A.20 An­gu­lar mo­men­tum of vec­tor par­ti­cles

This ad­den­dum is con­cerned with vec­tor par­ti­cles, par­ti­cles whose wave func­tions are vec­tors. To be sure, the wave func­tion of an elec­tron can also be writ­ten as a vec­tor, chap­ters 3.1 and 5.5.1:

$$\mbox{electron:} \quad \skew{-1}\vec\Psi({\skew0\vec r};t)... ...0\vec r};t) \\ \Psi_-({\skew0\vec r};t) \end{array} \right)$$

But that is not a nor­mal vec­tor. It is a two-di­men­sion­al vec­tor in three-di­men­sion­al space, and is known as a spinor. This ad­den­dum is con­cerned with wave func­tions that are nor­mal three-di­men­sion­al vec­tors. That is of im­por­tance for un­der­stand­ing, for ex­am­ple, the spin an­gu­lar mo­men­tum of pho­tons. A pho­ton is a vec­tor par­ti­cle, though a spe­cial one. It will be shown in this ad­den­dum that the spin of a vec­tor par­ti­cle is 1. The par­ity of such a par­ti­cle will also be dis­cussed.

To re­ally ap­pre­ci­ate this ad­den­dum, you may want to read the pre­vi­ous ad­den­dum {A.19} first. In any case, ac­cord­ing to that ad­den­dum an­gu­lar mo­men­tum is re­lated to what hap­pens to the wave func­tion un­der ro­ta­tion of the co­or­di­nate sys­tem. In par­tic­u­lar, the an­gu­lar mo­men­tum in the $z$-​di­rec­tion is re­lated to what hap­pens if the co­or­di­nate sys­tem is ro­tated around the $z$-​axis.

Con­sider first the sim­plest pos­si­ble vec­tor wave func­tion:

$$\vec\Psi = \skew3\vec A' f(r) \qquad \skew3\vec A' = \left(\begin{array}{c} A_x' \\ A_y' \\ A_z' \end{array}\right)$$

Here $\skew3\vec A'$ is a con­stant vec­tor. Also $r$ is the dis­tance from the ori­gin around which the an­gu­lar mo­men­tum is de­fined. Fi­nally, $f(r)$ is any ar­bi­trary func­tion of $r$. The big ques­tion is now what hap­pens to this wave func­tion if the co­or­di­nate sys­tem is ro­tated over some an­gle $\gamma$ around the $z$-​axis.

Fig­ure A.8: Ef­fect of ro­ta­tion of the co­or­di­nate sys­tem on a vec­tor. The vec­tor is phys­i­cally the same, but it has a dif­fer­ent math­e­mat­i­cal rep­re­sen­ta­tion, dif­fer­ent com­po­nents, in the two co­or­di­nate sys­tems.

The fac­tor $f(r)$ does not change un­der ro­ta­tions be­cause the dis­tance from the ori­gin does not. But the vec­tor $\skew3\vec A'$ does change. Fig­ure A.8 shows what hap­pens. The vec­tor in the ro­tated co­or­di­nate sys­tem $x,y,z$ has com­po­nents

$$A_x = \cos(\gamma) A_x' + \sin(\gamma) A_y' \qquad A_y = -\sin(\gamma) A_x' + \cos(\gamma) A_y' \qquad A_z = A_z' %$$ (A.82)

For ex­am­ple, the first re­la­tion ex­presses that $A_x'{\hat\imath}^{\,\prime}$ has a com­po­nent $\cos(\gamma)A_x'$ in the di­rec­tion of ${\hat\imath}$, while $A_y'{\hat\jmath}^{\,\prime}$ has a com­po­nent $\sin(\gamma)A_y'$ in that di­rec­tion. The sec­ond ex­pres­sion fol­lows sim­i­larly. The $z$-​com­po­nent does not change un­der the ro­ta­tion.

