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%%from  Rudro Rana Biswas
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bookmarks={false}, pdfauthor={John McGreevy}, pdftitle={Yay, physics!}]{hyperref}
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\begin{document}
\begin{center}

University of California at San Diego -- 
Department of Physics --
Prof.~John McGreevy

{\Large\bf  Physics 215B QFT Winter 2026}\\
{\Large\bf Assignment 5 \answer{--~~~ Solutions} }
\end{center}


\noindent
%Posted September 17, 2015 
\hfill {\bf Due  11:59pm Monday, February 9, 2026}




\bigskip
\hrule
\begin{enumerate}



\item {\bf Brain-warmer.} [optional]
%\begin{enumerate}
%\item 
%Show that we did the right thing in the numerator of the electron self-energy: 
%use the Clifford algebra to show that 
%$$  \gamma^\mu \( x \slashed{p} + m_0 \) \gamma_\mu = - 2  x \slashed{p}  + 4 m_0 . $$

%\item 
Prove the Gordon identities
$$ \bar u_2  \(q^\nu \sigma_{\mu\nu}   \) u_1 
= \ii \bar u_2  \( (p_1 + p_2)_\mu - (m_1 + m_2) \gamma_\mu \) u_1 
$$ 
and
$$ \bar u_2  \( (p_1+p_2)^\nu  \sigma_{\mu\nu}   \) u_1 
= \ii \bar u_2  \( {\cob(p_2 - p_1)}_\mu - {\cob (m_2 - m_{1})} \gamma_\mu \) u_1 
$$ 
where $q  \equiv {\cob p_2 - p_1} $ and $ \slashed{p}_1 u_1 = m_1 u_1, \bar u_2 \slashed{p}_2 = m_2 \bar u_2$,
using the definitions and the Clifford algebra.
%\end{enumerate}















\item {\bf Pauli-Villars practice.}

Consider a field theory of two scalar fields with
$$ \CL = - \half \phi \Box \phi - \half m^2 \phi^2 
- \half \Phi\Box  \Phi - \half M^2 \Phi^2 - g \phi \Phi^2+ \text{counterterms} . $$

Compute the one-loop contribution to the self-energy of $ \Phi$.
Use a Pauli-Villars regulator -- introduce a second copy of the $\phi$
field of mass $\Lambda$ with the wrong-sign propagator.




Determine the counterterms required to impose 
that the { $\Phi$} propagator has a pole 
at { $p^2 =M^2 $} with residue $1$.




\item {\bf Tadpole diagrams.}

\begin{enumerate}

\item{}
Why don't we worry about the following diagram 
\vspace{1em}
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	\draw[fermion] (0:1)--(0,0);
	% \node at (-140:1.2) {$e$};
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	% \node at (.5,.3) {$\gamma$};	
	\draw[fermion] (1.5,1) arc (360:0:.5);
%	\draw[scalarnoarrow] (2,0) arc (0:-180:.5);
	\draw[vector] (1,0) --(1,.5);
	\draw[fermionbar] (1,0) --(2,0);
	% \node at (2.5,.3) {$\gamma$};	
\end{tikzpicture}
as a correction to the electron self-energy in QED?  




\end{enumerate}



 For the remainder of the problem, we consider $\phi^3$ theory with a (small) mass:
$$ S = \int d^D x \( \half (\partial\phi)^2 - \half m^2 \phi^2 - {g \over 3!} \phi^3 \) .  $$

\begin{enumerate}[resume]

\item{} Notice that unlike $\phi^4$ theory (or QED), there is no symmetry that forbids a one-point function for the scalar.
Why don't we lose generality by not adding a term linear in $\phi$ to the Lagrangian?   



