### 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:

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 -​di­rec­tion is re­lated to what hap­pens if the co­or­di­nate sys­tem is ro­tated around the -​axis.

Con­sider first the sim­plest pos­si­ble vec­tor wave func­tion:

Here is a con­stant vec­tor. Also is the dis­tance from the ori­gin around which the an­gu­lar mo­men­tum is de­fined. Fi­nally, is any ar­bi­trary func­tion of . The big ques­tion is now what hap­pens to this wave func­tion if the co­or­di­nate sys­tem is ro­tated over some an­gle around the -​axis.

The fac­tor does not change un­der ro­ta­tions be­cause the dis­tance from the ori­gin does not. But the vec­tor does change. Fig­ure A.8 shows what hap­pens. The vec­tor in the ro­tated co­or­di­nate sys­tem has com­po­nents

 (A.82)

For ex­am­ple, the first re­la­tion ex­presses that has a com­po­nent in the di­rec­tion of , while has a com­po­nent in that di­rec­tion. The sec­ond ex­pres­sion fol­lows sim­i­larly. The -​com­po­nent does not change un­der the ro­ta­tion.

Now con­sider three very spe­cial vec­tors:

 (A.83)

If you plug into the re­la­tions (A.82) given above and use the Euler iden­tity (2.5), you get . So the vec­tor changes by a mere scalar fac­tor, the ex­po­nen­tial , un­der the ro­ta­tion of the co­or­di­nate sys­tem. Ac­cord­ing to the re­la­tion­ship be­tween ro­ta­tions and an­gu­lar mo­men­tum, {A.19}, that makes a state of def­i­nite an­gu­lar mo­men­tum in the -​di­rec­tion. Also, the mag­netic quan­tum num­ber of the mo­men­tum in the -​di­rec­tion is by de­f­i­n­i­tion the co­ef­fi­cient of in the ex­po­nen­tial fac­tor . Here that is 1, so 1. This value is shown as the su­per­script of the state . The ac­tual an­gu­lar mo­men­tum in the -​di­rec­tion is , so it is 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 has only a -​com­po­nent, so it does not change un­der the ro­ta­tion. Phrased dif­fer­ently, it changes by a unit fac­tor . That makes its mag­netic quan­tum num­ber zero, as the su­per­script in says. Then the an­gu­lar mo­men­tum in the -​di­rec­tion of this state is zero too. Fi­nally, the vec­tor changes by a fac­tor un­der ro­ta­tion, so it has 1, as its su­per­script says, and the an­gu­lar mo­men­tum in the -​di­rec­tion is .

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

Here is the op­er­a­tor that de­scribes the ef­fect of the ro­ta­tion of the co­or­di­nate sys­tem on the wave func­tion. Ap­plied on the vec­tor in the un­ro­tated co­or­di­nate sys­tem, that means

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

So the op­er­a­tor drops the -​com­po­nent and swaps the other two com­po­nents, chang­ing the sign of the first, and then adds a fac­tor . If the same op­er­a­tions are per­formed an­other time, the net re­sult is:

So the square op­er­a­tor just drops the -​com­po­nent and adds a fac­tor .

Of course, the op­er­a­tors and are de­fined sim­i­larly. There is noth­ing spe­cial about the -​axis. The op­er­a­tor of square spin an­gu­lar mo­men­tum is de­fined as

Since each of the op­er­a­tors in the right hand side drops a dif­fer­ent com­po­nent and adds a fac­tor to the other two, the to­tal for any vec­tor is,

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 , the spin, can by de­f­i­n­i­tion be found from equat­ing the co­ef­fi­cient of in the right hand side above to . The only non­neg­a­tive value that can sat­isfy this con­di­tion is 1.

That then means that the spin of vec­tor par­ti­cles is equal to 1. So a vec­tor par­ti­cle is a bo­son of spin 1. The sub­script on the spe­cial vec­tors in­di­cates their spin 1.

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

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

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

 (A.84)

Here and are the po­si­tion an­gles in spher­i­cal co­or­di­nates, and is a so-called spher­i­cal har­monic of or­bital an­gu­lar mo­men­tum, chap­ter 4.2.3. For the above wave func­tion to be prop­erly nor­mal­ized,

(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 -​di­rec­tion equal to in ad­di­tion to the spin an­gu­lar mo­men­tum . So the to­tal an­gu­lar mo­men­tum in the -​di­rec­tion is . To check that, note that un­der ro­ta­tions of the co­or­di­nate sys­tem, the vec­tor changes by a fac­tor while the spher­i­cal har­monic changes by an ad­di­tional fac­tor . That makes the to­tal change a fac­tor .

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

 (A.85)

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

That is a com­pli­cated is­sue best left to chap­ter 12. But a cou­ple of spe­cial cases are worth men­tion­ing al­ready. First, if 1 and , or al­ter­na­tively, if 1 and , then is sim­ply the sum of the spin and or­bital az­imuthal quan­tum num­bers .

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:

Here is the unit vec­tor stick­ing ra­di­ally out­ward away from the ori­gin. The fi­nal con­stant is the spher­i­cal har­monic . It is needed the sat­isfy the nor­mal­iza­tion re­quire­ment un­less you change the one on .

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 has Carte­sian com­po­nents

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

To check this, just plug in the ex­pres­sions for the of (A.83), and for the of ta­ble 4.3.

The bot­tom line is that by com­bin­ing states of unit spin 1, and unit or­bital an­gu­lar mo­men­tum 1, you can cre­ate a state of zero net an­gu­lar mo­men­tum, 0. Note also that in each of the three terms in the right hand side above, and add up to zero. A state of zero an­gu­lar mo­men­tum 0 must have 0 with­out un­cer­tainty. Fur­ther note that the val­ues of both the spin and or­bital an­gu­lar mo­men­tum in the -​di­rec­tion are un­cer­tain. Each of the two has mea­sur­able val­ues , 0, or with equal prob­a­bil­ity .

The above re­la­tion may be writ­ten more neatly in terms of “ket no­ta­tion.” In ket no­ta­tion, an an­gu­lar mo­men­tum state with az­imuthal quan­tum num­ber and mag­netic quan­tum num­ber is in­di­cated as . Us­ing this no­ta­tion, and drop­ping the com­mon fac­tor , the above re­la­tion can be writ­ten as

Here the sub­scripts , , and 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 to stand for spin, for or­bital, and 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 in clas­si­cal physics is a po­lar vec­tor. Each po­si­tion co­or­di­nate , , and changes sign, and there­fore so does the en­tire vec­tor. Sim­i­larly, a ve­loc­ity vec­tor 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, . 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 is odd, then the net par­ity is the op­po­site of the in­trin­sic one. If the quan­tum num­ber is even, then the net par­ity is the in­trin­sic one. The rea­son is that spher­i­cal har­mon­ics change sign un­der spa­tial in­ver­sion if is odd, but not when is even, {D.14}.

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