Now con­sider three very spe­cial vec­tors:

$$\fbox{$\displaystyle \vec Y_1^1 = \left(\begin{array}{c} ... ...om{-}{\rm i}/\sqrt{2} \\ \phantom{-}0 \end{array}\right) $} %$$ (A.83)

If you plug $\vec{A'}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vec{Y}_1^1$ into the re­la­tions (A.82) given above and use the Euler iden­tity (2.5), you get $\skew3\vec A$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{{\rm i}\gamma}\vec{Y}_1^1$. So the vec­tor $\vec{Y}_1^1$ changes by a mere scalar fac­tor, the ex­po­nen­tial $e^{{\rm i}\gamma}$, un­der the ro­ta­tion of the co­or­di­nate sys­tem. Ac­cord­ing to the re­la­tion­ship be­tween ro­ta­tions and an­gu­lar mo­men­tum, {A.19}, that makes $\vec{Y}_1^1$ a state of def­i­nite an­gu­lar mo­men­tum in the $z$-​di­rec­tion. Also, the mag­netic quan­tum num­ber $m_s$ of the mo­men­tum in the $z$-​di­rec­tion is by de­f­i­n­i­tion the co­ef­fi­cient of ${\rm i}\gamma$ in the ex­po­nen­tial fac­tor $e^{{\rm i}\gamma}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{1{\rm i}\gamma}$. Here that is 1, so $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. This value is shown as the su­per­script of the state $\vec{Y}_1^1$. The ac­tual an­gu­lar mo­men­tum in the $z$-​di­rec­tion is $m_s\hbar$, so it is $\hbar$ for this state. This an­gu­lar mo­men­tum should be called spin, in anal­ogy to the case for the elec­tron. It is due to the fact that the wave func­tion is a vec­tor.

The vec­tor $\vec{Y}_1^0$ has only a $z$-​com­po­nent, so it does not change un­der the ro­ta­tion. Phrased dif­fer­ently, it changes by a unit fac­tor $e^{0{\rm i}\gamma}$. That makes its mag­netic quan­tum num­ber $m_s$ zero, as the su­per­script in $\vec{Y}_1^0$ says. Then the an­gu­lar mo­men­tum in the $z$-​di­rec­tion of this state is zero too. Fi­nally, the vec­tor $\vec{Y}_1^{-1}$ changes by a fac­tor $e^{-1{\rm i}\gamma}$ un­der ro­ta­tion, so it has $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$1, as its su­per­script says, and the an­gu­lar mo­men­tum in the $z$-​di­rec­tion is $\vphantom{0}\raisebox{1.5pt}{$-$}$$\hbar$.

To get at square an­gu­lar mo­men­tum, first the op­er­a­tor ${\widehat S}_z$ of spin an­gu­lar mo­men­tum around the $z$-​axis is needed. The re­la­tion be­tween an­gu­lar mo­men­tum and ro­ta­tions shows that this op­er­a­tor takes the gen­eral form, {A.19} (A.76),

$${\widehat S}_z = \frac{\hbar}{{\rm i}} \lim_{\gamma\to 0}\frac{{\cal R}_{z,\gamma}-1}{\gamma}$$

Here ${\cal R}_{z,\gamma}$ is the op­er­a­tor that de­scribes the ef­fect of the ro­ta­tion of the co­or­di­nate sys­tem on the wave func­tion. Ap­plied on the vec­tor $\skew3\vec A'$ in the un­ro­tated co­or­di­nate sys­tem, that means

$${\widehat S}_z \skew3\vec A' = \frac{\hbar}{{\rm i}} \lim_{\gamma\to 0}\frac{\skew3\vec A- \skew3\vec A'}{\gamma}$$

Plug­ging in the com­po­nents of $\skew3\vec A$ as given ear­lier, (A.82), and tak­ing the lim­its us­ing l’Hôpital, that pro­duces

$${\widehat S}_z \left(\begin{array}{c} A_x' \\ A_y' \\ A_z' ... ... \phantom{-}A_y' \\ {-}A_x' \\ \phantom{-}0 \end{array}\right)$$

So the op­er­a­tor ${\widehat S}_z$ drops the $z$-​com­po­nent and swaps the other two com­po­nents, chang­ing the sign of the first, and then adds a fac­tor $\hbar$$\raisebox{.5pt}{$/$}$​${\rm i}$. If the same op­er­a­tions are per­formed an­other time, the net re­sult is:

$${\widehat S}_z^2 \left(\begin{array}{c} A_x' \\ A_y' \\ A_z... ...2 \left(\begin{array}{c} A_x' \\ A_y' \\ 0 \end{array}\right)$$

So the square op­er­a­tor just drops the $z$-​com­po­nent and adds a fac­tor $\hbar^2$.

