# Chapter 4: Spherical Harmonics

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The material covered in this chapter is partially presented in Boas Chapter 12, Section 10.

## 4.1 Associated Legendre Function

Chapter 3 was all about the Legendre polynomials $P_\ell(x)$. Here we build on these and introduce the associated Legendre functions $P_\ell^m(x)$ in the first part of the chapter, and the spherical harmonics $Y_\ell^m(\theta,\phi)$ in the second part. Spherical harmonics are used extremely widely in physics. You will see them soon enough in quantum mechanics, they are front and centre in advanced electromagnetism, and they will be among your best friends if you ever become a cosmologist. The presentation here will be fairly terse and dry: apologies! Applications will come in Chapter 10.

The associated Legendre functions are obtained by differentiating the Legendre polynomial $m$ times and multiplying by $(1-x^2)^{m/2}$,

\begin{equation} P_\ell^m(x) := (1-x^2)^{m/2} \frac{d^m}{dx^m} P_\ell(x). \tag{4.1} \end{equation}

By convention the label $m$ on the functions is indicated as a superscript. This label is not to be confused with a power: $P_\ell^m$ is not $P_\ell$ raised to the power $m$. The definition of Eq.(4.1) implies that $m \geq 0$, and it is understood that $P_\ell^0 \equiv P_\ell$. Because $P_\ell$ is a polynomial of degree $\ell$, taking $m$ derivatives reduces the largest power of $x$ to $\ell - m$, and there is nothing left to be differentiated once $m$ has reached its limiting value $m = \ell$. This implies that the associated Legendre functions are not defined for $m > \ell$, and Eq.(4.1) therefore comes with the condition $0 \leq m \leq \ell$.

While taking $m$ derivatives of a Legendre polynomial still returns a polynomial, the factor of $(1-x^2)^{m/2}$ inserted in the definition of $P_\ell^m$ implies that in general, these functions are not polynomials. They are still polynomials when $m$ is an even number, but they are not polynomials when $m$ is odd. The factor also implies that in general, the associated Legendre functions become imaginary when $x$ is smaller than $-1$ or larger than $+1$. To avoid this complication we shall restrict $x$ to the interval $(-1,1)$, and take the functions to be real.

It is a straightforward task to apply the definition of Eq.(4.1) and generate the first few associated Legendre functions. We find

\begin{align} P_1^1 &= (1-x^2)^{1/2}, \tag{4.2a}\\ P_2^1 &= 3x(1-x^2)^{1/2}, \tag{4.2b}\\ P_2^2 &= 3(1-x^2), \tag{4.2c}\\ P_3^1 &= \frac{3}{2}(5x^2-1)(1-x^2)^{1/2},\tag{4.2d}\\ P_3^2 &= 15x (1-x^2), \tag{4.2e}\\ P_3^3 &= 15(1-x^2)^{3/2}, \tag{4.2f}\\ P_4^1 &= \frac{5}{2}(7x^3-3x)(1-x^2)^{1/2}, \tag{4.2g}\\ P_4^2 &= \frac{15}{2}(7x^2-1)(1-x^2), \tag{4.2h}\\ P_4^3 &= 105x(1-x^2)^{3/2}, \tag{4.2i}\\ P_4^4 &= 105(1-x^2)^2. \tag{4.2j} \end{align}

Exercise 4.1: Verify that the listing of Eq.(4.2) follows from the definition of Eq.(4.1).

It is useful to extend the definition of the associated Legendre functions to negative values of $m$. This is achieved by the relation

\begin{equation} P_\ell^{-m}(x) := (-1)^m \frac{(\ell-m)!}{(\ell+m)!}\, P_\ell^m(x), \tag{4.3} \end{equation}

in which $m$ is taken to be positive. The definition indicates that except for an overall numerical factor, $P_\ell^{-m}$ is the same function of $x$ as $P_\ell^m$. With this extended definition, the parameter $m$ is now allowed to lie in the interval $-\ell \leq m \leq \ell$.

## 4.2 Associated Legendre Functions

We shall show that the associated Legendre functions satisfy the second-order differential equation

\begin{equation} (1-x^2) \frac{d^2 P_\ell^m}{dx^2} - 2 x \frac{d P_\ell^m}{dx} + \biggl[ \ell(\ell+1) - \frac{m^2}{1-x^2} \biggr] P_\ell^m = 0, \tag{4.4} \end{equation}

known as the associated Legendre equation. You may note that as it should, the equation reduces to Legendre's equation (3.31) when $m=0$. You may also note that the equation depends on $m^2$, and therefore does not care about the sign of $m$.

