12.6 Triplet and singlet states

With the ladder operators, you can determine how different angular momenta add up to net angular momentum. As an example, this section will examine what net spin values can be produced by two particles, each with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. They may be the proton and electron in a hydrogen atom, or the two electrons in the hydrogen molecule, or whatever. The actual result will be to rederive the triplet and singlet states described in chapter 5.5.6, but it will also be an example for how more complex angular momentum states can be combined.

The particles involved will be denoted as $a$ and $b$. Since each particle can have two different spin states $\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\...
...\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle $ and $\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\...
...\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle $, there are four different combined product states:

\begin{displaymath}
{\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scrip...
...\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b.
\end{displaymath}

In these product states, each particle is in a single individual spin state. The question is, what is the combined angular momentum of these four product states? And what combination states have definite net values for square and $z$ angular momentum?

The angular momentum in the $z$-​direction is simple; it is just the sum of those of the individual particles. For example, the $z$-​momentum of the ${\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/...
...rn-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b$ state follows from

\begin{eqnarray*}
\left({\mbox{${\widehat J}$}_z}_a + {\mbox{${\widehat J}$}_z...
...2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b
\end{eqnarray*}

which makes the net angular momentum in the $z$-​direction $\hbar$, or $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$ from each particle. Note that the $z$ angular momentum operators of the two particles simply add up and that ${\mbox{${\widehat J}$}_z}_a$ only acts on particle $a$, and ${\mbox{${\widehat J}$}_z}_b$ only on particle $b$ {N.28}. In terms of quantum numbers, the magnetic quantum number $m_{ab}$ is the sum of the individual quantum numbers $m_a$ and $m_b$; $m_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_a+m_b$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

The net total angular momentum is not so obvious; you cannot just add total angular momenta. To figure out the total angular momentum of ${\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/...
...rn-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b$ anyway, there is a trick: multiply it with the combined step-up operator

\begin{displaymath}
{\widehat J}^+_{ab} = {\widehat J}^+_a+{\widehat J}^+_b
\end{displaymath}

Each part returns zero: ${\widehat J}^+_a$ because particle $a$ is at the top of its ladder and ${\widehat J}^+_b$ because particle $b$ is. So the combined state ${\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/...
...rn-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b$ must be at the top of the ladder too; there is no higher rung. That must mean $j_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1; the combined state must be a $\big\vert 1\:1\big\rangle $ state. It can be defined it as the combination $\big\vert 1\:1\big\rangle $ state:
\begin{displaymath}
{\big\vert 1\:1\big\rangle }_{ab} \equiv {\big\vert\leavev...
...lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b %
\end{displaymath} (12.11)

You could just as well have defined ${\big\vert 1\:1\big\rangle }_{ab}$ as $-{\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
...
...rn-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b$ or ${\rm i}{\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern...
...rn-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b$, say. But why drag along a minus sign or ${\rm i}$ if you do not have to? The first triplet state has been found.

Here is another trick: multiply ${\big\vert 1\:1\big\rangle }_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/...
...rn-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b$ by ${\widehat J}^-_{ab}$: that will go one step down the combined states ladder and produce a combination state ${\big\vert 1\:0\big\rangle }_{ab}$:

\begin{eqnarray*}
{\widehat J}_{ab}^-{\big\vert 1\:1\big\rangle }_{ab}
& = &...
...2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b
\end{eqnarray*}

or

\begin{displaymath}
\hbar \sqrt{2} {\big\vert 1\:0\big\rangle }_{ab} =
\hbar...
...lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_{b}
\end{displaymath}

where the effects of the ladder-down operators were taken from (12.10). (Note that this requires that the individual particle spin states are normalized consistent with the ladder operators.) The second triplet state is therefore:
\begin{displaymath}
{\big\vert 1\:0\big\rangle }_{ab} \equiv
\sqrt{\leavevmo...
...wer.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_{b} %
\end{displaymath} (12.12)

But this gives only one $\big\vert j\:m\big\rangle $ combination state for the two product states ${\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/...
...-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_{b}$ and ${\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/...
...-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_{b}$ with zero net $z$-​momentum. If you want to describe unequal combinations of them, like ${\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/...
...-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_{b}$ by itself, it cannot be just a multiple of ${\big\vert 1\:0\big\rangle }_{ab}$. This suggests that there may be another ${\big\vert j\:0\big\rangle }_{ab}$ combination state involved here. How do you get this second state?

Well, you can reuse the first trick. If you construct a combination of the two product states that steps up to zero, it must be a state with zero $z$ angular momentum that is at the end of its ladder, a ${\big\vert\:0\big\rangle }_{ab}$ state. Consider an arbitrary combination of the two product states with as yet unknown numerical coefficients $C_1$ and $C_2$:

\begin{displaymath}
C_1 {\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\s...
...lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_{b}
\end{displaymath}

For this combination to step up to zero,

\begin{displaymath}
\begin{array}{r}
\displaystyle
\left({\widehat J}^+_a+...
...{\the\scriptfont0 2}\kern.05em\big\rangle }_{b}
\end{array}
\end{displaymath}

must be zero, which requires $C_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-C_1$, leaving $C_1$ undetermined. $C_1$ must be chosen such that the state is normalized, but that still leaves a constant of magnitude one undetermined. To fix it, $C_1$ is taken to be real and positive, and so the singlet state becomes
\begin{displaymath}
{\big\vert\:0\big\rangle }_{ab} =
\sqrt{\leavevmode \ker...
...er.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_{b}. %
\end{displaymath} (12.13)

To find the remaining triplet state, just apply ${\widehat J}^-_{ab}$ once more, to ${\big\vert 1\:0\big\rangle }_{ab}$ above. It gives:

\begin{displaymath}
{\big\vert 1\:\rule[2.5pt]{5pt}{.5pt}1\big\rangle }_{ab} =...
...wer.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_{b} %
\end{displaymath} (12.14)

Of course, the normalization factor of this bottom state had to turn out to be one; all three step-down operators produce only positive real factors.

Figure 12.3: Triplet and singlet states in terms of ladders
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\resizebox{5.6tr...
...pt]
\strut $j_{ab}=0$}}}
\end{picture}
}
\end{picture}}
\end{figure}

Figure 12.3 shows the results graphically in terms of ladders. The two possible spin states of each of the two electrons produce 4 combined product states indicated using up and down arrows. These product states are then combined to produce triplet and singlet states that have definite values for both $z$ and total net angular momentum, and can be shown as rungs on ladders.

Note that a product state like ${\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/...
...rn-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b$ cannot be shown as a rung on a ladder. In fact, from adding (12.12) and (12.13) it is seen that

\begin{displaymath}
{\big\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scrip...
...he\scriptfont0 2}\kern.05em}\;{\big\vert\:0\big\rangle }_{ab}
\end{displaymath}

which makes it a combination of the middle rungs of the triplet and singlet ladders, rather than a single rung.