# Why can we associate the rotation operator for a vector field to the angular momentum operator?

I've seen the operator that rotates a scalar field and its properties. Considering that the wave function of a particle (if I don't consider the spin) is described by a scalar field and that this operator must commute with the hamiltonian I associate it with the angular momentum operator. Then I've seen the operator that rotates a vector field and its properties and I thought it can be the angular momentum operator for a wave function that is described by a vector field. The only problem is that I don't know a quantum system that has a vector field as wave function. I read that the em field can be seen somehow as a quantum system with a vector field wave function but I don't understand what are the equations that define this system. Please answer with little math, I'm interested in the concept.

I'll talk about nonrelativistic quantum. The fundamental object at play here is the rotation group $$SO(3)$$. (If you want to talk about spin 1/2 objects, you need to talk about $$SU(2)$$, but let's start slowly.) You correctly described how a scalar field transforms, and you want to know about how vector fields transform.

For now, forget about fields of any kind, and let's just talk about scalars, vectors, etc. A scalar doesn't have any nontrivial transformation rule under rotations. However, a vector does transform nontrivially under rotations—just apply a rotation matrix!

We should talk about what a rotation matrix is: it's an orthogonal matrix with determinant 1. (Determinant 1 means I exclude reflections.) For good reason, I want to think about "infinitesimal rotation matrices". If I perform a very small rotation, the corresponding matrix should approximately look like $$1+i\epsilon L$$, where 1 is the identity matrix, $$L$$ is a 3x3 matrix, and $$\epsilon$$ should be thought of as small.

Now, $$L$$ isn't just any old matrix. For math reasons (that arent too difficult, but I don't want to get into it), $$L$$ has a very restricted form. If you consider the vector space of 3x3 matrices (with pure imaginary elements) as a 9-dimensional vector space, then our constraints above mean that the space of "allowed $$L$$ matrices" is a 3-dimensional subspace of that vector space. Importantly, we should be able to come up with 3 basis elements for that vector space. For other math reasons, those basis elements (Call them $$L_1, L_2, L_3$$) should satisfy the relations $$[L_1, L_2]=iL_3$$ and the analogous relations resulting from cyclically permitting the indices 1,2,3. These are called the $$\textit{commutation relations}$$.

One more piece of background knowledge that will be super important: Remember that scalars have 1 component, and vectors have 3. We figured out how to infinitesimally transform vectors by coming up with an appropriate basis of 3x3 matrices that, when applied onto a vector, infinitesimally rotate that vector. What if we wanted to do something similar for, say, 5- component vectors? As it turns out, we will need to come up with a basis of 3 5x5 matrices that satisfy the aforementioned commutation relations. This tells us how 5-component objects ought to transform under the rotation group SO(3).

This whole business is known as finding a $$\textit{representation}$$ of the rotation group (technically, its "Lie Algebra" because we talked about infinitesimal transforms), because we're "representing" (via a homomorphism) a group of 3x3 matrices with 5x5 matrices instead. Objects with (2n+1) components that transform accordingly under the rotation group are said to be in the "spin-n" representation of the rotation group. In life, some things are scalars, some things are vectors, and some things are spinors (which have n as a half-integer, and force us to look at a slightly different group called $$SU(2)$$, but that's for later. BTW, SU(2) is not isomorphic to SO(3), but their Lie algebras are ismorphic). For example, it just so happens that the wavefunction describing an electron is a spinor field, so we have to deal with that when we consider how that field transforms under rotations.

Now , I'll stop talking about math and start talking about some physics. Remember when you solved the Schrodinger equation for central potentials? (Hydrogen atom should come to mind.) You found that the eigenstates can be labeled by 3 principal quantum numbers, two of which are $$l$$ and $$m$$. Ignore the 3rd quantum number, it's not relevant.

Let's take $$l=1$$. The Subspace of the Hilbert space of wavefunctions with $$l=1$$ is 3 dimensional, because we have a state for each $$m=-1, 0, 1$$. Consider the matrix representation of the angular momentum operators (for a scalar field) in this 3-dimensional subspace. We have 3 operators $$L_x, L_y, L_z$$ that are all represented in this manner as 3x3 matrices on this subspace.

