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\begin{document}
\begin{center}

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

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


\noindent
%Posted September 17, 2015 
\hfill {\bf Due  11:59pm Monday, May 4, 2026}




\bigskip
\hrule
\begin{enumerate}

\item {\bf Correlators of composite operators made of free bosons in 1+1 dimensions.}
%The guts of the Veneziano amplitude.}

\def\del{\partial}
This problem is a continuation of the last problem on the previous homework.

Consider a collection of $n$ two-dimensional 
free bosons $X^\mu$ governed by the action
$$ S = -{ 1\over 4\pi g} \int d^2 \sigma \del_a X_\mu \del^a X^\mu. $$
%[The symbol $\alpha'$ is a convention from string theory.
[The coupling $g$ can be absorbed into the definition of $X$ if we prefer, 
but it is useful to leave this coupling constant arbitrary
since different physicists use different conventions for the normalization
and as you will see this affects the appearance of the final answer.]

Until further notice, we will assume that $X$ takes values on the real line.


\begin{enumerate}
\item{} Rotate $e^{\ii S}$ to Euclidean space ($d^2 \sigma = - \ii (d^2 \sigma)_E$)
and compute the Euclidean generating functional
$$Z[J] =  \vev{ e^{ \int (d^2\sigma)_E  J^\mu X_\mu }}
\equiv Z_0^{-1} \int [dX] e^{ \ii S} e^{ \int (d^2\sigma)_E J^\mu X_\mu }
$$
(where $Z_0^{-1} \equiv Z[J=0]$ but please don't worry too much about the normalization of the path integral).
%[NORMALIZATION]

[Hint: use the Green function from the previous problem, and Wick's theorem.
Or use our general formula for Gaussian integrals with sources.]

[Warning: In the problem at hand, even the euclidean kinetic operator has a kernel,
namely the zero-momentum mode.
You will need to do this integral separately.]

[Cultural remark 1: this field theory describes the propagation of featureless strings
in $n$-dimensional flat space $\IR^n$ -- think of $X^\mu(\sigma)$ as the parametrizing the position 
in $\IR^n$ to which the point $\sigma$ is mapped.

Cultural remark 2: this is an example of a {\it conformal field theory}.  
In particular recall that massless scalars in $D=2$ have engineering dimension zero.]



\item{} Show that  
\be\label{eq:veneziano} \vev{ \prod_{i=1}^N : e^{ - i \sqrt{ 2 \alpha^\prime} 
k_i \cdot X(\sigma^{(i)})   } 
:  }
= \delta^n\(\sum_i k_i^\mu\)  \prod_{i,j=1}^N | z_i - z_j |^{ {+} \alpha^\prime g k_i \cdot k_j} \ee
where $ \sigma^{(i)}$ label points in 2d Euclidean space, 
$z_i \equiv \sigma_1^{(i)} + i \sigma_2^{(i)}$, 
$\alpha^\prime$ is a parameter with dimensions of $[ X^2/g]$ 
(called the `Regge slope'), and $k_i^\mu$ are a set of arbitrary
$n$-vectors in the target space.
The $: ... : $ indicate the following prescription for {\it defining}
composite operators.   The prescription is simply
to leave out Wick contractions of objects within a pair of $: ... :$.  
Give a symmetry explanation of the delta function in $k$.


[Cultural remark: this calculation is the central ingredient
in the {\it Veneziano amplitude} for scattering of bosonic strings at tree level.]


\item{} Conclude that the composite operator $\CO_a \equiv  :e^{ \ii a X}:$
has {\it scaling dimension} $\Delta_a = { g a^2\over 2}$, in the sense that 
$$ \vev{ \CO_a(z) \CO_{b}^\dagger(0)} = \delta(a-b) { 1 \over |z|^{ 2 \Delta_a} } . $$

Notice that the correlation functions of these operators do not describe the propagation 
of {\it particles} in any sense.  The operator $ \CO$ produces
some power-law excitation of the CFT soup.


\item Suppose we have one field $(n=1)$ $X$ which takes values on the circle, that is,
we identify
$$ X \simeq X + 2 \pi R ~.$$
What values of $a$ label single-valued operators $ : e^{ \ii a X} :$ ? 
How should we modify \eqref{eq:veneziano}?

\end{enumerate}
  
  
  
  \item {\bf Order parameter exponent at the Wilson-Fisher fixed point.} [bonus problem]
In lecture we outlined the computation of $\eta$ using position space diagrams.  
Find the coefficient $c$ in $\eta = c \eps^2 + \CO(\eps^3)$ 
as a function of $n$ at the $\gO(n)$ Wilson-Fisher fixed point.
Check that the factors of $r_2$ drop out.

[Hint: The answer for the Ising model ($n=1$) is $\eta = {\eps^2\over 54}$.]


