Sub­sec­tions


D.14 The spher­i­cal har­mon­ics

This note de­rives and lists prop­er­ties of the spher­i­cal har­mon­ics.


D.14.1 De­riva­tion from the eigen­value prob­lem

This analy­sis will de­rive the spher­i­cal har­mon­ics from the eigen­value prob­lem of square an­gu­lar mo­men­tum of chap­ter 4.2.3. It will use sim­i­lar tech­niques as for the har­monic os­cil­la­tor so­lu­tion, {D.12}.

The im­posed ad­di­tional re­quire­ment that the spher­i­cal har­mon­ics $Y_l^m$ are eigen­func­tions of $L_z$ means that they are of the form $\Theta_l^m(\theta)e^{{{\rm i}}m\phi}$ where func­tion $\Theta_l^m(\theta)$ is still to be de­ter­mined. (There is also an ar­bi­trary de­pen­dence on the ra­dius $r$, but it does not have any­thing to do with an­gu­lar mo­men­tum, hence is ig­nored when peo­ple de­fine the spher­i­cal har­mon­ics.) Sub­sti­tu­tion into $\L ^2\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $L^2\psi$ with $\L ^2$ as in (4.22) yields an ODE (or­di­nary dif­fer­en­tial equa­tion) for $\Theta_l^m(\theta)$:

\begin{displaymath}
-\frac{\hbar^2}{\sin\theta}
\frac{\partial}{\partial \thet...
... \frac{\hbar^2 m^2}{\sin^2\theta} \Theta_l^m
= L^2 \Theta_l^m
\end{displaymath}

It is con­ve­nient de­fine a scaled square an­gu­lar mo­men­tum by $L^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2\lambda^2$ so that you can di­vide away the $\hbar^2$ from the ODE.

More im­por­tantly, rec­og­nize that the so­lu­tions will likely be in terms of cosines and sines of $\theta$, be­cause they should be pe­ri­odic if $\theta$ changes by $2\pi$. If you want to use power-se­ries so­lu­tion pro­ce­dures again, these tran­scen­den­tal func­tions are bad news, so switch to a new vari­able $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\cos\theta$. At the very least, that will re­duce things to al­ge­braic func­tions, since $\sin\theta$ is in terms of $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\cos\theta$ equal to $\sqrt{1-x^2}$. Con­vert­ing the ODE to the new vari­able $x$, you get

\begin{displaymath}
-(1-x^2)\frac{{\rm d}^2\Theta_l^m}{{\rm d}x^2}
+ 2x\frac{{...
...m d}x}
+ \frac{m^2}{1-x^2} \Theta_l^m
= \lambda^2 \Theta_l^m
\end{displaymath}

As you may guess from look­ing at this ODE, the so­lu­tions $\Theta_l^m$ are likely to be prob­lem­atic near $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm1$, (phys­i­cally, near the $z$-​axis where $\sin\theta$ is zero.) If you ex­am­ine the so­lu­tion near those points by defin­ing a lo­cal co­or­di­nate $\xi$ as in $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm(1-\xi)$, and then de­duce the lead­ing term in the power se­ries so­lu­tions with re­spect to $\xi$, you find that it is ei­ther $\xi^{m/2}$ or $\xi^{-m/2}$, (in the spe­cial case that $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, that sec­ond so­lu­tion turns out to be $\ln\xi$.) Ei­ther way, the sec­ond pos­si­bil­ity is not ac­cept­able, since it phys­i­cally would have in­fi­nite de­riv­a­tives at the $z$-​axis and a re­sult­ing ex­pec­ta­tion value of square mo­men­tum, as de­fined in chap­ter 4.4.3, that is in­fi­nite. You need to have that $\Theta_l^m$ be­haves as $\xi^{m/2}$ at each end, so in terms of $x$ it must have a fac­tor $(1-x)^{m/2}$ near $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $(1+x)^{m/2}$ near $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$1. The two fac­tors mul­ti­ply to $(1-x^2)^{m/2}$ and so $\Theta_l^m$ can be writ­ten as $(1-x^2)^{m/2}f_l^m$ where $f_l^m$ must have fi­nite val­ues at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$1.

