Renormalization in Two Dimensions

[MathematicalPhysicist]

Homework Statement

I am looking for references for the scalar field theory in one-space, one-time dimension defined by:

$$\mathcal{L}=-\frac{1}{2}(\partial_{\mu}\phi)^2-\frac{1}{2}m_{2,0}^2\phi^2-\frac{1}{4!}m_{4,0}^2\phi^4-\frac{1}{6!}m_{6,0}^2\phi^6$$

That explains why the only divergences arise from closed loops which start and end on the same vertex.
And basically does renormalization in two dimensions on this Lagrangian density.

The full question is from Brown's textbook on QFT, on page 277 question 2.
I'll type it here:

The Lagrangian is the above one, for scalar field theory in one-space, one-time dimension.

a) Explain why the only divergences arise from closed loops which start and end on the same vertex.
In working out this problem be sure to draw graphs to explain what you are doing.

b) First work out the renormalization associated with the $\phi^6$ interaction.
Suppose four external lines or four lines from other vertices or some combination of external lines and lines from other vertices are connected to this interaction.
They can do so in $6!/2!$ ways. The remaining $\phi^2$ can form a single loop.
This gives a divergent term $\frac{1}{2}m_{6,0}^2<\phi(0)^2>$.
Continue to $n$ dimensions and write: $m_{6,0}^2=m_6^2\mu^{4-2n}$, where $m_6$ is a finite six-leg coupling constant with the dimensions of mass.
Why is $m_6$ finite?
Show that:$$<\phi(0)^2>=m^{n-2}\frac{\Gamma(1-n/2)}{(4\pi)^{n/2}},$$
where $m$ is the mass used for the propagator. We shall see that the particular value of this mass is irrelevant.
Thus the divergent term is given by $$-\frac{1}{2}m_6^2\mu^{2-n}\frac{1}{4\pi}\frac{1}{1-n/2},$$
and this divergence is removed by the renormalization: $$m_{4,0}^2=\mu^{2-n}\bigg[m_4^2+\frac{m_6^2}{4\pi}\frac{1}{n-2}\bigg].$$
Suppose now that two "external" lines are connected to the $\phi^6$ vertex.
They can do so in $6\cdot 5$ ways.
The remaining $\phi^4$ can form two closed loops.
This gives a divergent term $$-\frac{1}{4!}m_{6,0}^23<\phi(0)^2>^2.$$
But the two external lines can also connect to the $\phi^4$ interaction in $4\cdot 3$ ways and the remaining $\phi^2$ can form a closed loop.
This gives a divergent term:$$-\frac{1}{2}m_{4,0}^2<\phi(0)^2>.$$
Show that the divergences are removed by by the renormalization $$m_{2,0}^2=m_2^2+\frac{m_4^2}{4\pi}\frac{1}{n-2}+\frac{1}{2}\frac{m_6^2}{(4\pi)^2}\frac{1}{(n-2)^2}.$$
Note that this is independent of $m$ as advertised; there are no terms of the form $\log (m/\mu) \frac{1}{n-2}$.
Note also the triangular counter term structure.

c) Check this result by computing $$\mu dm_6^2/d\mu , \mu dm_4^2/d\mu , \mu dm_2^2/d\mu$$
and showing that they are finite.

d) check the book by Brown.

I can only do part c) of this question.
We have from part b) that $m_6^2\propto \mu^{2n-4}$ so $\mu dm_6^2/d\mu \propto \mu^{2n-4}=O(1)$ so this is finite since we take $n=2$ since we have two dimensions case here.
$m_4^2 \propto \mu^{n-2}$ so $\mu dm_4^2/d\mu \propto \mu^{n-2} = O(1)$.

Assuming I answered correctly item c), how to answer the other items here?
I am not sure I answered it correctly, though.

Homework Equations

The Attempt at a Solution

 

[nrqed]

Homework Statement

I am looking for references for the scalar field theory in one-space, one-time dimension defined by:

$$\mathcal{L}=-\frac{1}{2}(\partial_{\mu}\phi)^2-\frac{1}{2}m_{2,0}^2\phi^2-\frac{1}{4!}m_{4,0}^2\phi^4-\frac{1}{6!}m_{6,0}^2\phi^6$$

That explains why the only divergences arise from closed loops which start and end on the same vertex.
And basically does renormalization in two dimensions on this Lagrangian density.

