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Disclaimer.
Peeter’s lecture notes from class. May not be entirely coherent.
Recap.
Recall our table
First column
Let’s start with computation of the kets in the lowest position of the first column, which we will obtain by successive application of the lowering operator to the state
Recall that our lowering operator was found to be (or defined as)
so that application of the lowering operator gives us
Proceeding iteratively would allow us to finish off this column.
Second column
Moving on to the second column, the top most element in the table
can only be made up of 
with 
. There are two possibilities
So for some 
and 
to be determined we must have
Observe that these are the same kets that we ended up with by application of the lowering operator on the topmost element of the first column in our table. Since 
and 
are orthogonal, we can construct our ket for the top of the second column by just seeking such an orthonormal superposition. Consider for example
With 
we find that 
, so we have
So we find, for real 
and 
that
for any orthonormal pair of kets 
and 
. Using this we find
This will work, although we could also multiply by any phase factor if desired. Such a choice of phase factors is essentially just a convention.
The Clebsch-Gordon convention
This is the convention we will use, where we
\begin{itemize}
\item choose the coefficients to be real.
\item require the coefficient of the 
term to be 
\end{itemize}
This gives us the first state in the second column, and we can proceed to iterate using the lowering operators to get all those values.
Moving on to the third column
can only be made up of with 
. There are now three possibilities
and 2 orthogonality conditions, plus conventions. This is enough to determine the ket in the third column.
We can formally write
where
and
are the Clebsch-Gordon coefficients, sometimes written as
Properties
\begin{enumerate}
\item 
only if 
This is sometimes called the triangle inequality, depicted in figure (\ref{fig:qmTwoL18:qmTwoL18fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL18fig1}
\caption{Angular momentum triangle inequality.}
\end{figure}
\item only if 
.
\item Real (convention).
\item 
positive (convention again).
\item Proved in the text. If follows that
\end{enumerate}
Note that the 
are all real. So, they can be assembled into an orthogonal matrix. Example
Example. Electrons
Consider the special case of an electron, a spin 
particle with 
and 
where we have
possible values of 
are 
Our table representation is then
Here 
can only have contributions from
from the same two. So using this and conventions we can work out (in section 28 page 524, of our text [1]).
Tensor operators
section 29 of the text.
Recall how we characterized a rotation
Here we are using an active rotation as depicted in figure (\ref{fig:qmTwoL18:qmTwoL18fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL18fig2}
\caption{active rotation}
\end{figure}
Suppose that
so that
rotates in the same way. Rotating a ket as in figure (\ref{fig:qmTwoL18:qmTwoL18fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL18fig3}
\caption{Rotating a wavefunction.}
\end{figure}
Rotating a ket
using the prescription
and write
![\begin{aligned}{\left\lvert {\psi'} \right\rangle} = U[M] {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.29)](http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D%7B%5Cleft%5Clvert+%7B%5Cpsi%27%7D+%5Cright%5Crangle%7D+%3D+U%5BM%5D+%7B%5Cleft%5Clvert+%7B%5Cpsi%7D+%5Cright%5Crangle%7D%5Cend%7Baligned%7D+%5Chspace%7B%5Cstretch%7B1%7D%7D%283.29%29&bg=fafcff&fg=2a2a2a&s=0 "\begin{aligned}{\left\lvert {\psi'} \right\rangle} = U[M] {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.29)")
Now look at
and compare with
![\begin{aligned}{\left\langle {\psi'} \right\rvert} \mathcal{O} {\left\lvert {\psi'} \right\rangle}={\left\langle {\psi} \right\rvert} \underbrace{U^\dagger[M] \mathcal{O} U[M]}_{{*}} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.31)](http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D%7B%5Cleft%5Clangle+%7B%5Cpsi%27%7D+%5Cright%5Crvert%7D+%5Cmathcal%7BO%7D+%7B%5Cleft%5Clvert+%7B%5Cpsi%27%7D+%5Cright%5Crangle%7D%3D%7B%5Cleft%5Clangle+%7B%5Cpsi%7D+%5Cright%5Crvert%7D+%5Cunderbrace%7BU%5E%5Cdagger%5BM%5D+%5Cmathcal%7BO%7D+U%5BM%5D%7D_%7B%7B%2A%7D%7D+%7B%5Cleft%5Clvert+%7B%5Cpsi%7D+%5Cright%5Crangle%7D%5Cend%7Baligned%7D+%5Chspace%7B%5Cstretch%7B1%7D%7D%283.31%29&bg=fafcff&fg=2a2a2a&s=0 "\begin{aligned}{\left\langle {\psi'} \right\rvert} \mathcal{O} {\left\lvert {\psi'} \right\rangle}={\left\langle {\psi} \right\rvert} \underbrace{U^\dagger[M] \mathcal{O} U[M]}_{{*}} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.31)")
We’ll be looking in more detail at 
.
References
[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.
each with components 
, 
, 
, 
, 
can be thought of as “length”s of the respective angular momentum vectors.
, and 
implies that 
.
(singlet).
. Triplet: 
.
pairs our partition function is
is
cancellation, is
)
term will dominate.
