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--- notes:physics:quantum-mechanics:angular-momentum [2026/09/30 20:22] – [First Property] chris
+++ notes:physics:quantum-mechanics:angular-momentum [2026/10/01 19:57] (current) – [Example calculations of different quantities] chris
@@ Line 91: / Line 91: @@
 ==== Second Property ====
+$$\begin{align}
+\left[\hat L_z,\hat L_\pm\right]
+&= \left[\hat L_z,\hat L_x\pm i\hat L_y\right] \\
+&= \left[\hat L_z,\hat L_x\right]\pm i\left[\hat L_z,\hat L_y\right] \\
+&= \left[\hat L_z,\hat L_x\right]\mp i\left[\hat L_y,\hat L_z\right] \\
+&= i\hbar\left(\hat L_y\mp i\hat L_x\right) \\
+&= \hbar\left(i\hat L_y\pm\hat L_x\right) \\
+&= \pm\hbar\left(\hat L_x\pm i\hat L_y\right) \\
+&= \pm\hbar\hat L_\pm
+\end{align}$$
+The quantum number $m$ is the eigenvalue of $\hat L_z$, so in order to see if
+$\hat L_\pm$ indeed takes us from state $\left|l,m\right>$ to state
+$\left|l,m\pm1\right>$ we can act with it on state $\hat L_\pm\left|l,m\right>$.
+$$\begin{align}
+\hat L_z\left(\hat L_\pm\left|l,m\right>\right)
+&= \left(\hat L_z\hat L_\pm\right)\left|l,m\right> \\
+&= \left(\hat L_\pm\hat L_z\pm\hbar\hat L_\pm\right)\left|l,m\right> \\
+&= \hat L_\pm\left(\left(\hat L_z\pm\hbar\right)\left|l,m\right>\right) \\
+&= \hat L_\pm\left(\hbar m\pm\hbar\right)\left|l,m\right> \\
+&= \hbar\left(m\pm1\right)\left(\hat L_\pm\left|l,m\right>\right)
+\end{align}$$
+The eigenvalue is $\hbar\mu=\hbar\left(m\pm1\right)$, so the operator $\hat L_\pm$ indeed
+takes us from state $\left|l,m\right>$ to state $\left|,m\pm1\right>$.
 ==== Normalization ====
-==== Accepted Values ====
+We have shown that there is a relation between the two states, $\hat L_\pm\left|l,m\right>\propto\left|l,m\pm1\right>$.
+In order to find an exact equation we can use the fact that wavefunctions are normalized.
+But first we need to calculate two operators that show up later on during the process.
+$$\begin{align}
+\hat L_-\hat L_+
+&= \left(\hat L_x-i\hat L_y\right)\left(\hat L_x+i\hat L_y\right)
+ = \hat L_x^2+\hat L_y^2+i\left(\hat L_x\hat L_y-\hat L_y\hat L_x\right) \\
+&= \hat L^2-\hat L_z^2+i\left[\hat L_x,\hat L_y\right]
+ = \hat L^2-\hat L_z^2+i\left(i\hbar\hat L_z\right) \\
+&= \hat L^2-\hat L_z^2-\hbar\hat L_z
+ = \hat L^2-\hat L_z\left(\hat L_z+\hbar\right)
+\end{align}$$
+$$\begin{align}
+\hat L_+\hat L_-
+&= \left(\hat L_x+i\hat L_y\right)\left(\hat L_x-i\hat L_y\right)
+ = \hat L_x^2+\hat L_y^2-i\left(\hat L_x\hat L_y-\hat L_y\hat L_x\right) \\
+&= \hat L^2-\hat L_z^2-i\left[\hat L_x,\hat L_y\right]
+ = \hat L^2-\hat L_z^2-i\left(i\hbar\hat L_z\right) \\
+&= \hat L^2-\hat L_z^2+\hbar\hat L_z
+ = \hat L^2-\hat L_z\left(\hat L_z-\hbar\right)
+\end{align}$$
+If we write $\hat L_\pm\left|l,m\right>=c_{lm}^\pm\left|l,m\pm1\right>$, we then can find the
+constant by taking the square magnitudes, since $\left\|\left|l,m\pm1\right>\right\|^2=1$
+by definition.
+$$\begin{align}
+\left|c_{lm}^\pm\right|^2
+&= \left\|\hat L_\pm\left|l,m\right>\right\|^2 \\
+&= \left(\hat L_\pm\left|l,m\right>\right)^\dagger\left(\hat L_\pm\left|l,m\right>\right) \\
+&= \left< l,m\left|\hat L_\mp\hat L_\pm\right|l,m\right> \\
+&= \left< l,m\left|\hat L^2-\hat L_z\left(\hat L_z\pm\hbar\right)\right|l,m\right> \\
+&= \left< l,m\left|\hbar^2\left[l(l+1)\right]-\hbar m\left(\hbar m\pm\hbar\right)\right|l,m\right> \\
+&= \hbar^2\left[l(l+1)-m(m\pm1)\right]\left< l,m|l,m\right> \\
+&= \hbar^2\left[l(l+1)-m(m\pm1)\right]
+\end{align}$$
+We now have fully defined how the angular momentum raising and lower operators act on a state.
