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 $S_z=\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\hbar$ 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 ${\downarrow}=\big\vert\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}... ...\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle$:

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

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 $2\times2$ 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 $a=b=1$,

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

so $a=b=1$ 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 $z$-axes, the $x$ or $y$ axis in the system you are looking at now, {D.70}.