7.4: Eigenvalues of Lz

It seems reasonable to attempt to write the eigenstate \(Y_{l,m}(\theta,\phi)\) in the separable form

\[ Y_{l,m}(\theta,\phi) = {\mit\Theta}_{l,m}(\theta)\,{\mit\Phi}_m(\phi). \label{e8.34} \]

We can satisfy the orthonormality constraint \ref{e8.31} provided that

\[\begin{align} \int_{0}^\pi {\mit\Theta}^{\,\ast}_{l',m'}(\theta)\,{\mit\Theta}_{l,m}(\theta)\,\sin\theta\,d\theta &= \delta_{ll'},\\[4pt] \int_0^{2\pi}{\mit\Phi}^{\,\ast}_{m'}(\phi)\,{\mit\Phi}_{m}(\phi)\,d\phi &= \delta_{mm'}.\label{e8.36}\end{align} \]

Note, from Equation \ref{e8.26}, that the differential operator which represents \(L_z\) only depends on the azimuthal angle \(\phi\), and is independent of the polar angle \(\theta\). It therefore follows from Equations \ref{e8.26}, \ref{e8.29}, and \ref{e8.34} that

\[-{\rm i}\,\hbar\,\frac{d{\mit\Phi}_m}{d\phi} = m\,\hbar\,{\mit\Phi}_m. \nonumber \]

The solution of this equation is

\[ {\mit\Phi}_m(\phi)\sim \rm e^{\,{\rm i}\,m\,\phi}. \label{e8.38} \]

Here, the symbol \(\sim\) just means that we are neglecting multiplicative constants.

Our basic interpretation of a wavefunction as a quantity whose modulus squared represents the probability density of finding a particle at a particular point in space suggests that a physical wavefunction must be single-valued in space. Otherwise, the probability density at a given point would not, in general, have a unique value, which does not make physical sense. Hence, we demand that the wavefunction \ref{e8.38} be single-valued: that is, \({\mit\Phi}_m(\phi+2\,\pi)= {\mit\Phi}_m(\phi)\) for all \(\phi\). This immediately implies that the quantity \(m\) is quantized. In fact, \(m\) can only take integer values. Thus, we conclude that the eigenvalues of \(L_z\) are also quantized, and take the values \(m\,\hbar\), where \(m\) is an integer. [A more rigorous argument is that \({\mit\Phi}_m(\phi)\) must be continuous in order to ensure that \(L_z\) is an Hermitian operator, because the proof of hermiticity involves an integration by parts in \(\phi\) that has canceling contributions from \(\phi=0\) and \(\phi=2\pi\).]

Finally, we can easily normalize the eigenstate \ref{e8.38} by making use of the orthonormality constraint \ref{e8.36}. We obtain

\[{\mit\Phi}_m(\phi) = \frac{\rm e^{\,{\rm i}\,m\,\phi}}{\sqrt{2\pi}}. \nonumber \]

This is the properly normalized eigenstate of \(L_z\) corresponding to the eigenvalue \(m\,\hbar\).

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