N.10 A less fishy story

This note gives a sim­ple model for the emis­sion of a par­ti­cle like a pho­ton. It is as­sumed that the emit­ted par­ti­cle has a typ­i­cal quan­tum wave length $\lambda$ that is large com­pared to the typ­i­cal size $R$ of the atom or nu­cleus that does the emit­ting. The pur­pose of the model is to show that in that case, the par­ti­cle will very likely come out with zero or­bital an­gu­lar mo­men­tum but has some prob­a­bil­ity of nonzero an­gu­lar mo­men­tum.

First, pho­ton wave func­tions are messy and not that easy to make sense of, {A.21.7}. The pho­ton would be much sim­pler if it did not have spin and was non­rel­a­tivis­tic. A rea­son­able wave func­tion for a hy­po­thet­i­cal spin­less non­rel­a­tivis­tic pho­ton com­ing out of the cen­ter of the emit­ter with typ­i­cal wave length $\lambda$ would be

$$\psi = \frac{1}{\lambda^{3/2}}f\left(\frac{r^2}{\lambda^2}\right)$$

where $r$ is the dis­tance from the cen­ter. (The var­i­ous fac­tors $\lambda$ have been added to make the func­tion $f$ in­de­pen­dent of the pho­ton wave length $\lambda$ de­spite the cor­re­spond­ing spa­tial scale and the nor­mal­iza­tion re­quire­ment.)

The above wave func­tion has no pre­ferred di­rec­tion in the emis­sion, mak­ing it spher­i­cally sym­met­ric. It de­pends only on the dis­tance $r$ from the cen­ter of the emit­ter. That means that the wave func­tion has zero or­bital an­gu­lar mo­men­tum. Re­call that zero an­gu­lar mo­men­tum cor­re­sponds to the spher­i­cal har­monic $Y^0_0$, which is in­de­pen­dent of the an­gu­lar po­si­tion, chap­ter 4.2.

There are var­i­ous rea­sons to give why you would want the wave func­tion of a par­ti­cle com­ing out of the ori­gin to have zero an­gu­lar mo­men­tum. For one, since it comes out of a fea­ture­less point, there should not be a pre­ferred di­rec­tion. Or in terms of clas­si­cal physics, if it had an­gu­lar mo­men­tum then it would have to have in­fi­nite ve­loc­ity at the ori­gin. The sim­i­lar quan­tum idea is that the rel­e­vant wave func­tions for a par­ti­cle mov­ing away from the ori­gin, the Han­kel func­tions of the first kind, blow up very strongly at the ori­gin if they have an­gu­lar mo­men­tum, {A.6}. But it is re­ally bet­ter to de­scribe the emit­ted par­ti­cle in terms of the Bessel func­tions of the first kind. These have zero prob­a­bil­ity of the par­ti­cle be­ing at the ori­gin if the an­gu­lar mo­men­tum is not zero. And a par­ti­cle should not be cre­ated at a point where it has zero prob­a­bil­ity of be­ing.

Of course, a spher­i­cally sym­met­ric quan­tum wave func­tion also means that the par­ti­cle is mov­ing away from the emit­ter equally in all di­rec­tions. Fol­low­ing the stated ideas of quan­tum me­chan­ics, this will be true un­til the po­si­tion of the par­ti­cle is mea­sured. Any macro­scopic sur­round­ings can­not rea­son­ably re­main un­com­mit­ted to ex­actly where the out­go­ing par­ti­cle is for very long.

Now con­sider the same sort of emis­sion, but from a point in the emit­ter a bit away from the cen­ter. For sim­plic­ity, as­sume the emis­sion point to be at $R{\hat k}$, where $R$ is the typ­i­cal size of the emit­ter and ${\hat k}$ is the unit vec­tor along the cho­sen $z$-​axis. In that case the wave func­tion is

$$\psi = \frac{1}{\lambda^{3/2}}f\left(\frac{({\skew0\vec r}-R{\hat k})^2}{\lambda^2}\right)$$

Us­ing Tay­lor se­ries ex­pan­sion, that be­comes

$$\psi = \frac{1}{\lambda^{3/2}}f\left(\frac{r^2}{\lambda^2}... ...}{\lambda^2}\right) 2 \frac{r}{\lambda} \frac{z}{r} + \ldots$$

In the sec­ond term, $z$$\raisebox{.5pt}{$/$}$​$r$ is the spher­i­cal har­monic $Y_1^0$, ta­ble 4.3. This term has an­gu­lar mo­men­tum quan­tum num­ber $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. So there is now un­cer­tainty in mo­men­tum. And fol­low­ing the stated ideas of quan­tum me­chan­ics, the prob­a­bil­ity for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is given by the square mag­ni­tude of the co­ef­fi­cient of the (nor­mal­ized) eigen­func­tion.

That makes the prob­a­bil­ity for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 pro­por­tional to $(R/\lambda)^2$. If you car­ried out the Tay­lor se­ries to the next or­der, you would end up with a $(z/r)^2$ term, which, com­bined with a spher­i­cally sym­met­ric con­tri­bu­tion, makes up the spher­i­cal har­monic $Y_2^0$. It then fol­lows that the prob­a­bil­ity for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 is of or­der $(R/\lambda)^4$. And so on. Un­der the as­sumed con­di­tion that the emit­ter size $R$ is much less than the quan­tum wave length $\lambda$ of the emit­ted par­ti­cle, the prob­a­bil­i­ties for nonzero an­gu­lar mo­men­tum are small and de­crease rapidly even fur­ther with in­creas­ing $l$.