d \mu} \phi + Z \frac{\mu}{\sqrt{Z}} \frac{d \sqrt{Z}}{d \mu} \frac{\partial \phi}{\partial \sqrt{Z}}=0$
$\frac{\partial \phi}{\partial \sqrt{Z}} = - \frac{\phi}{\sqrt{Z}}$
$\log{\phi} = - \log{\sqrt{Z}}$
$\phi = \sqrt{Z}^{-1}$
That looks a bit off?

You can add a source term $$j x$$ to the "action" in that integral and then mimic the functional methods used in scalar field theory to derive Feynman rules. They're not necessary here, because we can actually do all the integrals, but this toy example is illustrative of the method.

So let's say I define

$I(J)=\int_{-\infty}^\infty e^{-m^2 / 2 x^2 + Jx} ( 1- \frac{x^4}{4!}\lambda + \dots)$
First of all, should I remove the taylor expansion for the timebeing? It looks like the maths will be easier if I keep things as
$I(J)=\int_{-\infty}^\infty e^{-m^2 / 2 x^2 + Jx - \lambda \frac{x^4}{4!}}$

However, to get the Feynman rules for the propagator we want to consider
$\langle \phi(x_1) \phi(x_2) \rangle = \frac{1}{I(0)} \frac{\partial}{\partial J(x_1)} \frac{\partial}{\partial J(x_2)} \vline_{J=0}$

There is a slight problem now though because
a) I don't think I should have called it $\langle \phi(x_1) \phi(x_2) \rangle$ as there certainly isn't a scalar field present but I don't know what else to put? What woudl you recommend? Is $\langle x_1 x_2 \rangle$ fine?
b) Is my definition of the propagator in terms of derivatives ok? The next step would be to complete the square but because of the $x^4$ term, this is proving problematic - any advice?

Thanks a lot.

 

[fzero]

Adding in the terms I missed I find
$\sqrt{Z} \gamma_\phi ( sqrt{Z} ) \phi + \sqrt{Z} \sqrt{Z} \gamma_\phi \phi = 0 \Rightarrow \sqrt{Z} \frac{\mu}{Z} \frac{d \sqrt{Z}}{d \mu} \phi + Z \frac{\mu}{\sqrt{Z}} \frac{d \sqrt{Z}}{d \mu} \frac{\partial \phi}{\partial \sqrt{Z}}=0$
$\frac{\partial \phi}{\partial \sqrt{Z}} = - \frac{\phi}{\sqrt{Z}}$
$\log{\phi} = - \log{\sqrt{Z}}$
$\phi = \sqrt{Z}^{-1}$
That looks a bit off?

I was a bit off before, $$\gamma_\phi = \mu d(\log \phi)/d\mu$$. This is the definition that results in

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

So let's say I define

$I(J)=\int_{-\infty}^\infty e^{-m^2 / 2 x^2 + Jx} ( 1- \frac{x^4}{4!}\lambda + \dots)$
First of all, should I remove the taylor expansion for the timebeing? It looks like the maths will be easier if I keep things as
$I(J)=\int_{-\infty}^\infty e^{-m^2 / 2 x^2 + Jx - \lambda \frac{x^4}{4!}}$

However, to get the Feynman rules for the propagator we want to consider
$\langle \phi(x_1) \phi(x_2) \rangle = \frac{1}{I(0)} \frac{\partial}{\partial J(x_1)} \frac{\partial}{\partial J(x_2)} \vline_{J=0}$

There is a slight problem now though because
a) I don't think I should have called it $\langle \phi(x_1) \phi(x_2) \rangle$ as there certainly isn't a scalar field present but I don't know what else to put? What woudl you recommend? Is $\langle x_1 x_2 \rangle$ fine?
b) Is my definition of the propagator in terms of derivatives ok? The next step would be to complete the square but because of the $x^4$ term, this is proving problematic - any advice?

Thanks a lot.

You want to establish the analogue of the formula at the bottom of page 20 of those notes. The propagator is going to be just $$1/m^2$$. So something like

$$I_0[J] = \int dx e^{-m^2 x^2/2 + x J}\sim e^{ J^2/(4m^2) }$$

$$I_\lambda[J] = e^{- (\lambda/4!) (\delta/\delta J)^4 } I_0[J]$$

But that's also just formality. If you already know the Feynman rules for scalar field theory, you can guess that we have simpler versions of them here. We can explicitly compute the coefficients in the perturbation expansion of $$I$$, we just need to explain them from vacuum diagrams (bubbles). It turns out to just be combinatorics of how many ways we can connect the legs of our propagators and vertices.

 

[latentcorpse]

I was a bit off before, $$\gamma_\phi = \mu d(\log \phi)/d\mu$$. This is the definition that results in

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

Ok. so what's wrong with my above calculation? I used the $\gamma_\phi = \frac{\mu}{\sqrt{Z}} \frac{\partial \sqrt{Z}}{\partial \mu}$ definition.

You want to establish the analogue of the formula at the bottom of page 20 of those notes. The propagator is going to be just $$1/m^2$$. So something like

$$I_0[J] = \int dx e^{-m^2 x^2/2 + x J}\sim e^{ J^2/(4m^2) }$$

$$I_\lambda[J] = e^{- (\lambda/4!) (\delta/\delta J)^4 } I_0[J]$$

But that's also just formality. If you already know the Feynman rules for scalar field theory, you can guess that we have simpler versions of them here. We can explicitly compute the coefficients in the perturbation expansion of $$I$$, we just need to explain them from vacuum diagrams (bubbles). It turns out to just be combinatorics of how many ways we can connect the legs of our propagators and vertices.