Of course, the op­er­a­tors ${\widehat S}_x^2$ and ${\widehat S}_y^2$ are de­fined sim­i­larly. There is noth­ing spe­cial about the $z$-​axis. The op­er­a­tor of square spin an­gu­lar mo­men­tum is de­fined as

$${\widehat S}^2 \equiv {\widehat S}_x^2 + {\widehat S}_y^2 + {\widehat S}_z^2$$

Since each of the op­er­a­tors in the right hand side drops a dif­fer­ent com­po­nent and adds a fac­tor $\hbar^2$ to the other two, the to­tal for any vec­tor $\skew3\vec A'$ is,

$${\widehat S}^2 \skew3\vec A' = 2 \hbar^2 \skew3\vec A'$$

So the square spin op­er­a­tor al­ways pro­duces a sim­ple mul­ti­ple of the orig­i­nal vec­tor. That makes any vec­tor an eigen­vec­tor of square spin an­gu­lar mo­men­tum. Also, the az­imuthal quan­tum num­ber $s$, the spin, can by de­f­i­n­i­tion be found from equat­ing the co­ef­fi­cient $2\hbar^2$ of $\skew3\vec A'$ in the right hand side above to $s(s+1)\hbar^2$. The only non­neg­a­tive value $s$ that can sat­isfy this con­di­tion is $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

That then means that the spin $s$ of vec­tor par­ti­cles is equal to 1. So a vec­tor par­ti­cle is a bo­son of spin 1. The sub­script on the spe­cial vec­tors $\vec{Y}_1^{m_s}$ in­di­cates their spin $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

You can write the most gen­eral vec­tor wave func­tion in the form

$$\skew{-1}\vec\Psi({\skew0\vec r};t) = \Psi_1({\skew0\vec r}... ...vec r};t)\vec Y_1^1 + \Psi_{-1}({\skew0\vec r};t)\vec Y_1^{-1}$$

Then you can put the co­ef­fi­cients in a vec­tor much like the wave func­tion of the elec­tron, but now three-di­men­sion­al:

$$\mbox{vector boson:}\quad \skew{-1}\vec\Psi({\skew0\vec r}... ...ec r};t) \\ \Psi_{-1}({\skew0\vec r};t) \end{array} \right)$$

Like the elec­tron, the vec­tor par­ti­cle can of course also have or­bital an­gu­lar mo­men­tum. That is due to the co­ef­fi­cients $\Psi_{m_s}$ in the wave func­tion above. So far it has been as­sumed that these co­ef­fi­cients only de­pended on the dis­tance $r$ from the ori­gin. How­ever, con­sider the fol­low­ing more gen­eral com­po­nent of a vec­tor wave func­tion:

$$\vec Y_1^{m_s} f(r) Y_l^{m_l}(\theta,\phi) %$$ (A.84)

Here $\theta$ and $\phi$ are the po­si­tion an­gles in spher­i­cal co­or­di­nates, and $Y_l^{m_l}$ is a so-called spher­i­cal har­monic of or­bital an­gu­lar mo­men­tum, chap­ter 4.2.3. For the above wave func­tion to be prop­erly nor­mal­ized,

$$\int_{r=0}^\infty r^2 f(r){\,\rm d}r = 1$$

(To check this, take a dot, or rather in­ner, prod­uct of the wave func­tion with it­self. Then in­te­grate over all space us­ing spher­i­cal co­or­di­nates and the or­tho­nor­mal­ity of the spher­i­cal har­mon­ics.)