If you sub­sti­tute $\Theta_l^m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(1-x^2)^{m/2}f_l^m$ into the ODE for $\Theta_l^m$, you get an ODE for $f_l^m$:

\begin{displaymath}
-(1-x^2)\frac{{\rm d}^2 f_l^m}{{\rm d}x^2}
+ 2(1+m)x\frac{{\rm d}f_l^m}{{\rm d}x}
+ (m^2+m) f_l^m
= \lambda^2 f_l^m
\end{displaymath}

Plug in a power se­ries, $f_l^m$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\sum}c_px^p$, to get, af­ter clean up,

\begin{displaymath}
\sum p(p-1)c_p x^{p-2}
= \sum \left[(p+m)(p+m+1)-\lambda^2\right] c_p x^p
\end{displaymath}

Us­ing sim­i­lar ar­gu­ments as for the har­monic os­cil­la­tor, you see that the start­ing power will be zero or one, lead­ing to ba­sic so­lu­tions that are again odd or even. And just like for the har­monic os­cil­la­tor, you must again have that the power se­ries ter­mi­nates; even in the least case that $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the se­ries for $f_l^m$ at $\vert x\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is like that of $\ln(1-x^2)$ and will not con­verge to the fi­nite value stip­u­lated. (For rigor, use Gauss’s test.)

To get the se­ries to ter­mi­nate at some fi­nal power $p$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$, you must have ac­cord­ing to the above equa­tion that $\lambda^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(n+m)(n+m+1)$, and if you de­cide to call $n+m$ the az­imuthal quan­tum num­ber $l$, you have $\lambda^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l(l+1)$ where $l$ $\raisebox{-.5pt}{$\geqslant$}$ $m$ since $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n+m$ and $n$, like any power $p$, is greater or equal to zero.

The rest is just a mat­ter of ta­ble books, be­cause with $\lambda^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l(l+1)$, the ODE for $f_l^m$ is just the $m$-​th de­riv­a­tive of the dif­fer­en­tial equa­tion for the $L_l$ Le­gendre poly­no­mial, [40, 28.1], so the $f_l^m$ must be just the $m$-​th de­riv­a­tive of those poly­no­mi­als. In fact, you can now rec­og­nize that the ODE for the $\Theta_l^m$ is just Le­gendre's as­so­ci­ated dif­fer­en­tial equa­tion [40, 28.49], and that the so­lu­tions that you need are the as­so­ci­ated Le­gendre func­tions of the first kind [40, 28.50].

To nor­mal­ize the eigen­func­tions on the sur­face area of the unit sphere, find the cor­re­spond­ing in­te­gral in a ta­ble book, like [40, 28.63]. As men­tioned at the start of this long and still very con­densed story, to in­clude neg­a­tive val­ues of $m$, just re­place $m$ by $\vert m\vert$. There is one ad­di­tional is­sue, though, the sign pat­tern. In or­der to sim­plify some more ad­vanced analy­sis, physi­cists like the sign pat­tern to vary with $m$ ac­cord­ing to the so-called lad­der op­er­a­tors. That re­quires, {D.64}, that start­ing from $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the spher­i­cal har­mon­ics for $m$ $\raisebox{.3pt}{$>$}$ 0 have the al­ter­nat­ing sign pat­tern of the lad­der-up op­er­a­tor, and those for $m$ $\raisebox{.3pt}{$<$}$ 0 the un­vary­ing sign of the lad­der-down op­er­a­tor. Physi­cists will still al­low you to se­lect your own sign for the $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 state, bless them.

The fi­nal so­lu­tion is

\begin{displaymath}
Y_l^m(\theta,\phi)=
(-1)^{\max(m,0)}
\sqrt{\frac{2l+1}{4\...
...t m\vert)!}}
P_l^{\vert m\vert}(\cos\theta)e^{{\rm i}m\phi} %
\end{displaymath} (D.5)

where the prop­er­ties of the as­so­ci­ated Le­gendre func­tions of the first kind $P_l^{\vert m\vert}$ can be found in ta­ble books like [40, pp. 162-166]. This uses the fol­low­ing de­f­i­n­i­tion of the as­so­ci­ated Le­gendre poly­no­mi­als:

\begin{displaymath}
P_l^m(x) \equiv (1-x^2)^{m/2} \frac{{\rm d}^m P_l(x)}{{\rm d}x^m}
\end{displaymath}

where $P_l$ is the nor­mal Le­gendre poly­no­mial. Need­less to say, some other au­thors use dif­fer­ent de­f­i­n­i­tions, po­ten­tially putting in a fac­tor $(-1)^m$.


