$$\require{cancel}$$

# 7.4: Eigenvalues of Lz

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

• { {template.ContribFitzpatrick()}}