Exercise 4.2: Verify that the functions of Eq.(4.2) are solutions to the associated Legendre equation.

To establish Eq.(4.4) it is useful to introduce the auxiliary functions

\begin{equation} U^m_\ell(x) := \frac{d^m}{dx^m} P_\ell(x), \tag{4.5} \end{equation}

so that $P_\ell^m = (1-x^2)^{m/2} U^m_\ell$. In these equations and the manipulations below, it is understood that $m \geq 0$. We know that the functions $U^0_\ell \equiv P_\ell$ are solutions to Legendre's equation,

\begin{equation} (1-x^2) \frac{d^2 U_\ell^0}{dx^2} - 2 x \frac{d U_\ell^0}{dx} + \ell(\ell+1) U_\ell^0 = 0, \tag{4.6} \end{equation}

and differentiating this $m$ times reveals that $U_\ell^m$ satisfies

\begin{equation} (1-x^2) \frac{d^2 U_\ell^m}{dx^2} - 2(m+1) x \frac{d U_\ell^m}{dx} + \bigl[ \ell(\ell+1) - m(m+1) \bigr] U_\ell^m = 0. \tag{4.7} \end{equation}

You are encouraged to verify this statement, but be warned: it is not at all pleasant to differentiate Legendre's equation $m$ times, because of the factor $1-x^2$ in front of the second derivative, and the factor $2x$ in front of the first derivative. Fortunately there is an easier (and more interesting) way to prove the validity of Eq.(4.7). We shall proceed by induction.

Let us assume that Eq.(4.7) is known to be valid for some value of $m$, denoted $m^*$. We will show that it must then also be valid for the next value of $m$, $m^* + 1$. So if the equation is valid for $m=m^*$ it is also valid for $m=m^*+1$, and resetting $m^*$ to $m^*+1$, we also have validity for $m=m^*+2$, and so on, all the way to $m=\ell$. The conclusion thus far is that if the equation is valid for $m = m^*$, it must also be valid for all values of $m$ larger than $m^*$. But we know that Eq.(4.7) is valid for $m=0$, because it then reduces to Legendre's equation. We can therefore set $m^* = 0$, and we have established the validity of Eq.(4.7) for all values of $m$.

The inductive argument rests on the essential fact that if Eq.(4.7) is valid for $m=m^*$, it must also be valid for $m = m^*+1$. This we can prove by differentiating the equation with respect to $x$ and noting that by virtue of its definition,

\begin{equation} U^{m^*+1}_\ell = \frac{d U^{m^*}_\ell}{dx}. \tag{4.8} \end{equation}

So set $m=m^*$ in Eq.(4.7) and differentiate to get

\begin{equation} (1-x^2) \frac{d^3 U^{m^*}_\ell}{dx^3} - 2(m^*+2) x \frac{d^2 U^{m^*}_\ell}{dx^2} + \bigl[ \ell(\ell+1) - (m^*+1)(m^*+2) \bigr] \frac{d U^{m^*}_\ell}{dx} = 0 \tag{4.9} \end{equation}

after simplification. This can re-expressed as

\begin{equation} (1-x^2) \frac{d^2 U^{m^*+1}_\ell}{dx^2} - 2(m^*+2) x \frac{d U^{m^*+1}_\ell}{dx} + \bigl[ \ell(\ell+1) - (m^*+1)(m^*+2) \bigr] U^{m^*+1}_\ell = 0, \tag{4.10} \end{equation}

which is just Eq.(4.7) again with $m=m^*+1$. This derivation, combined with the inductive argument detailed in the preceding paragraph, provides a complete proof of the validity of Eq.(4.7) for any value of $m$.

Exercise 4.3: Reproduce the steps that lead to Eq.(4.10). Take the time to think through the inductive proof of Eq.(4.7). Proof by induction is a wonderful mathematical method, but it can be elusive and difficult to understand at first.

With Eq.(4.7) established it is now straightforward to prove the validity of Eq.(4.4). It is simply a matter of turning Eq.(4.1) into

\begin{equation} U^m_\ell(x) = (1-x^2)^{-m/2} P_\ell^m(x) \tag{4.11} \end{equation}

by virtue of Eq.(4.5), and making the substitution in Eq.(4.7). A single differentiation yields

\begin{equation} \frac{d U^m_\ell}{dx} = (1-x^2)^{-m/2} \frac{d P^m_\ell}{dx} + m x (1-x^2)^{-m/2-1} P^m_\ell, \tag{4.12} \end{equation}

and two differentiations give

\begin{align} \frac{d^2 U^m_\ell}{dx^2} &= (1-x^2)^{-m/2} \frac{d^2 P^m_\ell}{dx^2} + 2mx(1-x^2)^{-m/2-1} \frac{d P^m_\ell}{dx} \nonumber \\ & \quad text{} + m(1-x^2)^{-m/2-2} \bigl[ 1 + (m+1) x^2 \bigr] P^m_\ell \tag{4.13} \end{align}

after simplification. Inserting this within Eq.(4.7), multiplying through by $(1-x^2)^{m/2}$, and simplifying, we finally arrive at Eq.(4.4).