\item{}
Now think about the following contribution to the scalar self-energy: 
%\parbox{1.3pt}{
\vspace{1em}
\begin{tikzpicture}[line width=1.3 pt, scale=1]
	\draw[scalarnoarrow] (0:1)--(0,0);
	% \node at (-140:1.2) {$e$};
	% \node at (140:1.2) {$e$};
	% \node at (.5,.3) {$\gamma$};	
	\draw[scalarnoarrow] (1.5,1) arc (360:0:.5);
%	\draw[scalarnoarrownoarrow] (2,0) arc (0:-180:.5);
	\draw[scalarnoarrow] (1,0) --(1,.5);
	\draw[scalarnoarrow] (1,0) --(2,0);
	% \node at (2.5,.3) {$\gamma$};	
\end{tikzpicture}
%} \\
Show that in the limit $m\to 0$ there is an IR divergence.
By thinking about the significance for the scalar potential of this part of the diagram
\vspace{1em}
\begin{tikzpicture}[line width=1.3 pt, scale=.5]
%	\draw[scalarnoarrow] (0:1)--(0,0);
	% \node at (-140:1.2) {$e$};
	% \node at (140:1.2) {$e$};
	% \node at (.5,.3) {$\gamma$};	
	\draw[scalarnoarrow] (1.5,1) arc (360:0:.5);
%	\draw[scalarnoarrownoarrow] (2,0) arc (0:-180:.5);
	\draw[scalarnoarrow] (1,0) --(1,.5);
%	\draw[scalarnoarrow] (1,0) --(2,0);
	% \node at (2.5,.3) {$\gamma$};	
\end{tikzpicture}
explain the meaning of this divergence.


\end{enumerate}


\item {\bf Scale invariance in QFT in $D=0+0$, part 3.}  [I got this problem from Frederik Denef.]

We continue our study of QFT in $D=0+0$ with two fields:
$$ Z = \int dP_X dP_Y dX dY e^{ - H/ T} .$$
Let's start by considering again
\be \label{eq:3.3} H = \half P_X^2 + \half P_Y^2 + V(X,Y),~~~~V(X,Y) = aX^4 + b Y^8 \ee
for some nonzero constants $a,b$.   

A generic relevant deformation of \eqref{eq:3.3} will flow to a Gaussian fixed point $V(X,Y) \sim X^2 + Y^2$ in the IR.  
Some other, more fine-tuned deformations will flow to other fixed points.  
For example, $\delta V(X,Y) = \eps Y^4$ will flow to 
$V(X,Y) = X^4 + Y^4$.  But something more interesting happens for $\delta V(X,Y) = \eps X^2 Y^2$.  
This deformation is a relevant perturbation of \eqref{eq:3.3} in the sense that $\delta V(\lambda^{1/4} X, \lambda^{1/8} Y) = \lambda^\kappa V(X,Y)$ 
with $\kappa = 3/4 < 1$.  But it is not true that the model simply flows to a fixed point with $V \propto X^2 Y^2 $ in the IR.  
That's because the model with such a potential has a divergent partition function: 
$ \int_{-\infty}^\infty dX \int_{-\infty}^\infty dY e^{ - \eps X^2 Y^2 /T} \propto \sqrt{T\over \eps} \int { dX \over |X|} = \infty $.  We cannot throw away the higher-order terms 
because they regulate the large-$X$ and large-$Y$ behavior of the integral.  Thus, in this model, the UV does not completely decouple from the IR.
As a consequence, naive scaling arguments break down, and the partition function develops ``anomalous" logarithmic dependence on $T$ for small $T$.  


\begin{enumerate}
\item \label{part-n} Compute the partition function for the model \eqref{eq:3.3} deformed by $\delta V(X,Y) = \eps X^2 Y^2 $ analytically using Mathematica 
or some other symbolic software.  This will give a horrible mess of hypergeometric functions.  Expand it at small $T$ and you should find something of the form
\be\label{eq:power-law} Z = Z_0 T^c \log { \Lambda \over T} \ee
up to corrections suppressed by positive powers of $\sqrt{T/\Lambda}$.
Find the constants $Z_0, c, \Lambda$.  
Plot the function and compare with your approximate expression in the small-$T$ regime.
The overall normalization $Z_0$ does not mean anything in classical statistical mechanics.  


