Sub­sec­tions


5.5 Mul­ti­ple-Par­ti­cle Sys­tems In­clud­ing Spin

Spin will turn out to have a ma­jor ef­fect on how quan­tum par­ti­cles be­have. There­fore, quan­tum me­chan­ics as dis­cussed so far must be gen­er­al­ized to in­clude spin. Just like there is a prob­a­bil­ity that a par­ti­cle is at some po­si­tion ${\skew0\vec r}$, there is the ad­di­tional prob­a­bil­ity that it has spin an­gu­lar mo­men­tum $S_z$ in an ar­bi­trar­ily cho­sen $z$-​di­rec­tion and this must be in­cluded in the wave func­tion. This sec­tion dis­cusses how.


5.5.1 Wave func­tion for a sin­gle par­ti­cle with spin

The first ques­tion is how spin should be in­cluded in the wave func­tion of a sin­gle par­ti­cle. If spin is ig­nored, a sin­gle par­ti­cle has a wave func­tion $\Psi({\skew0\vec r};t)$, de­pend­ing on po­si­tion ${\skew0\vec r}$ and on time $t$. Now, the spin $S_z$ is just some other scalar vari­able that de­scribes the par­ti­cle, in that re­spect no dif­fer­ent from say the $x$-​po­si­tion of the par­ti­cle. The “every pos­si­ble com­bi­na­tion” idea of al­low­ing every pos­si­ble com­bi­na­tion of states to have its own prob­a­bil­ity in­di­cates that $S_z$ needs to be added to the list of vari­ables. So the com­plete wave func­tion $\Psi$ of the par­ti­cle can be writ­ten out fully as:

\begin{displaymath}
\fbox{$\displaystyle
\Psi \equiv \Psi({\skew0\vec r},S_z;t)
$}
\end{displaymath} (5.16)

The value of $\vert\Psi({\skew0\vec r},S_z;t)\vert^2{ \rm d}^3{\skew0\vec r}$ gives the prob­a­bil­ity of find­ing the par­ti­cle within a vicin­ity ${\rm d}^3{\skew0\vec r}$ of ${\skew0\vec r}$ and with spin an­gu­lar mo­men­tum in the $z$-​di­rec­tion $S_z$.

But note that there is a big dif­fer­ence be­tween the spin co­or­di­nate and the po­si­tion co­or­di­nates: while the po­si­tion vari­ables can take on any value, the val­ues of $S_z$ are highly lim­ited. In par­tic­u­lar, for the elec­tron, pro­ton, and neu­tron, $S_z$ can only be $\frac12\hbar$ or $-\frac12\hbar$, noth­ing else. You do not re­ally have a full $S_z$ axis, just two points.

As a re­sult, there are other mean­ing­ful ways of writ­ing the wave func­tion. The full wave func­tion $\Psi({\skew0\vec r},S_z;t)$ can be thought of as con­sist­ing of two parts $\Psi_+$ and $\Psi_-$ that only de­pend on po­si­tion:

\begin{displaymath}
\fbox{$\displaystyle
\Psi_+({\skew0\vec r};t) \equiv \Psi(...
...equiv \Psi({\skew0\vec r},-{\textstyle\frac{1}{2}}\hbar;t)
$}
\end{displaymath} (5.17)

These two parts can in turn be thought of as be­ing the com­po­nents of a two-di­men­sion­al vec­tor that only de­pends on po­si­tion:

\begin{displaymath}
\skew{-1}\vec\Psi({\skew0\vec r};t)
\equiv
\left(
\begin...
...0\vec r};t) \\
\Psi_-({\skew0\vec r};t)
\end{array} \right)
\end{displaymath}

Re­mark­ably, Dirac found that the wave func­tion for par­ti­cles like elec­trons has to be a vec­tor, if it is as­sumed that the rel­a­tivis­tic equa­tions take a guessed sim­ple and beau­ti­ful form, like the Schrö­din­ger and all other ba­sic equa­tions of physics are sim­ple and beau­ti­ful. Just like rel­a­tiv­ity re­veals that par­ti­cles should have build-in en­ergy, it also re­veals that par­ti­cles like elec­trons have build-in an­gu­lar mo­men­tum. A de­scrip­tion of the Dirac equa­tion is in chap­ter 12.12 if you are cu­ri­ous.