Exercise 4.4: Verify that Eq.(4.7) leads to the associated Legendre equation (4.4).

## 4.3 Recursion Relations

The associated Legendre functions satisfy a number of recursion relations, which can be found in the standard textbooks. A useful one is

\begin{equation} P_\ell^m = 2(m-1)x (1-x^2)^{-1/2}\, P_\ell^{m-1} - (\ell+m-1)(\ell-m+2) P_\ell^{m-2}. \tag{4.14} \end{equation}

It can be used to compute the functions efficiently, once the functions with low values of $m$ have been obtained from the definition.

Exercise 4.5: Knowing that $P_4 = \frac{1}{8} (35x^4 - 30x^2 + 3)$ and $P_4^1 = \frac{5}{2} (7x^3-3x)(1-x^2)^{1/2}$, compute $P_4^2$, $P_4^3$, and $P_4^4$ by exploiting the recursion relation. Make sure your results agree with the listing of Eq.(4.2).

The recursion relation follows from Eq.(4.7), which we write in the equivalent form

\begin{equation} (1-x^2) U_\ell^{m+2} - 2(m+1) x\, U_\ell^{m+1} + (\ell+m+1) (\ell-m) U_\ell^m = 0. \tag{4.15} \end{equation}

We put $m \to m - 2$ in this equation, multiply through by $(1-x^2)^{m/2 - 1}$, recall the definition of Eq.(4.5), and arrive directly at Eq.(4.14).

Exercise 4.6: Reproduce the steps that take us from Eq.(4.7) to Eq.(4.14).

## 4.4 Orthogonality

Like the Legendre polynomials, the associated Legendre functions form a set of orthogonal functions on the interval $(-1,1)$,

\begin{equation} \int_{-1}^1 P_\ell^m(x) P_{\ell'}^m(x)\, dx = 0 \tag{4.16} \end{equation}

when $\ell \neq \ell'$. Notice that $m$ is the same for both functions; the statement would not be true if the integrand were replaced by $P_\ell^m P_{\ell'}^{m'}$ with $m \neq m'$. The proof of Eq.(4.16) follows exactly the same steps as those involved in establishing the orthogonality of the Legendre polynomials. We shall not go over this again, but simply note that the proof begins with Eq.(4.4) multiplied by $P_{\ell'}^m$, which is subtracted from a second copy of the equation in which $\ell$ is interchanged with $\ell'$. With $m$ the same in both copies of the equation, the terms involving $m^2/(1-x^2)$ cancel out after subtraction, and the proof proceeds just as in Sec.3.7.

Exercise 4.7: Fill in the steps and provide a complete derivation of Eq.(4.16).

We define the norm of the associated Legendre functions by setting $\ell'$ equal to $\ell$ inside the integral. We shall show that this is given by

\begin{equation} N_\ell^m := \int_{-1}^1 \bigl[ P_\ell^m(x) \bigr]^2\, dx =\frac{2}{2\ell+1} \frac{(\ell+m)!}{(\ell-m)!}. \tag{4.17} \end{equation}

The norm evidently reduces to Eq.(3.47) when $m = 0$. Equations (4.16) and (4.17) can be combined into the single statement

\begin{equation} \int_{-1}^1 P_\ell^m(x) P_{\ell'}^m(x)\, dx = \frac{2}{2\ell+1} \frac{(\ell+m)!}{(\ell-m)!}\, \delta_{\ell\ell'}, \tag{4.18} \end{equation}

which involves the Kronecker delta $\delta_{\ell\ell'}$.