Ok so I find the power of m to be 4 not 2:

$I_0[J]=\int dx e^{-\frac{m^2x^2}{2} + xJ} = \int d \tilde{x} e^{-\frac{m^2 \tilde{x}^2}{2} + \frac{J^2}{4m^4}}$ where $\tilde{x}=x-\frac{J}{2m^2}$

This means we have $I_0[J]=I_0[0] e^{\frac{J^2}{4m^2}}$
and $I_\lambda[J]=e^{-\frac{\lambda}{4!} ( \frac{\delta}{\delta J})^4 } I_0[J]$

Now to work out the propagator i need to evaluate the 2 point greens function right? Is this my next step?

 

[fzero]

Ok. so what's wrong with my above calculation? I used the $\gamma_\phi = \frac{\mu}{\sqrt{Z}} \frac{\partial \sqrt{Z}}{\partial \mu}$ definition.

I thought the extra factor of $$\phi$$ floating around was giving you trouble.

Ok so I find the power of m to be 4 not 2:

$I_0[J]=\int dx e^{-\frac{m^2x^2}{2} + xJ} = \int d \tilde{x} e^{-\frac{m^2 \tilde{x}^2}{2} + \frac{J^2}{4m^4}}$ where $\tilde{x}=x-\frac{J}{2m^2}$

It's $$m^2$$ because of the overall factor of $$m^2$$ in the first term.

This means we have $I_0[J]=I_0[0] e^{\frac{J^2}{4m^2}}$
and $I_\lambda[J]=e^{-\frac{\lambda}{4!} ( \frac{\delta}{\delta J})^4 } I_0[J]$

Now to work out the propagator i need to evaluate the 2 point greens function right? Is this my next step?

You could compute the 2pt function easily, or just read the propagator off of the free part of the action for $$x$$ as $$1/m^2$$. At some point you also need to compute the first few terms in the $$\lambda$$ expansion of the original integral $$I$$.

 

[latentcorpse]

I thought the extra factor of $$\phi$$ floating around was giving you trouble.

Yeah but if we take $\mu \frac{d}{d \mu} (\sqrt{Z} \phi) =0$
this means $\sqrt{Z} \gamma_\phi (\sqrt{Z} \phi)=0 \Rightarrow \gamma_\phi (\sqrt{Z} \phi)=0$
$\frac{\mu}{\sqrt{Z}} \frac{d \mu}{\d \sqrt{Z}} ( \frac{\partial \sqrt{Z}}{\partial \sqrt{Z}} \phi + \sqrt{Z} \frac{\partial \phi}{\partial \sqrt{Z}} )=0$
$\Rightarrow \frac{\partial \sqrt{Z}}{\partial \sqrt{Z}} \phi + \sqrt{Z} \frac{\partial \phi}{\partial \sqrt{Z}} =0$
$\frac{\partial \phi}{\partial \sqrt{Z}} =-\phi$
so there is still a $\phi$ left over?

It's $$m^2$$ because of the overall factor of $$m^2$$ in the first term.

I don't get this.

we have $\frac{m^2}{2}x^2 - Jx$ in the exponent
$\tilde{x} = x- \frac{J}{2m}$
would give $\frac{m^2}{2} (x^2 - \frac{Jx}{m} + \frac{J^2}{4m^2})$
so when we multiply this out, the cross term doesn't give us what we want, does it?

You could compute the 2pt function easily, or just read the propagator off of the free part of the action for $$x$$ as $$1/m^2$$. At some point you also need to compute the first few terms in the $$\lambda$$ expansion of the original integral $$I$$.

Well I don't really understand why we can just read of 1/m^2?
Nonetheless, I think for the sake of practice I'd like to compute this explicitly!
The only thing is,
$\frac{I_\lambda[J]}{I_0[0]} = e^{-\frac{\lambda}{4!} ( \frac{\delta}{\delta J})^4 } e^{\frac{J^2}{4m^2}}$
So now should I expand the taylor series?
that would give
$(1-\lambda / 4! (\frac{\delta}{\delta J})^4 + \dots ) e^{\frac{J^2}{4m^2}}$

Is this looking ok?
[/itex]

 

[fzero]

Yeah but if we take $\mu \frac{d}{d \mu} (\sqrt{Z} \phi) =0$
this means $\sqrt{Z} \gamma_\phi (\sqrt{Z} \phi)=0 \Rightarrow \gamma_\phi (\sqrt{Z} \phi)=0$
$\frac{\mu}{\sqrt{Z}} \frac{d \mu}{\d \sqrt{Z}} ( \frac{\partial \sqrt{Z}}{\partial \sqrt{Z}} \phi + \sqrt{Z} \frac{\partial \phi}{\partial \sqrt{Z}} )=0$
$\Rightarrow \frac{\partial \sqrt{Z}}{\partial \sqrt{Z}} \phi + \sqrt{Z} \frac{\partial \phi}{\partial \sqrt{Z}} =0$
$\frac{\partial \phi}{\partial \sqrt{Z}} =-\phi$
so there is still a $\phi$ left over?

Computing

$\mu \frac{d}{d \mu} (\sqrt{Z} \phi) =0$

only leads to the equation

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

Also

$$\mu \frac{d}{d \mu} (\sqrt{Z} \phi) \neq \sqrt{Z} \gamma_\phi (\sqrt{Z} \phi),$$

so the result that's confusing you is not even correct.

Well I don't really understand why we can just read of 1/m^2?