The two-di­men­sion­al vec­tor is called a “spinor” to in­di­cate that its com­po­nents do not change like those of or­di­nary phys­i­cal vec­tors when the co­or­di­nate sys­tem is ro­tated. (How they do change is of no im­por­tance here, but will even­tu­ally be de­scribed in de­riva­tion {D.68}.) The spinor can also be writ­ten in terms of a mag­ni­tude times a unit vec­tor:

\begin{displaymath}
\skew{-1}\vec\Psi({\skew0\vec r};t)
= \Psi_m({\skew0\vec r...
...\vec r};t) \\
\chi_2({\skew0\vec r};t)
\end{array} \right).
\end{displaymath}

This book will just use the scalar wave func­tion $\Psi({\skew0\vec r},S_z;t)$; not a vec­tor one. But it is of­ten con­ve­nient to write the scalar wave func­tion in a form equiv­a­lent to the vec­tor one:

\begin{displaymath}
\Psi({\skew0\vec r},S_z;t) =
\Psi_+({\skew0\vec r};t) {\uparrow}(S_z) + \Psi_-({\skew0\vec r};t) {\downarrow}(S_z).
\end{displaymath} (5.18)

The square mag­ni­tude of func­tion $\Psi_+$ gives the prob­a­bil­ity of find­ing the par­ti­cle near a po­si­tion with spin-up. That of $\Psi_-$ gives the prob­a­bil­ity of find­ing it with spin-down. The spin-up func­tion ${\uparrow}(S_z)$ and the spin-down func­tion ${\downarrow}(S_z)$ are in some sense the equiv­a­lent of the unit vec­tors ${\hat\imath}$ and ${\hat\jmath}$ in nor­mal vec­tor analy­sis; they have by de­f­i­n­i­tion the fol­low­ing val­ues:

\begin{displaymath}
\fbox{$\displaystyle
{\uparrow}({\textstyle\frac{1}{2}}\hb...
...ar)=0\quad
{\downarrow}(-{\textstyle\frac{1}{2}}\hbar)=1.
$}
\end{displaymath}

The func­tion ar­gu­ments will usu­ally be left away for con­cise­ness, so that

\begin{displaymath}
\fbox{$\displaystyle
\Psi = \Psi_+ {\uparrow}+ \Psi_- {\downarrow}
$}
\end{displaymath}

is the way the wave func­tion of, say, an elec­tron will nor­mally be writ­ten out.


Key Points
$\begin{picture}(15,5.5)(0,-3)
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\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Spin must be in­cluded as an in­de­pen­dent vari­able in the wave func­tion of a par­ti­cle with spin.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Usu­ally, the wave func­tion $\Psi({\skew0\vec r},S_z;t)$ of a sin­gle par­ti­cle with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ will be writ­ten as

\begin{displaymath}
\Psi = \Psi_+ {\uparrow}+ \Psi_- {\downarrow}
\end{displaymath}

where $\Psi_+({\skew0\vec r};t)$ de­ter­mines the prob­a­bil­ity of find­ing the par­ti­cle near a given lo­ca­tion ${\skew0\vec r}$ with spin up, and $\Psi_-({\skew0\vec r};t)$ the one for find­ing it spin down.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The func­tions ${\uparrow}(S_z)$ and ${\downarrow}(S_z)$ have the val­ues

\begin{displaymath}
{\uparrow}({\textstyle\frac{1}{2}}\hbar)=1\quad
{\uparrow}...
...}\hbar)=0\quad
{\downarrow}(-{\textstyle\frac{1}{2}}\hbar)=1
\end{displaymath}

and rep­re­sent the pure spin-up, re­spec­tively spin-down states.

