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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!

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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 $\mathbb{N}$.

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  • $\begingroup$ 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? $\endgroup$ – toby Dec 18 '19 at 3:46
  • $\begingroup$ @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. $\endgroup$ – Aaron Stevens Dec 18 '19 at 5:39
  • $\begingroup$ @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. $\endgroup$ – toby Dec 18 '19 at 7:50
  • $\begingroup$ I missed the information that your equation was only the azimutal part of the full Schrödinger equation. I have updated my answer. $\endgroup$ – Christophe Dec 18 '19 at 18:31
  • $\begingroup$ Thank you so much! $\endgroup$ – toby Dec 20 '19 at 6:47
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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.

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

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