\item {\bf OPE.} [bonus problem]
Consider the Gaussian fixed point with $\gO(n)$ symmetry.  
Compute the OPE coefficients for the operators $ \CO_2 \equiv :\phi_a \phi_a:$, 
$\CO_4 \equiv :( \phi_a \phi_a)^2:$, and the identity operator (here $a=1..n$ and the repeated index is summed).
Use this information to compute the beta function, 
find the Wilson-Fisher fixed point and the correlation length critical exponent $\nu$
there.





\item {\bf RG analysis of less-symmetric spin systems.} 


Suppose that we break the rotation symmetry of the $\gO(n)$ model
to the subgroup of $\pi/2$ rotations, \ie~the cubic symmetry,
(for example, for $n=2$, $ (s_1, s_2) \to (s_2, -s_1) $.)
If the spins live on the cubic lattice, a spin-orbit coupling
could do this.

\begin{enumerate}

\item {\cor Don't look at the next part of the problem yet!}
What functions of an $n$-vector $\phi_a$ are invariant under $ \pi/2$ rotations, but not general
rotations?


$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$


$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$

$$\vdots$$



\vfill\eject
\item Show that 
in addition to the usual  $\gO(n)$-symmetric interaction
$$\int d^d x ~ \sum_{a,b=1}^n u \phi_a^2\phi_b^2, $$
the LG free energy should include a term of the form
$$\int d^d x ~ \sum_{a=1}^n v \phi_a^4 .$$
Argue that this is the only new term 
(preserving cubic symmetry but not the full $\gO(n)$ symmetry)
which can be a relevant perturbation of 
the Gaussian fixed point near $ d=4$.

\item 
Treating $ \CO(u) = \CO(v) = \CO(\eps)$, 
redo the analysis of the running couplings in $d=4-\eps$ dimensions to derive 
beta functions for $ u $ and $v$ up to corrections of order $ \CO(u^3) = \CO(v^2u) = ... = \CO(\eps^3) $.

\item Your answer to the previous part will be of the form
\bea
- s \partial_s u &=&  -\eps u + A_1 u^2 + A_2  uv + A_3 v^2 + \CO(u^3 ) \cr 
- s \partial_s v &=& - \eps v + B_1 u^2 + B_2  uv + B_3 v^2 + \CO(u^3 ) .
\eea
You should find that $ A_3= B_1 = 0$.
Find four fixed points: 
\begin{itemize}
\item The gaussian fixed point.
\item A fixed point where only $u\neq 0$.  
\item A fixed point where only $v\neq 0$.  Describe the physics of this fixed point.  (Hint: the action is a sum of $n$ terms.)
\item A fixed point where both $(u,v)$ are nonzero.
\end{itemize}
(In every case, the assumption of $ u \sim v \sim \eps$ is self-consistent.)



\item Analyze the stability of these fixed points 
(by computing the matrix of derivatives of the beta functions at each fixed point).
Draw the phase diagram. 
Which fixed point dominates the critical behavior?
You will want to consider different cases depending on whether $n>4$ or $n<4$.

\item When $ n>4$ you may find that $v$ wants to become negative.  This means that 
the effective potential for $m$ becomes unbounded,  within our approximation.  
What have we left out that will restore sanity?  
What does this mean for the order of the phase transition?  
(Notice that mean field theory predicts a continuous transition, 
so any change in this conclusion is a dramatic effect of the fluctuations, 
more dramatic than just changing the values of critical exponents by a little.)


\end{enumerate}


\end{enumerate}
\end{document}


  
  \item{\bf Topological terms in QM.}
\answer{[from Abanov]}

The purpose of this problem is to demonstrate that 
%(even in this simple QM example)
total derivative terms in the action  (like the $\theta$ term in QCD) do affect the physics.


The euclidean path integral for a particle on a ring with
magnetic flux $\theta = \int \vec B \cdot \dd\vec a $ through the ring is given by
$$ Z = \int [D\phi] e^{ - \int_0^\beta \dd\tau \( { m\over 2 } \dot \phi^2 - \ii { \theta \over 2\pi} \dot \phi\) } ~.$$
Here 
\be\label{eq:ident} \phi \equiv \phi + 2\pi\ee is a coordinate on the ring.
Because of the identification \eqref{eq:ident}, $\phi$ need not be a single-valued function of $\tau$ --
it can wind around the ring.
On the other hand, $ \dot \phi$ is single-valued and periodic and hence has an ordinary Fourier decomposition.
This means that we can expand the field as 
\be\label{eq:fourier} \phi(\tau) = { 2\pi \over \beta} Q \tau + \sum_{\ell \in \IZ {\setminus 0}} \phi_\ell e^{ \ii { 2\pi \over \beta} \ell \tau} .\ee


\begin{enumerate}

\item Show that the $\dot \phi$ term in the action
does not affect the classical equations of motion.  
In this sense, it is a topological term.

