# 7.4: Eigenvalues of Lz

- Page ID
- 15767

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

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

We can satisfy the orthonormality constraint ([e8.31]) provided that

\[\begin{aligned} \int_{0}^\pi {\mit\Theta}^{\,\ast}_{l',m'}(\theta)\,{\mit\Theta}_{l,m}(\theta)\,\sin\theta\,d\theta &= \delta_{ll'},\\[0.5ex] \int_0^{2\pi}{\mit\Phi}^{\,\ast}_{m'}(\phi)\,{\mit\Phi}_{m}(\phi)\,d\phi &= \delta_{mm'}.\label{e8.36}\end{aligned}\]

Note, from Equation ([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 ([e8.26]), ([e8.29]), and ([e8.34]) that \[-{\rm i}\,\hbar\,\frac{d{\mit\Phi}_m}{d\phi} = m\,\hbar\,{\mit\Phi}_m.\] The solution of this equation is \[\label{e8.38} {\mit\Phi}_m(\phi)\sim {\rm e}^{\,{\rm i}\,m\,\phi}.\] 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 ([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 ([e8.38]) by making use of the orthonormality constraint ([e8.36]). We obtain \[{\mit\Phi}_m(\phi) = \frac{ {\rm e}^{\,{\rm i}\,m\,\phi}}{\sqrt{2\pi}}.\] This is the properly normalized eigenstate of \(L_z\) corresponding to the eigenvalue \(m\,\hbar\).

## Contributors and Attributions

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