# Intuitive interpretation of angular momentum in a macroscopic system

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?

• Why is $\psi=0$ for $|x|\geq\epsilon$ and same for $\L_z\psi$? I dont think this is true. – Daniel Aug 22 '15 at 12:37
• If $L_z$ is a 90 degree rotation around the z axis, then it would be true, but as the answers point out, this is not true. – Bass Aug 23 '15 at 10:54

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.

• Thank you, I understand now that there's a difference between the generator of rotations and the 90-degree-rotation, even thought their matrices look the same. But your second paragraph confuses me further. I always thought the rotations $R\in SO(3)$ operate on $\psi\in\mathcal{H}$ by $(R(\psi))(x)=\psi(R^{-1}x)$. So there is a well-defined, natural operation of the rotation matrix on the wave function, even though the wave function is not vector-valued. How can I apply the matrix to the domain of the wavefunction? – Bass Aug 23 '15 at 11:02
• @BastianTreichler You have the general idea. Wave functions have configuration space as their domain not physical space so it isn't as simple as just a a single 3x3 matrix multiply inside the argument. You need to apply the whole transformation to the whole configuration space. But if you want to learn QM for a single unconstrained particle in an xyz basis and then relearn everything the first time you have a constraint or more than one particle or even just a different coordinate system that is another option (one where I've seen people develop misconceptions that they never ever resolve). – Timaeus Aug 23 '15 at 17:05
• thanks for the hint on the configuration space. Is it correct that if we're talking about rotations $R\in SO(3)$, then surely the configuration space must have some structure that allows $R$ to act on it? For example, a rotation of some particles' location, and for other particles just the identity.. And for configuration spaces where $\psi(R^{-1}x)$ does not make sense, we would have other groups representing things like angular momentum. Do I understand your remarks about configuration spaces correctly? – Bass Aug 23 '15 at 19:48
• @BastianTreichler You have to adjust spin states in one way, momentum space representations another, position space representations in another, and configurations of constrained systems in another (they could have fewer than three variables per particle). Hopefully it is obvious. However don't forget that the electromagnetic field also stores angular momentum. Then you can worry about gauge freedom, canonical momentum versus conserved momentum, etc. I was trying to make sure you don't walk away with an oversimplified view of angular momentum in general. For a simple situation it can be simple – Timaeus Aug 23 '15 at 20:22

$L_z$ 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}