# 30.8 Quantum numbers and rules  (Page 2/10)

 Page 2 / 10
${L}_{z}={m}_{l}\frac{h}{2\pi }\phantom{\rule{1.00em}{0ex}}\left({m}_{l}=-l,-l+1, ...,\phantom{\rule{0.25em}{0ex}}-1, 0, 1, ...\phantom{\rule{0.25em}{0ex}}l-1,\phantom{\rule{0.25em}{0ex}}l\right)\text{,}$

where ${L}_{z}$ is the $z$ -component of the angular momentum and ${m}_{l}$ is the angular momentum projection quantum number. The rule in parentheses for the values of ${m}_{l}$ is that it can range from $-l$ to $l$ in steps of one. For example, if $l=2$ , then ${m}_{l}$ can have the five values –2, –1, 0, 1, and 2. Each ${m}_{l}$ corresponds to a different energy in the presence of a magnetic field, so that they are related to the splitting of spectral lines into discrete parts, as discussed in the preceding section. If the $z$ -component of angular momentum can have only certain values, then the angular momentum can have only certain directions, as illustrated in [link] .

## What are the allowed directions?

Calculate the angles that the angular momentum vector $\mathbf{\text{L}}$ can make with the $z$ -axis for $l=1$ , as illustrated in [link] .

Strategy

[link] represents the vectors $\mathbf{\text{L}}$ and ${\mathbf{\text{L}}}_{z}$ as usual, with arrows proportional to their magnitudes and pointing in the correct directions. $\mathbf{\text{L}}$ and ${\mathbf{\text{L}}}_{z}$ form a right triangle, with $\mathbf{L}$ being the hypotenuse and ${\mathbf{\text{L}}}_{z}$ the adjacent side. This means that the ratio of ${\mathbf{L}}_{z}$ to $\mathbf{L}$ is the cosine of the angle of interest. We can find $\mathbf{\text{L}}$ and ${\mathbf{\text{L}}}_{z}$ using $L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$ and ${L}_{z}=m\frac{h}{2\pi }$ .

Solution

We are given $l=1$ , so that ${m}_{l}$ can be +1, 0, or −1. Thus $L$ has the value given by $L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$ .

$L=\frac{\sqrt{l\left(l+1\right)}h}{2\pi }=\frac{\sqrt{2}h}{2\pi }$

${L}_{z}$ can have three values, given by ${L}_{z}={m}_{l}\frac{h}{2\pi }$ .

${L}_{z}={m}_{l}\frac{h}{2\pi }=\left\{\begin{array}{cccc}\phantom{\rule{0.25em}{0ex}}\frac{h}{2\pi },& \phantom{\rule{0.25em}{0ex}}{m}_{l}& =& +1\phantom{\rule{0.25em}{0ex}}\\ \phantom{\rule{0.25em}{0ex}}0,\phantom{\rule{0.25em}{0ex}}& {m}_{l}& =& 0\phantom{\rule{0.25em}{0ex}}\\ -\frac{h}{2\pi },& \phantom{\rule{0.25em}{0ex}}{m}_{l}& =& -1\end{array}$

As can be seen in [link] , $cos\phantom{\rule{0.25em}{0ex}}\theta ={\text{L}}_{z}\text{/L,}$ and so for ${m}_{l}=+1$ , we have

$\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}=\frac{{L}_{Z}}{L}=\frac{\frac{h}{2\pi }}{\frac{\sqrt{2}h}{2\pi }}=\frac{1}{\sqrt{2}}=0\text{.}\text{707.}$

Thus,

${\theta }_{1}={\text{cos}}^{-1}\text{0.707}=\text{45}\text{.}0º.$

Similarly, for ${m}_{l}=0$ , we find $\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}=0$ ; thus,

${\theta }_{2}={\text{cos}}^{-1}0=\text{90}\text{.}0º.$

And for ${m}_{l}=-1$ ,

$\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{3}=\frac{{L}_{Z}}{L}=\frac{-\frac{h}{2\pi }}{\frac{\sqrt{2}h}{2\pi }}=-\frac{1}{\sqrt{2}}=-0\text{.}\text{707,}$

so that

${\theta }_{3}={\text{cos}}^{-1}\left(-0\text{.}\text{707}\right)=\text{135}\text{.}0º.$

Discussion

The angles are consistent with the figure. Only the angle relative to the $z$ -axis is quantized. $L$ can point in any direction as long as it makes the proper angle with the $z$ -axis. Thus the angular momentum vectors lie on cones as illustrated. This behavior is not observed on the large scale. To see how the correspondence principle holds here, consider that the smallest angle ( ${\theta }_{\text{1}}$ in the example) is for the maximum value of ${m}_{l}=0$ , namely ${m}_{l}=l$ . For that smallest angle,

$\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\frac{{L}_{z}}{L}=\frac{l}{\sqrt{l\left(l+1\right)}}\text{,}$

which approaches 1 as $l$ becomes very large. If $\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =1$ , then $\theta =0º$ . Furthermore, for large $l$ , there are many values of ${m}_{l}$ , so that all angles become possible as $l$ gets very large.

## Intrinsic spin angular momentum is quantized in magnitude and direction

There are two more quantum numbers of immediate concern. Both were first discovered for electrons in conjunction with fine structure in atomic spectra. It is now well established that electrons and other fundamental particles have intrinsic spin , roughly analogous to a planet spinning on its axis. This spin is a fundamental characteristic of particles, and only one magnitude of intrinsic spin is allowed for a given type of particle. Intrinsic angular momentum is quantized independently of orbital angular momentum. Additionally, the direction of the spin is also quantized. It has been found that the magnitude of the intrinsic (internal) spin angular momentum    , $S$ , of an electron is given by

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hour glass, pendulum clock, atomic clock?
S.M
tnks
David
how did they solve for "t" after getting 67.6=.5(Voy + 0)t
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no it is derived
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no
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