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Eigenvalues of $L_z$

It seems reasonable to attempt to write the eigenstate $Y_{l,m}(\theta,\phi)$ in the separable form
\begin{displaymath}
Y_{l,m}(\theta,\phi) = \Theta_{l,m}(\theta) \Phi_m(\phi).
\end{displaymath} (570)

We can satisfy the orthonormality constraint (558) provided that
$\displaystyle \int_{-\pi}^\pi \Theta^{ \ast}_{l',m'}(\theta) \Theta_{l,m}(\theta) \sin\theta d\theta$ $\textstyle =$ $\displaystyle \delta_{ll'},$ (571)
$\displaystyle \int_0^{2\pi}\Phi^{ \ast}_{m'}(\phi) \Phi_{m}(\phi) d\phi$ $\textstyle =$ $\displaystyle \delta_{mm'}.$ (572)

Note, from Eq. (553), 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 Eqs. (553), (556), and (570) that

\begin{displaymath}
-{\rm i} \hbar\frac{d\Phi_m}{d\phi} = m \hbar \Phi_m.
\end{displaymath} (573)

The solution to this equation is
\begin{displaymath}
\Phi_m(\phi)\sim {\rm e}^{ {\rm i} m \phi}.
\end{displaymath} (574)

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

Now, 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 (574) be single-valued: i.e., $\Phi_m(\phi+2 \pi)= \Phi_m(\phi)$ 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 $\Phi_m(\phi)$ must be continuous in order to ensure that $L_z$ is an Hermitian operator, since 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 (574) by making use of the orthonormality constraint (572). We obtain

\begin{displaymath}
\Phi_m(\phi) = \frac{{\rm e}^{ {\rm i} m \phi}}{\sqrt{2\pi}}.
\end{displaymath} (575)

This is the properly normalized eigenstate of $L_z$ corresponding to the eigenvalue $m \hbar$.


next up previous
Next: Eigenvalues of Up: Orbital Angular Momentum Previous: Eigenstates of Angular Momentum
Richard Fitzpatrick 2010-07-20