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.

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