Because you know what the propagator is for a scalar field, so you can dimensionally reduce to get the propagator for quantum mechanics.

Nonetheless, I think for the sake of practice I'd like to compute this explicitly!
The only thing is,
$\frac{I_\lambda[J]}{I_0[0]} = e^{-\frac{\lambda}{4!} ( \frac{\delta}{\delta J})^4 } e^{\frac{J^2}{4m^2}}$
So now should I expand the taylor series?
that would give
$(1-\lambda / 4! (\frac{\delta}{\delta J})^4 + \dots ) e^{\frac{J^2}{4m^2}}$

Is this looking ok?
[/itex]

It should let you compute the integral. Just go ahead and keep going with it. Also go back and do the regular Gaussian integrals so that you have something to compare with. This is all elementary calculations, so there's no reason to stop and ask me about it.

 

[latentcorpse]

Computing

$\mu \frac{d}{d \mu} (\sqrt{Z} \phi) =0$

only leads to the equation

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

Also

$$\mu \frac{d}{d \mu} (\sqrt{Z} \phi) \neq \sqrt{Z} \gamma_\phi (\sqrt{Z} \phi),$$

so the result that's confusing you is not even correct.

How do we work out $\gamma_\phi$ then? To work out $\beta_\lambda , \gamma_{m^2}$ we took their defining equation e.g. $\mu^{-\epsilon}=\lambda+F_\lambda(\lambda,\epsilon)$ and computed $\mu \frac{d}{d \mu}$ of both sides which is what I was trying to do here.

Because you know what the propagator is for a scalar field, so you can dimensionally reduce to get the propagator for quantum mechanics.

Can you explain this a bit simpler please? I don't know how we know this propagator is for a scalar field? Is there even a field present? It looks like just $x$ 's rather than $\phi$'s to me?
And what does dimensionally reduce mean?
And I still don't see why it isn't an m^4?

It should let you compute the integral. Just go ahead and keep going with it. Also go back and do the regular Gaussian integrals so that you have something to compare with. This is all elementary calculations, so there's no reason to stop and ask me about it.

Can you tell me what the integral I actually want to calculate is though? I assume that as per our usual procedure, to find the propagator we want to compute the 2 point green's function, yes?

But the 2 point green's function should have 2 x's in it? But using our differential operator at first order will bring down 4 x's!

Also, is the 2 point green's function equal to $\frac{I_\lambda[J]}{I_0[0]}$? Because that doesn't have an integral in it?

This "toy" model is kind of frustrating me because I thought I actually understood this stuff fairly well but this is just confusing me loads!

 

[fzero]

How do we work out $\gamma_\phi$ then? To work out $\beta_\lambda , \gamma_{m^2}$ we took their defining equation e.g. $\mu^{-\epsilon}=\lambda+F_\lambda(\lambda,\epsilon)$ and computed $\mu \frac{d}{d \mu}$ of both sides which is what I was trying to do here.

It's an analogous procedure where you write

$$Z = 1 + \sum_{n>0} \frac{a_n(\lambda)}{\epsilon^n}.$$

You substitute this in

$Z \gamma_\phi = \frac{\mu}{2} \frac{d Z}{d \mu}$

and collect powers of $$\epsilon$$.

Can you explain this a bit simpler please? I don't know how we know this propagator is for a scalar field? Is there even a field present? It looks like just $x$ 's rather than $\phi$'s to me?
And what does dimensionally reduce mean?
And I still don't see why it isn't an m^4?

Functional integrals are generalizations of ordinary integrals

$$\int dx F(x)$$

where we promote the coordinates to scalar fields $$x(\sigma)$$ and the measure to one on the space of paths $$\mathcal{D}x$$:

$$\int \mathcal{D}x F[x(\sigma)].$$

The scalar field $$x(\sigma)$$ lives in $$d$$ dimensions, but the ordinary coordinate is a scalar field in 0 dimensions. Dimensional reduction is just the statement that quantities in $$d$$ dimensions can be related to $$d-1$$ dimensional quantities by ignoring a coordinate in an appropriate way. For the propagator, in $$d$$ dimensions, we have $$1/(p^2+m^2)$$. In zero dimensions, there is no momentum, so it becomes $$1/m^2$$.

Can you tell me what the integral I actually want to calculate is though?

The problem asks you to compute the integral $$I$$ in a Taylor expansion. That's the one.

I assume that as per our usual procedure, to find the propagator we want to compute the 2 point green's function, yes?

You can obtain it from the 2pt function if you need to.

But the 2 point green's function should have 2 x's in it? But using our differential operator at first order will bring down 4 x's!

Also, is the 2 point green's function equal to $\frac{I_\lambda[J]}{I_0[0]}$? Because that doesn't have an integral in it?

The leading order 2pt function appears in $$I_0$$. You could compute it from the leading part of

$$\left( \frac{\partial}{\partial J}\right)^2 I[J]$$.

This "toy" model is kind of frustrating me because I thought I actually understood this stuff fairly well but this is just confusing me loads!

Well hopefully you understand it better after working through the toy model.

 

[latentcorpse]

Thanks for the reply.

It's an analogous procedure where you write

$$Z = 1 + \sum_{n>0} \frac{a_n(\lambda)}{\epsilon^n}.$$

You substitute this in

$Z \gamma_\phi = \frac{\mu}{2} \frac{d Z}{d \mu}$

and collect powers of $$\epsilon$$.
[/itex]
Ok. Where did $Z \gamma_\phi = \frac{\mu}{2} \frac{d Z}{d \mu}$ come from? I tried doing $\mu \frac{d}{d \mu}$ of $\phi_B=Z^{1/2} \phi$ as usual but this gives
$\frac{\mu}{Z^{1/2}} Z^{1/2} \frac{d Z^{1/2}}{d \mu} \frac{\partial}{\partial \mu} ( Z^{1/2} \phi) = 0$
using product rule we get $\mu \phi \frac{d Z^{1/2}}{d \mu} = Z \gamma_\phi \phi$
So you see I am missing the 2 amongst other things that you had before I am able to start subbing in and collecting powers.