5.5.1 Re­view Ques­tions
1.

What is the nor­mal­iza­tion re­quire­ment of the wave func­tion of a spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cle in terms of $\Psi_+$ and $\Psi_-$?

So­lu­tion com­plexsa-a


5.5.2 In­ner prod­ucts in­clud­ing spin

In­ner prod­ucts are im­por­tant: they are needed for find­ing nor­mal­iza­tion fac­tors, ex­pec­ta­tion val­ues, un­cer­tainty, ap­prox­i­mate ground states, etcetera. The ad­di­tional spin co­or­di­nates add a new twist, since there is no way to in­te­grate over the few dis­crete points on the spin axis. In­stead, you must sum over these points.

As an ex­am­ple, the in­ner prod­uct of two ar­bi­trary elec­tron wave func­tions $\Psi_1({\skew0\vec r},S_z;t)$ and $\Psi_2({\skew0\vec r},S_z;t)$ is

\begin{displaymath}
\langle\Psi_1\vert\Psi_2\rangle
=
\sum_{S_z = \pm\frac12\...
...S_z;t) \Psi_2({\skew0\vec r},S_z;t) { \rm d}^3 {\skew0\vec r}
\end{displaymath}

or writ­ing out the two-term sum,

\begin{displaymath}
\langle\Psi_1\vert\Psi_2\rangle
=
\int_{{\rm all} {\skew...
...r},-{\textstyle\frac{1}{2}}\hbar;t) { \rm d}^3 {\skew0\vec r}
\end{displaymath}

The in­di­vid­ual fac­tors in the in­te­grals are by de­f­i­n­i­tion the spin-up com­po­nents $\Psi_{1+}$ and $\Psi_{2+}$ and the spin down com­po­nents $\Psi_{1-}$ and $\Psi_{2-}$ of the wave func­tions, so:

\begin{displaymath}
\langle\Psi_1\vert\Psi_2\rangle
=
\int_{{\rm all} {\skew...
... r};t)
\Psi_{2-}({\skew0\vec r};t) { \rm d}^3 {\skew0\vec r}
\end{displaymath}

In other words, the in­ner prod­uct with spin eval­u­ates as

\begin{displaymath}
\fbox{$\displaystyle
\langle
\Psi_{1+}{\uparrow}+ \Psi_{1...
...si_{2+}\rangle
+
\langle\Psi_{1-}\vert\Psi_{2-}\rangle
$} %
\end{displaymath} (5.19)

It is spin-up com­po­nents to­gether and spin-down com­po­nents to­gether.

An­other way of look­ing at this, or maybe re­mem­ber­ing it, is to note that the spin states are an or­tho­nor­mal pair,

\begin{displaymath}
\fbox{$\displaystyle
\langle{\uparrow}\vert{\uparrow}\rang...
... \qquad
\langle{\downarrow}\vert{\downarrow}\rangle = 1
$} %
\end{displaymath} (5.20)

as can be ver­i­fied di­rectly from the de­f­i­n­i­tions of those func­tions as given in the pre­vi­ous sub­sec­tion. Then you can think of an in­ner prod­uct with spin as mul­ti­ply­ing out as:

\begin{eqnarray*}
\lefteqn{
\langle\Psi_{1+}{\uparrow}+ \Psi_{1-}{\downarrow}\...
...}\vert\Psi_{2+}\rangle
+
\langle\Psi_{1-}\vert\Psi_{2-}\rangle
\end{eqnarray*}


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
In in­ner prod­ucts, you must sum over the spin states.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
For spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cles:

\begin{displaymath}
\langle
\Psi_{1+}{\uparrow}+ \Psi_{1-}{\downarrow}
\vert
...
...ert\Psi_{2+}\rangle
+
\langle\Psi_{1-}\vert\Psi_{2-}\rangle
\end{displaymath}

which is spin-up com­po­nents to­gether plus spin-down com­po­nents to­gether.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The spin-up and spin-down states ${\uparrow}$ and ${\downarrow}$ are an or­tho­nor­mal pair.

