12.4 Possible values of angular momentum

The fact that the angular momentum ladders of the previous section must have a top and a bottom rung restricts the possible values that angular momentum can take. This section will show that the azimuthal quantum number $j$ can either be a nonnegative whole number or half of one, but nothing else. And it will show that the magnetic quantum number $m$ must range from $-j$ to $+j$ in unit increments. In other words, the bosonic and fermionic example ladders in figures 12.1 and 12.2 are representative of all that is possible.

To start, in order for a ladder to end at a top rung $m_{\rm {max}}$, ${\widehat J}^+\big\vert l\:m\big\rangle $ has to be zero for $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {max}}$. More specifically, its magnitude $\left\vert{\widehat J}^+\big\vert j\:m\big\rangle \right\vert$ must be zero. The square magnitude is given by the inner product with itself:

\begin{displaymath}
\left\vert{\widehat J}^+\big\vert j\:m\big\rangle \right\v...
...rt{\widehat J}^+\big\vert j\:m\big\rangle \bigg\rangle
= 0.
\end{displaymath}

Now because of the complex conjugate that is used in the left hand side of an inner product, (see chapter 2.3), ${\widehat J}^+$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\widehat J}_x+{\rm i}{\widehat J}_y$ goes to the other side of the product as ${\widehat J}^-$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\widehat J}_x-{\rm i}{\widehat J}_y$, and you must have

\begin{displaymath}
\left\vert{\widehat J}^+\big\vert j\:m\big\rangle \right\v...
...ehat J}^-{\widehat J}^+\big\vert j\:m\big\rangle \bigg\rangle
\end{displaymath}

That operator product can be multiplied out:

\begin{displaymath}
{\widehat J}^-{\widehat J}^+\equiv({\widehat J}_x-{\rm i}{...
...{\widehat J}_x{\widehat J}_y - {\widehat J}_y{\widehat J}_x),
\end{displaymath}

but ${\widehat J}_x^2+{\widehat J}_y^2$ is the square angular momentum ${\widehat J}^2$ except for ${\widehat J}_z^2$, and the term within the parentheses is the commutator $[{\widehat J}_x,{\widehat J}_y]$ which is according to the fundamental commutation relations equal to ${\rm i}\hbar{\widehat J}_z$, so
\begin{displaymath}
{\widehat J}^-{\widehat J}^+ = {\widehat J}^2 - {\widehat J}_z^2 - \hbar{\widehat J}_z %
\end{displaymath} (12.5)

The effect of each of the operators in the left hand side on a state $\big\vert j\:m\big\rangle $ is known and the inner product can be figured out:
\begin{displaymath}
\left\vert{\widehat J}^+\big\vert j\:m\big\rangle \right\vert^2 =
j(j+1)\hbar^2 - m^2 \hbar^2 - m\hbar^2 %
\end{displaymath} (12.6)

The question where angular momentum ladders end can now be answered:

\begin{displaymath}
j(j+1)\hbar^2 - m_{\rm max}^2 \hbar^2 - m_{\rm max}\hbar^2 = 0
\end{displaymath}

There are two possible solutions to this quadratic equation for $m_{\rm {max}}$, to wit $m_{\rm {max}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j$ or $-m_{\rm {max}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j+1$. The second solution is impossible since it already would have the square $z$ angular momentum exceed the total square angular momentum. So unavoidably,

\begin{displaymath}
m_{\rm max} = j
\end{displaymath}

That is one of the things this section was supposed to show.

The lowest rung on the ladder goes the same way; you get

\begin{displaymath}
{\widehat J}^+{\widehat J}^- = {\widehat J}^2 - {\widehat J}_z^2 + \hbar{\widehat J}_z %
\end{displaymath} (12.7)

and then
\begin{displaymath}
\left\vert{\widehat J}^-\big\vert j\:m\big\rangle \right\vert^2 =
j(j+1)\hbar^2 - m^2 \hbar^2 + m\hbar^2 %
\end{displaymath} (12.8)

and the only acceptable solution for the lowest rung on the ladders is

\begin{displaymath}
m_{\rm min} = -j
\end{displaymath}

It is nice and symmetric; ladders run from $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$j$ up to $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j$, as the examples in figures 12.1 and 12.2 already showed.

And in fact, it is more than that; it also limits what the quantum numbers $j$ and $m$ can be. For, since each step on a ladder increases the magnetic quantum number $m$ by one unit, you have for the total number of steps up from bottom to top:

\begin{displaymath}
\mbox{total number of steps } = m_{\rm max} - m_{\rm min} = 2j
\end{displaymath}

But the number of steps is a whole number, and so the azimuthal quantum $j$ must either be a nonnegative integer, such as 0, 1, 2, ..., or half of one, such as $\frac12$, $\frac32$, ...

Integer $j$ values occur, for example, for the spherical harmonics of orbital angular momentum and for the spin of bosons like photons. Half-integer values occur, for example, for the spin of fermions such as electrons, protons, neutrons, and $\Delta$ particles.

Note that if $j$ is a half-integer, then so are the corresponding values of $m$, since $m$ starts from $\vphantom0\raisebox{1.5pt}{$-$}$$j$ and increases in unit steps. See again figures 12.1 and 12.2 for some examples. Also note that ladders terminate just before $z$-​momentum would exceed total momentum.

It may also be noted that ladders are distinct. It is not possible to go up one ladder, like the first $Y^m_3$ one in figure 12.1 with ${\widehat J}^+$ and then come down the second one using ${\widehat J}^-$. The reason is that the states $\big\vert j\:m\big\rangle $ are eigenstates of the operators ${\widehat J}^-{\widehat J}^+$, (12.5), and ${\widehat J}^+{\widehat J}^-$, (12.7), so going up with ${\widehat J}^+$ and then down again with ${\widehat J}^-$, or vice-versa, returns to the same state. For similar reasons, if the tops of two ladders are orthonormal, then so is the rest of their rungs.