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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. 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 ${\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em ... ...n-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}$ and ${\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em ... ...n-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}$, there are four dif­fer­ent com­bined prod­uct states:

$${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scri... ...ower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b.$$

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 ${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em... ...2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b$ state fol­lows from

$$\begin{eqnarray*} \left({\mbox{${\widehat J}$}_z}_a + {\mbox{${\widehat J}$}_z}... ...m\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b \end{eqnarray*}$$

which makes the net an­gu­lar mo­men­tum in the $z$-​di­rec­tion $\hbar$, or $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$ 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 ${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em... ...2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b$ any­way, there is a trick: mul­ti­ply it with the com­bined step-up op­er­a­tor

$${\widehat J}^+_{ab} = {\widehat J}^+_a+{\widehat J}^+_b$$

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 ${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em... ...2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b$ 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 ${\left\vert 1\:1\right\rangle}$ state. It can be de­fined it as the com­bi­na­tion ${\left\vert 1\:1\right\rangle}$ state:

$${{\left\vert 1\:1\right\rangle}}_{ab} \equiv {{\left\vert\l... ...wer.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b %$$ (12.11)

You could just as well have de­fined ${{\left\vert 1\:1\right\rangle}}_{ab}$ as $-{{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2e... ...2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b$ or ${\rm i}{{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\ke... ...2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b$, 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 ${{\left\vert 1\:1\right\rangle}}_{ab}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em... ...2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b$ by ${\widehat J}^-_{ab}$: that will go one step down the com­bined states lad­der and pro­duce a com­bi­na­tion state ${{\left\vert 1\:0\right\rangle}}_{ab}$:

$$\begin{eqnarray*} {\widehat J}_{ab}^-{{\left\vert 1\:1\right\rangle}}_{ab} & =... ...m\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b \end{eqnarray*}$$

or

$$\hbar \sqrt{2} {{\left\vert 1\:0\right\rangle}}_{ab} = \hb... ...wer.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_{b}$$

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:

$${{\left\vert 1\:0\right\rangle}}_{ab} \equiv \sqrt{\leavev... ...r.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_{b} %$$ (12.12)

But this gives only one ${\left\vert j\:m\right\rangle}$ com­bi­na­tion state for the two prod­uct states ${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em... ...m /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_{b}$ and ${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em... ...m /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_{b}$ with zero net $z$-​mo­men­tum. If you want to de­scribe un­equal com­bi­na­tions of them, like ${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em... ...m /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_{b}$ by it­self, it can­not be just a mul­ti­ple of ${{\left\vert 1\:0\right\rangle}}_{ab}$. This sug­gests that there may be an­other ${{\left\vert j\:0\right\rangle}}_{ab}$ 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 ${{\left\vert\:0\right\rangle}}_{ab}$ state. Con­sider an ar­bi­trary com­bi­na­tion of the two prod­uct states with as yet un­known nu­mer­i­cal co­ef­fi­cients $C_1$ and $C_2$:

$$C_1 {{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\... ...wer.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_{b}$$

For this com­bi­na­tion to step up to zero,

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

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

$${{\left\vert\:0\right\rangle}}_{ab} = \sqrt{\leavevmode \k... ....56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_{b}. %$$ (12.13)

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

$${{\left\vert 1\:\rule[2.5pt]{5pt}{.5pt}1\right\rangle}}_{ab... ...r.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_{b} %$$ (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

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 ${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em... ...2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}}_b$ can­not be shown as a rung on a lad­der. In fact, from adding (12.12) and (12.13) it is seen that

$${{\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scri... ...scriptfont0 2}\kern.05em}\;{{\left\vert\:0\right\rangle}}_{ab}$$

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