To establish Eq.(4.17) we return\footnote{The following derivation cannot be found in the standard textbooks. It was extracted from Sec.~II.1.8 of Equations of Mathematical Physics by A.N.\ Tikhonov and A.A.\ Samarskii (Dover, 1990).} once more to Eq.(4.7), which we write in the alternative form

\begin{equation} \frac{d}{dx} \bigl[ (1-x^2)^{m+1} U_\ell^{m+1} \bigr] + (\ell+m+1)(\ell-m) (1-x^2)^m U_\ell^m = 0. \tag{4.19} \end{equation}

In this we put $m \to m-1$ and obtain

\begin{equation} \frac{d}{dx} \bigl[ (1-x^2)^m U_\ell^m \bigr] = -(\ell+m)(\ell-m+1) (1-x^2)^{m-1} U_\ell^{m-1}. \tag{4.20} \end{equation}

Now, Eq.(4.17) involves $[P^m_\ell]^2 = (1-x^2)^m [U^m_\ell]^2$, which we write as

\begin{align} \bigl[ P^m_\ell \bigr]^2 &= (1-x^2)^m U^m_\ell \frac{d U^{m-1}_\ell}{dx} \nonumber \\ &= \frac{d}{dx} \bigl[ (1-x^2)^m U^m_\ell U^{m-1}_\ell \bigr] - U^{m-1}_\ell \frac{d}{dx} \bigl[ (1-x^2)^m U^m_\ell \bigr] \nonumber \\ &= \frac{d}{dx} \bigl[ (1-x^2)^m U^m_\ell U^{m-1}_\ell \bigr] + (\ell+m)(\ell-m+1) (1-x^2)^{m-1} \bigl[ U_\ell^{m-1} \bigr]^2 \nonumber \\ &= \frac{d}{dx} \bigl[ (1-x^2)^{1/2} P^m_\ell P^{m-1}_\ell \bigr] + (\ell+m)(\ell-m+1) \bigl[ P_\ell^{m-1} \bigr]^2; \tag{4.21} \end{align}

in the last step we invoked the definition of Eq.(4.5). Integrating this from $x=-1$ to $x=1$, we see that the total derivative produces vanishing boundary terms, and we are left with

\begin{equation} N_\ell^m = (\ell+m)(\ell-m+1) N_\ell^{m-1}. \tag{4.22} \end{equation}

This is a recursion relation for the norm of the associated Legendre functions.

Exercise 4.8: Make sure that you can reproduce all the steps that lead to Eq.(4.22).

It is now a simple matter to solve Eq.(4.22) for $N^m_\ell$. With $m=1$ the equation becomes $N^1_\ell = (\ell+1) \ell\, N^0_\ell$. With $m=2$ we get $N^2_\ell = (\ell+2)(\ell-1) N^1_\ell = (\ell+2)(\ell+1) \ell (\ell-1) N^0_\ell$, and with $m=3$ we obtain $N^3_\ell = (\ell+3)(\ell-2) N^2_\ell = (\ell+3)(\ell+2)(\ell+1) \ell (\ell-1) (\ell-2) N^0_\ell$. A clear pattern emerges, and after $m$ iterations of the recursion relation we get

\begin{equation} N^m_\ell = (\ell+m)(\ell+m-1)(\ell+m-2) \cdots (\ell-m+1) N^0_\ell. \tag{4.23} \end{equation}

The factor in front of $N^0_\ell$ can be written neatly as $(\ell+m)!/(\ell-m)!$, because

\begin{align} \frac{(\ell+m)!}{(\ell-m)!} &= \frac{ (\ell+m)(\ell+m-1) \cdots (\ell-m+1) (\ell-m)(\ell-m-1) \cdots (1)} {(\ell-m)(\ell-m-1) \cdots (1)} \nonumber \\ &= (\ell+m)(\ell+m-1) \cdots (\ell-m+1). \tag{4.24} \end{align}

And because we know that $N^0_\ell = 2/(2\ell+1)$ --- this is the norm of the Legendre polynomials --- we have arrived at Eq.(4.17).

Exercise 4.9: Make sure that you can reproduce all the steps that took us from the recursion relation to Eq.(4.17).

## 4.5 A Change of Variables

In many applications, the associated Legendre functions are expressed in terms of an angle $\theta$ instead of the original variable $x$. The transformation is

\begin{equation} x = \cos\theta, \tag{4.25} \end{equation}

and the functions are now denoted $P^m_\ell(\cos\theta)$. The interval $-1 \leq x \leq 1$ is covered by letting $\theta$ range from $\theta = 0$ to $\theta = \pi$; notice that $x$ decreases as $\theta$ increases. The ubiquitous factor of $(1-x^2)^{1/2}$ in the associated Legendre functions becomes the more compelling $\sin\theta$ when we make the change of variables; notice that $\sin\theta$ is everywhere positive in the interval $0 \leq \theta \leq \pi$.

