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--- notes:physics:quantum-mechanics:angular-momentum [2026/10/01 16:36] – [Solving the Eigenvalue Problems] chris
+++ notes:physics:quantum-mechanics:angular-momentum [2026/10/01 19:57] (current) – [Example calculations of different quantities] chris
@@ Line 350: / Line 350: @@
 ==== Using Operators ====
+The raising and lower operators $\hat L_\pm$ obey all physical constraints for the
+spherical harmonics as they have been defined. So an easier way to find said
+harmonics is to solve the differential equation for one of the termination conditions
+and find all other states one by one.
+$$\hat L_+Y_{l,l}(\theta,\phi)=0,\quad \hat L_-Y_{l,-l}(\theta,\phi)=0$$
+First, we have to find formulas of $\hat L_x, \hat L_y$ with respect to $\theta, \phi$ the same
+way we found $\hat L_z=-i\hbar\partial_\phi$.
+$$\hat L_x = i\hbar\left(\sin\phi\frac{\partial}{\partial\theta} + \frac{\cos\phi}{\tan\theta}\frac{\partial}{\partial\phi}\right),
+\quad
+\hat L_y = i\hbar\left(-\cos\phi\frac{\partial}{\partial\theta}+\frac{\sin\phi}{\tan\theta}\frac{\partial}{\partial\phi}\right)$$
+Calculating the raising and lowering operators through $\hat L_\pm=\hat L_x\pm i\hat L_y$ we find:
+$$\hat L_+ =\hbar e^{i\phi}\left(\frac{\partial}{\partial\theta}+i\frac{1}{\tan\theta}\frac{\partial}{\partial\phi}\right),
+\quad
+\hat L_-=\hbar e^{-i\phi}\left(-\frac{\partial}{\partial\theta}+i\frac{1}{\tan\theta}\frac{\partial}{\partial\phi}\right)$$
+The lowest state will have the form $Y_{l,-l}=\Theta_l(\theta)e^{-il\phi}$. Solving the termination
+condition we have:
+$$\hat L_-Y_{l,-l}(\theta,\phi)=0$$
+$$\hbar e^{-i\phi}\left(-e^{-il\phi}\frac{d\Theta}{d\theta}+i\frac{\Theta(\theta)}{\tan\theta}(-il)e^{-il\phi}\right)=0$$
+$$-e^{il\phi}\left(\frac{d\Theta}{d\theta}-\frac{l}{\tan\theta}\Theta(\theta)\right)=0$$
+We then perform a $u$ substitution, $u=\sin\theta, du=\cos\theta d\theta$.
+$$\cos\theta\frac{d\Theta}{du}-\frac{l\cos\theta}{u}\Theta(u)=0$$
+$$\frac{d\Theta}{du}-\frac{l}{u}\Theta(u)=0$$
+$$\Theta(u)=N_lu^l$$
+$$\Theta(\theta)=N_l\sin^l\!\theta$$
+Due to normalization we require that:
+$$\oint\left|Y_{lm}(\theta,\phi)\right|^2\ d\Omega=1$$
+$$\require{cancel}\oint N_l^2\sin^{2l}\!\theta\cancelto{1}{\left|e^{im\phi}\right|^2}\ d\Omega = 1$$
+$$N_l=\frac{1}{2^ll!}\sqrt{\frac{(2l+1)!}{4\pi}}$$
+The lowest state is:
+$$\boxed{Y_{l,-l}(\theta,\phi)=(-1)^l\frac{1}{2^ll!}\sqrt{\frac{(2l+1)!}{4\pi}}\ \sin^l\!\theta\ e^{-il\phi}}$$
+and the rest can be found through the raising operator.
+$$\boxed{Y_{l,m+1}(\theta,\phi)=\frac{1}{\hbar\sqrt{l(l+1)-m(m+1)}}\hat L_+Y_{l,m}(\theta,\phi)}$$
+Below are listed the first few spherical harmonics:
+$$Y_{00}(\theta,\phi)=\frac{1}{\sqrt{4\pi}}$$
+$$Y_{10}(\theta,\phi)=\sqrt{\frac{3}{4\pi}}\ \cos\theta$$
+$$Y_{1,\pm1}(\theta,\phi)=\mp\sqrt{\frac{3}{8\pi}}\ \sin\theta\ e^{\pm i\phi}$$
+$$Y_{20}(\theta,\phi)=\sqrt{\frac{5}{16\pi}}\ (3\cos^2\!\theta-1)$$
+$$Y_{2,\pm1}(\theta,\phi)=\mp\sqrt{\frac{15}{8\pi}}\ \cos\theta\sin\theta\ e^{\pm i\phi}$$
+$$Y_{2,\pm2}(\theta,\phi)=\sqrt{\frac{15}{32\pi}}\ \sin^2\!\theta\ e^{\pm i2\phi}$$
 ===== Example calculations of different quantities =====
+Calculating quantities is trivial if their operator can be described as
+a function of the raising and lowering operators. Take for example the
+$x$ and $y$ components of the angular momentum.
+$$\hat L_x=\frac{\hat L_++\hat L_-}{2},\quad\hat L_y=\frac{\hat L_+-\hat L_-}{2}$$
+For example below are the calculations for
+$\left<\hat L_x\right>,\left<\hat L_y\right>,\left<\hat L_x^2\right>, \left<\hat L_y^2\right>$.
+$$\begin{align}
+\left<\hat L_x\right>
+&= \left< l,m\left|\hat L_x\right|l,m\right> \\
+&\propto \left< l,m\left|\hat L_+\right|l,m\right> + \left< l,m\left|\hat L_-\right|l,m\right> \\
+&\propto \left< l,m|l,m+1\right> + \left< l,m|l,m-1\right> \\
+&= 0
+\end{align}$$
+The system is $\phi$-independent, since it has a central potential. This means symmetry around
+the $z$ axis and thus $\left<\hat L_x\right>=\left<\hat L_y\right> =0$. Similarly
+$\left<\hat L_x^2\right>=\left<\hat L_y^2\right>$. Then:
+$$
+\left<\hat L^2\right>
+= \left<\hat L_x^2+\hat L_y^2+\hat L_z^2\right>
+= \left<\hat L_x^2\right> + \left<\hat L_y^2\right> + \left<\hat L_z^2\right>
+= 2\left<\hat L_x^2\right> + \left<\hat L_z^2\right>
+$$
+$$\begin{align}
+\left<\hat L_x^2\right>
+= \left<\hat L_y^2\right>
+&= \frac12\left(\left<\hat L^2\right> - \left<\hat L_z^2\right> \right) \\
+&= \frac12\left(\left< l,m\left|\hat L^2\right|l,m\right> -\left< l,m\left|\hat L_z^2\right|l,m\right>\right) \\
+&= \frac12\Big(l(l+1)\left< l,m|l,m\right> - m^2\left< l,m|l,m\right>\Big) \\
+&= \frac{l(l+1)-m^2}{2}
+\end{align}$$
+This way we can find the dispersion around the expected value.
+$$\require{cancel}
+\Delta L_x
+= \Delta L_y
+= \sqrt{\left<\hat L_x^2\right> - \cancelto{0}{\left<\hat L_x\right>^2}}
+= \sqrt{\frac{l(l+1)-m^2}{2}}
+$$

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