5.5.2 Re­view Ques­tions
1.

Show that the nor­mal­iza­tion re­quire­ment for the wave func­tion of a spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cle in terms of $\Psi_+$ and $\Psi_-$ re­quires its norm $\sqrt{\langle\Psi\vert\Psi\rangle}$ to be one.

So­lu­tion com­plex­sai-a

2.

As­sume that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are nor­mal­ized spa­tial wave func­tions. Now show that a com­bi­na­tion of the two like $\left(\psi_{\rm {l}}{\uparrow}+\psi_{\rm {r}}{\downarrow}\right)$$\raisebox{.5pt}{$/$}$$\sqrt 2$ is a nor­mal­ized wave func­tion with spin.

So­lu­tion com­plex­sai-b


5.5.3 Com­mu­ta­tors in­clud­ing spin

There is no known in­ter­nal phys­i­cal mech­a­nism that gives rise to spin like there is for or­bital an­gu­lar mo­men­tum. For­tu­nately, this lack of de­tailed in­for­ma­tion about spin is to a con­sid­er­able amount made less of an is­sue by knowl­edge about its com­mu­ta­tors.

In par­tic­u­lar, physi­cists have con­cluded that spin com­po­nents sat­isfy the same com­mu­ta­tion re­la­tions as the com­po­nents of or­bital an­gu­lar mo­men­tum:

\begin{displaymath}
\fbox{$\displaystyle
[{\widehat S}_x,{\widehat S}_y] = {\r...
...idehat S}_z,{\widehat S}_x] = {\rm i}\hbar{\widehat S}_y
$} %
\end{displaymath} (5.21)

These equa­tions are called the “fun­da­men­tal com­mu­ta­tion re­la­tions.” As will be shown in chap­ter 12, a large amount of in­for­ma­tion about spin can be teased from them.

Fur­ther, spin op­er­a­tors com­mute with all func­tions of the spa­tial co­or­di­nates and with all spa­tial op­er­a­tors, in­clud­ing po­si­tion, lin­ear mo­men­tum, and or­bital an­gu­lar mo­men­tum. The rea­son why can be un­der­stood from the given de­scrip­tion of the wave func­tion with spin. First of all, the square spin op­er­a­tor ${\widehat S}^2$ just mul­ti­plies the en­tire wave func­tion by the con­stant $\hbar^2s(s+1)$, and every­thing com­mutes with a con­stant. And the op­er­a­tor ${\widehat S}_z$ of spin in an ar­bi­trary $z$-​di­rec­tion com­mutes with spa­tial func­tions and op­er­a­tors in much the same way that an op­er­a­tor like $\partial$$\raisebox{.5pt}{$/$}$$\partial{x}$ com­mutes with func­tions de­pend­ing on $y$ and with $\partial$$\raisebox{.5pt}{$/$}$$\partial{y}$. The $z$-​com­po­nent of spin cor­re­sponds to an ad­di­tional axis sep­a­rate from the $x$, $y$, and $z$ ones, and ${\widehat S}_z$ only af­fects the vari­a­tion in this ad­di­tional di­rec­tion. For ex­am­ple, for a par­ti­cle with spin one half, ${\widehat S}_z$ mul­ti­plies the spin-up part of the wave func­tion $\Psi_+$ by the con­stant $\frac12\hbar$ and $\Psi_-$ by $-\frac12\hbar$. Spa­tial func­tions and op­er­a­tors com­mute with these con­stants for both $\Psi_+$ and $\Psi_-$ hence com­mute with ${\widehat S}_z$ for the en­tire wave func­tion. Since the $z$-​di­rec­tion is ar­bi­trary, this com­mu­ta­tion ap­plies for any spin com­po­nent.


Key Points
$\begin{picture}(15,5.5)(0,-3)
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\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
While a de­tailed mech­a­nism of spin is miss­ing, com­mu­ta­tors with spin can be eval­u­ated.

