Angular momentum | Quantum mechanics
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- Written by: Germán Fernández
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Consider a moving particle of mass m, whose position vector is $\vec{r}=x\vec{i}+y\vec{j}+z\vec{k}$. Let the linear momentum of the particle be $\vec{p}=p_{x}\vec{i}+p_{y}\vec{j}+p_{z}\vec{k}$
The angular momentum of the particle, $\vec{l}$ is defined: \begin{equation} \vec{l}=\vec{r}\times\vec{p}=\left(yp_{z}-zp_ {y}\right)\vec{i}-\left(xp_{z}-zp_{x}\right)\vec{j}+\left(xp_{y}-yp_{x}\right)\vec {k} \end{equation} \begin{eqnarray} l_{x} &=& yp_z - zp_y\\ l_{y} &=& zp_x - zp_z\\ l_{z} &=& xp_y - yp_x \end{eqnarray}
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- Written by: Germán Fernández
- Category: Angular Moment | Quantum mechanics
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In quantum mechanics there are two types of angular momentum:
- Orbital angular momentum, refers to the movement of the particle in space and is analogous to classical mechanics.
- Spin angular momentum is an intrinsic property of microscopic particles and has no classical analogue.
We are going to build the mechanical-classical operators of angular momentum.
We start from the classical expression $l_x = yp_z - zp_y$ and replace Cartesian coordinates and linear moments with the corresponding quantum operators:
\begin{equation} \hat{l}_x = \hat{y}\left(-i\hbar\frac{\partial}{\partial z}\right)-\hat{z}\left(-i\hbar\frac{\partial}{\partial y}\right)=-i\hbar\left(\hat {y}\frac{\partial}{\partial z}-\hat{z}\frac{\partial}{\partial y}\right) \end{equation}
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Next, we will calculate the commutation relations between the different angular momentum operators, to determine which of them can be known simultaneously.
\begin{equation} [\hat{l}_x , \hat{l}_y]=\hat{l}_x \hat{l}_y - \hat{l}_y \hat{l}_x \end{equation}
So that expressions are not too long, let's calculate the two terms separately
\begin{equation} \hat{l}_x\hat{l}_y f=\left[-i\hbar\left(y\frac{\partial} {\partial z}-z\frac{\partial}{\partial y}\right)\right]\left[-i\hbar\left(z\frac{\partial f}{\partial x}-x\frac{\partial f}{\partial z}\right)\right]= \end{equation}
\begin{equation} =-\hbar^2\left(y\frac{\partial f}{\partial x} +yz\frac{\partial^2 f}{\partial z \partial x}-yz\frac{\partial^2 f}{\partial z^2}-z^2\frac{\partial^2 f} {\partial y \partial x}+zx\frac{\partial^2 f}{\partial y \partial z}\right) \end{equation}
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The symmetry of the atom forces us to work in spherical polar coordinates. In Cartesian coordinates the components of angular momentum are given by the following equations: \begin{eqnarray} \hat{l}_x &=& -i\hbar\left[\hat{y}\frac{\partial}{\partial z}-\hat{z}\frac{\partial}{\partial y}\right]\\ \hat{l}_y &=& -i\hbar\left[\hat{z}\frac{\partial }{\partial x}-\hat{x}\frac{\partial}{\partial z}\right]\\ \hat{l}_z &=& -i\hbar\left[\hat{x}\ frac{\partial}{\partial y}-\hat{y}\frac{\partial}{\partial x}\right]\label{1} \end{eqnarray}
Relationships between Cartesian and spherical coordinates:
\begin{eqnarray} x &=& rsen\theta\cos\varphi \\ y &=& rsen\theta sin\varphi \\ z &=& r\cos\theta \\ r^2 &=& x^2 +y^2+z^2 \\ \cos\theta &=& \frac{z}{r}=\frac{z}{\left(x^2+y^2+z^2\right)^ {1/2}}\\ \tan\varphi &=& \frac{y}{x} \label{2} \end{eqnarray}
Let the function $f(r,\theta,\varphi)=g(x,y,z)$ be the function, we will calculate the partial derivatives of the function $g$ with respect to the variables $x,y,z$ taking into account the following dependency: $r=r(x,y,z)$, $\theta=\theta(x,y,z)$ and $\varphi=\varphi(x,y,z)$. Applying the The chain rule to the function $g(x,y,z)$ gives us:
