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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

Now look at

and compare with

We’ll be looking in more detail at .

# References

[1] BR Desai. *Quantum mechanics with basic field theory*. Cambridge University Press, 2009.