The full question is from Brown's textbook on QFT, on page 277 question 2.
I'll type it here:

The Lagrangian is the above one, for scalar field theory in one-space, one-time dimension.

a) Explain why the only divergences arise from closed loops which start and end on the same vertex.
In working out this problem be sure to draw graphs to explain what you are doing.

b) First work out the renormalization associated with the $\phi^6$ interaction.
Suppose four external lines or four lines from other vertices or some combination of external lines and lines from other vertices are connected to this interaction.
They can do so in $6!/2!$ ways. The remaining $\phi^2$ can form a single loop.
This gives a divergent term $\frac{1}{2}m_{6,0}^2<\phi(0)^2>$.
Continue to $n$ dimensions and write: $m_{6,0}^2=m_6^2\mu^{4-2n}$, where $m_6$ is a finite six-leg coupling constant with the dimensions of mass.
Why is $m_6$ finite?
Show that:$$<\phi(0)^2>=m^{n-2}\frac{\Gamma(1-n/2)}{(4\pi)^{n/2}},$$
where $m$ is the mass used for the propagator. We shall see that the particular value of this mass is irrelevant.
Thus the divergent term is given by $$-\frac{1}{2}m_6^2\mu^{2-n}\frac{1}{4\pi}\frac{1}{1-n/2},$$
and this divergence is removed by the renormalization: $$m_{4,0}^2=\mu^{2-n}\bigg[m_4^2+\frac{m_6^2}{4\pi}\frac{1}{n-2}\bigg].$$
Suppose now that two "external" lines are connected to the $\phi^6$ vertex.
They can do so in $6\cdot 5$ ways.
The remaining $\phi^4$ can form two closed loops.
This gives a divergent term $$-\frac{1}{4!}m_{6,0}^23<\phi(0)^2>^2.$$
But the two external lines can also connect to the $\phi^4$ interaction in $4\cdot 3$ ways and the remaining $\phi^2$ can form a closed loop.
This gives a divergent term:$$-\frac{1}{2}m_{4,0}^2<\phi(0)^2>.$$
Show that the divergences are removed by by the renormalization $$m_{2,0}^2=m_2^2+\frac{m_4^2}{4\pi}\frac{1}{n-2}+\frac{1}{2}\frac{m_6^2}{(4\pi)^2}\frac{1}{(n-2)^2}.$$
Note that this is independent of $m$ as advertised; there are no terms of the form $\log (m/\mu) \frac{1}{n-2}$.
Note also the triangular counter term structure.

c) Check this result by computing $$\mu dm_6^2/d\mu , \mu dm_4^2/d\mu , \mu dm_2^2/d\mu$$
and showing that they are finite.

d) check the book by Brown.

I can only do part c) of this question.
We have from part b) that $m_6^2\propto \mu^{2n-4}$ so $\mu dm_6^2/d\mu \propto \mu^{2n-4}=O(1)$ so this is finite since we take $n=2$ since we have two dimensions case here.
$m_4^2 \propto \mu^{n-2}$ so $\mu dm_4^2/d\mu \propto \mu^{n-2} = O(1)$.

Assuming I answered correctly item c), how to answer the other items here?
I am not sure I answered it correctly, though.

Homework Equations

The Attempt at a Solution

For part a), consider looking at the powers of loop momenta in a typical Feynman diagram. What does the propagator look like? What is the measure? For what diagrams will you have ultraviolet divergences?

 

[MathematicalPhysicist]

For part a), consider looking at the powers of loop momenta in a typical Feynman diagram. What does the propagator look like? What is the measure? For what diagrams will you have ultraviolet divergences?

I looked at the appendix of Peskin&Schrodinger and there's no mention of Feynman rules for $\phi^6$, so I don't know how do they look like?

Perhaps in Brown's they appear?

 

[nrqed]

I looked at the appendix of Peskin&Schrodinger and there's no mention of Feynman rules for $\phi^6$, so I don't know how do they look like?

Perhaps in Brown's they appear?

There is no derivative in that term so the Feynman run is simply that these vertices are accompanied by the coefficient $m^2_{0,6}$. So it is like the $\lambda \Phi^4/4!$ term in $\phi^4$ theory.

 

[MathematicalPhysicist]

There is no derivative in that term so the Feynman run is simply that these vertices are accompanied by the coefficient $m^2_{0,6}$. So it is like the $\lambda \Phi^4/4!$ term in $\phi^4$ theory.

But we have already a $\phi^4/4!$ in this Lagrangian.

So how is it like the $\phi^4$ theory?

 

[nrqed]

But we have already a $\phi^4/4!$ in this Lagrangian.

So how is it like the $\phi^4$ theory?

You were asking what the Feynman rule was, so I meant that the Feynman rule is just a constant times the vertex. To prove that, you follow exactly the same step as proving the Feynman rule for the phi^4 vertex. This is what I meant. Of course this vertex has six scalar lines connected to it.

 

[nrqed]

But we have already a $\phi^4/4!$ in this Lagrangian.

So how is it like the $\phi^4$ theory?

You are probably done with this problem, but I will just say add a few words in case someone else is interested in working this out.

Now that we know that there are no momenta in the Feynman rules of the vertices, the only source of momenta in the numerator of Feynman diagram comes from the measure. The momenta in the denominators come from the propagators, as usual. Now do you see why only closed loops starting and ending on the same vertex will produce a UV divergence?