Functional integrals are generalizations of ordinary integrals

$$\int dx F(x)$$

where we promote the coordinates to scalar fields $$x(\sigma)$$ and the measure to one on the space of paths $$\mathcal{D}x$$:

$$\int \mathcal{D}x F[x(\sigma)].$$

The scalar field $$x(\sigma)$$ lives in $$d$$ dimensions, but the ordinary coordinate is a scalar field in 0 dimensions. Dimensional reduction is just the statement that quantities in $$d$$ dimensions can be related to $$d-1$$ dimensional quantities by ignoring a coordinate in an appropriate way. For the propagator, in $$d$$ dimensions, we have $$1/(p^2+m^2)$$. In zero dimensions, there is no momentum, so it becomes $$1/m^2$$.

So how do we know that the propagator is $1/(p^2+m^2)$. This isn't always the case. Or do you just mean we recognise that the propagator is usually proportional to this factor when we work in d dimensions?
Surely the particle itself is in one dimension though so it can still have some momentum can't it?

The problem asks you to compute the integral $$I$$ in a Taylor expansion. That's the one.



You can obtain it from the 2pt function if you need to.



The leading order 2pt function appears in $$I_0$$. You could compute it from the leading part of

$$\left( \frac{\partial}{\partial J}\right)^2 I[J]$$.



Well hopefully you understand it better after working through the toy model.

Ok.
It doesn't actually say that i need to evaluate the integral although presumably that's going to be useful...
So we need to get the feynman rules. the propagator is exactly the 2 point greens function
you say this comes from [itex]\left( \frac{\partial}{\partial J}\right)^2 I[J] \vline_{J=0}[/tex] but wouldn't that just give $x^2I[0]$

Is that correct? A propagator gets a factor of $x^2I[0]$?

And then we can see the vertex term from looking at the interaction action $V^{(4)}(0)=\lambda$ so we can expect that a vertex picks up a $-\lambda$ right?

How is that?

A couple of other things?
(i) Why do we take $-V^{(k)}(0)$ for the vertex and not +?
(ii) Where did your expression for the two point greens function come from?
Usually we'd have something like $\langle \phi(x_1) \phi(x_2) \rangle = - \frac{1}{Z[0]} \frac{\delta^2}{\delta J(x_1) \delta J(x_2)}Z[J] \vline_{J=0}$
Why do we not have something of this form here? And if we do - can youe xplain why they are of the same form?

Thanks very much as usual!

 

[fzero]

Thanks for the reply.

Ok. Where did $Z \gamma_\phi = \frac{\mu}{2} \frac{d Z}{d \mu}$ come from? I tried doing $\mu \frac{d}{d \mu}$ of $\phi_B=Z^{1/2} \phi$ as usual but this gives
$\frac{\mu}{Z^{1/2}} Z^{1/2} \frac{d Z^{1/2}}{d \mu} \frac{\partial}{\partial \mu} ( Z^{1/2} \phi) = 0$
using product rule we get $\mu \phi \frac{d Z^{1/2}}{d \mu} = Z \gamma_\phi \phi$
So you see I am missing the 2 amongst other things that you had before I am able to start subbing in and collecting powers.

As I said several posts ago, the correct definition of the anomalous dimension is (I had a sign wrong)

$$\gamma_\phi = - \mu \frac{d(\log \phi)}{d\mu}.$$

We compute

$$0 = \mu \frac{d}{d\mu}(\sqrt{Z}\phi) = \phi \mu \frac{d\sqrt{Z}}{d\mu} + \sqrt{Z} \mu \frac{d\phi)}{d\mu} = \left( \frac{\mu}{\sqrt{Z}} \frac{d\sqrt{Z}}{d\mu} - \gamma_\phi \right) \sqrt{Z}\phi,$$

from which we conclude that

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

If we use the chain rule on $$\sqrt{Z}$$, we obtain

$$\gamma_\phi = \frac{\mu}{2Z} \frac{dZ}{d\mu} .$$

So how do we know that the propagator is $1/(p^2+m^2)$. This isn't always the case. Or do you just mean we recognise that the propagator is usually proportional to this factor when we work in d dimensions?

What else do you propose it to be?

Surely the particle itself is in one dimension though so it can still have some momentum can't it?

As a scalar field theory, $$x$$ is in zero dimensions. One dimension would mean that the coordinate was a function of, for example, time.

Ok.
It doesn't actually say that i need to evaluate the integral although presumably that's going to be useful...
So we need to get the feynman rules. the propagator is exactly the 2 point greens function
you say this comes from [itex]\left( \frac{\partial}{\partial J}\right)^2 I[J] \vline_{J=0}[/tex] but wouldn't that just give $x^2I[0]$

Is that correct? A propagator gets a factor of $x^2I[0]$?

$$I[J]$$ is an integral over $$x$$, so there cannot be any $$x$$ dependence in the functional expression above.

And then we can see the vertex term from looking at the interaction action $V^{(4)}(0)=\lambda$ so we can expect that a vertex picks up a $-\lambda$ right?

You also have to keep track of the numerical factors, since the Feynman rules have to reproduce them.