\begin{equation}\label{10} \left(\frac{\partial g}{\partial x}\right)_{ y,z}=\left(\frac{\partial f}{\partial r}\right)_{\theta,\varphi}\cdot\left(\frac{\partial r}{\partial x}\right )_{y,z} +\left(\frac{\partial f}{\partial \theta}\right)_{r,\varphi}\cdot\left(\frac{\partial\theta}{\partial x}\right)_{y,z}+\left(\frac{\partial f}{\partial\varphi}\right)_{r,\theta}\cdot\left(\frac{\partial\varphi }{\partial x}\right)_{y,z} \end{equation}
\begin{equation}\label{11} \left(\frac{\partial g}{\partial y}\right)_{ x,z}=\left(\frac{\partial f}{\partial r}\right)_{\theta,\varphi}\cdot\left(\frac{\partial r}{\partial y}\right)_{x,z} +\left(\frac{\partial f}{\partial \theta}\right)_{r,\varphi} \cdot\left(\frac{\partial\theta}{\partial y}\right)_{x,z}+\left(\frac{\partial f}{\partial\varphi}\right)_{r ,\theta}\cdot\left(\frac{\partial\varphi}{\partial y}\right)_{x,z} \end{equation}
\begin{equation}\label{12} \left(\frac{\partial g}{\partial z}\right)_{x,y}=\left(\frac{\partial f}{\partial r}\right)_{\theta,\varphi}\cdot\left(\frac{\partial r}{\partial z}\right)_{x,y} +\left(\frac{\partial f}{\partial \theta}\right)_{r,\varphi} \cdot\left(\frac{\partial\theta}{\partial z}\right)_{x,y}+\left(\frac{\partial f}{\partial\varphi}\right)_{r ,\theta}\cdot\left(\frac{\partial\varphi}{\partial z}\right)_{x,y}\end{equation}
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The eigenfunctions of the angular momentum operators are the spherical harmonics $Y=Y(\theta , \varphi)$. The eigenvalue equation is given by: \begin{equation} \hat{l}_z Y(\theta , \varphi)=bY(\theta , \varphi) \end{equation} Where b is the eigenvalue of $\ hat{l}_z$. Substituting $\hat{l}_z = -i\hbar\frac{\partial}{\partial\varphi}$ into the eigenvalue equation \begin{equation} -i\hbar\frac{\partial}{\partial \varphi}Y(\theta , \varphi)=bY(\theta , \varphi) \end{equation} Splitting the spherical harmonic into a product of two functions dependent on a single variable $Y(\theta , \varphi)=S (\theta)T(\varphi)$ \begin{equation} -i\hbar\frac{\partial}{\partial\varphi}\left[S(\theta)T(\varphi)\right]=bS( \theta)T(\varphi) \end{equation}
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\begin{equation} \hat{l}^2 Y(\theta , \varphi)=cY(\theta , \varphi) \end{equation}
Substituting the operator $\hat{l}^2$ into sphericals and writing the spherical harmonic as a product $S(\theta)T(\varphi)$ \begin{equation} -\hbar^2\left(\frac{\partial^2}{\partial\theta^2}+cotg\theta \frac{\partial}{\partial\theta}+\frac{1}{sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right)\left(S(\ theta)\frac{1}{\sqrt{2\pi}}e^{im\varphi}\right)=cS(\theta)\frac{1}{\sqrt{2\pi}}e^{im \varphi} \end{equation}
Acting with the operator on the function and simplifying:
\begin{equation} -\hbar^2\left(\frac{\partial^2S(\theta)}{\partial\theta^2 }+cotg\theta\frac{\partial S}{\partial\theta}-\frac{m^2}{sin^2\theta}S(\theta)\right)\cancel{\frac{1}{ \sqrt{2\pi}}e^{im\varphi}}=cS(\theta)\cancel{\frac{1}{\sqrt{2\pi}}e^{im\varphi}} \end{equation}
\begin{equation} \frac{\partial^2S(\theta)}{\partial\theta^2}+cotg\theta\frac{\partial S}{\partial\theta}-\frac{m^ 2}{sin^2\theta}S(\theta)=-\frac{c}{\hbar^2}S(\theta) \end{equation}
Solving this differential equation gives us the eigenvalue, c, and the eigenfunction $S(\theta)$.
\begin{equation} c=\hbar^2l(l+1) \end{equation} \begin{equation} S_{l,m}(\theta)=(-1)^{\frac{m+|m| }{2}}sin^{|m|}\theta\sum_{j=0,(1),2,(3)..}^{l-|m|}a_j cos^{j}\theta \end{equation}