\item Using the decomposition \eqref{eq:fourier}, write the partition function  
as a sum over topological sectors labelled by the {\it winding number} $Q \in \IZ$
and calculate it explicitly. 

[Hint: use the Poisson resummation formula
$$ \sum_n f(n) = \sum_l \hat f(2\pi l) $$
where $\hat f(p) = \int dx e^{ - \ii p x } f(x) $ is the fourier transform of $f$.]


\answer{Applied to our case, the Poisson resummation formula says
$$ \sum_{n \in \IZ} e^{ - \half t n^2 + \ii z n } 
= \sqrt{ 2 \pi \over t} \sum_{\ell \in \IZ} e^{ - { 1\over 2 t} \( z - 2 \pi \ell\)^2 } .~~$$
}

%\answer{
%I should have mentioned that more generally the Poisson resummation formula says 
%$$ \sum_n f(n) = \sum_l \hat f(2\pi l) $$
%where $\hat f(p) = \int dx e^{ - \ii p x } f(x) $ is the fourier transform of $f$.
%}


\answer{
Using the given mode expansion and $ \int_0^\beta dt e^{ 2 \pi \ii (l-l') \tau \over \beta}= \beta \delta_{l, l'} $ 
the action is
$$ S[\phi]=  \ii \theta Q + { m (2 \pi Q)^2 \over 2 \beta} + \sum_{\ell \neq 0} { ( 2 \pi \ell)^2 m \over 2 \beta } \phi_\ell \phi_{-\ell} $$
where $ \phi_\ell = \phi_{-\ell}^\star $.  
Thus
\begin{align} Z & =   \sum_{Q\in \IZ} e^{  - \ii \theta Q + { m (2 \pi Q)^2 \over 2 \beta}  } 
\prod_{\ell \neq 0} \int d^2 \phi_\ell e^{ { ( 2 \pi \ell)^2 m \over 2 \beta } \phi_\ell \phi_{\ell}^\star }   
\\ & 
=   \sum_{Q\in \IZ} e^{  - \ii \theta Q + { m (2 \pi Q)^2 \over 2 \beta}  } 
\prod_{\ell \neq 0} \( \beta \over 2 \pi \ell^2 m \) 
\\ 
& \propto  
\sum_{n \in \IZ} e^{ - \beta { 1 \over 2m (2\pi)^2 } ( \theta - 2 \pi n )^2 }
= \sum_{n \in \IZ} e^{ - \beta { 1 \over 2m } \( n -{ \theta \over 2 \pi} \)^2 } 
\end{align} 
where in the last step we used the above Poisson summation formula with 
$ z =  \theta$ and $ t = {m (2\pi)^2 \over \beta}$.
}

\item  Use the result from the previous part to determine the energy spectrum as a function of $\theta$.


\answer{ After the Poisson resummation, this is manifestly the partition function 
of a system with energies $E_n = {1\over 2m} (n - {\theta \over 2\pi} )^2 $.
}



\item{} Derive the canonical momentum and Hamiltonian from the action above
and verify the spectrum.

\answer{
Note that the action given above is the {\it Euclidean} action.  The real time action 
(from which we should derive the hamiltonian) is
$$ S = \int dt  \( \half m \dot \phi^2   + \dot \phi {\theta \over 2 \pi}  \) . $$
This gives $ p = { \partial L \over \partial \dot \phi} = m \dot \phi + { \theta \over  2\pi } $, 
and hence 
$$ H = {\(p - {\theta\over 2 \pi } \)^2 \over 2m} .$$
Now, since $\phi \equiv \phi + 2\pi$, its canonical momentum is quantized, $p \in \IZ$, so
$$ E_n = {1\over 2m } \( n - {\theta \over 2\pi} \)^2 $$
as above.
We find the following spectrum for various $\theta$ (I am plotting the energies of the states with wavenumbers $ n \in [-3, 2]$):\\
\includegraphics[width=.5\textwidth]{fig-pset07-theta-spectrum.png}\\
(In the axis label, $I $ is the moment of inertia of the rotor.)
Notice that when $ \theta=\pi$, the groundstate becomes doubly degenerate.
}


\item Consider what happens in the limit $ m \to0, \theta \to \pi$ with
$ X \equiv { \theta - \pi \over m} \sim \beta^{-1} $ fixed.
Interpret the result as the partition function for a spin 1/2 particle.
What is the meaning of the ratio  $X$ in this interpretation?

\answer{In this limit, the higher bands of energies go off to $\infty$, 
and we are left with a two-state system.   $X$ is a Zeeman field
splitting the two states.
}

\end{enumerate}
%-- particle on a ring.  topological terms 



\end{enumerate}
\end{document}
