[NOTES/QM-16001] Angular Momentum Algebra — Coordinate Representation

                                                                   Contents

1.  Eigenvalues and Eigenvectors .
2.  Separation of Variables .
3.  Solution of φ equation . 
4.  Solution of θ equation. 

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The orbital angular momentum of a particle is given by $\vec{L}=\vec{r}\times \vec{p}$ and the components of the angular momentum operator in coordinate representation are \begin{eqnarray} \hat{L}_x &=& -i\hbar\Big(\hat{y} \frac{\partial}{\partial z} - \hat{z} \frac{\partial}{\partial y} \Big)\label{eq01}\\ \hat{L}_y &=& -i\hbar\Big(\hat{z} \frac{\partial}{\partial x} - \hat{x} \frac{\partial}{\partial z} \Big)\label{eq02}\\ \hat{L}_z &=& -i\hbar\Big(\hat{x} \frac{\partial}{\partial y} - \hat{y} \frac{\partial}{\partial x} \Big)\label{eq03} \end{eqnarray} Here \(\hat{A}\) means operator corresponding to the dynamical variable \(A\). In terms of spherical polar coordinates these expressions take the form \begin{eqnarray} \hat{L}_x &=& i\hbar \left(\sin\phi \frac{\partial}{\partial \theta} +\cot\theta\cos\phi\frac{\partial}{\partial \phi} \right)\label{eq04}\\ \hat{L}_y &=& i\hbar \left(-\cos\phi \frac{\partial}{\partial \theta} +\cot\theta\sin\phi\frac{\partial}{\partial \phi} \right)\label{eq05}\\ \hat{L}_z &= & i\hbar \frac{\partial }{\partial \phi}\label{eq06} \end{eqnarray} The operator $\vec{L}^2$ given by \begin{equation} \vec{L}^2 = \hat{L}_x^2 +\hat{L}_y^2 + \hat{L}_z^2 \label{eq07} \end{equation} takes the form \begin{equation}\label{eq08} \vec{L}^2 = -\hbar^2\left[ \frac{1}{\sin\theta} \frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial\theta} \right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} \right] \end{equation} % The components of orbital angular momentum satisfy commutation relations % $$ [L_x,L_y]= i\hbar L_z;\quad [L_y,L_z]= i\hbar L_x;\quad[L_z,L_x]= i\hbar L_y; $$
1. Eigenvalues and Eigenvectors
These commutation relations of angular momentum imply that $\vec{L}^2$ commutes with $\vec{n}\cdot\hat{L}$ for all numerical$\hat{n}$. Hence we can find simultaneous eigenfunctions of $\vec{L}^2$ and a component of $\vec{L}$. along any direction $\vec{n}$. Taking $\hat{n}$ to be along $z-$ axis the eigenvalue equations \begin{eqnarray} \vec{L}^2 Y(\theta,\phi)&=&\lambda \hbar^2 Y(\theta,\phi)\label{eq09} \\ L_z Y(\theta,\phi)&= &\mu\hbar Y(\theta,\phi)\label{eq10} \end{eqnarray} become differential equations \begin{equation}\label{eq11} \left[ \frac{1}{\sin\theta} \frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial\theta} \right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} \right]Y(\theta,\phi) + \lambda Y(\theta,\phi)=0 \end{equation} and \begin{equation}\label{eq12} -i\frac{\partial }{\partial \phi}Y(\theta,\phi)= \mu Y(\theta,\phi) \end{equation} We shall now show that acceptable solutions exist only for \begin{equation} \label{eq13} \lambda =\ell(\ell+1) ; \qquad\qquad \mu=m \end{equation} where $\ell$ can take only positive integral values $0,1,2,\cdots$ and $m$ must satisfy the relation \(( -\ell\le m \le\ell )\), taking values in steps of 1: \begin{equation}\label{eq14} m= \ell,\ell-1,\cdots,-\ell+1, -\ell. \qquad \end{equation} There are $(2\ell+1)$ eigenvalues of $L_z$ for a fixed $\vec{L}^2$ and the spherical harmonics $Y_{\ell m}{\theta,\phi}$ will be seen to be the corresponding eigenfunctions. These results on eigenvalues and eigenfunctions of $\vec{L}^2$ and $L_z$ will be proved by solving the differential equations by the method of separation of variables.
2. Separation of Variables