D.14.2 Par­ity

One spe­cial prop­erty of the spher­i­cal har­mon­ics is of­ten of in­ter­est: their “par­ity.” The par­ity of a wave func­tion is 1, or even, if the wave func­tion stays the same if you re­place ${\skew0\vec r}$ by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. The par­ity is $\vphantom{0}\raisebox{1.5pt}{$-$}$1, or odd, if the wave func­tion stays the same save for a sign change when you re­place ${\skew0\vec r}$ by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. It turns out that the par­ity of the spher­i­cal har­mon­ics is $(-1)^l$; so it is $\vphantom{0}\raisebox{1.5pt}{$-$}$1, odd, if the az­imuthal quan­tum num­ber $l$ is odd, and 1, even, if $l$ is even.

To see why, note that re­plac­ing ${\skew0\vec r}$ by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$ means in spher­i­cal co­or­di­nates that $\theta$ changes into $\pi-\theta$ and $\phi$ into $\phi+\pi$. Ac­cord­ing to trig, the first changes $\cos\theta$ into $\vphantom{0}\raisebox{1.5pt}{$-$}$$\cos\theta$. That leaves $P_l(\cos\theta)$ un­changed for even $l$, since $P_l$ is then a sym­met­ric func­tion, but it changes the sign of $P_l$ for odd $l$. So the sign change is $(-1)^l$. The value of $m$ has no ef­fect, since while the fac­tor $e^{{{\rm i}}m\phi}$ in the spher­i­cal har­mon­ics pro­duces a fac­tor $(-1)^{\vert m\vert}$ un­der the change in $\phi$, $m$ also puts $\vert m\vert$ de­riv­a­tives on $P_l$, and each de­riv­a­tive pro­duces a com­pen­sat­ing change of sign in $P_l^{\vert m\vert}(\cos\theta)$.


D.14.3 So­lu­tions of the Laplace equa­tion

The Laplace equa­tion is

\begin{displaymath}
\nabla^2 u = 0
\end{displaymath}

So­lu­tions $u$ to this equa­tion are called “har­monic func­tions.” In spher­i­cal co­or­di­nates, the Laplace equa­tion has so­lu­tions of the form

\begin{displaymath}
r^l Y_l^m(\theta\phi)
\end{displaymath}

This is a com­plete set of so­lu­tions for the Laplace equa­tion in­side a sphere. Any so­lu­tion $u$ of the Laplace equa­tion in­side a sphere is a lin­ear com­bi­na­tion of these so­lu­tions.

As you can see in ta­ble 4.3, each so­lu­tion above is a power se­ries in terms of Carte­sian co­or­di­nates.

For the Laplace equa­tion out­side a sphere, re­place $r^l$ by 1$\raisebox{.5pt}{$/$}$$r^{l+1}$ in the so­lu­tions above. Note that these so­lu­tions are not ac­cept­able in­side the sphere be­cause they blow up at the ori­gin.

To check that these are in­deed so­lu­tions of the Laplace equa­tion, plug them in, us­ing the Lapla­cian in spher­i­cal co­or­di­nates given in (N.5). Note here that the an­gu­lar de­riv­a­tives can be sim­pli­fied us­ing the eigen­value prob­lem of square an­gu­lar mo­men­tum, chap­ter 4.2.3.


