Intuitive interpretation of angular momentum in a macroscopic system

Question

My professor defined the angular momentum operators in quantum mechanics as the infinitesimal rotations in the Hilbert space of quantum states.

I'm looking for a qualitative, intuitive explanation of why these operators would detect the angular momentum in a way that is analogous to the classical angular momentum.

Imagine a thin vertical metal plate of thickness $2\epsilon$ that is situated at the origin. At $t=0$ the x axis is orthogonal to the metal plate's plane. It rotates around the z axis, so after some time, its normal vector will point to the y axis.

The expectation value $\langle L_z\rangle$ of the angular momentum operator $L_z$ should be positive in this macroscopic system (if the metal plate rotates in positive direction).

The infinitesimal rotation around z is

$$ L_z = \left( \begin{matrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) $$

so $L_z |\psi\rangle$ is the metal plate rotated around the z axis by 90 degree.

$$ \langle L_z \rangle = \langle\psi | L_z \psi\rangle = \int d\vec{x} \psi^* L_z\psi $$

However, the function under the integral is zero nearly everywhere, because $\psi$ is zero everywhere except at $-\epsilon<x<\epsilon$, and $L_z\psi$ is zero everywhere except at $-\epsilon<y<\epsilon$.

So, for a small $\epsilon$, $\langle L_z \rangle$ would be nearly zero, which is obviously wrong.

Where's my error?

Answer 1

My professor defined the angular momentum operators in quantum mechanics as the infinitesimal rotations in the Hilbert space of quantum states.

Your professor likely did not mean what you think they meant. Rotations in the Hilbert space are performed by unitary transformations. If you choose the position representation then your Hilbert space is a set of L2 functions whose domain is the configuration space of your system. You can consider rotations of your domain and that it what angular momentum is all about.

So in particular your $\Psi$ is a function of $x,$ $y,$ and $z.$ it is not vector valued and multiplying it by a matrix makes no sense whatsoever. You should apply the matrix to the domain of the wavefunction.

And finally you need to distinguish a generator of a rotation from a finite rotation. Take the generator multiply it by a small scalar and then you can exponentiate the matrix to get a small rotation. And $e^{aA}$ for a matrix A is the limit of the following sum (where 1 is the identity matrix):

$$e^{aA}=1+\frac{a^1}{1!}A^1+\frac{a^2}{2!}A^2+\frac{a^3}{3!}A^3+\dots$$

In your case you get a rotation by $\phi$ if you consider the rotation $e^{i\phi\hat L_z}.$ And have that act on the domain of the wave.

Answer 2

$L_z$ is related to an infinitesimal rotation, and you seem to think you have rotated you coordinates by $90^\circ$.

A rotation around the $z$ axis by a small angle $\phi$ is given by $$ R(\phi) = 1 + \imath \phi L_z + O(\phi^2) $$

To see how $L_z$ acts on a wavefunction we can apply $R$ to it, expand and compare terms \begin{align} R(\phi)\psi(\mathbf{r}) &= \psi(\mathbf{r}) + \imath \phi L_z \psi(\mathbf{r}) + O(\phi^2)\\ &=\psi(\mathbf{r}+\phi\delta\mathbf{r}). \end{align} As you correctly say in your post $\delta\mathbf{r} = (-y, x, 0)$, so we can taylor expand \begin{align} R(\phi)\psi(\mathbf{r}) &=\psi(\mathbf{r}) + \imath \phi L_z \psi(\mathbf{r}) + O(\phi^2)\\ &= \psi(\mathbf{r}) + \phi\left[x\frac{\partial \psi}{\partial y} - y \frac{\partial \psi}{\partial x}\right] + O(\phi^2) \end{align}

Comparing terms we find that \begin{align} \imath L_z \psi= x\frac{\partial \psi}{\partial y} - y \frac{\partial \psi}{\partial x}\\ L_z = \frac{1}{\hbar}\hat{\mathbf{z}}\cdot (\mathbf{x}\times\mathbf{p}) \end{align}