It is easy to verify that the functions listed in Eq.(4.2) become

\begin{align} P_0 &= 1, \tag{4.26a}\\ P_1^0 &= \cos\theta, \tag{4.26b}\\ P_1^1 &= \sin\theta, \tag{4.26c}\\ P_2 &= \frac{1}{2} (3\cos^2\theta-1), \tag{4.26d}\\ P_2^1 &= 3\sin\theta \cos\theta, \tag{4.26e}\\ P_2^2 &= 3\sin^2\theta, \tag{4.26f}\\ P_3 &= \frac{1}{2} (5\cos^3\theta - 3\cos\theta), \tag{4.26g}\\ P_3^1 &= \frac{3}{2} \sin\theta (5\cos^2\theta - 1), \tag{4.26h}\\ P_3^2 &= 15\sin^2\theta \cos\theta, \tag{4.26i}\\ P_3^3 &= 15\sin^3\theta, \tag{4.26j}\\ P_4 &= \frac{1}{8} (35\cos^4\theta - 30\cos^2\theta + 3), \tag{4.26k}\\ P_4^1 &= \frac{5}{2} \sin\theta (7\cos^3\theta - 3\cos\theta), \tag{4.26l}\\ P_4^2 &= \frac{15}{2} \sin^2\theta (7\cos^2\theta-1), \tag{4.26m}\\ P_4^3 &= 105\sin^3\theta \cos\theta, \tag{4.26n}\\ P_4^4 &= 105\sin^4\theta \tag{4.26o} \end{align}

when expressed in terms of $\theta$. For completeness we have included the Legendre polynomials $P_\ell \equiv P_\ell^0$ in the list, also expressed in terms of the angular variable.

Exercise 4.10: Verify that the Legendre polynomials of Eq.(3.8) and the associated functions of Eq.(4.2) take the forms listed in Eq.(4.26) when the transformation $x = \cos\theta$ is implemented.

The differential equation of Eq.(4.4) becomes

\begin{equation} \frac{d^2 P^m_\ell}{d\theta^2} + \frac{\cos\theta}{\sin\theta} \frac{dP^m_\ell}{d\theta} + \biggl[ \ell(\ell+1) - \frac{m^2}{\sin^2\theta} \biggr] P^m_\ell = 0 \tag{4.27} \end{equation}

when expressed in terms of the new variable. To arrive at this is simply a matter of applying the chain rule to the derivatives. We have

\begin{align} \frac{d P^m_\ell}{dx} &= \frac{d P^m_\ell}{d\theta} \frac{d\theta}{dx} = -\frac{1}{\sin\theta} \frac{d P^m_\ell}{d\theta}, \tag{4.28a}\\ \frac{d^2 P^m_\ell}{dx^2} &= -\frac{1}{\sin\theta} \frac{d}{d\theta} \biggl( -\frac{1}{\sin\theta }\frac{d P^m_\ell}{d\theta} \biggr) = \frac{1}{\sin^2\theta} \frac{d^2 P^m_\ell}{d\theta^2} - \frac{\cos\theta}{\sin^3\theta} \frac{d P^m_\ell}{d\theta}, \tag{4.29} \end{align}

and inserting these expressions within Eq.(4.4) returns Eq.(4.27) after simplification. The equation can also be put in the alternative form

\begin{equation} \frac{1}{\sin\theta} \frac{d}{d\theta} \biggl( \sin\theta \frac{dP^m_\ell}{d\theta} \biggr) + \biggl[ \ell(\ell+1) - \frac{m^2}{\sin^2\theta} \biggr] P^m_\ell = 0. \tag{4.30} \end{equation}

Exercise 4.11: Verify Eqs.(4.27) and (4.30).

The orthonormality relation of Eq.(4.18) becomes

\begin{equation} \int_0^\pi P_\ell^m(\cos\theta) P^m_{\ell'}(\cos\theta) \sin\theta\, d\theta = \frac{2}{2\ell+1} \frac{(\ell+m)!}{(\ell-m)!}\, \delta_{\ell\ell'} \tag{4.31} \end{equation}

in the new variable. To get this we simply implement the change of variables. With $x = \cos\theta$ we have that $dx = -\sin\theta\, d\theta$, and we see that the lower bound $x=-1$ is mapped to $\theta = \pi$, while the upper bound $x = 1$ is mapped to $\theta = 0$. The integral is initially expressed as $\int_\pi^0 \cdots$, but rewritten as $-\int_0^\pi \cdots$ by reversing the direction of integration. The minus sign in front of the integral is finally combined with the minus sign in $dx = -\sin\theta\, d\theta$, and we arrive at Eq.(4.31).

