D.14 Derivation of the spherical harmonics

This analysis will use similar techniques as for the harmonic oscillator solution, {D.12}. The 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 $\L ^2\psi=L^2\psi$ with $\L ^2$ as in (4.22) yields an ODE (ordinary differential equation) for $\Theta_l^m(\theta)$:

$$-\frac{\hbar^2}{\sin\theta} \frac{\partial}{\partial \th... ...frac{\hbar^2 m^2}{\sin^2\theta} \Theta_l^m = L^2 \Theta_l^m$$

It is convenient define a scaled square angular momentum by $L^2=\hbar^2\lambda^2$ 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 $x=\cos\theta$. At the very least, that will reduce things to algebraic functions, since $\sin\theta$ is in terms of $x=\cos\theta$ equal to $\sqrt{1-x^2}$. Converting the ODE to the new variable $x$, you get

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

As you may guess from looking at this ODE, the solutions $\Theta_l^m$ are likely to be problematic near $x=\pm1$, (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 $x=\pm(1-\xi)$, 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 $m=0$, 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 $x=1$ and $(1+x)^{m/2}$ near $x=-1$. 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 $x=1$ and $x=-1$.

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

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

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

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

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 $m=0$, the series for $f_l^m$ at $\vert x\vert=1$ 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 $p=n$, you must have according to the above equation that $\lambda^2=(n+m)(n+m+1)$, and if you decide to call $n+m$ the azimuthal quantum number $l$, you have $\lambda^2=l(l+1)$ where $l\ge m$ since $l=n+m$ and $n$, like any power $p$, is greater or equal to zero.

The rest is just a matter of table books, because with $\lambda^2=l(l+1)$, the ODE for $f_l^m$ is just the $m$-th derivative of the differential equation for the $L_l$ Legendre polynomial, [39, 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 [39, 28.49], and that the solutions that you need are the associated Legendre functions of the first kind [39, 28.50].

To normalize the eigenfunctions on the surface area of the unit sphere, find the corresponding integral in a table book, like [39, 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.66}, that starting from $m=0$, the spherical harmonics for $m>0$ have the alternating sign pattern of the “ladder-up operator,” and those for $m<0$ the unvarying sign of the “ladder-down operator.” Physicists will still allow you to select your own sign for the $m=0$ state, bless them.

The final solution is

$$Y_l^m(\theta,\phi)= (-1)^{\max(m,0)} \sqrt{\frac{2l+1}... ...\vert)!}} P_l^{\vert m\vert}(\cos\theta)e^{{\rm i}m\phi} %$$ (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 [39, pp. 162-166]. This uses the following definition of the associated Legendre polynomials:

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

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

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 $-{\skew0\vec r}$. The parity is $-$1, or odd, if the wave function stays the same save for a sign change when you replace ${\skew0\vec r}$ by $-{\skew0\vec r}$. It turns out that the parity of the spherical harmonics is $(-1)^l$; so it is $-$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 $-{\skew0\vec r}$ means in spherical coordinates that $\theta$ changes into $\pi-\theta$ and $\phi$ into $\phi+\pi$. According to trig, the first changes $\cos\theta$ into $-\cos\theta$. 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 $P_l^{\vert m\vert}(\cos\theta)$.

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

$$\sum_{\alpha+\beta+\gamma=l} c_{\alpha\beta\gamma} x^\alpha y^\beta z^\gamma$$

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

$$\sum_{2a+b=l-m} c_{ab} u^{a+m}v^a z^b \qquad a,b,m\ge 0$$

$$\sum_{2a+b=l-\vert m\vert} c_{ab} u^av^{a+\vert m\vert} z^b \qquad a,b,-m\ge 0$$

where the coordinates $u=x+{\rm i}y$ and $v=x-{\rm i}y$ 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.

The simplest way of getting the spherical harmonics is probably the one given in derivation {D.66}.