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 and Attributions

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