Exercise 4.12: Go through the steps leading to Eq.(4.31).

## 4.6 Spherical Harmonics

The spherical harmonics $Y^m_\ell(\theta,\phi)$ are functions of two angles, $\theta$ and $\phi$. They are defined by taking the associated Legendre functions $P^m_\ell(\cos\theta)$, which depend on $\theta$ only, and multiplying them by

\begin{equation} e^{im\phi} = \cos(m\phi) + i \sin(m\phi), \tag{4.32} \end{equation}

a complex function of the second angle. To make things official we insert some strange numerical factors, and define the spherical harmonics as

\begin{equation} Y^m_{\ell}(\theta,\phi) := (-1)^m \sqrt{ \frac{2\ell+1}{4\pi} \frac{(\ell-m)!}{(\ell+m)!} }\, P^m_\ell(\cos\theta)\, e^{im\phi}. \tag{4.33} \end{equation}

It is understood that $\theta$ lies in the interval $(0,\pi)$, that $\phi$ is confined to $(0, 2\pi)$, and that $m$ is an integer between $-\ell$ and $\ell$. The reason for the square-root factor will be made clear below. The factor $(-1)^m$ is conventional, and you should be warned that the convention can depend on the author. In applications, $\theta$ and $\phi$ are interpreted as the familiar angles in the spherical coordinates $(r, \theta, \phi)$. These, we recall from Sec.1.8 are defined by the relations $x= r\sin\theta\cos\phi$, $y = r\sin\theta\sin\phi$, and $z = r\cos\theta$. The angles specify the position of a point on the surface of a sphere $r = \text{constant}$, and any function of $\theta$ and $\phi$ can therefore be thought of as a function on the surface of a sphere. This shall be our interpretation of the spherical harmonics: they constitute a set of functions on the surface of a sphere.

Equation (4.33) reduces to

\begin{equation} Y^0_\ell = \sqrt{ \frac{2\ell+1}{4\pi} } P_\ell(\cos\theta) \tag{4.34} \end{equation}

when $m=0$, and we see that $Y^0_\ell$ is simply a rescaled version of the Legendre polynomials; these functions depend on $\theta$ only. The spherical harmonics with negative values of $m$ can be obtained directly from those with positive values by exploiting the identity

\begin{equation} Y_\ell^{-m} = (-1)^m \bigl( Y^m_\ell \bigr)^*, \tag{4.35} \end{equation}

with the asterisk indicating complex conjugation. This property follows directly from Eq.(4.33) after we make use of Eq.(4.3) and recall that $e^{-im\phi} = (e^{im\phi})^*$.

Exercise 4.13: Prove the validity of Eq.(4.35).

The first few spherical harmonics are given by

\begin{align} Y_0^0 &= \frac{1}{\sqrt{4\pi}}, \tag{4.36a}\\ Y_1^0 &= \sqrt{\frac{3}{4\pi}}\, \cos\theta, \tag{4.36b}\\ Y_1^1 &= -\sqrt{\frac{3}{8\pi}}\, \sin\theta\, e^{i\phi},\tag{4.36c} \\ Y_2^0 &= \sqrt{\frac{5}{16\pi}}\, (3 \cos^2\theta - 1), \tag{4.36d}\\ Y_2^1 &= -\sqrt{\frac{15}{8\pi}}\, \sin\theta \cos\theta\, e^{i\phi},\tag{4.36e} \\ Y_2^2 &= \sqrt{\frac{15}{32\pi}}\, \sin^2\theta\, e^{2i\phi}, \tag{4.36f}\\ Y_3^0 &= \sqrt{\frac{7}{16\pi}}(5\cos^3\theta - 3\cos\theta), \tag{4.36g}\\ Y_3^1 &= -\sqrt{\frac{21}{64\pi}} \sin\theta (5\cos^2\theta - 1)\, e^{i\phi}, \tag{4.36h}\\ Y_3^2 &= \sqrt{\frac{105}{32\pi}} \sin^2\theta \cos\theta\, e^{2i\phi}, \tag{4.36i}\\ Y_3^3 &= -\sqrt{\frac{35}{64\pi}} \sin^3\theta\, e^{3i\phi}. \tag{4.36j} \end{align}

## 4.7 Orthonormality

We established in Sec.4.4 that the functions $P_\ell^m$ and $P_{\ell'}^m$ are orthogonal to each other, in the sense provided by Eq.(4.31),

\begin{equation} \int_0^\pi P_\ell^m(\cos\theta) P^m_{\ell'}(\cos\theta) \sin\theta\, d\theta =\frac{2}{2\ell+1} \frac{(\ell+m)!}{(\ell-m)!}\, \delta_{\ell\ell'}. \tag{4.37} \end{equation}

We have already noted that both functions must share the same value of $m$ for this statement to be true. On the other hand, the functions $e^{im\phi}$ and $e^{im'\phi}$ are also orthogonal to each other, in the sense provided by

\begin{equation} \int_0^{2\pi} ( e^{im\phi} )^* e^{im'\phi}\, d\phi = 2\pi \delta_{mm'}. \tag{4.38} \end{equation}

Recall the notion of orthogonality of complex functions introduced back in Sec.3.6.