The wave func­tion (A.84) above has or­bital an­gu­lar mo­men­tum in the $z$-​di­rec­tion equal to $m_l\hbar$ in ad­di­tion to the spin an­gu­lar mo­men­tum $m_s\hbar$. So the to­tal an­gu­lar mo­men­tum in the $z$-​di­rec­tion is $(m_s+m_l)\hbar$. To check that, note that un­der ro­ta­tions of the co­or­di­nate sys­tem, the vec­tor changes by a fac­tor $e^{m_s{\rm i}\gamma}$ while the spher­i­cal har­monic changes by an ad­di­tional fac­tor $e^{m_l{\rm i}\gamma}$. That makes the to­tal change a fac­tor $e^{(m_s+m_l){\rm i}\gamma}$.

In gen­eral, then, the mag­netic quan­tum num­ber $m_j$ of the net an­gu­lar mo­men­tum is sim­ply the sum of the spin and or­bital ones,

$$\fbox{$\displaystyle m_j = m_s + m_l $} %$$ (A.85)

How­ever, the sit­u­a­tion for the az­imuthal quan­tum num­ber $j$ of the net an­gu­lar mo­men­tum is not so sim­ple. In gen­eral the wave func­tion (A.84) above will have un­cer­tainty in the value of $j$. Com­bi­na­tions of wave func­tions of the form (A.84) are usu­ally needed to get states of def­i­nite $j$.

That is a com­pli­cated is­sue best left to chap­ter 12. But a cou­ple of spe­cial cases are worth men­tion­ing al­ready. First, if $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $m_l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l$, or al­ter­na­tively, if $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$1 and $m_l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$l$, then $j$ is sim­ply the sum of the spin and or­bital az­imuthal quan­tum num­bers $1+l$.

The other spe­cial case is that there is zero net an­gu­lar mo­men­tum. Zero net an­gu­lar mo­men­tum means that the wave func­tion is ex­actly the same re­gard­less how the co­or­di­nate sys­tem is ro­tated. And that only hap­pens for a vec­tor wave func­tion if it is purely ra­dial:

$${\hat\imath}_r f(r) \frac{1}{\sqrt{4\pi}}$$

Here ${\hat\imath}_r$ is the unit vec­tor stick­ing ra­di­ally out­ward away from the ori­gin. The fi­nal con­stant is the spher­i­cal har­monic $Y_0^0$. It is needed the sat­isfy the nor­mal­iza­tion re­quire­ment un­less you change the one on $f$.

The above state has zero net an­gu­lar mo­men­tum. The ques­tion of in­ter­est is what can be said about its spin and or­bital an­gu­lar mo­men­tum. To an­swer that, it must be rewrit­ten in terms of Carte­sian com­po­nents. Now the unit vec­tor ${\hat\imath}_r$ has Carte­sian com­po­nents

$${\hat\imath}_r = \frac{x}{r} {\hat\imath}+ \frac{y}{r} {\hat\jmath}+ \frac{z}{r} {\hat k}$$

The spa­tial fac­tors in this ex­pres­sion can be writ­ten in terms of the spher­i­cal har­mon­ics $Y_1^{m_l}$, chap­ter 4.2.3. That gives the state of zero net an­gu­lar mo­men­tum as

$${\hat\imath}_r f(r) Y_0^0 = \left( {\textstyle\sqrt{\leav... ...e\scriptfont0 3}\kern.05em}} \vec Y_1^{-1} Y_1^1 \right) f(r)$$

To check this, just plug in the ex­pres­sions for the $\vec{Y}_1^{m_s}$ of (A.83), and for the $Y_1^{m_l}$ of ta­ble 4.3.

The bot­tom line is that by com­bin­ing states of unit spin $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, and unit or­bital an­gu­lar mo­men­tum $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, you can cre­ate a state of zero net an­gu­lar mo­men­tum, $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Note also that in each of the three terms in the right hand side above, $m_s$ and $m_l$ add up to zero. A state of zero an­gu­lar mo­men­tum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 must have $m_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 with­out un­cer­tainty. Fur­ther note that the val­ues of both the spin and or­bital an­gu­lar mo­men­tum in the $z$-​di­rec­tion are un­cer­tain. Each of the two has mea­sur­able val­ues $\hbar$, 0, or $\vphantom{0}\raisebox{1.5pt}{$-$}$$\hbar$ with equal prob­a­bil­ity $\frac13$.