D.14.4 Or­thog­o­nal in­te­grals

The spher­i­cal har­mon­ics are or­tho­nor­mal on the unit sphere:

\begin{displaymath}
\int_{\rm all} Y_{{\underline l}}^{{\underline m}}\strut^* ...
...ad {\rm d}\Omega \equiv sin\theta {\,\rm d}\theta{\rm d}\phi %
\end{displaymath} (D.6)

Here $\delta_{l{\underline l}}$ is de­fined to be 0 if ${\underline l}$ and $l$ are dif­fer­ent, and 1 if they are equal, and sim­i­lar for $\delta_{m{\underline m}}$. In other words, the in­te­gral above is 1 if $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline l}$ and $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline m}$, and 0 in every other case. This ex­presses phys­i­cally that the spher­i­cal har­mon­ics, as eigen­func­tions of the Her­mit­ian $z$ and square an­gu­lar mo­men­tum op­er­a­tors, are or­tho­nor­mal. Math­e­mat­i­cally, it al­lows you to in­te­grate each spher­i­cal har­monic sep­a­rately and quickly when you are find­ing $\int\vert\psi\vert^2{\,\rm d}^3{\skew0\vec r}$ for a wave func­tion $\psi$ ex­pressed in terms of spher­i­cal har­mon­ics.

Fur­ther

\begin{displaymath}
\int_{\rm all}
\left(
\frac{Y_{{\underline l}}^{{\underli...
...ga = l(l+1) \delta_{{\underline l}l}\delta_{{\underline m}m} %
\end{displaymath} (D.7)

This ex­pres­sion sim­pli­fies your life when you are find­ing the $\int\vert\nabla\psi\vert^2{\,\rm d}^3{\skew0\vec r}$ for a wave func­tion $\psi$ ex­pressed in terms of spher­i­cal har­mon­ics.

See the no­ta­tions for more on spher­i­cal co­or­di­nates and $\nabla$.

To ver­ify the above ex­pres­sion, in­te­grate the first term in the in­te­gral by parts with re­spect to $\theta$ and the sec­ond term with re­spect to $\phi$ to get

\begin{displaymath}
- \int \bar Y \left(\frac{1}{\sin\theta}(Y\sin\theta)_\theta
+ \frac{1}{\sin^2\theta} Y_{\phi\phi}\right){\,\rm d}\Omega
\end{displaymath}

and then ap­ply the eigen­value prob­lem of chap­ter 4.2.3.


D.14.5 An­other way to find the spher­i­cal har­mon­ics

There is a more in­tu­itive way to de­rive the spher­i­cal har­mon­ics: they de­fine the power se­ries so­lu­tions to the Laplace equa­tion. In par­tic­u­lar, each $r^lY_l^m$ is a dif­fer­ent power se­ries so­lu­tion $P$ of the Laplace equa­tion $\nabla^2P$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 in Carte­sian co­or­di­nates. Each takes the form

\begin{displaymath}
\sum_{\alpha+\beta+\gamma=l}
c_{\alpha\beta\gamma} x^\alpha y^\beta z^\gamma
\end{displaymath}

where the co­ef­fi­cients $c_{\alpha\beta\gamma}$ are such as to make the Lapla­cian zero.

Even more specif­i­cally, the spher­i­cal har­mon­ics are of the form

\begin{displaymath}
\sum_{2a+b=l-m} c_{ab} u^{a+m}v^a z^b \qquad a,b,m\mathrel{\raisebox{-1pt}{$\geqslant$}}0
\end{displaymath}


\begin{displaymath}
\sum_{2a+b=l-\vert m\vert} c_{ab} u^av^{a+\vert m\vert} z^b \qquad a,b,-m\mathrel{\raisebox{-1pt}{$\geqslant$}}0
\end{displaymath}

where the co­or­di­nates $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x+{{\rm i}}y$ and $v$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x-{{\rm i}}y$ serve to sim­plify the Lapla­cian. That these are the ba­sic power se­ries so­lu­tions of the Laplace equa­tion is read­ily checked.

To get from those power se­ries so­lu­tions back to the equa­tion for the spher­i­cal har­mon­ics, one has to do an in­verse sep­a­ra­tion of vari­ables ar­gu­ment for the so­lu­tion of the Laplace equa­tion in a sphere in spher­i­cal co­or­di­nates (com­pare also the de­riva­tion of the hy­dro­gen atom.) Also, one would have to ac­cept on faith that the so­lu­tion of the Laplace equa­tion is just a power se­ries, as it is in 2D, with no ad­di­tional non­power terms, to set­tle com­plete­ness. In other words, you must as­sume that the so­lu­tion is an­a­lytic.


D.14.6 Still an­other way to find them

The sim­plest way of get­ting the spher­i­cal har­mon­ics is prob­a­bly the one given later in de­riva­tion {D.64}.