# Why is representation theory important in physics?

Given a certain group we can find many representations of it. And If I'm not wrong a representation is a group itself. For example, given the group of the unitary 2x2 matrices with determinant 1 $$SU(2)$$, its three-dimensional representation is a subsets of 3x3 matrices that is a group itself. Since a representation is a homomorphism I expect that the group and its representation can be different groups. So it looks like there exist many groups that are related to a certain group because they are its representation. Why is it important to know the representation of a group? is there a physically important property that the representation group inherit from the group it represents?

If I'm wrong feel free to correct me (I'm pretty new to the topic)

I've recently wrote a related post but there were too many questions inside, maybe it's useful somehow Some clarifications about the ideas of representation of a group

• A representation is not a group, but rather a map from the group into the endomorphism group of a vector space. Thus it takes an element of the group and gives you something which can act on the vector space, thereby representing how the group acts on that space. However, physicists will often refer to the representation by this vector space, but always keep in mind we must have a map as well. Commented Apr 27, 2020 at 15:24
• Possible duplicates: physics.stackexchange.com/q/3/2451 and links therein. Commented Apr 28, 2020 at 4:10

One can give many examples of where specific aspects of representation theory are useful in physics (see the current other answers to this question) but the fact of the matter is simply that you cannot do physics without having representations, whether you call them that or not:

Don't think about representations as "a group and a different group". Even faithful (but different) representations are relevant. A representation is a pair - it consists of both a vector space $$V_\rho$$ and a representation map $$\rho : G \to \mathrm{GL}(V_\rho)$$ that represerves the group structure, i.e. is a group homomorphism. Without a representation, the group $$G$$ remains abstract and acts on nothing.

Whenever we ask a question like "How does X transform under rotations?" (or with "rotations" replaced by any other transformation), this is - if X lives in a vector space, as it often does, e.g. when it is any sort of number or array of numbers - the same as asking "In which representation of $$\mathrm{SO}(3)$$ (the rotation group) does $$X$$ transform?". You cannot have transformations that form a group acting on vectors without having representations. Most of physics is literally impossible to do without having a representation somewhere, since the ideas of transformations and symmetries are fundamental to all fields of physics. And questions like "If I multiply X and Y, how does their product transform?" are so natural that it is mostly unavoidable to have more than one representation.

You might as well ask "Why are groups important?", because without their representations, groups aren't very interesting from a physical perspective at all (this, coincidentally, is why you'll often hear physicists say "group theory" to what mathematicians would consider "representation theory")!

• this Is very interesting, but I don't understand how it's possible to say which group is a group without referring to a representation of it Commented Apr 27, 2020 at 16:14
• @SimoBartz The mathematical definition of a group makes no reference to any representations. Often one will define a group as a subset of matrices on some space so it comes with a natural representation, but e.g. I could define the 2d rotation group (essentially just an angle) as the quotient group $\mathbb{R}/\mathbb{Z}$ without any reference to 2d vectors. Commented Apr 27, 2020 at 16:19
• when you say "How does X transform under rotations? is the same as: In which representation of SO(3)(the rotation group) does X transform?" do you mean that in non relativistic physics every rotation operator is a representation of SO(3)? if yes why do we need SU(2)? Commented Apr 27, 2020 at 16:29
• @SimoBartz It's just an example for a classical 3d theory, I wasn't trying to make any general statement at all. But see my Q&A at physics.stackexchange.com/q/203944/50583 for a (highly technical) explanation of how SU(2) emerges from SO(3). Commented Apr 27, 2020 at 16:44

1) The physical states of a theory(the 'particles', so to speak), lie in a vector space $$\mathcal{H}$$- the Hilbert space of the theory. This is quantum mechanics.

2) The notion of a physical state $$|k\rangle$$ evolving into another physical state $$|k'\rangle$$ can be implemented by means of a linear operator on $$\mathcal{H}$$ that takes the former to the latter. This is a concrete operator(a matrix, if you will)-that is somehow "applying" an abstract transformation to these vectors.

