I was reading about the solution to the Schrodinger Equation in spherical coordinates with a radially symmetric potential, $V(r)$, and the book split the wavefunction into two parts: an angular part and a radial part. When dealing with the angular part of $\psi$, the book claims that the angular part "must" be an eigenfunction of $L^2$ (the square of the angular momentum operator) and since the eigenfunctions of $L^2$ are the spherical harmonics $Y_{lm}$, then the angular part of $\psi$ is just the spherical harmonic equations.
I am confused as to why the angular part of $\psi$ must be an eigenfunction of $L^2$.
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asked Apr 18, 2013 at 20:32
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1$\begingroup$ what book are you reading? $\endgroup$– PrasttApr 18, 2013 at 20:37
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$\begingroup$ The eigenfunctions of $L^2$ form a complete basis, and furthermore $H$ commutes with $L^2$ (if $V(r)$ is spherically symmetric). So you can always choose a basis of energy eigenstates that are also eigenstates of $L^2$. This is very convenient in practice, because you get states with a well-defined energy and spin! $\endgroup$– VibertApr 18, 2013 at 21:54
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2 Answers
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It has to do with the Laplacian $(\nabla^2)$. When trying a separable solution
$$\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$$
You will get two ODE's, a radial and an angular equation. They must be equal to separation constants. Turns out the angular part is the $L^2$ operator. You can rewrite $$H \psi = E \psi$$ as $$\left( \frac{-\hbar^2}{2 m} \nabla^2 + V(r) \right) \psi = E \psi $$ But $$\nabla^2 = \left(\frac{1}{r^2}\partial_r ( r^2 \partial_r ) - \frac{1}{\sin{\theta}} \partial_{\theta} (\sin{\theta}\, \partial_{\theta}) + \frac{1}{\sin^2{\theta}} \partial_{\phi \phi} \right) = \left(\frac{1}{r^2}\partial_r ( r^2 \partial_r ) + \frac{L^2}{\hbar^2} \right) $$ Now by inserting the above relation for $\psi$ into our Schrodinger equation and finding separation constants for the ODE's you will get a second order harmonic oscillator ODE for $\Phi(\phi)$ which proves that you will have integer values of m and the associated Legendre equation that will yield $P^m_l (\cos{\theta})$. Combined these will give you the spherical harmonics which must be solutions to this angular equation, and conversely eigenfunctions of $L^2$ with eigenvalues $l(l+1)$
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The operator $L^2$ is a symmetry of the Hamiltonian, which means that $[H,L^2] = 0$. This means that we can find simultaneous eigenfunctions to $H$ and $L^2$ since $HL^2 | \psi \rangle = L^2 H | \psi \rangle = \ell (\ell + 1) E | \psi \rangle$. The operator $L^2$ does not depend on the radial coordinate, thus the angular part of the wavefunction for hydrogen can be chosen to be an eigenfunciton of $L^2$. This is useful as it allows to label states with quantum numbers corresponding to the energy and angular momentum of the state.
In general, whenever you have a spherically symmetric potential (as in the case of the hydrogen atom) the spherical harmonics will show up as eigenfunctions of the angular part.
