# Why is the azimuthal part of spherical harmonics in the form of $Ae^{im\phi}$, but not $Ae^{im\phi}+Be^{-im\phi}$?

The general solution to the azimuthal equation for a quantum-mechanical rigid rotor (spherical harmonics) $$\frac{d^2}{d\phi^2}\psi(\phi)=n^2\psi(\phi)$$ ($$n^2$$ is the separation constant) is given by $$\psi(\phi)=Ae^{in\phi}+Be^{-in\phi}$$ With the boundary condition $$\psi(\phi)=\psi(\phi+2\pi)$$, we have $$n = ... -2, -1, 0, +1, +2 ...$$ and $$\psi_n(\phi)=Ae^{in\phi}+Be^{-in\phi},\,\,\,\,\,\,n = ... -2, -1, 0, +1, +2 ...$$ However, the complete wave function (spherical harmonics) is expressed as $$Y(\theta,\phi)=\Theta(\theta)\psi(\phi)=\Theta(\theta)(Ce^{in\phi})$$ but not $$Y(\theta,\phi)=\Theta(\theta)(Ae^{in\phi}+Be^{-in\phi})$$ It is very strange to me because $$Ce^{in\phi}$$ is not the general solution to the azimuthal equation, or in general $$Ce^{in\phi} \neq Ae^{in\phi}+Be^{-in\phi}$$ How can we use $$Ce^{in\phi}$$ here instead of $$Ae^{in\phi}+Be^{-in\phi}$$? Thank you for your help in advance!

The complete set of solutions of the azimutal equation can be written under of the form $$\psi_m(\phi)=A_me^{im\phi}+B_me^{-im\phi},\quad m\in\mathbb{N}$$ or equivalently under the form $$\psi_m'(\phi)=C_ne^{im\phi},\quad m\in\mathbb{Z}$$ Note that $$\psi_m\ne\psi_m'$$ but, in both cases, the general solution is $$\psi(\phi)=\sum_{m\in\mathbb{Z}}C_me^{im\phi}$$ with $$C_m=A_m$$ if $$m>0$$, $$C_m=B_{-m}$$ if $$m<0$$, and $$C_0=A_0+B_0$$.

Similarly, the solutions of the full Schrödinger equation are of the form $$\psi_{n,l,m}(r,\theta,\phi)=R_{nl}(r)P_{lm}(\cos\theta)e^{im\phi}$$ so the general solution is $$\psi(r,\theta,\phi)=\sum_{n,l,m} D_{n,l,m}R_{nl}(r)P_{lm}(\cos\theta)e^{im\phi}$$ where the sum over $$m$$ extends over $$\mathbb{Z}$$. Since $$P_{lm}(\cos\theta)$$ is proportionnal to $$P_{l,-m}(\cos\theta)$$ (https://en.wikipedia.org/wiki/Associated_Legendre_polynomials), you can also find some coefficients $$F_{n,l,m}$$ and $$G_{n,l,m}$$ such that $$\psi(r,\theta,\phi)=\sum_{n,l,m} R_{nl}(r)P_{lm}(\cos\theta)\big[F_{n,l,m}e^{im\phi}+G_{n,l,m}e^{-im\phi}\big]$$ where the sum over $$m$$ extends now only over $$\mathbb{N}$$.

• So the general solution is $$\psi(\phi)=\sum_n C_n e^{in\phi}$$ and the sum of C_n e^{in\phi} over all $n$ here seems to play an important role because without the summation, $\psi(\phi)$ cannot be the general solution. However, in spherical harmonics, the complete wave functions are expressed in the form of $$Y(\theta,\phi)=\Theta(\theta)\psi(\phi)=\Theta(\theta)(C_n e^{in\phi})$$ but not $$Y(\theta,\phi)=\Theta(\theta)(\sum_n C_n e^{in\phi})$$ Why is a particular solution for some $n$ used here instead of the general solution? – toby Dec 18 '19 at 3:46
• @toby Typically the general expression is given for the solution as just one term with the understanding that the entire general solution is a sum over those functions whose "amplitudes" depend on the initial conditions. – Aaron Stevens Dec 18 '19 at 5:39
• @Stevens 1) Isn't it too early to say about the amplitude since $\psi(\phi)$ is not a wave function but just a part of it? 2) There's no initial condition because this equation is time-independent. – toby Dec 18 '19 at 7:50
• I missed the information that your equation was only the azimutal part of the full Schrödinger equation. I have updated my answer. – Christophe Dec 18 '19 at 18:31
• Thank you so much! – toby Dec 20 '19 at 6:47

It turns out that the energies depend only on $$n^2$$ so any combination of the $$+n$$ and $$-n$$ eigenstates will be solution of the Schrödinger equation for this energy.

The actual value of the coefficients $$A$$ and $$B$$ depend on the initial condition. For instance, if $$\psi(0)=1$$ then you need $$A=B=1/2$$.

Note that you can also use Euler's formula and combine the exponentials so that your general solution is now of the form $$\psi(\phi)= a \cos(\phi)+b\sin(\phi)$$ where again $$a$$ and $$b$$ would be determined by the initial conditions. If again you have $$\psi(0)=1$$ then your solution would be $$\psi(\phi)=\cos(\phi)$$: you can verify that this actually satisfies your original differential equation.

• Comments are not for extended discussion; this conversation has been moved to chat. – Chris Dec 18 '19 at 7:34