To solve the differential equations we substitute \begin{equation}\label{eq15} Y(\theta,\phi)=Q(\theta)E(\phi) \end{equation} in eq.\eqref{eq11} and \eqref{eq12} and divide by $Y(\theta,\phi)=Q(\theta)E(\phi)$. This gives \begin{equation}\label{eq16} -i\frac{dE(\phi)}{d\phi} = \mu E(\phi) \end{equation} Similarly, \eqref{eq11} gives \begin{equation}\label{eq17} \left[\frac{1}{Q(\theta)} \frac{1}{\sin\theta} \frac{d}{d\theta}\left(\sin\theta \frac{d}{d\theta} Q(\theta) \right) + \frac{1}{\sin^2\theta}\frac{1}{E(\phi)}\frac{d^2 E(\phi)}{d \phi^2} \right]+\lambda=0 \end{equation} On using eq.\eqref{eq16} in \eqref{eq17} we get \begin{equation} \sin^2\theta \left\{\frac{1}{Q(\theta)} \frac{1}{\sin\theta} \frac{d}{d\theta}\left(\sin\theta \frac{d}{d\theta} Q(\theta) \right) \right\} + \lambda \sin^2\theta = - \frac{1}{E(\phi)} \frac{d^2E(\phi)}{d\phi^2} \end{equation} While the left hand side of the above equation is a function of $\theta$, the right hand side is a function of $\phi$ alone. Hence each side must be a constant, from eq.\eqref{eq16} this constant is $\mu$. Thus we get \begin{equation}\label{eq19} \frac{1}{\sin\theta} \frac{d}{d\theta}\left(\sin\theta \frac{d}{d\theta} Q(\theta) \right) + \Big(\lambda - \frac{\mu^2}{\sin^2\theta} \Big) Q(\theta) =0 \end{equation}
3. Solution of $\phi$ equation
General solution of eq.\eqref{eq16} is \begin{equation} E(\phi) = \begin{cases} A \exp(i\sqrt{\mu} \phi)+ B \exp(-i\sqrt{\mu}\phi), & \text{ if } \mu \ne 0\\ C + D \phi, & \text{ if } \mu=0 \end{cases} \end{equation} A wave function must be single valued function. For a fixed $r,\theta, \phi$ the values of $\phi$ and $\phi+2\pi$ correspond to the same point. Hence the solution should have the same value for $\phi$ and $\phi+2\pi$. Thus we demand that $E(\phi)$ must satisfy \begin{equation} E(\phi+2\pi)= E(\phi) \end{equation} for all $\phi$. For \(\mu=0\) this implies that \(D=0\).\\ Next, when \(\mu \ne 0\) we must have \begin{equation} A \exp(i\sqrt{\mu}( \phi+2\pi))+ B \exp(-i\sqrt{\mu}(\phi+2\pi))=A \exp(i\sqrt{\mu} \phi)+ B \exp(-i\sqrt{\mu}\phi) \end{equation} or \begin{equation}\label{eq23} A \exp(i\sqrt{\mu}\phi)\exp(2\pi i)+ B \exp(-i\sqrt{\mu}\phi)\exp(2\pi i))=A \exp(i\sqrt{\mu} \phi)+ B \exp(-i\sqrt{\mu}\phi). \end{equation} For $\mu\ne 0$, the linear independence of the $\exp(\pm{i}\sqrt{\mu} \phi)$ implies that the corresponding coefficients must be separately equal implying that $m$ is an integer. Thus the solutions of eq.\eqref{eq16} are \begin{equation} E(\phi) = \exp(i m\phi),\qquad m =0,\pm1,\pm2,\cdots \end{equation}
4. Solution of $\theta$ equation
If we substitute $w=\cos\theta$ in eq.\eqref{eq19} takes the form \begin{equation}\label{eq26} \frac{d}{dw}(1-w^2)\frac{dP(w)}{dw} +\left(\lambda -\frac{m^2}{1-w^2}\right)P(w)=0 \end{equation} where we have introduced $P(w)\equiv Q(\cos\theta)$ and have used $$ \frac{d P(w)}{d\theta} = \frac{dP(w)}{dw} \cdot \frac{dw}{d\theta}= -\sin\theta \frac{dP(w)}{dw} $$ The equation \eqref{eq26} is known as associated Legendre equation. This equation can be solved by the method of series solution. Since \eqref{eq26} is a second order differential equation, there are two linearly independent solutions of this equation. For general values of $\lambda$ both the solutions become infinite at $w=\pm 1$ corresponding to $\theta=0,\pi$ These solutions are therefore unacceptable. For special values $\lambda=\ell(\ell+1)$, where $\ell$ is a positive integer, and with $ |m|\le \ell$, one solution remains finite , but not the other solution. Thus we fix \begin{equation} \lambda = \ell(\ell+1) \qquad |m| \le \ell \end{equation} For the above choice, the non singular solution for $P(w)$ is known as the associated Legendre function and has the form \begin{equation} P^\ell_m(w) = (1-w^2)^{|m|/2} \frac{d^{|m|}}{d w^{|m|}} P_\ell(w) \end{equation} where $P_\ell(w)$ is Legendre polynomial of degree $\ell$.Thus the eigenfunctions of $\vec{L}^2$ and $L_z$ are the \begin{equation}\label{eq29} Y_{\ell m}(\theta,\phi) = N P^\ell_m(\cos\theta)e^{im\phi} , \qquad\qquad m=\ell,\ell-1,\cdots, \ell \end{equation} The normalization is fixed by demanding \begin{equation} \int_0^{2\pi}d\phi \int_0^\pi Y_{\ell m}^*(\theta,\phi)Y_{\ell m}(\theta,\phi)\sin\theta {}d\theta =1 \end{equation} The functions \(Y_{\ell m}(\theta,\phi)\) in eq.\eqref{eq29} are known as spherical harmonics.