$\begin{picture}(15,5.5)(0,-3)
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\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The com­po­nents of spin sat­isfy the same mu­tual com­mu­ta­tion re­la­tions as the com­po­nents of or­bital an­gu­lar mo­men­tum.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Spin com­mutes with spa­tial func­tions and op­er­a­tors.

5.5.3 Re­view Ques­tions
1.

Are not some com­mu­ta­tors miss­ing from the fun­da­men­tal com­mu­ta­tion re­la­tion­ship? For ex­am­ple, what is the com­mu­ta­tor $[{\widehat S}_y,{\widehat S}_x]$?

So­lu­tion com­plexsac-a


5.5.4 Wave func­tion for mul­ti­ple par­ti­cles with spin

The ex­ten­sion of the ideas of the pre­vi­ous sub­sec­tions to­wards mul­ti­ple par­ti­cles is straight­for­ward. For two par­ti­cles, such as the two elec­trons of the hy­dro­gen mol­e­cule, the full wave func­tion fol­lows from the every pos­si­ble com­bi­na­tion idea as

\begin{displaymath}
\fbox{$\displaystyle
\Psi = \Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2};t)
$}
\end{displaymath} (5.22)

The value of $\vert\Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2};t)\vert^2{ \rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}_2$ gives the prob­a­bil­ity of si­mul­ta­ne­ously find­ing par­ti­cle 1 within a vicin­ity ${\rm d}^3{\skew0\vec r}_1$ of ${\skew0\vec r}_1$ with spin an­gu­lar mo­men­tum in the $z$-​di­rec­tion $S_{z1}$, and par­ti­cle 2 within a vicin­ity ${\rm d}^3{\skew0\vec r}_2$ of ${\skew0\vec r}_2$ with spin an­gu­lar mo­men­tum in the $z$-​di­rec­tion $S_{z2}$.

Re­strict­ing the at­ten­tion again to spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cles like elec­trons, pro­tons and neu­trons, there are now four pos­si­ble spin states at any given point, with cor­re­spond­ing spa­tial wave func­tions

\begin{displaymath}
\fbox{$\displaystyle
\begin{array}{c}
\displaystyle
\Psi...
...kew0\vec r}_2,-{\textstyle\frac{1}{2}}\hbar;t)
\end{array} $}
\end{displaymath} (5.23)

For ex­am­ple, $\vert\Psi_{+-}({\skew0\vec r}_1,{\skew0\vec r}_2;t)\vert^2{ \rm d}^3{\skew0\vec r}_1{ \rm d}^3{\skew0\vec r}_2$ gives the prob­a­bil­ity of find­ing par­ti­cle 1 within a vicin­ity ${\rm d}^3{\skew0\vec r}_1$ of ${\skew0\vec r}_1$ with spin up, and par­ti­cle 2 within a vicin­ity ${\rm d}^3{\skew0\vec r}_2$ of ${\skew0\vec r}_2$ with spin down.

The wave func­tion can be writ­ten us­ing purely spa­tial func­tions and purely spin func­tions as

\begin{eqnarray*}
\Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2};t)
& = ...
..._1,{\skew0\vec r}_2;t) {\downarrow}(S_{z1}) {\downarrow}(S_{z2})
\end{eqnarray*}

As you might guess from this multi-line dis­play, usu­ally this will be writ­ten more con­cisely as

\begin{displaymath}
\fbox{$\displaystyle
\Psi =
\Psi_{++} {\uparrow}{\uparrow...
...\downarrow}{\uparrow}+
\Psi_{-} {\downarrow}{\downarrow}
$}
\end{displaymath}

by leav­ing out the ar­gu­ments of the spa­tial and spin func­tions. The un­der­stand­ing is that the first of each pair of ar­rows refers to par­ti­cle 1 and the sec­ond to par­ti­cle 2.

