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%%from  Rudro Rana Biswas
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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 1 \answer{--~~~ Solutions} }
\end{center}


\noindent
%Posted September 17, 2015 
\hfill {\bf Due  11:59pm Monday, January 12, 2025}

\begin{itemize}
\item
Please hand in your homework electronically via Canvas.
The preferred option is to typeset your homework.   
It is easy to do and you need to learn to do it anyway as a practicing scientist. 
I'll provide LaTeX template file with each of the problem sets.
%\htmladdnormallink{here}{https://mcgreevy.physics.ucsd.edu/w21/tex-template.tar.gz}.
If you need help getting set up or have any other questions please email me.  I am happy to give TeX advice.

\item
To hand in your homework, please 
submit a pdf file through the course's Canvas website,
under the assignment labelled hw 01.

\item I will post solutions as soon as everyone has submitted their homework.
%email a pdf file
%Please put the filename in the format:
%\begin{verbatim}
%220-hw01-YourLastName-YourFirstName.pdf
%\end{verbatim}
% to mcgreevy@physics.ucsd.edu
% with the subject line:
%\begin{verbatim}
%212C-hw01
%\end{verbatim}
\end{itemize}

 Thanks in advance for following these guidelines.  Please ask me by email if you have any trouble.



%I will add another problem later today.
\bigskip
\hrule
\begin{enumerate}



\item {\bf Brain-warmer.}  
Convince yourself that 
$$ \(\partial_g\)^n e^{ - 1 / g } |_{g=0} = 0 ~~\forall n .$$
This means that a function of the form $ e^{ - 1/g} $ does not have a useful series expansion
about $g=0$.


\answer{If you put your answer inside the `answer' command, it will show up in blue.}

\item {\bf Scale invariant quantum mechanics.}

Consider the action for one quantum variable $r$ with $ r>0 $ and 
% \in \IR^+$ 
%$$ S[r] = \int dt \( \half \dot r^2 - V(r) \) ,~~~~V(r) = { \lambda \over r^2}~.$$
$$ S[r] = \int dt \( \half m \dot r^2 - V(r) \) ,~~~~V(r) = { \lambda \over r^2}~.$$


\begin{enumerate}


\item 
%\item 
Show that the (non-relativistic) mass parameter $m$ can be eliminated by a multiplicative redefinition of the field $r$ or of the time $t$.
As a result, convince yourself that the physics of interest here should only depend on 
the combination $m\lambda$.
Show that the coupling $m\lambda$ is dimensionless: $ [m\lambda]=0$.


\item Show that this action is {\it scale invariant}, i.e.~show that 
the transformation 
%\cor
\be\label{eq:scale} r(t) \to s^\alpha \cdot r(st)~~~\ee
%\be\label{eq:scale} r(t) \to r_s(t_s) \equiv s^\alpha \cdot r(t)~~~, ~~~~ t \to t_s = t/s \ee
%\cob
(for some $\alpha$ which you must determine),
(with $s \in \IR^+$) is a symmetry.  %$$ \delta r(t) = a \alpha r(t) . $$


\item Find the associated Noether charge $ \CD$.
For this last step, it will be useful to note that the infinitesimal version of 
\eqref{eq:scale} is ($ s = e^{a}, a \ll 1 $)
%\cor 
$$ \delta r(t) = a \(  \alpha + t {d \over dt} \) r(t) . $$


\begin{tcolorbox}[breakable]
[Hint: In field theory the best way to find the Noether current is the following.
If we know that under a transformation $ \phi \to \phi_\eps$ with
parameter $ \eps$ constant in spacetime, 
the action does not change: $ S[\phi] = S[\phi_\eps] $
then if we allow $\eps = \eps(x)$ (infinitesimal) 
then the variation must be proportional to derivatives of $\eps$: 
\be \delta S \equiv S[\phi_{\eps(x) } ] - S[\phi] = \int d^D x \partial_\mu \eps j^\mu(x) \label{eq:variation}\ee
for some functional of the fields $j^\mu$.
The RHS is $ \delta S = - \int d^D x  \eps \partial_\mu j^\mu $
by integration by parts (we assume no boundary).
$j^\mu$ is the conserved Noether current, which is conserved 
$ \partial_\mu j^\mu = 0 $
when evaluated on a solution of the equations of motion.
This is because at a solution of the equations of motion, the action is invariant under {\it any} variation $\delta S = 0$, 
including $\phi \to \phi_\eps$ spacetime-dependent $\epsilon$, which given \eqref{eq:variation} requires $ \partial_\mu j^\mu=0$.
As a result, its time component, integrated over space, is time independent: 
$$ { d \over dt} Q \equiv  { d \over dt}  \int d^{D-1} \vec x j^0  = - \int \nabla \cdot j = 0 $$
(again we ignore boundary terms, in space).
This method is superior to formulae you may remember from classical mechanics
(like $ Q = { \partial L \over \partial \dot r } \delta r + F^0 $ where $\delta L = \dot F^0$)
because it makes no assumptions about the dependence of the Lagrangian 
on $\dot q$ and it doesn't require remembering anything.]
\end{tcolorbox}