3) Such transformations are known to have a group structure; the Lorentz transformations for example are $$SO(3,1)$$. But at this stage, we don't have a way to concretely "apply" these group elements in a vector space. What we need, is a map from the group to the operators on a vector space, in such a way that the map preserves the group composition structure(this is what homomorphism means).

4) This is exactly what a representation is. For example, consider rotations in a plane. The group here, is only an abstract set that can be labelled by elements $$\{\theta\}$$, following composition laws such as $$\theta_1\cdot\theta_2=\theta_1+\theta_2$$. We have not yet said what these $$\theta$$ are-moreover, it is unclear how is one supposed to ACT them on a vector you wish to rotate. It is just an abstract symbol.

5) So we look for concrete quantities that 'behave' exactly like the abstract $$\theta$$. An example is the $$2\times2$$ rotation matrix. This will now allow us to "apply" the behavior of our group onto vectors. Another crucial point-when we say 'vectors', we implicitly mean the vector space $$R^2$$. Is there a way to implement this group on vectors in $$R^3$$? An arbitrary vector space?

6) So we realise that it is not enough to define the map to operators-we must simultaneously define the vector space over which these operator act. This amounts to choosing our Hilbert space. The group will, in general, act in different ways over different vector spaces-there are different representations of the group.

7) In particle physics, elementary particles are synonyms to (irreducible unitary) representations the Poincare group. This is why photons are different from electrons-they transform under different representations, of the same group. The laws governing their transformations are the same-Lorentz transforms-except that they are implemented in different ways.

8) As an aside, in case it seemed like a lot of guesswork goes in, these representations are furnished by looking at the eigenvalues of a quantity called the Casimir of the algebra corresponding to the group. Which of these are physically realised ofcourse is a different matter.

• Can we say something similar for classical physics? I am bit stuck because the states are not necessarily elements of a linear space. Commented Nov 11, 2020 at 4:13

First seeing everything in terms of morphism is formally correct but not needed to understand the basic idea: one starts with an abstract set of operations and represents this set in an explicit way (in physics, usually by matrices), keeping the relation between the abstract elements. Thus, multiplication of the matrices associated with two elements of the group will produce the matrix of the correct third element of a group: $$\Gamma(a)\Gamma(b)=\Gamma(c)$$ if $$a\cdot b=c$$ and $$\Gamma$$ is the representation. Note that mathematicians make a difference between representations, modules and other technical terms that physicists lump together in a slight abuse of notation (no distinction is needed in most cases).

There are several (sometimes infinitely many) representations for (the elements of) a group, and in physics most often we deal with matrix representations. Closely related are representations of the algebra of a continuous group, and we see that even more often in physics.

The simplest examples are representations of dimension $$2j+1$$ of the angular momentum operators, or alternatively representations of dimension $$2j+1$$ of $$\mathfrak{su}(2)$$. The connection between the algebra $$\mathfrak{su}(2)$$ and the group $$SU(2)$$ is through exponentiation.

Thus the Pauli matrices are a 2-dimension representation of $$\mathfrak{su}(2)$$. At the level of the algebra, the $$2\times 2$$ matrices for $$\sigma_{x,y,z}$$ have the same (matrix) commutator as the abstract elements. It is possible of course to construct $$3\times 3$$ matrices for angular momentum operators, which still have the correct (matrix) commutators. Those are two inequivalent representations: it's not possible to from one to the other by a similarity transformation (obvious since they are of different dimensions).

The basic blocks are irreducible representations, one for which no similarity transformation will simultaneously bring a representation of all the elements to a block diagonal form. Since pretty much every representation in physics can be written in terms of irreducible bits, the irreducibles thus function as "elementary" representations.

One reason we use this is to block diagonalize the Hilbert space of states. If one has an $$SO(3)$$-invariant operator, it cannot connect blocks with different $$j$$ values. Thus one can work within each representation and thus in general within a smaller subspace.

Another reason is that the group properties impose constraints on the representations. The matrix elements of $$\hat L_{x,y,z}$$ for instance are not random numbers, but must be so that the matrices produce the right commutation relations. In particular they must all have the same eigenvalues. Group theory in this way is a power way to relate quantities related by some symmetry operations.