The above re­la­tion may be writ­ten more neatly in terms of “ket no­ta­tion.” In ket no­ta­tion, an an­gu­lar mo­men­tum state with az­imuthal quan­tum num­ber $j$ and mag­netic quan­tum num­ber $m_j$ is in­di­cated as ${\left\vert j\:m_j\right\rangle}$. Us­ing this no­ta­tion, and drop­ping the com­mon fac­tor $f(r)$, the above re­la­tion can be writ­ten as

$${{\left\vert\:0\right\rangle}}_j = {\textstyle\sqrt{\leave... ...pt}{.5pt}1\right\rangle}}_s {{\left\vert 1\:1\right\rangle}}_l$$

Here the sub­scripts $j$, $s$, and $l$ in­di­cate net, spin, and or­bital an­gu­lar mo­men­tum, re­spec­tively.

There is a quicker way to get this re­sult than go­ing through the above al­ge­braic mess. You can sim­ply read off the co­ef­fi­cients in the ap­pro­pri­ate col­umn of the bot­tom-right tab­u­la­tion in fig­ure 12.6. (In this fig­ure take $a$ to stand for spin, $b$ for or­bital, and $ab$ for net an­gu­lar mo­men­tum.) Fig­ure 12.6 also has the co­ef­fi­cients for many other net spin states that you might need. A de­riva­tion of the fig­ure must wait un­til chap­ter 12.

The par­ity of vec­tor wave func­tions is also im­por­tant. Par­ity is what hap­pens to a wave func­tion if you in­vert the pos­i­tive di­rec­tion of all three Carte­sian axes. What hap­pens to a vec­tor wave func­tion un­der such an in­ver­sion can vary. A nor­mal, or “po­lar,” vec­tor changes sign when you in­vert the axes. For ex­am­ple, a po­si­tion vec­tor ${\skew0\vec r}$ in clas­si­cal physics is a po­lar vec­tor. Each po­si­tion co­or­di­nate $x$, $y$, and $z$ changes sign, and there­fore so does the en­tire vec­tor. Sim­i­larly, a ve­loc­ity vec­tor $\vec{v}$ is a po­lar vec­tor; it is just the time de­riv­a­tive of po­si­tion. A par­ti­cle with a vec­tor wave func­tion that be­haves like a nor­mal vec­tor has neg­a­tive in­trin­sic par­ity. The sign of the wave func­tion flips over un­der axes in­ver­sion. Par­ti­cles of this type turn out to in­clude the pho­ton.

But now con­sider an ex­am­ple like a clas­si­cal an­gu­lar mo­men­tum vec­tor, ${\skew0\vec r}$ $\times$ ${m}\vec{v}$. Since both the po­si­tion and the ve­loc­ity change sign un­der spa­tial in­ver­sion, a clas­si­cal an­gu­lar mo­men­tum vec­tor stays the same. A vec­tor that does not change un­der axes in­ver­sion is called a “pseudovec­tor” or “ax­ial” vec­tor. A par­ti­cle whose wave func­tion be­haves like a pseudovec­tor has pos­i­tive in­trin­sic par­ity.

Note how­ever that the or­bital an­gu­lar mo­men­tum of the par­ti­cle also has an ef­fect on the net par­ity. In par­tic­u­lar, if the quan­tum num­ber of or­bital an­gu­lar mo­men­tum $l$ is odd, then the net par­ity is the op­po­site of the in­trin­sic one. If the quan­tum num­ber $l$ is even, then the net par­ity is the in­trin­sic one. The rea­son is that spher­i­cal har­mon­ics change sign un­der spa­tial in­ver­sion if $l$ is odd, but not when $l$ is even, {D.14}.

Par­ti­cles of all types of­ten have def­i­nite par­ity. Such a par­ti­cle may still have un­cer­tainty in $l$. But if par­ity is def­i­nite, the mea­sur­able val­ues of $l$ will need to be all even or all odd.