The in­ner prod­uct now eval­u­ates as

\begin{eqnarray*}
\lefteqn{\langle\Psi_1\vert\Psi_2\rangle =} \\
&&\!\!
\sum...
...2};t)
{ \rm d}^3 {\skew0\vec r}_1 { \rm d}^3 {\skew0\vec r}_2
\end{eqnarray*}

This can be writ­ten in terms of the purely spa­tial com­po­nents as
\begin{displaymath}
\fbox{$\displaystyle
\langle\Psi_1\vert\Psi_2\rangle =
\l...
...si_{2-+}\rangle +
\langle\Psi_{1-}\vert\Psi_{2-}\rangle
$}
\end{displaymath} (5.24)

It re­flects the fact that the four spin ba­sis states ${\uparrow}{\uparrow}$, ${\uparrow}{\downarrow}$, ${\downarrow}{\uparrow}$, and ${\downarrow}{\downarrow}$ are an or­tho­nor­mal quar­tet.


Key Points
$\begin{picture}(15,5.5)(0,-3)
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...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The wave func­tion of a sin­gle par­ti­cle with spin gen­er­al­izes in a straight­for­ward way to mul­ti­ple par­ti­cles with spin.

$\begin{picture}(15,5.5)(0,-3)
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\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The wave func­tion of two spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cles can be writ­ten in terms of spa­tial com­po­nents mul­ti­ply­ing pure spin states as

\begin{displaymath}
\Psi =
\Psi_{++} {\uparrow}{\uparrow}+
\Psi_{+-} {\uparro...
...} {\downarrow}{\uparrow}+
\Psi_{-} {\downarrow}{\downarrow}
\end{displaymath}

where the first ar­row of each pair refers to par­ti­cle 1 and the sec­ond to par­ti­cle 2.

$\begin{picture}(15,5.5)(0,-3)
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\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
In terms of spa­tial com­po­nents, the in­ner prod­uct $\langle\Psi_1\vert\Psi_2\rangle$ eval­u­ates as in­ner prod­ucts of match­ing spin com­po­nents:

\begin{displaymath}
\langle\Psi_{1++}\vert\Psi_{2++}\rangle +
\langle\Psi_{1+-...
...t\Psi_{2-+}\rangle +
\langle\Psi_{1-}\vert\Psi_{2-}\rangle
\end{displaymath}

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The four spin ba­sis states ${\uparrow}{\uparrow}$, ${\uparrow}{\downarrow}$, ${\downarrow}{\uparrow}$, and ${\downarrow}{\downarrow}$ are an or­tho­nor­mal quar­tet.

5.5.4 Re­view Ques­tions
1.

As an ex­am­ple of the or­tho­nor­mal­ity of the two-par­ti­cle spin states, ver­ify that $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ is zero, so that ${\uparrow}{\uparrow}$ and ${\downarrow}{\uparrow}$ are in­deed or­thog­o­nal. Do so by ex­plic­itly writ­ing out the sums over $S_{z1}$ and $S_{z2}$.

So­lu­tion com­plexsb-a

2.

A more con­cise way of un­der­stand­ing the or­tho­nor­mal­ity of the two-par­ti­cle spin states is to note that an in­ner prod­uct like $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ equals $\langle{\uparrow}\vert{\downarrow}\rangle\langle{\uparrow}\vert{\uparrow}\rangle$, where the first in­ner prod­uct refers to the spin states of par­ti­cle 1 and the sec­ond to those of par­ti­cle 2. The first in­ner prod­uct is zero be­cause of the or­thog­o­nal­ity of ${\uparrow}$ and ${\downarrow}$, mak­ing $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ zero too.

To check this ar­gu­ment, write out the sums over $S_{z1}$ and $S_{z2}$ for $\langle{\uparrow}\vert{\downarrow}\rangle\langle{\uparrow}\vert{\uparrow}\rangle$ and ver­ify that it is in­deed the same as the writ­ten out sum for $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ given in the an­swer for the pre­vi­ous ques­tion.

