12.10 Pauli spin ma­tri­ces

This sub­sec­tion re­turns to the sim­ple two-rung spin lad­der (dou­blet) of an elec­tron, or any other spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cle for that mat­ter, and tries to tease out some more in­for­ma­tion about the spin. While the analy­sis so far has made state­ments about the an­gu­lar mo­men­tum in the ar­bi­trar­ily cho­sen $z$-​di­rec­tion, you of­ten also need in­for­ma­tion about the spin in the cor­re­spond­ing $x$ and $y$ di­rec­tions. This sub­sec­tion will find it.

But be­fore get­ting at it, a mat­ter of no­ta­tions. It is cus­tom­ary to in­di­cate an­gu­lar mo­men­tum that is due to spin by a cap­i­tal $S$. Sim­i­larly, the az­imuthal quan­tum num­ber of spin is in­di­cated by $s$. This sub­sec­tion will fol­low this con­ven­tion.

Now, sup­pose you know that the par­ti­cle is in the spin-up state with $S_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$ an­gu­lar mo­men­tum in a cho­sen $z$ di­rec­tion; in other words that it is in the ${\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em ... ...n-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}$, or ${\uparrow}$, state. You want the ef­fect of the ${\widehat S}_x$ and ${\widehat S}_y$ op­er­a­tors on this state. In the ab­sence of a phys­i­cal model for the mo­tion that gives rise to the spin, this may seem like a hard ques­tion in­deed. But again the faith­ful lad­der op­er­a­tors ${\widehat S}^+$ and ${\widehat S}^-$ clam­ber up and down to your res­cue!

As­sum­ing that the nor­mal­iza­tion fac­tor of the ${\downarrow}$ state is cho­sen in terms of the one of the ${\uparrow}$ state con­sis­tent with the lad­der re­la­tions (12.9) and (12.10), you have:

$${\widehat S}^+ {\uparrow}= ({\widehat S}_x+{\rm i}{\widehat... ...dehat S}_x-{\rm i}{\widehat S}_y){\uparrow}= \hbar{\downarrow}$$

By adding or sub­tract­ing the two equa­tions, you find the ef­fects of ${\widehat S}_x$ and ${\widehat S}_y$ on the spin-up state:

$${\widehat S}_x{\uparrow}= {\textstyle\frac{1}{2}}\hbar{\dow... ...}_y{\uparrow}= {\textstyle\frac{1}{2}}{\rm i}\hbar{\downarrow}$$

It works the same way for the spin-down state ${\downarrow}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\left\vert\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em ... ...n-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em\right\rangle}$:

$${\widehat S}_x{\downarrow}= {\textstyle\frac{1}{2}}\hbar{\u... ..._y{\downarrow}= -{\textstyle\frac{1}{2}}{\rm i}\hbar{\uparrow}$$

You now know the ef­fect of the $x$ and $y$ an­gu­lar mo­men­tum op­er­a­tors on the $z$-​di­rec­tion spin states. Chalk one up for the lad­der op­er­a­tors.

Next, as­sume that you have some spin state that is an ar­bi­trary com­bi­na­tion of spin-up and spin-down:

$$a{\uparrow}+ b{\downarrow}$$

Then, ac­cord­ing to the ex­pres­sions above, ap­pli­ca­tion of the $x$ spin op­er­a­tor ${\widehat S}_x$ will turn it into:

$${\widehat S}_x \left(a{\uparrow}+ b{\downarrow}\right) = a... ...t({\textstyle\frac{1}{2}}\hbar{\uparrow}+ 0{\downarrow}\right)$$

while the op­er­a­tor ${\widehat S}_y$ turns it into

$${\widehat S}_y \left(a{\uparrow}+ b{\downarrow}\right) = a... ...tstyle\frac{1}{2}}\hbar{\rm i}{\uparrow}+ 0{\downarrow}\right)$$

And of course, since ${\uparrow}$ and ${\downarrow}$ are the eigen­states of ${\widehat S}_z$,

$${\widehat S}_z\left(a{\uparrow}+ b{\downarrow}\right) = a ... ...t(0{\uparrow}- {\textstyle\frac{1}{2}}\hbar{\downarrow}\right)$$