+$$\boxed{\hat L_+\left|l,m\right> = \hbar\sqrt{l(l+1)-m(m+1)}\left|l,m+1\right>}$$
+$$\boxed{\hat L_-\left|l,m\right> = \hbar\sqrt{l(l+1)-m(m-1)}\left|l,m-1\right>}$$
+==== Accepted Values of $m$ and $l$ ====
+There's a natural limit for the eigenvalue $m$. It isn't possible for the projection
+of the angular momentum on the $z$ axis to be of greater magnitude than the total
+angular momentum. Thus $m$ is bounded, and its minimum and maximum values are given
+by requiring the normalization term to be zero in the two formulas above.
+$$\hat L_+\left|l,m_{max}\right> = 0\Rightarrow m_{max}=l$$
+$$\hat L_-\left|l,m_{min}\right> = 0\Rightarrow m_{min}=-l$$
+By proving that $\left[\hat L_z,\hat L_\pm\right]=\hbar\hat L_\pm$ we showed that the
+eigenvalues $\hbar m$ of $\hat L_z$ are equally spaced multiples of $\hbar$ and that
+the number $m$ is **integer**. The allowed values are all integer numbers in the interval
+$\left[-l,l\right]$. This interval has length $2l$, which also integer. So $l$ can
+be any integer divided by $2$, meaning an **integer or half-integer**.
+The two paragraphs above is the reason we write the eigenvalue $\lambda$ from the problem
+$\hat L^2\left|l,m\right>=\lambda\left|l,m\right>$ as $\lambda=l(l+1)$. Without this
+formulation, finding a relation between the limits of $m$ and $\lambda$ would be much
+more difficult and less usable.
+How is it possible for $l$ to theoretically be able to take half-integer values, while
+it's forced to be strictly integer since $m$ is strictly integer? During the analysis
+for $l$ and $\hat L^2$ above we silently assumed that by angular momentum we mean orbital
+angular momentum, as an equivalent to [[..:classical-dynamics|classical angular momentum]].
+We can follow the exact same procedure for a more general angular momentum $j$, which
+include both the orbital angular momentum with $m=m_l$ and spin $m_s$.
+The quantum number of spin $m_s$ can accept half-integer values, and ensures that all
+theoretically possible values of $j$ can be reached. Another impressive property is
+that it splits up all elementary particles in two categories, //fermions// and //bosons//,
+depending on if they have half-integer or strictly integer values of spin.
 ===== Constructing the Eigenfunctions =====
-==== Solving the Eigenvalue Problem ====
+The momentum operator $\hat p^2=-\hbar^2\nabla^2$ represents the total momentum.
+When written in spherical coordinates, it is easy to spot a radial and an angular
+part.
+$$\hat p^2
+=-\hbar^2\nabla^2
+=\underbrace{
+  -\hbar^2\left(\frac{\partial ^2}{\partial r^2}+\frac2r\frac{\partial}{\partial r}\right)
+}_{\large\hat p_r^2}
++ \frac{1}{r^2} \underbrace{
+\left[-\hbar^2\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)
+  +\frac{1}{\sin^2\!\theta}\frac{\partial^2}{\partial\phi^2}\right)\right]
+}_{\large\hat L^2}$$
+This is how we can //guess// the form of $\hat L^2$. We can also try to guess how $\hat L_z$
+looks like. Rotation around the $z$ axis is described exclusively from the
+azimuthal angle $\phi$. Taking the $\phi$ dependence from the formula above and ignoring
+the $\sin\theta$ scaling factor of the spherical Laplacian, we have a //guess//.
+$$\hat L_z=-i\hbar\frac{\partial}{\partial\phi}$$
+One can prove that this is indeed the operator if one starts from $\hat L_z=-i\hbar(x\partial_y-y\partial_x)$
+and applies the chain rule on $\partial_x,\partial_y$ to end up with derivatives
+with respect to $r,\theta$ and $\phi$.
+It was mentioned before that the two eigenvalue problems, $\hat L^2\left|l,m\right> =\hbar^2l(l+1)\left|l,m\right>$
+and $\hat L_z\left|l,m\right> =\hbar m\left|l,m\right>$ must be solved at the same time. Thus, they have
+common eigenfunctions $Y_{lm}(\theta,\phi)$, which are called **Spherical Harmonics**.
+==== Solving the Eigenvalue Problems ====
+In order to start solving the differential equations we initially try
+separating the variables and writing $Y_{lm}(\theta,\phi)=\Theta(\theta)\Phi(\phi)$ and
+start from the second eigenvalue problem.
