# Spectrum of the angular momentum and angular momentum squared

I am trying to understand how to build the spectrum of the angular momentum; of course since different components of the angular momentum do not commute with each other we must chose only one component to focus on: let's say we choose the $$L_z$$ component, and so we want to find the spectrum of $$L_z$$. We also know incidentally that every component of the angular momentum commute with the square of the angular momentum $$\vec{L}^2$$, so we can diagonalize $$L_z$$ and $$\vec{L}^2$$ simultaneously. Problem is: why we should care? Why do we care about having the spectrum of eigenfunction of both $$L_z$$ and $$\vec{L}^2$$? Coulden't we just determine the spectrum of $$L_z$$? This is the first part of my question.

But let's say that we care for some reason: to determine the eigenfunctions I would expect a system like: $$L_z|m\rangle=a_{L_z}|m\rangle$$ $$\vec{L}^2|m\rangle=a_{L^2}|m\rangle$$ instead in my lecture notes the following system is present: $$L_z|l \ \ m\rangle=\hbar m|l \ \ m\rangle$$ $$\vec{L}^2|l \ \ m\rangle=\lambda _l|l \ \ m\rangle$$ of course $$a_{L_z},a_{L^2}$$ are arbitrary names for the eigenvalues and we can replace them with whatever we want, including $$\hbar m$$, we can do this, but why? Why is there an $$\hbar$$?. Why is putting an $$\hbar$$ there useful? Doesen't it just create more confusion?

And secondly but most importantly: why are the eigenvectors labeled with the two letters $$l,m$$? Usually when we see something like this, for example $$|+ \ \ -\rangle$$ it means that we are dealing with two particles (or maybe that we are in 2D); why is the double index present here since we are talking about the same collection of eigenvectors for both the operators? Is it simply to show that the same eigenvector is correlated with both a $$L_z$$ value and a $$\vec{L}^2$$ value and dimensionality has nothing to do with it?

• Representation theory of SU(2) Commented Oct 27, 2020 at 15:25
• Unfortunately your link is unhelpful to me in this moment. I am not supposed to study this kind of math right now, and I will prefer a more low level and direct response Commented Oct 27, 2020 at 16:00

Why do we care about having the spectrum of eigenfunction of both $$L_z$$ and $$L^2$$? Couldn't we just determine the spectrum of $$L_z$$?

Contrary to how it may appear, having more constraints makes the problem easier to solve, not harder. For example, a generic eigenstate of $$L_z$$ with eigenvalue $$0$$ is of the form

$$|\psi\rangle = \sum_{\ell=0}^\infty c_\ell |\ell,0\rangle$$

which ranges over all possible values of $$\ell$$. There is a tremendous amount of degeneracy here, and trying to find a solution with an infinity of undetermined constants is an exercise in totally unnecessary masochism.

Instead, we can also demand that our eigenstate of $$L_z$$ is an eigenstate of $$L^2$$ as well. Doing this eliminates all of our unwanted freedom, because the simultaneous eigenstates of $$L_z$$ and $$L^2$$ are unique (up to multiplication by a constant, of course).

of course aLz,aL2 are arbitrary names for the eigenvalues and we can replace them with whatever we want, including ℏm, we can do this, but why? Why is there an ℏ?. Why is putting an ℏ there useful? Doesen't it just create more confusion?

The eigenvalues of $$L_z$$ are integer multiples of $$\hbar$$, and the eigenvalues of $$L^2$$ are of the form $$\ell(\ell+1)\hbar^2$$ where $$\ell$$ is a non-negative integer. You can label a simultaneous eigenstate of $$L^2$$ and $$L_z$$ with the eigenvalues if you'd like, but that would lead to expressions like

$$|6\hbar^2,2\hbar \rangle + |2\hbar^2,\hbar\rangle$$

rather than

$$|3,2\rangle + |1,1\rangle$$ where we've instead labeled the eigenstates by the integers $$\ell$$ and $$m$$. If you want to write $$\hbar$$'s all over the place and constantly work out what $$\ell$$ is when you've only written down $$\ell(\ell+1)$$, then you're welcome to it, but it's not the standard convention and nobody will want to decipher your work.

And secondly but most importantly: why are the eigenvectors labeled with the two letters l,m?

Because the eigenvectors are simultaneous eigenvectors of both $$L^2$$ and $$L_z$$. If you only write $$|m\rangle$$, how am I supposed to know what value of $$\ell$$ that state corresponds to?

Usually when we see something like this, for example |+ −⟩ it means that we are dealing with two particles (or maybe that we are in 2D).

That is not the case here, nor is it usually the case. One of the things you must get used to if you're to be successful with advanced physics and mathematics is that notation means no more or less than what we define it to mean.

Somewhere in your text/lesson, the author/your instructor said something like "we will label a simultaneous eigenvector of $$L^2$$ and $$L_z$$ with two integers, $$\ell$$ and $$m$$, to reflect the fact that $$L^2|\ell,m\rangle = \ell(\ell+1)\hbar^2 |\ell,m\rangle$$ and $$L_z|\ell,m\rangle = m\hbar|\ell,m\rangle$$" so that's what it means. If you aren't familiar with some bit of notation, then you need to go check where it is defined, but you are now past the stage when you can expect notation to serve as a definition. The ideas you are working with are too subtle and complicated to be condensed into a few symbols, so you need to keep the definitions in your head (or refer to a reference when necessary).

Is it simply to show that the same eigenvector is correlated with both a $$L_z$$ value and an $$L^2$$ value and dimensionality has nothing to do with it?

$$\uparrow$$ Yep.

• Thanks a lot. It's all clear now. Regarding the last part of my question unfortunately my lectures do not contain any explanation of the notation, they simply implemented it without any warning. Pretty frustrating when you are trying to understand a new subject. Commented Oct 27, 2020 at 17:45