Exercise 4.14: Verify Eq.(4.38). Now may be a good time to refresh your understanding of the complex exponential. We will make heavy use of it in Chapters 7, 9, and 10.

Because the spherical harmonics result from the merger of the associated Legendre functions with the complex exponentials, we might consider merging the two statements of orthogonality. We express this as

\begin{equation} \int_0^{2\pi} \int_0^\pi \bigl[ P^m_\ell(\cos\theta) e^{im\phi} \bigr]^* \bigl[ P^{m'}_{\ell'} e^{im'\phi} \bigr]\, \sin\theta d\theta d\phi = \frac{4\pi}{2\ell+1} \frac{(\ell+m)!}{(\ell-m)!}\, \delta_{\ell\ell'} \delta_{mm'}, \tag{4.39} \end{equation}

and conclude that the double integral vanishes unless $\ell' = \ell$ and $m' = m$. Both conditions must be met; equality between $m'$ and $m$ is required by the $\phi$-integral, and with this first condition satisfied, equality between $\ell'$ and $\ell$ is required by the $\theta$-integral. Notice that the $\theta$-integral never has to be performed with $m' \neq m$, because the $\phi$-integral always vanishes under such circumstances.

The merged statement of orthogonality can be re-expressed in terms of the spherical harmonics by making use of the definition of Eq.(4.33). Because the double integral vanishes unless $\ell'= \ell$ and $m' = m$, the numerical factor on the right-hand side of Eq.(4.39) can be factorized as

$(-1)^m \sqrt{ \frac{4\pi}{2\ell+1} \frac{(\ell+m)!}{(\ell-m)!} } \cdot (-1)^{m'} \sqrt{ \frac{4\pi}{2\ell'+1} \frac{(\ell'+m')!}{(\ell'-m')!}}$

and transferred to the left-hand side. You may worry about the overall factor of $(-1)^{m+m'}$, which was not present in Eq.(4.39). This factor is innocuous, because it becomes $(-1)^{2m}$ when $m'=m$, and since $2m$ is necessarily an even number, $(-1)^{2m} = 1$. The factor of $(-1)^m$ and the first square root are then naturally combined with $P_\ell^m(\cos\theta) e^{im\phi}$ inside the integral, to give rise to the spherical harmonic $Y_\ell^m(\theta,\phi)$. Absorbing $(-1)^{m'}$ and the second square root in a similar way to get $Y_{\ell'}^{m'}(\theta,\phi)$, we arrive at

\begin{equation} \int_0^{2\pi} \int_0^\pi \bigl[ Y^m_\ell(\theta,\phi) \bigr]^* \bigl[ Y^{m'}_{\ell'}(\theta,\phi) \bigr]\, \sin\theta d\theta d\phi = \delta_{\ell\ell'} \delta_{mm'}, \tag{4.40} \end{equation}

the statement of orthogonality for the spherical harmonics. Because these functions depend on two variables, $\theta$ and $\phi$, this is a two-dimensional notion of orthogonality, defined in terms of an integral over the surface of a sphere. We emphasize that the integral returns zero unless $\ell'=\ell$ and $m=m'$. When both conditions are met, the integral becomes

\begin{equation} \int_0^{2\pi} \int_0^\pi \bigl| Y^m_\ell(\theta,\phi) \bigr|^2\, \sin\theta d\theta d\phi = 1, \tag{4.41} \end{equation}

and we say that the norm of the spherical harmonics is equal to one. Alternatively, we might say that the spherical harmonics are normalized functions of $\theta$ and $\phi$. This is why we inserted the strange numerical factors in the definition of Eq.(4.33): we wanted to ensure that the spherical harmonics are normalized.

Exercise 4.15: Verify explicitly by performing the double integral that the functions $Y^2_2(\theta,\phi)$ and $Y^1_3(\theta,\phi)$ listed in Eq.(4.36) are normalized.