A couple of other things?
(i) Why do we take $-V^{(k)}(0)$ for the vertex and not +?

You have to pick the sign for the vertex that reproduces the sign from the functional expansion.

(ii) Where did your expression for the two point greens function come from?
Usually we'd have something like $\langle \phi(x_1) \phi(x_2) \rangle = - \frac{1}{Z[0]} \frac{\delta^2}{\delta J(x_1) \delta J(x_2)}Z[J] \vline_{J=0}$
Why do we not have something of this form here? And if we do - can youe xplain why they are of the same form?

Thanks very much as usual!

You should probably normalize the expression by $$I[0]$$, but the $$x$$ and $$J$$ are just numbers, not functions, so the rest of that expression isn't needed.

I would hope that things would get clearer if you would actually compute the first few terms in the original $$I$$ and what terms your Feynman rules actually have to reproduce.

 

[latentcorpse]

As I said several posts ago, the correct definition of the anomalous dimension is (I had a sign wrong)

$$\gamma_\phi = - \mu \frac{d(\log \phi)}{d\mu}.$$

We compute

$$0 = \mu \frac{d}{d\mu}(\sqrt{Z}\phi) = \phi \mu \frac{d\sqrt{Z}}{d\mu} + \sqrt{Z} \mu \frac{d\phi)}{d\mu} = \left( \frac{\mu}{\sqrt{Z}} \frac{d\sqrt{Z}}{d\mu} - \gamma_\phi \right) \sqrt{Z}\phi,$$

from which we conclude that

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

If we use the chain rule on $$\sqrt{Z}$$, we obtain

$$\gamma_\phi = \frac{\mu}{2Z} \frac{dZ}{d\mu} .$$

So this gets us to $\gamma_\phi(1+A) = - \frac{1}{2} \lambda \frac{\partial}{\partial \lambda} a_1(\lambda)$
I'm meant to demonstrate that $\gamma_\phi = \mathcal{O}(\lambda^2)$ I think. How do I get that out of this mess?

As a scalar field theory, $$x$$ is in zero dimensions. One dimension would mean that the coordinate was a function of, for example, time.

So is it because there is no time dependence that there is no momentum?

$$I[J]$$ is an integral over $$x$$, so there cannot be any $$x$$ dependence in the functional expression above.



You also have to keep track of the numerical factors, since the Feynman rules have to reproduce them.



You have to pick the sign for the vertex that reproduces the sign from the functional expansion.
[/itex]
So basically if the interaction term in the action has a minus sign in front of it, the vertex will be defined with a minus sign? Is this also true for why we get factors of i in position space feynman rules but not in momentum space?

You should probably normalize the expression by $$I[0]$$, but the $$x$$ and $$J$$ are just numbers, not functions, so the rest of that expression isn't needed.

I would hope that things would get clearer if you would actually compute the first few terms in the original $$I$$ and what terms your Feynman rules actually have to reproduce.

I get
$I(J) = \int dx e^{-\frac{\lambda}{4!} ( \frac{\partial}{\partial J})^4} e^{-\frac{m^2x^2}{2}+Jx} = \int dx (1-\frac{\lambda}{4!} \left( \frac{\partial}{\partial J} \right)^4 + \frac{\lambda^2}{4!^2} ( \frac{\partial}{\partial J})^8 + \dots ) e^{-\frac{m^2x^2}{2}+Jx}$
$=\int dx (1-\frac{\lambda x^4}{4!} + \frac{\lamdba^2 x^8}{4!^2} + \dots) e^{-\frac{m^2x^2}{2} + Jx}$
Now how do you compute this integral??

 

[fzero]

So this gets us to $\gamma_\phi(1+A) = - \frac{1}{2} \lambda \frac{\partial}{\partial \lambda} a_1(\lambda)$
I'm meant to demonstrate that $\gamma_\phi = \mathcal{O}(\lambda^2)$ I think. How do I get that out of this mess?

The problem states that there is no field renormalization at order $$\lambda$$.

So is it because there is no time dependence that there is no momentum?

Yes, $$x$$ is not a function of any other coordinates.

So basically if the interaction term in the action has a minus sign in front of it, the vertex will be defined with a minus sign? Is this also true for why we get factors of i in position space feynman rules but not in momentum space?

Yes. Some of the factors of $$i$$ will be different depending on whether you're in Euclidean space or not. You can always work out a consistent set of conventions from the functional approach.

I get
$I(J) = \int dx e^{-\frac{\lambda}{4!} ( \frac{\partial}{\partial J})^4} e^{-\frac{m^2x^2}{2}+Jx} = \int dx (1-\frac{\lambda}{4!} \left( \frac{\partial}{\partial J} \right)^4 + \frac{\lambda^2}{4!^2} ( \frac{\partial}{\partial J})^8 + \dots ) e^{-\frac{m^2x^2}{2}+Jx}$
$=\int dx (1-\frac{\lambda x^4}{4!} + \frac{\lamdba^2 x^8}{4!^2} + \dots) e^{-\frac{m^2x^2}{2} + Jx}$
Now how do you compute this integral??

These are standard Gaussian integrals and there are a number of methods to deal with them. You can make a substitution to write them in terms of Gamma functions. You can also compute

$$M = \int_{-\infty}^\infty dx e^{-\frac{m^2x^2}{2} + Jx}$$

by completing the square. Then

$$\int_{-\infty}^\infty dx ~ x^{2n} e^{-\frac{m^2x^2}{2} + Jx} = c_n \left( \frac{d}{dJ}\right)^{2n} M$$

for some appropriate coefficient $$c_n$$. This is method that you actually use the in the generating function approach to correlation functions.

