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In Landau & Lifshitz's QM book, p86, a derivation of the eigenvalues of the angular momentum operator is given by taking its expression in polar form in one component (say $z$): $$-i\frac{\partial{\psi}}{\partial\phi}=l_z\psi,$$ solving for $\psi$ to get $$\psi = f(r,\theta)e^{il_z\phi},$$ and then making the argument

"If the function $\psi$ is to be single valued, it must be periodic in $\phi$ with period $2\pi$. Hence we find $l_z=m$, where $m=0,+-1,+-2,...$ Thus the eigenvalues $l_z$ are the positive and negative integers including 0."

I don't understand the argument in quotations. In complex analysis to make a complex exponential single valued we choose a branch, which would translate here to imposing the condition $0\leq l_z\phi \leq 2\pi$. I don't see how this translates into forcing $l_z$ to be an integer.

Qmechanic
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summersfreezing
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3 Answers3

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The argument is that (single particle) wavefunctions should satisfy $\psi(\phi)=\psi(\phi+2\pi)$ since rotating a distribution by $2\pi$ about the $z$ axis should leave the distribution invariant.

Thus, $e^{im\phi}=e^{im \phi}e^{2i \pi m}$ implies $m$ should be an integer.

ZeroTheHero
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If you don't make a branch cut, you have a function that wraps around from $0$ to $2\pi$ and keeps going. It may well have different values at $\phi$ and $\phi + 2\pi$. But if $l_z$ is an integer, the values will be the same.

mmesser314
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There is an interesting argument based on manipulating the operators into suitable forms and relating them to the quantum mechanical harmonic oscillator, which does not require the single-valued-ness of the wavefunction. As far as I am concerned, this proof has already been outlined here, but I shall provide a more dimensionally consistent version of the proof.

First we adopt the non-dimensionalised operators $$ \xi_{\{x,y\}} = \sqrt{\frac{m\omega}{\hbar}}\{x,y\} ,\quad \pi_{\{x,y\}}=\frac{\{p_x,p_y\}}{\sqrt{m\omega\hbar}}.$$

Now the azimuthal angular momentum becomes $$ L_z = \hbar\left(\xi_x\pi_y-\xi_y\pi_x\right). $$ Then we make the following transformation: $$ Q_1 = \frac{\xi_x+\pi_y}{\sqrt{2}},\quad Q_2 = \frac{\xi_x+\pi_y}{\sqrt{2}},\quad P_1 = \frac{\pi_x-\xi_y}{\sqrt{2}},\quad P_2 = \frac{\pi_x+\xi_y}{\sqrt{2}},$$ and note that the variables here obey the canonical commutation relation $$ \left[Q_i,Q_j\right] = \left[P_i,P_j\right] = 0,\quad \left[Q_i,P_j\right] = i\hbar\delta_{ij}. $$

This means we can treat them as formal position and momentum operators, leading to the quantum harmonic oscillator Hamiltonians $$ \mathcal{H}_1 = \frac{\hbar\omega}{2}\left(Q_1^2+P_1^2\right),\quad \mathcal{H}_2 = \frac{\hbar\omega}{2}\left(Q_2^2+P_2^2\right). $$

Calculate their difference to get $$ \mathcal{H}_1-\mathcal{H}_2 = \hbar\omega\left(\xi_x\pi_y-\xi_y\pi_x\right) = \omega L_z . $$ But $$ \mathcal{H}_1-\mathcal{H}_2 = \hbar\omega\left(n_1+\frac{1}{2}\right) - \hbar\omega\left(n_2+\frac{1}{2}\right) = \hbar\omega\left(n_1-n_2\right) , $$

so $$ L_z = \hbar\left(n_1-n_2\right) = m\hbar,\quad m\in\mathbb{Z}. $$

Jono94
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