12.10 Pauli spin matrices

This subsection returns to the simple two-rung spin ladder (doublet) of an electron, or any other spin $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ 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 $S_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$ angular momentum in a chosen $z$ direction; in other words that it is in the $\big\vert\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\... ...\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle$, 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:

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

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

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

It works the same way for the spin-down state ${\downarrow}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\big\vert\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\... ...\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle$:

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

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:

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

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

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

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

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

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”:

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

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

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.69}.