 

[latentcorpse]

The problem states that there is no field renormalization at order $$\lambda$$.

What does that mean? That a1=0?

Yes. Some of the factors of $$i$$ will be different depending on whether you're in Euclidean space or not. You can always work out a consistent set of conventions from the functional approach.

We covered the procedure for getting propagators quite extensively. What is the procedure for getting the vertex term. I have pretty much been using the method of read off the coefficient of the interaction term in the action?

These are standard Gaussian integrals and there are a number of methods to deal with them. You can make a substitution to write them in terms of Gamma functions. You can also compute

$$M = \int_{-\infty}^\infty dx e^{-\frac{m^2x^2}{2} + Jx}$$

by completing the square. Then

$$\int_{-\infty}^\infty dx ~ x^{2n} e^{-\frac{m^2x^2}{2} + Jx} = c_n \left( \frac{d}{dJ}\right)^{2n} M$$

for some appropriate coefficient $$c_n$$. This is method that you actually use the in the generating function approach to correlation functions.

This gets hard very fast. Consider $\int dx x^4 e^{-\frac{m^2x^2}{2}+Jx}$
$=\int dx x^4 e^{-\frac{m^2}{2} \tilde{x}^2 +\frac{J^2}{m^4}}$ (I'm sorry but I'm convinced its an m^4 rather than an m^2 on the bottom lol!)

But before we can even begin making substitutions to get it into the form of a gamma function we're going to need everything in terms of $\tilde{x}$.

The measure is fine as $dx=d \tilde{x}$

But the $x^4=(\tilde{x}+\frac{J}{m^2})^4$ This is going to produce horrendous cross terms!

 

[fzero]

What does that mean? That a1=0?

It looks like it means that $$a_1 = O(\lambda^2)$$.

We covered the procedure for getting propagators quite extensively. What is the procedure for getting the vertex term. I have pretty much been using the method of read off the coefficient of the interaction term in the action?

That should work.

This gets hard very fast. Consider $\int dx x^4 e^{-\frac{m^2x^2}{2}+Jx}$
$=\int dx x^4 e^{-\frac{m^2}{2} \tilde{x}^2 +\frac{J^2}{m^4}}$ (I'm sorry but I'm convinced its an m^4 rather than an m^2 on the bottom lol!)

But before we can even begin making substitutions to get it into the form of a gamma function we're going to need everything in terms of $\tilde{x}$.

The measure is fine as $dx=d \tilde{x}$

But the $x^4=(\tilde{x}+\frac{J}{m^2})^4$ This is going to produce horrendous cross terms!

The actual integrals that you want to compute have $$J=0$$. Even if you kept J in, the integrals over the odd cross-terms vanish.

 

[latentcorpse]

It looks like it means that $$a_1 = O(\lambda^2)$$.

Ok so this means that $\gamma_\phi=\mathcal{O}(\lambda^2)$

So back to the problem of using a running coupling to solve the CS/RG eqn.

Is the $\gamma(g(\mu))$ that now appears in the CS eqn just $\gamma_\phi$?

In that case we would get the solution to be $G_n( \{ p_i \} ; g(\mu) , \mu) =e^{-\int_{\mu_0}^\mu \frac{ds}{s} \gamma(g(s))} G_n( \{ p_i \} ; g(\mu_0) , \mu_0)$

substituting $\gamma(g(s))=0$ since $\gamma_\phi = \mathcal{O}(\lambda^2)$ would give
$G_n( \{ p_i \} ; g(\mu) , \mu) =e^C G_n( \{ p_i \} ; g(\mu_0) , \mu_0)$
for some constant $C$

Is that correct for the next bit of the question?

Then for the final parts:
(i) how is your answer changed in the presence of a non zero $\gamma_\phi$
Perhaps I'm missing something but wouldn't that just be saying that rather than having a constant green's function we have one varies exponentially?
(ii)Working with the same one-loop formula for $\beta_\lambda$ (and $m^2 = 0$), determine the new solution for $\gamma_\phi=c\lambda^2$
Well I think this is just substituting in $c \lambda^2$ so we'd have
$G_n(p_i,\mu)=e^{-c\int_{\mu_0}^\mu \frac{d \lambda}{\lambda} \lambda^2} G_n(p_i,\mu_0) = e^{-c (\mu-\mu_0)}G_n(p_i,\mu_0)$

The actual integrals that you want to compute have $$J=0$$. Even if you kept J in, the integrals over the odd cross-terms vanish.

I get $I=\frac{\sqrt{2 \pi}}{m} - \lambda \frac{4}{\sqrt{2} 4! m^5} \Gamma(\frac{5}{2}) + \lambda^2 \frac{16}{\sqrt{2}8!m^9} \Gamma(\frac{9}{2}) + \dots$Cheers.

 

[fzero]

Ok so this means that $\gamma_\phi=\mathcal{O}(\lambda^2)$

So back to the problem of using a running coupling to solve the CS/RG eqn.

Is the $\gamma(g(\mu))$ that now appears in the CS eqn just $\gamma_\phi$?

In that case we would get the solution to be $G_n( \{ p_i \} ; g(\mu) , \mu) =e^{-\int_{\mu_0}^\mu \frac{ds}{s} \gamma(g(s))} G_n( \{ p_i \} ; g(\mu_0) , \mu_0)$

substituting $\gamma(g(s))=0$ since $\gamma_\phi = \mathcal{O}(\lambda^2)$ would give
$G_n( \{ p_i \} ; g(\mu) , \mu) =e^C G_n( \{ p_i \} ; g(\mu_0) , \mu_0)$
for some constant $C$

Is that correct for the next bit of the question?

