# Rotational Symmetry and the Conservation of Angular Momentum in Quantum Mechanics

Definition. A quantum mechanical system is said to be rotationally invariant if the system still obeys Schrödinger’s equation after a rotation.

Let us assume that the quantum mechanical system under consideration is rotationally invariant. Consider an infinitesimal rotation of a point $(x,y,z)\in\mathbb{R}^3$ about the $z$-axis. The point $(x,y,z)$ may be written as
$$x=r\cos\theta,\ y=r\sin\theta,\ z=z$$ where $r$ and $z$ are held fixed. Then
\begin{align*}
dx&=-r\sin\theta d\theta=-yd\theta,\\
dy&=r\cos\theta d\theta=x d\theta,\\
dz&=0
\end{align*}
Hence, by an infinitesimal rotation the point $(x,y,z)$ changes to $(x’,y’,z’)$ as
\begin{align*}
x’&=x+dx=x-yd\theta,\\
y’&=y+dy=y+xd\theta,\\
z’&=z+dz=z
\end{align*}
The infinitesimal rotation about the $z$-axis by the angle $d\theta$ can be given by the transformation matrix
$$R_z=I+\begin{pmatrix} 0 & -d\theta & 0\\ d\theta & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$$ where $I$ is the $3\times 3$ identity matrix. In terms of $R_z$, $(x’,y’,z’)$ can be written as
$$\begin{pmatrix} x’\\ y’\\ z’ \end{pmatrix}=\begin{pmatrix} x\\ y\\ z \end{pmatrix}+\begin{pmatrix} 0 & -d\theta & 0\\ d\theta & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x\\ y\\ z \end{pmatrix}$$

Let $H\psi(x,y,z)=E\psi(x,y,z)$. Then by the assumption
$$H\psi(x-yd\theta,y+xd\theta,z)=E\psi(x-yd\theta,y+xd\theta,z)\ \ \ \ \ (1)$$
Recall Taylor expansion
$$f(x+dx,y+dy)\approx f(x,y)+f_x(x,y)dx+f_y(x,y)dy$$ for infinitesimal $dx$ and $dy$. If we Taylor expand $\psi(x-yd\theta,y+xd\theta,z)$, we obtain
\begin{align*}
\psi(x-yd\theta,y+xd\theta,z)&\approx \psi(x,y,z)-y\psi_x(x,y,z)d\theta+x\psi_y(x,y,z)d\theta\\
&=\psi(x,y,z)+\left(x\frac{\partial\psi}{\partial y}-y\frac{\partial\psi}{\partial x}\right)d\theta
\end{align*}
The Schrödinger equation (1) is then written as
$$H\psi+H\left(x\frac{\partial\psi}{\partial y}-y\frac{\partial\psi}{\partial x}\right)d\theta=E\psi+E\left(x\frac{\partial\psi}{\partial y}-y\frac{\partial\psi}{\partial x}\right)d\theta$$
Since $H\psi=E\psi$ and $E$ commutes with the operator $\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)$, we obtain
$$\left[H,\frac{\hbar}{i}\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)\right]=0$$
Recall that $\frac{\hbar}{i}\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)$ is $L_z$, the $z$-component of the angular momentum operator $L$. That is, we have $[H,L_z]=0$. By a similar manner, we also obtain $[H,L_x]=[H,L_y]=0$. Hence, $[H,L]=0$ i.e. the angular momentum operator commutes with the hamiltonian $H$. This means that the angular momentum is conserved. This is the reason why rotational invariance is one of the fundamental assumptions of quantum mechanics.