\item Find the position-space Hamiltonian $\HH$ governing the dynamics of $r$.  
Show that the Schr\"odinger equation is Bessel's equation
$$ \( - 
%\cor \half \cob 
{\partial_r^2\over 2m} + { \lambda \over r^2 } \) \psi_E(r) = E \psi_E(r) .$$
%\cor 
Show that the Noether charge associated $\CD$ with scale transformations ($\equiv$ dilatations) satisfies:
 $ [ \CD, \HH] =  { -} \ii \HH$.  This equation says that the Hamiltonian has a 
 definite scaling dimension, i.e.~that its scale tranformation is
 $ \delta \HH = \ii a [ \CD , \HH ] = { +}  a \HH$. \cob
 Note that you should not need to use arcane facts about Bessel functions,
 only the asymptotic analysis of the equation, in subsequent parts of the problem.

 
\item Describe the behavior of the solutions to this equation 
as $ r \to 0$.
[Hint: in this limit you can ignore the RHS.  Make a power-law ansatz: $\psi(r) \sim r^{\Delta}$ and find $\Delta$.]


\item What happens if $  2m \cob \lambda < - {1\over 4} $ ?
It looks like there is a continuum of negative-energy solutions (boundstates).
This is another example of a {\it too-attractive} potential.





\item  A hermitian operator has orthogonal eigenvectors.
We will show next that to make $\HH$ hermitian when $ 2m \cob \lambda < - {1\over 4}$,
we must impose a constraint on the wavefunctions: 
\be\label{eq:self-adjoint} \( \psi^\star_E \partial_r \psi_E - \psi_E \partial_r \psi_E^\star \)|_{r=\eps}= 0 ~,\ee
where $\epsilon$ is an infinitesimal cutoff.
There are two useful perspectives on this condition: one is that 
the LHS is the probability current passing through the point $r=\eps$.  
If it is nonzero, it means probability is leaking into a hole in the world at $r=\eps$.  

The other perspective is the following.  
Consider two eigenfunctions: $$ \HH \psi_E = E \psi_E ,~~\HH \psi_{E'} = E'\psi_{E'} . $$
Multiply the first equation by $\psi_{E'}^\star$ and integrate;
multiply the second by $\psi_{E}^\star$ and integrate;
take the difference (maybe take the complex conjugate of the second term).  Show that the result is a boundary term which must vanish when $ E = E'$.



\item Show that the condition 
\eqref{eq:self-adjoint} is empty for $ 2m \cob \lambda > -{1\over 4}$.
We must impose a further condition
on the eigenfunctions for $ 2m \cob \lambda <  - {1\over 4}$ in order to satisfy \eqref{eq:self-adjoint}.  
[Really, we want to impose a boundary condition at $r=0$, but in order to actually do this, we must impose the condition at $r=\eps$ for some UV cutoff $\eps$.
Notice that this regulator explicitly breaks scale invariance.]
Show that the resulting spectrum of boundstates 
has a {\it discrete} scale invariance.

[Cultural remark: For some reason I don't know, restricting the Hilbert space in this way is called a
{\it self-adjoint extension}.]



\item{} [Extra credit] Consider instead a 
particle moving in $\IR^d$ with a central $1/r^2$ potential, $r^2 \equiv \vec x \cdot \vec x$,
$$ S[\vec x] = \int dt  \(  
 \half m  \cob
%\half
\dot{\vec x}  \cdot \dot{\vec x}  - {\lambda \over r^2} \) .$$
Show that the same analysis applies 
(e.g.~to the s-wave states)
with minor modifications.  

[A useful intermediate result is the following representation of (minus) the laplacian in $\IR^d$:
$$ \vec p^2 =- { 1\over r^{d-1} } \partial_r \(r^{d-1} \partial_r \) + {\hat L^2 \over r^2} , ~~~
\hat L^2 \equiv \half \hat L_{ij} \hat L_{ij}, ~~ L_{ij} = - \ii \( x_i \partial_j - x_j \partial_i \) , $$
where $ r^2 \equiv x^i x^i$.
By `s-wave states' I mean those annihilated by $ \hat L^2$.]


\end{enumerate}

\end{enumerate}




%\end{enumerate}
\end{document}