The un­der­ly­ing math­e­mat­i­cal prin­ci­ple is that sums of prod­ucts can be fac­tored into sep­a­rate sums as in:

\begin{displaymath}
\sum_{{\rm all} S_{z1}} \sum_{{\rm all} S_{z2}} f(S_{z1}) ...
...S_{z1})\right] \left[\sum_{{\rm all} S_{z2}} g(S_{z2})\right]
\end{displaymath}

This is sim­i­lar to the ob­ser­va­tion in cal­cu­lus that in­te­grals of prod­ucts can be fac­tored into sep­a­rate in­te­grals:

\begin{eqnarray*}\lefteqn{\int_{{\rm all} {\skew0\vec r}_1} \int_{{\rm all} {\...
...ec r}_2} g({\skew0\vec r}_2) { \rm d}^3 {\skew0\vec r}_2\right]
\end{eqnarray*}

So­lu­tion com­plexsb-b


5.5.5 Ex­am­ple: the hy­dro­gen mol­e­cule

As an ex­am­ple, this sec­tion con­sid­ers the ground state of the hy­dro­gen mol­e­cule. It was found in sec­tion 5.2 that the ground state elec­tron wave func­tion must be of the ap­prox­i­mate form

\begin{displaymath}
\psi_{{\rm gs},0} =
a
\left[
\psi_{\rm {l}}({\skew0\vec ...
...r}}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2)
\right]
\end{displaymath}

where $\psi_{\rm {l}}$ was the elec­tron ground state of the left hy­dro­gen atom, and $\psi_{\rm {r}}$ the one of the right one; $a$ was just a nor­mal­iza­tion con­stant. This so­lu­tion ex­cluded all con­sid­er­a­tion of spin.

In­clud­ing spin, the ground state wave func­tion must be of the gen­eral form

\begin{displaymath}
\psi_{{\rm gs}} =
\psi_{++} {\uparrow}{\uparrow}+ \psi_{+-...
...+} {\downarrow}{\uparrow}+ \psi_{-} {\downarrow}{\downarrow}.
\end{displaymath}

As you might guess, in the ground state, each of the four spa­tial func­tions $\psi_{++}$, $\psi_{+-}$, $\psi_{-+}$, and $\psi_{-}$ must be pro­por­tional to the no-spin so­lu­tion $\psi_{{\rm {gs}},0}$ above. Any­thing else would have more than the low­est pos­si­ble en­ergy, {D.24}.

So the ap­prox­i­mate ground state in­clud­ing spin must take the form

\begin{displaymath}
\psi_{{\rm gs}} =
a
\left[
\psi_{\rm {l}}({\skew0\vec r}...
...downarrow}{\uparrow}+
a_{-}{\downarrow}{\downarrow}
\right]
\end{displaymath} (5.25)

where $a_{++}$, $a_{+-}$, $a_{-+}$, and $a_{-}$ are con­stants.


Key Points
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\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The elec­tron wave func­tion $\psi_{{\rm {gs}},0}$ for the hy­dro­gen mol­e­cule de­rived pre­vi­ously ig­nored spin.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
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...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
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In the full elec­tron wave func­tion, each spa­tial com­po­nent must sep­a­rately be pro­por­tional to $a(\psi_{\rm {l}}\psi_{\rm {r}}+\psi_{\rm {r}}\psi_{\rm {l}})$.

5.5.5 Re­view Ques­tions
1.

Show that the nor­mal­iza­tion re­quire­ment for $\psi_{\rm {gs}}$ means that

\begin{displaymath}
\vert a_{++}\vert^2 + \vert a_{+-}\vert^2 + \vert a_{-+}\vert^2 + \vert a_{-}\vert^2 = 1
\end{displaymath}

So­lu­tion com­plexsc-a


5.5.6 Triplet and sin­glet states

In the case of two par­ti­cles with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, it is of­ten more con­ve­nient to use slightly dif­fer­ent ba­sis states to de­scribe the spin states than the four ar­row com­bi­na­tions ${\uparrow}{\uparrow}$, ${\uparrow}{\downarrow}$, ${\downarrow}{\uparrow}$, and ${\downarrow}{\downarrow}$. The more con­ve­nient ba­sis states can be writ­ten in ${\left\vert s\:m\right\rangle}$ ket no­ta­tion, and they are:

\begin{displaymath}
\fbox{$\displaystyle
\underbrace{
{\left\vert 1\:1\right\...
...ownarrow}{\uparrow}\right)
}_{\mbox{the singlet state}}
$} %
\end{displaymath} (5.26)