Then for the final parts:
(i) how is your answer changed in the presence of a non zero $\gamma_\phi$
Perhaps I'm missing something but wouldn't that just be saying that rather than having a constant green's function we have one varies exponentially?
(ii)Working with the same one-loop formula for $\beta_\lambda$ (and $m^2 = 0$), determine the new solution for $\gamma_\phi=c\lambda^2$
Well I think this is just substituting in $c \lambda^2$ so we'd have
$G_n(p_i,\mu)=e^{-c\int_{\mu_0}^\mu \frac{d \lambda}{\lambda} \lambda^2} G_n(p_i,\mu_0) = e^{-c (\mu-\mu_0)}G_n(p_i,\mu_0)$

I thought that

$$\int_{\mu_0}^\mu \frac{ds}{s} \gamma(g(s)) = \int_{\lambda(\mu_0)}^{\lambda(\mu)} d\lambda \frac{\gamma(\lambda)}{\beta(\lambda)} ,$$

which is not what you have.

I get $I=\frac{\sqrt{2 \pi}}{m} - \lambda \frac{4}{\sqrt{2} 4! m^5} \Gamma(\frac{5}{2}) + \lambda^2 \frac{16}{\sqrt{2}8!m^9} \Gamma(\frac{9}{2}) + \dots$

You'll want to clean this up a bit. I found that

$$I = \sqrt{\frac{2\pi}{m}} \left( 1 - \frac{3\lambda}{4! m^4} + \frac{105 \lambda^2}{2(4!)^2 m^8} + \cdots \right).$$

 

[latentcorpse]

I thought that

$$\int_{\mu_0}^\mu \frac{ds}{s} \gamma(g(s)) = \int_{\lambda(\mu_0)}^{\lambda(\mu)} d\lambda \frac{\gamma(\lambda)}{\beta(\lambda)} ,$$

which is not what you have.
You'll want to clean this up a bit. I found that

$$I = \sqrt{\frac{2\pi}{m}} \left( 1 - \frac{3\lambda}{4! m^4} + \frac{105 \lambda^2}{2(4!)^2 m^8} + \cdots \right).$$

I thought the running coupling gets rid of the $\beta$ term.

We have $(\mu \frac{\partial}{\partial \mu} + \beta(g) \frac{\partial}{\partial g} + n\gamma(g))G_n=0$ (Should the n still be there?)

Anyway if we introduce a running coupling, it obeys
$\mu \frac{d}{d \mu} g(\mu) = \beta(g(\mu))$

So wouldn't that mean $\beta(g(\mu)) \frac{\partial}{\partial g} = \mu \frac{\partial}{\partial \mu}$ and so we can combine it with the first term in the RG equation.

This appears to be the technique we used when we solve for the case of $\gamma=0$. In this instance we combine the first two terms and the third obviously vanishes and we end up with

$\mu \frac{d}{d \mu} G_n( \{ p_i \} ; g( \mu) , \mu) =0 \Rightarrow G_n( \{ p_i \} ; g(\mu) , \mu) = G_m ( \{ p_i \} ; g( \mu_0) , \mu_0)$

So shouldn't the $\beta$ term get absorbed? And what about the n?

Secondly id you could take a look at the thread i made about minus signs in the feynman rules for tree level diagrams that would be awesome.

Thanks a lot!

 

[fzero]

My remark was about your equation for $$\gamma\neq 0$$. The integral you write down in

$G_n(p_i,\mu)=e^{-c\int_{\mu_0}^\mu \frac{d \lambda}{\lambda} \lambda^2} G_n(p_i,\mu_0) = e^{-c (\mu-\mu_0)}G_n(p_i,\mu_0)$

is wrong.

 

[latentcorpse]

My remark was about your equation for $$\gamma\neq 0$$. The integral you write down in

$G_n(p_i,\mu)=e^{-c\int_{\mu_0}^\mu \frac{d \lambda}{\lambda} \lambda^2} G_n(p_i,\mu_0) = e^{-c (\mu-\mu_0)}G_n(p_i,\mu_0)$

is wrong.

Ok. I think we're getting too far ahead here.

(i) In the CS equation once we've introduced a running coupling, should there still be an n? i.e. is the last term $n \gamma(g(s))$ or just $\gamma(g(s))$?

(ii) If so, does that mean $\gamma(g(s))=n\gamma_\phi$. Why?

(iii) When I'm trying the first part of the question i.e. to solve to 1 loop, I wrote down the formula which appears to be the 3rd from the bottom on p60 of those notes. I don't see what's wrong with it though!

Cheers.

 

[fzero]

Ok. I think we're getting too far ahead here.

(i) In the CS equation once we've introduced a running coupling, should there still be an n? i.e. is the last term $n \gamma(g(s))$ or just $\gamma(g(s))$?

(ii) If so, does that mean $\gamma(g(s))=n\gamma_\phi$. Why?

Yes, it looks like they set $\gamma(g(s))=n\gamma_\phi$ to simplify the notation.

(iii) When I'm trying the first part of the question i.e. to solve to 1 loop, I wrote down the formula which appears to be the 3rd from the bottom on p60 of those notes. I don't see what's wrong with it though!

Cheers.

The integral over $$s$$ corresponds to integrating over the RG scale $$\mu$$. To write this as an integral over the coupling constant means changing the measure according to

$$\frac{d\mu}{\mu} = \frac{d\lambda}{\beta(\lambda)},$$

as well as the limits of integration.