%\answer{I'm going to wait to post this solution until next week.
%\end{enumerate}\end{enumerate}\end{document}
%}




\item Using \eqref{eq:power-law}, compute the dimensionless quantities $U/T$ and $C$ and plot them as a function of $T$ at small $T$.  (Without the logarithmic dependence on $T$, these would be equal.)  Check that in the strict limit $T\to 0$, you get the values for $U/T$ and $C$ that you would have guessed based on naive scaling arguments for $V \propto X^2 Y^2$.  Note that a logarithm varies more slowly than the $T^{1/2}$ corrections that we threw away. 


\item To what extent does the IR physics depend on the UV completion of the $V\propto X^2 Y^2 $ model?  We could have started with $ V = a X^8 + b Y^8+ \eps X^2 Y^2 $ instead.  This model would have different high-temperature physics.  Redo part \label{part-n} for this potential. 
You'll find an equally-horrendous, but different combination of hypergeometric functions.  Which of the parameters $Z_0, c, \Lambda$ are the same? 



\item The result of the previous part remains true for any other UV completion of the $V \propto X^2 Y^2$ model, as long as $\delta V = \eps X^2 Y^2$ 
remains a relevant deformation.  In fact, we could equally well just take $ V = \eps X^2 Y^2$ and impose a hard cutoff on the $X$ and $Y$ integrals at some fixed values $|X| \leq X_0, |Y|\leq Y_0$ (this is like $V = X^n + Y^n$ with $n \to \infty$).  Check that this again reduces to \eqref{eq:power-law}.  



\item \label{parte} In view of this apparent universality of \eqref{eq:power-law} at low $T$, it is desirable to have a way of deriving it without having to take the detour involving the horrendous hypergeometric functions.  Here is one way.  We use the hard cutoff $|X|\leq  L, |Y|\leq L$, so that the position-space factor is
\be\label{eq:cutoffZ} Z_V(T, L) = \int_{-L}^L dX \int_{-L}^L dY e^{ - X^2 Y^2 /T } \ee
where we've set $\eps = 1$ by a choice of temperature units.  
A rescaling of the integration variables $(X,Y) \to (T^{1/4} X, T^{1/4} Y) $ shows that $Z_V(T,L) = \sqrt{T} F(T^{-1/4} L) $ for some function $F$ of one variable.  To find $F$, compute $L \partial_L Z_V$ directly from \eqref{eq:cutoffZ}.  
By another suitable rescaling, show that $L\partial_L Z$ is finite and easily computable for $L^4 /T \to \infty$.  
Infer from this the dependence on the cutoff $L$ in the regime $T \ll L^4$ and thus the function $F$ in this regime.  This reproduces \eqref{eq:power-law}.  

\item We conclude that even when some kind of UV completion is required to give finite answers, the observable low-energy physics remains essentially independent of the UV completion.  The infinite number of possible UV completions all flow in the IR to a partition function of the same form \eqref{eq:power-law}, with the details of the UV completion all lumped into a single scale parameter $\Lambda$.  
In fact, in the absence of other reference scales that can be used to fix a unit of temperature, the parameter $\Lambda$ does not really label physically distinct models, since we can always choose units with $\Lambda=1$.  Equivalently, only dimensionless quantities (and relations between them) are physically meaningful.  Examples of such dimensionless quantities are $C$ and $u \equiv U/T$.  Show that $C$ and $u$ obey a universal relation $C = f(u)$ with $f(u)$ independent of $T$ and $\Lambda$, and thus independent of the UV completion of the $X^2Y^2 $ model.  In the same spirit, show that the function $g(u)$ in the flow equation $T\partial_T u = g(u) $ is independent of the UV completion.


\item Show that on the other hand $f(u)$ and $g(u)$ {\it do} depend on the IR part of the potential, for example by comparing the IR potential $V = X^2Y^2$ considered above to another IR potential such as $V = X^6 Y^6$.  


\end{enumerate}






\end{enumerate}
\end{document}