A state ${\left\vert s\:m\right\rangle}$ has net spin $s$, giv­ing a net square an­gu­lar mo­men­tum $s(s+1)\hbar^2$, and has net an­gu­lar mo­men­tum in the $z$-​di­rec­tion $m\hbar$. For ex­am­ple, if the two par­ti­cles are in the state ${\left\vert 1\:1\right\rangle}$, the net square an­gu­lar mo­men­tum is $2\hbar^2$, and their net an­gu­lar mo­men­tum in the $z$-​di­rec­tion is $\hbar$.

The ${\uparrow}{\downarrow}$ and ${\downarrow}{\uparrow}$ states can be writ­ten as

\begin{displaymath}
{\uparrow}{\downarrow}= \frac{1}{\sqrt2}\left({\left\vert 1...
...\vert 1\:0\right\rangle} - {\left\vert\:0\right\rangle}\right)
\end{displaymath}

This shows that while they have zero an­gu­lar mo­men­tum in the $z$-​di­rec­tion; they do not have a value for the net spin: they have a 50/50 prob­a­bil­ity of net spin 1 and net spin 0. A con­se­quence is that ${\uparrow}{\downarrow}$ and ${\downarrow}{\uparrow}$ can­not be writ­ten in ${\left\vert s\:m\right\rangle}$ ket no­ta­tion; there is no value for $s$. (Re­lated to that, these states also do not have a def­i­nite value for the dot prod­uct of the two spins, {A.10}.

In­ci­den­tally, note that $z$ com­po­nents of an­gu­lar mo­men­tum sim­ply add up, as the New­ton­ian anal­ogy sug­gests. For ex­am­ple, for ${\uparrow}{\downarrow}$, the $\frac12\hbar$ spin an­gu­lar mo­men­tum of the first elec­tron adds to the $-\frac12\hbar$ of the sec­ond elec­tron to pro­duce zero. But New­ton­ian analy­sis does not al­low square an­gu­lar mo­menta to be added to­gether, and nei­ther does quan­tum me­chan­ics. In fact, it is quite a messy ex­er­cise to ac­tu­ally prove that the triplet and sin­glet states have the net spin val­ues claimed above. (See chap­ter 12 if you want to see how it is done.)

The spin states ${\uparrow}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
...
...n-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}$ and ${\downarrow}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
...
...n-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}$ that ap­ply for a sin­gle spin-${\textstyle\frac{1}{2}}$ par­ti­cle are of­ten re­ferred to as the “dou­blet” states, since there are two of them.


Key Points
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...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The set of spin states ${\uparrow}{\uparrow}$, ${\uparrow}{\downarrow}$, ${\downarrow}{\uparrow}$, and ${\downarrow}{\downarrow}$ are of­ten bet­ter re­placed by the triplet and sin­glet states ${\left\vert 1\:1\right\rangle}$, ${\left\vert 1\:0\right\rangle}$, ${\left\vert 1\:\rule[2.5pt]{5pt}{.5pt}1\right\rangle}$, and ${\left\vert\:0\right\rangle}$.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The triplet and sin­glet states have def­i­nite val­ues for the net square spin.

5.5.6 Re­view Ques­tions
1.

Like the states ${\uparrow}{\uparrow}$, ${\uparrow}{\downarrow}$, ${\downarrow}{\uparrow}$, and ${\downarrow}{\downarrow}$; the triplet and sin­glet states are an or­tho­nor­mal quar­tet. For ex­am­ple, check that the in­ner prod­uct of ${\left\vert 1\:0\right\rangle}$ and ${\left\vert\:0\right\rangle}$ is zero.

So­lu­tion com­plexse-a