 

[latentcorpse]

Yes, it looks like they set $\gamma(g(s))=n\gamma_\phi$ to simplify the notation.



The integral over $$s$$ corresponds to integrating over the RG scale $$\mu$$. To write this as an integral over the coupling constant means changing the measure according to

$$\frac{d\mu}{\mu} = \frac{d\lambda}{\beta(\lambda)},$$

as well as the limits of integration.

So I can live with that change of measure - it just comes from the definition of the running coupling!

So we would have $G_n ( \{ p_i \} ; \lambda(\mu) , \mu) = e^{-\int_{\lambda(\mu_0)^{\lambda(\mu)} \frac{d \lambda}{\beta(\lambda)} \gamma(\lambda(\mu))}G_n( \{p_i \} ; \lambda(\mu_0),\mu_0)$


So for this part where i am asked to solve to one loop accuracy, i find that

$G_n ( \{ p_i \} ; \lambda(\mu) , \mu) = e^{-\int_{\lambda(\mu_0)^{\lambda(\mu)} \frac{d \lambda}{\beta(\lambda)} \times 0}G_n( \{p_i \} ; \lambda(\mu_0),\mu_0)$

Since $\gamma(\lambda(s))=n \gamma_\phi$ and we saw $\gamma_\phi = \mathcal{O}(\lambda^2)$ and so can be considered negligible.

Therefore the thing we integrate in the exponent is 0 anyway so what was the point in changing variables?

 

[latentcorpse]

I thought that

$$\int_{\mu_0}^\mu \frac{ds}{s} \gamma(g(s)) = \int_{\lambda(\mu_0)}^{\lambda(\mu)} d\lambda \frac{\gamma(\lambda)}{\beta(\lambda)} ,$$

which is not what you have.



You'll want to clean this up a bit. I found that

$$I = \sqrt{\frac{2\pi}{m}} \left( 1 - \frac{3\lambda}{4! m^4} + \frac{105 \lambda^2}{2(4!)^2 m^8} + \cdots \right).$$

and why don't you have any $\Gamma$;s in here?

 

[fzero]

So I can live with that change of measure - it just comes from the definition of the running coupling!

So we would have $G_n ( \{ p_i \} ; \lambda(\mu) , \mu) = e^{-\int_{\lambda(\mu_0)^{\lambda(\mu)} \frac{d \lambda}{\beta(\lambda)} \gamma(\lambda(\mu))}G_n( \{p_i \} ; \lambda(\mu_0),\mu_0)$


So for this part where i am asked to solve to one loop accuracy, i find that

$G_n ( \{ p_i \} ; \lambda(\mu) , \mu) = e^{-\int_{\lambda(\mu_0)^{\lambda(\mu)} \frac{d \lambda}{\beta(\lambda)} \times 0}G_n( \{p_i \} ; \lambda(\mu_0),\mu_0)$

Since $\gamma(\lambda(s))=n \gamma_\phi$ and we saw $\gamma_\phi = \mathcal{O}(\lambda^2)$ and so can be considered negligible.

Therefore the thing we integrate in the exponent is 0 anyway so what was the point in changing variables?

In the last part of the problem you're told to consider $$\gamma_\phi\neq 0$$.

and why don't you have any $\Gamma$;s in here?

Their values have been evaluated.

 

[latentcorpse]

In the last part of the problem you're told to consider $$\gamma_\phi\neq 0$$.



Their values have been evaluated.

I'm still on the bit before that where we're just told to solve to one loop accuracy. For this part, we aren't given any values so should I just leave the final result

$G_n( \{ p_i \} ; \lambda(\mu), \mu) = e^{-\int_{\mu_0}^\mu \frac{ds}{s} \gamma(\lambda(s))} G_n( \{ p_i \} ; \lambda(\mu_0) , \mu_0)$?

Then for the final bit where we are given the value I find

$G_n(\mu) = e^{-\int_{\lambda(\mu_0)}^{\lambda(\mu)} \frac{d \lambda}{\beta(\lambda)} nc \lambda^2 } G_n(\mu_0)$

But we know $\beta_\lambda = \frac{3 \lambda^2}{16 \pi^2}$ from earlier so

$G_n(\mu) = e^{-\int_{\lambda(\mu_0)}^{\lambda(\mu)} \frac{16 \pi^2 nc}{3} d \lambda}G_n(\mu_0)$
$G_n(\mu) = e^{- \frac{16 \pi^2nc}{3} (\lambda(\mu)-\lambda(\mu_0)} G_n(\mu_0)$

So can I simplify this any further?

Finally, what would your comment be about how the solution changes when $\gamma_\phi \neq 0$? I can't really think of much else to say other than that it is no longer constant??

Thanks.

 

[fzero]

I don't think the problem is asking for more than the result that you've derived. You could determine the asymptotic behavior of $$\lambda$$ as a function of scale and the resulting asymptotics of the the correlation functions depending on the sign of c, but the problem isn't asking for that.

 

[latentcorpse]

I don't think the problem is asking for more than the result that you've derived. You could determine the asymptotic behavior of $$\lambda$$ as a function of scale and the resulting asymptotics of the the correlation functions depending on the sign of c, but the problem isn't asking for that.

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?

(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?

(iii) Is it possible to show that your definition $$\gamma_\phi = - \mu \frac{d(\log \phi)}{d\mu}.$$ 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.

 

[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 $$\gamma_\phi = - \mu \frac{d(\log \phi)}{d\mu}.$$ 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.