+$$\begin{align}
+\hat L_zY_{lm}(\theta,\phi)
+&=
+\hbar mY_{lm}(\theta,\phi)\\
+-i\hbar\frac{\partial}{\partial\phi}\left(\Theta(\theta)\Phi(\phi)\right)
+&= \hbar m\Theta(\theta)\Phi(\phi) \\
+\Theta(\theta)\frac{\partial\Phi}{\partial\phi}
+&= im\Theta(\theta)\Phi(\phi) \\
+\frac{\partial\Phi}{\partial\phi}
+&=im\Phi(\phi)
+\end{align}$$
+$$\boxed{\Phi(\phi)=e^{im\phi}}$$
+Now, for the first eigenvalue problem:
+$$
+\hat L^2Y_{lm}(\theta,\phi)
+=
+\hbar^2l(l+1)Y_{lm}(\theta,\phi)
+$$
+$$\begin{align}
+-\hbar^2\left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)
++\frac{1}{\sin^2\!\theta}\frac{\partial^2}{\partial\phi^2}\right)
+\left(\Theta(\theta)e^{im\phi}\right)
+&=
+\hbar^2l(l+1)\Theta(\theta)e^{im\phi}
+\\
+-\left(\frac{e^{im\phi}}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)
++\frac{\Theta(\theta)}{\sin^2\!\theta}\frac{d^2\left(e^{im\phi}\right)}{d\phi^2}\right)
+&=
+l(l+1)\Theta(\theta)e^{im\phi}
+\\
+\frac{m^2}{\sin^2\!\theta}\Theta(\theta)-\frac{1}{\sin\theta}\frac{d}{d\theta}
+\left(\sin\theta\frac{d\Theta}{d\theta}\right)
+&=
+l(l+1)\Theta(\theta)
+\end{align}$$
+By doing a variable change of $\xi=\cos\theta, d\xi=-\sin\theta d\theta$ we can simplify the equation.
+$$
+\frac{m^2}{1-\xi^2}\Theta(\xi)
+-\frac{d}{d\xi}\left(\left(1-\xi^2\right)\frac{d\Theta}{d\xi}\right)
+=
+l(l+1)\Theta(\xi)
+$$
+$$
+\frac{d}{d\xi}\left(\left(1-\xi^2\right)\frac{d\Theta}{d\xi}\right)
++\left(l(l+1)-\frac{m^2}{1-\xi^2}\right)\Theta(\xi)
+=
+$$
+This is the **General** (or **Associated**) **Legendre Equation**, and some of its solutions
+are the **Associated Legendre Polynomials** $P_l^m(x)$. From all the solutions, only the
+Associated Legendre Polynomials can describe this physical system. They are defined with
+respect to the normal, run of the mill Legendre Polynomials $P_l(x)$((The Associated Legendre
+Polynomials are not actually polynomials. They are functions, and got this name because
+they are generated from the Legendre Polynomials)).
+This is their historical definition, and only later on they were characterized as solutions
+to the differential equation above.
+$$P_l^m(x)=(-1)^m\left(1-x^2\right)^{m/2}\frac{d^m}{dx^m}P_l(x)$$
+<WRAP tip round center 85%>
+**The Condon-Shortley phase**
+The $(-1)^m$ term above is called the Condon-Shortley phase and it neither required for
+solving the Legendre differential equations, nor for solving the angular momentum eigenvalue
+problem. Its addition helps on calculations with the raising and lowering operators by making
+the algebra simpler. Depending on the physics textbook, it is usually added either directly
+in the definition of the Associated Legendre Polynomials (like above), or in the definition
+of the spherical harmonics.
+</WRAP>
+The Legendre Polynomials are the solutions to the Legendre differential equation,
+$$\frac{d}{dx}\left(\left(1-x^2\right)\frac{dP_l}{dx}\right) + l(l+1)P_l(x) = 0$$
+and are generated by the Rodrigues formula.
+$$P_l(x)=\frac{1}{2^ll!}\frac{d^l}{dx^l}\left[\left(x^2-1\right)^l\right]$$
+By replacing the Rodrigues formula on the definition of the associated polynomials we get:
+$$P_l^m(x)=(-1)^m\frac{1}{2^ll!}\left(1-x^2\right)^{m/2}\frac{d^{l+m}}{dx^{l+m}}\left[\left(x^2-1\right)^l\right]$$
+The formula above stands only for positive values of $m$. It can expanded to negative values through the formula
+below:
+$$P_l^{-m}(x)=(-1)^m\frac{(l-m)!}{(l+m)!}P_l^m,\quad m>0$$
+After taking all of the above into account, and //some// calculations, one can find the
+exact form of the spherical harmonics.
+$$\boxed{
+Y_{lm}(\theta,\phi)
+= (-1)^m\frac{1}{2^ll!}
+\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}
+\left(1-\cos^2\!\theta\right)^{\left|m\right|/2}
+e^{im\phi}
+\left(\frac{d}{d(\cos\theta)}\right)^{l+\left|m\right|}
+\left[\left(\cos^2\!\theta-1\right)^l\right]
+}$$
+The maximum degree of the polynomial $\left(\cos^2\!\theta-1\right)^l$ is $2l$, so
+the constraint regarding the values of $m$ is also obeyed. For any $\left|m\right|>l$ there
+will be the differentiation of a constant which will immediately zero the spherical harmonic.
 ==== 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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