# I Are representations of Rotation groups in non-physical spaces "responses" to actual physical rotations?

#### Blub

I have questions concerning group theory, esprecially Rotation groups. The first is: Are rotations groups f.ex. SO(2) defined for rotations in the actual physical 2 dimensional plane or are general rotations in any 2 dimensional space included?
Someone wrote that "the action of an element of SO(3) corresponds to a physical rotation. The response of the (point!) particle is a rotation in the internal space (the representation space) of the particle". Is this true? I thought that the representation of a rotation in a vector space V is a rotation in the vector space V --- the basis in V is rotated, without the physical space to rotate. Can someone clear my confusion. I would be so grateful.

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#### PeroK

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I have questions concerning group theory, esprecially Rotation groups. The first is: Are rotations groups f.ex. SO(2) defined for rotations in the actual physical 2 dimensional plane or are general rotations in any 2 dimensional space included?
Someone wrote that "the action of an element of SO(3) corresponds to a physical rotation. The response of the (point!) particle is a rotation in the internal space (the representation space) of the particle". Is this true? I thought that the representation of a rotation in a vector space V is a rotation in the vector space V --- the basis in V is rotated, without the physical space to rotate. Can someone clear my confusion. I would be so grateful.
You can do it either way. Rotating the physical system and rotating the coordinate system are equal and opposite. You need to know which one you are doing and be consistent.

These are sometimes called "active" and "passive" rotations.

I don't think there is a standard.

#### Antarres

Rotation groups $SO(n)$ are groups of orthogonal transformations on $n$ dimensional Euclidean space, which can be represented by orthogonal matrices with determinant equal to 1. Orthogonal transformations are transformations that preserve the scalar product(in case of a real space). This means that lengths and angles between vectors are preserved. The extra condition that determinant is equal to 1, means that we also preserve the orientation of the system(a.k.a. no reflections included).

So there's a theorem by Euler for example that every element of $SO(3)$ corresponds to a rotation in three dimensional space. In general every $SO(n)$ element is element of the group of symmetries of a sphere in $n$ dimensional space, with all types of reflections excluded. Which means that it could be taken that it is a rotation group, that is, that the elements are indeed rotations.

Your question then asks, whether it can be a 'rotation' in some other space, not just Euclidean(I think that's what you meant by your first question). Well, since it's a group of transformations, the most abstract way to define it is as a group of operators on some vector space, which is orthogonal, and preserves vector orientation. It is common that it's defined with orthogonal matrices on Euclidean space, but that doesn't mean that the same group cannot be represented in some other space. It's properties are defined in terms of the scalar product, so you need the scalar product of vectors defined in order to have orthogonal operators.

Also regarding the question in the title, no, they are not responses to actual rotations in coordinate space. It just happens that internal space is invariant under same transformations, so they can be described with this group.

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#### vanhees71

Gold Member
I'm a bit confused by the question in #1. What do you mean by "non-physical spaces".

The rotation group SO(3) enters Newtonian and special-relativistic physics as a symmetry group of physical space. The symmetry is also called "isotropy of space", basically saying that there's no preferred direction in physical space within Newtonian and special-relativistic physics (in the latter case for what an inertial observer calls "space").

In quantum theory you have the Hilbert space, and rotations as a symmetry must be realized by a unitary representation of the rotation group (or its covering group, SU(2), because not the vectors represent pure states but rays in Hilbert space), i.e., any rotation $R \in \mathbb{SO}(3)$ is represented by a unitary transformation $\hat{U}(R)$ such that for two rotations $\hat{U}(R_1 R_2)=\hat{U}(R_1) \hat{U}(R_2)$.

One example are the realization of the Hilbert space in wave mechanics of spin-0 particles, where the vectors are square-integrable functions $\psi(\vec{x})$, and the unitary transformation acting on the wave function is given by the usual rule how to transform fields, i.e., with $\vec{r}'=R \vec{r}$ with $R \in \mathrm{SO}(3)$ you have
$$\psi'(\vec{r}')=\psi(\vec{r})=\psi(R^{-1} \vec{r}').$$
It's very easy and a good exercise to show that this is a unitary transformation in Hilbert space and that this unitary transformation obeys the rule for a group representation, i.e., $\hat{U}(R_1 R_2)=\hat{U}(R_1) \hat{U}(R_2)$!

• PeroK

#### Antarres

@vanhees71 Actually, rotations form a projective representation on the Hilbert space, right? So the multiplication rule depends on the total angular momentum. I think by 'non-physical space' he meant any space that isn't our usual configuration space, so I identified it with internal space(isospin, lepton number etc all those internal symmetries). Internal spaces are obviously not 'non-physical', but that's how I figured he meant it according to the knowledge that is expected in an 'I' thread.

Either way, we'll wait for him to react to the answers, I guess.

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#### vanhees71

Gold Member
You can represent rotations as unitary representations of the covering group $\mathrm{SU}(2)$. A rotation is generated in the sense of a Lie group by the total-angular-momentum operators.

• Antarres

#### Antarres

You can represent rotations as unitary representations of the covering group $\mathrm{SU}(2)$. A rotation is generated in the sense of a Lie group by the total-angular-momentum operators.
Yeah, by switching to the representation of the covering group you get a normal unitary representation. I was just adding that, in case OP decided to check the representation rule for $SO(3)$ in which case projective phase factor would necessarily appear(in case he didn't go for the representation of $SU(2)$, just followed the general procedure for $SO(3)$). So yeah, you're obviously right, I just wanted to make clear to him, that in case you're looking at the representations of $SO(3)$ they would necessarily be projective, while if you look at the covering group $SU(2)$ the phase factor from projective representations is eliminated and we have a normal unitary representation. It just depends on how well OP understands this difference, so that he doesn't run into a dead end trying to prove something with $SO(3)$ that should be worked in $SU(2)$ or so(treating them as if they were the same thing).

• vanhees71

#### vanhees71

Gold Member
Yes, that's indeed a very important point. The pure states are not represented by the Hilbert-space vector ("state ket") itself but by the corresponding projection operator as the statistical operator (or equivalently the ray in Hilbert space represented by this state ket), i.e., symmetries are realized by unitary ray representations. A single symmetry transformation can then always be represented by a unitary or antiunitary transformation (in practice only time-reversal symmetry is realized by an anti-unitary ray representation). For a Lie group it's always a unitary representation, but the observables are defined via Noether through the "generators" of the Lie group, i.e., by the Lie-algebra elements, and that leads finally to the realization of the symmetry through unitary representations of the covering group of the original symmetry group. For rotations you thus get the representations of SU(2) and not only those of SO(3), which of course is very important since this includes the possibility of half-integer representations of total angular momentum and thus spin 1/2 for leptons and quarks.

For the Galilei group it's also very important, because with proper unitary representations you get no useful quantum dynamics. In this case you realize Galilei symmetry by a unitary representation of the central extension of its covering group. The Galilei algebra has one non-trivial central charge, which turns out to be the mass of the particle (implying a superselection rule for mass for non-relativistic particles) and the covering group is simply considered by using SU(2) instead of SO(3) in the semidirect product of spacetime translations, rotations, and Galilei boosts.

• Antarres