If you put the co­ef­fi­cients in the for­mula above, ex­cept for the com­mon fac­tor ${\textstyle\frac{1}{2}}\hbar$, in lit­tle 2 $\times$ 2 ta­bles, you get the so-called Pauli spin ma­tri­ces:

$$\sigma_x = \left(\begin{array}{rr} 0 & 1\ 1 & 0\end{array... ... = \left(\begin{array}{rr} 1 & 0\ 0 & -1\end{array}\right) %$$ (12.15)

where the con­ven­tion is that $a$ mul­ti­plies the first col­umn of the ma­tri­ces and $b$ the sec­ond. Also, the top rows in the ma­tri­ces pro­duce the spin-up part of the re­sult and the bot­tom rows the spin down part. In lin­ear al­ge­bra, you also put the co­ef­fi­cients $a$ and $b$ to­gether in a vec­tor:

$$a{\uparrow}+ b{\downarrow}\equiv \left(\begin{array}{c} a\ b\end{array}\right)$$

You can now go fur­ther and find the eigen­states of the ${\widehat S}_x$ and ${\widehat S}_y$ op­er­a­tors in terms of the eigen­states ${\uparrow}$ and ${\downarrow}$ of the ${\widehat S}_z$ op­er­a­tor. You can use the tech­niques of lin­ear al­ge­bra, or you can guess. For ex­am­ple, if you guess $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $b$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1,

$${\widehat S}_x \left(\begin{array}{c} 1\ 1\end{array}\rig... ...c{1}{2}} \hbar \left(\begin{array}{c} 1\ 1\end{array}\right)$$

so $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $b$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is an eigen­state of ${\widehat S}_x$ with eigen­value ${\textstyle\frac{1}{2}}\hbar$, call it a $\rightarrow$, spin-right, state. To nor­mal­ize the state, you still need to di­vide by $\sqrt2$:

$$\rightarrow\; = \frac1{\sqrt2}{\uparrow}+ \frac1{\sqrt2}{\downarrow}$$

Sim­i­larly, you can guess the other eigen­states, and come up with:

$$\rightarrow\; = \frac1{\sqrt2}{\uparrow}+ \frac1{\sqrt2}{\... ...\frac1{\sqrt2}{\uparrow}- \frac{{\rm i}}{\sqrt2}{\downarrow} %$$ (12.16)

Note that the square mag­ni­tudes of the co­ef­fi­cients of the states are all one half, giv­ing a 50/50 chance of find­ing the $z$-​mo­men­tum up or down. Since the choice of the axis sys­tem is ar­bi­trary, this can be gen­er­al­ized to mean that if the spin in a given di­rec­tion has an def­i­nite value, then there will be a 50/50 chance of the spin in any or­thog­o­nal di­rec­tion turn­ing out to be ${\textstyle\frac{1}{2}}\hbar$ or $-{\textstyle\frac{1}{2}}\hbar$.

You might won­der about the choice of nor­mal­iza­tion fac­tors in the spin states (12.16). For ex­am­ple, why not leave out the com­mon fac­tor ${\rm i}$ in the $\leftarrow$, (neg­a­tive $x$ spin, or spin-left), state? The rea­son is to en­sure that the $x$-​di­rec­tion lad­der op­er­a­tor ${\widehat S}_y\pm{\rm i}{\widehat S}_z$ and the $y$-​di­rec­tion one ${\widehat S}_z\pm{\rm i}{\widehat S}_x$, as ob­tained by cyclic per­mu­ta­tion of the ones for $z$, pro­duce real, pos­i­tive mul­ti­pli­ca­tion fac­tors. This al­lows re­la­tions valid in the $z$-​di­rec­tion (like the ex­pres­sions for triplet and sin­glet states) to also ap­ply in the $x$ and $y$ di­rec­tions. In ad­di­tion, with this choice, if you do a sim­ple change in the la­bel­ing of the axes, from $xyz$ to $yzx$ or $zxy$, the form of the Pauli spin ma­tri­ces re­mains un­changed. The $\rightarrow$ and $\otimes$ states of pos­i­tive $x$-​, re­spec­tively $y$-​mo­men­tum were cho­sen a dif­fer­ent way: if you ro­tate the axis sys­tem 90$\POW9,{\circ}$ around the $y$ or $x$ axis, these are the spin-up states along the new $z$-​axis, the $x$-​axis or $y$-​axis in the sys­tem you are look­ing at now, {D.68}.