Now for the big reveal: those matrices turn out to be a suitable basis for the allowed $$L$$ matrices from before! Thus, we can approximate small rotations via the angular momentum operators! So in some sense, the rotation operator for a vector object is associated to some sort of angular momentum operator—but that angular momentum operator isn't a differential operator, it's a matrix. That's because I restricted the discussion to vectors, not vector fields.

This is no accident that angular momentum operators can give you rotation operators. In (relativistic) classical field theory, we can write down objects like scalars, vectors, spinors (maybe more on this later) as part of a Lagrangian that describes the dynamics (e.g. Klein-Gordon or Dirac equations). These Lagrangians have a symmetry under Lorentz transforms (the full symmetry group is the Poincaré group, which includes spacetime translations). By Noether's theorem, there are conserved quantities. Linear energy-momentum is one, resulting from translation invariance. Angular momentum is another, resulting from rotational invariance. But what do we mean by angular momentum?

As it turns out, for fields, this depends strongly on whether the object you're describing transforms like a scalar, or a vector, or a spinor. If you remember the derivation of Noether's theorem (if you don't, study it, it will make everything make sense), the expression for the conserved quantity (Noether charge) depends crucially on how the field transforms under the rotation. If we have a scalar field, we only need to take into account how the field coordinate transforms under rotations, and we get a Noether charge from that—call it orbital angular momentum. If we have something other than a scalar field, we not only have a transformation on the coordinate that gets us orbital angular momentum, but we have a transformation that mixes field components into each other under the rotation, and this adds an additional term to the Noether charge—a different type of angular momentum. We call that spin. This is the (mathematical) origin of that mysterious form of intrinsic angular momentum that doesn't result from anything physically spinning about.

Now, the answer to your actual question. Suppose we write down a field theory for your favorite type of object, be it a scalar, a vector, a spinor, etc. It has rotational symmetry (or the more general Lorentz symmetry). Thus, your object is in some $$\textit{representation}$$ of the rotation group (or Lorentz group, if you want relativistic stuff). We have our Noether charge, i.e. "angular momentum" (which we now know incorporates both orbital and spin angular momentum), which came from the rotational symmetry.

Upon a formal procedure called "canonical quantization", the Noether charges become operators that act both on the spacetime coordinate of the field, as well as on the components of the object itself (e.g. a Pauli matrix operating on a ket state represented as a 2-component vector). Those quantized operators turn out to obey the operator commutation relations that defined the rotation group $$SO(3)$$ (or $$SU(2)$$). That means, we can use them to define infinitesimal rotations acting on your field (scalar, vector, spinor, or whatever)!

Tl;dr: The reason why rotation operators acting on various types of objects correspond to angular momentum operators is because angular momentum is a Noether charge corresponding to rotational symmetry, so upon quantization the angular momentum operators will obey the same commutation relations as the Lie algebra of the rotation group, hence the angular momentum operators "form a representation of the rotation group" and hence can be used to transform objects under rotations.

I've left out a great deal of technical detail and subtleties and said next to nothing about spinors, which are the real fun here. I suggest you look at Peskin & Schroeder: Introduction to Quantum Field Theory for a more organized and motivated treatment of all the stuff I've mentioned here. If you need something clarified, feel free to ask, because this is admittedly jumbled.

• "What if we wanted to do something similar for, say, 5- component vectors? As... " do you mean that any set of operators that follow the rule of the 3x3 $L_x, L_y, L_z$ matrices is a representation of the same matrices because we can map one into the other? So if we find 3 operators that follow those rules we can map them into 3x3 matrices and study them in the same way? Mar 26, 2020 at 11:42
• Essentially yes: the formal definition is here: en.m.wikipedia.org/wiki/Lie_algebra_representation Basically the Lie algebra is defined as a vector space with n basis vectors (For SO(3) and SU(2), n=3) that has an associated Lie Bracket (its like commutation relations, but more abstract). A representation means you find a linear map from that abstract vector space to a vector space of your choosing, with the stipulation that the image (under the map) of the Lie Bracket of 2 elements is the commutator (in the sense of ordinary matrices) of the image. Mar 27, 2020 at 3:05