## 4.8 Differential Equations

The spherical harmonics are also proportional to $e^{im\phi}$, and for fixed $\theta$ they must satisfy

\begin{equation} \frac{\partial^2 Y^m_\ell}{\partial \phi^2} = -m^2 Y^m_\ell. \tag{4.42} \end{equation}

This is the associated Legendre equation (4.30), which we re-expressed in terms of partial derivatives to make it clear that $\phi$ is held fixed while differentiating.

The spherical harmonics are also proportional to $e^{im\phi}$, and for fixed $\theta$ they must satisfy

\begin{equation} \frac{\partial^2 Y^m_\ell}{\partial \phi^2} = -m^2 Y^m_\ell. \tag{4.43} \end{equation}

Again this was expressed in terms of partial derivatives, to emphasize that $\theta$ is now held fixed while differentiating.

The second equation reveals that the $m^2$ term in the first equation can be equivalently written in terms of partial derivatives with respect to $\phi$. This gives us

\begin{equation} \frac{1}{\sin\theta} \frac{\partial}{\partial \theta} \biggl( \sin\theta \frac{\partial Y^m_\ell}{\partial \theta} \biggr) + \frac{1}{\sin^2\theta} \frac{\partial^2 Y^m_\ell}{\partial \phi^2} + \ell(\ell+1) Y^m_\ell = 0. \tag{4.44} \end{equation}

This is a second-order, partial differential equation for the spherical harmonics.

The relevance of this equation comes from the fact that it is closely associated with the Laplacian operator in spherical coordinates. We saw back in Sec.1.8 --- refer to Eq.(1.86) --- that $\nabla^2 f$ is expressed as

\begin{equation} \nabla^2 f = \frac{1}{r^2} \biggl[ \frac{\partial}{\partial r} \biggl( r^2 \frac{\partial f}{\partial r} \biggr) + \frac{1}{\sin\theta} \frac{\partial}{\partial \theta} \biggl( \sin\theta \frac{\partial f}{\partial \theta} \biggr) + \frac{1}{\sin^2\theta} \frac{\partial^2 f}{\partial \phi^2} \biggr] \tag{4.45} \end{equation}

in spherical coordinates. You will recognize the last two terms if your compare this with Eq.(4.44). This similarity is not an accident, and in fact it provides the main motivation for defining and studying the spherical harmonics. We'll pick this thread up again in Chapter 10.

## 4.9 Practice Problems

1. Show by evaluating the integral explicitly that $P_2^1$ and $P_3^1$ are orthogonal on the interval $(-1,1)$.

2. Calculate $\int_{-1}^1 (P_2^2)^2\, dx$ explicitly and show that your result agrees with Eq.(4.18).

3. Using the definition of Eq.(4.1), show that $P^m_\ell(-x)$ is proportional to $P^m_\ell(x)$, with a numerical factor that is either $+1$ or $-1$. How does this numerical factor depend on $\ell$ and $m$? Verify that your result agrees with the listing of Eq.(4.2).

4. Complete the listing of Eq.(4.36) by including the spherical harmonics with negative values of $m$.

5. Express $r Y_1^0$, $r^2 Y_2^1$, and $r^3 Y_3^3$ in Cartesian coordinates.

## 4.10 Challenge Problems

1. a) Derive the recursion relation

$(2\ell+1) (1-x^2)^{1/2}\, P_\ell^m = P^{m+1}_{\ell+1}- P^{m+1}_{\ell-1}$

satisfied by the associated Legendre functions.

b) Use this recursion relation to find an explicit expression for $P^\ell_\ell(x)$. Verify your result with the special cases $\ell = 1, 2, 3, 4$.

2. a)Prove that acting on the spherical harmonic $Y^m_\ell$ with the differential operator

$L_+ := e^{i\phi} \biggl( \frac{\partial}{\partial \theta}+ i \frac{\cos\theta}{\sin\theta} \frac{\partial}{\partial \phi}\biggr)$

returns $Y^{m+1}_\ell$ up to a numerical factor. That is, prove that $L_+ Y^m_\ell = \alpha^m_\ell Y^{m+1}_\ell$ and find an expression for the constant $\alpha^m_\ell$. Simplify your expression as much as possible.

b) Recycle your result in part (a) and prove that acting on the spherical harmonic $Y^m_\ell$ with the differential operator

$L_- := e^{-i\phi} \biggl( -\frac{\partial}{\partial \theta}+ i \frac{\cos\theta}{\sin\theta} \frac{\partial}{\partial \phi}\biggr)$

returns $Y^{m-1}_\ell$ up to a numerical factor.