Why are physicists so interested in irreps if in their non-block-diagonal form they mix all components of a vector?

Question

Consider a group $\{G,\circ\}$, with elements $e,g_1,g_2,...$, represented by the matrices $\{D(e), D(g_1), D(g_2)...\}$. If all the matrices can be brought to block diagonal forms by a similarity transformation, such a representation is reducible.

In this block-diagonalized form, any vector on which the representative matrices act, do not mix all components of the vector. So we conclude that there exists more than one nontrivial (except the null space and whole space) invariant subspaces.

I cannot fully understand the importance of this. Irreps are not always block-diagonal to start with. But can be made block-diagonal. If they are not block-diagonal, they would mix all components of a vector. Is this not a problem?

Let me give a simple example. The 2-dimensional representation of SO(2): $$D=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}$$ is reducible to $$D\to D^\prime= SDS^{-1}=\begin{pmatrix}e^{i\theta} & 0\\ 0 & e^{-i\theta}\end{pmatrix}$$

In this first case, the basis vectors are $(1~0)^T$ and $(0~1)^T$. In this case, $A_x$ and $A_y$ mix with each other.

In the second case, the basis vectors are $2^{-1/2}(1~~i)^T$ and $2^{-1/2}(1-i)^T$ and S is given by $$S=2^{-1/2}\begin{pmatrix}1 & i\\ 1 & -i\end{pmatrix}.$$ It is easy to see that $A_{\pm}=A_x\pm iA_y$ do not mix with each other.

So if I am getting it correct, this example shows that there are two invariant 1-dimensional subspaces in this case - one spanned by $2^{-1/2}(1~~i)^T$ and the other by $2^{-1/2}(1-i)^T$. So finding irreps mean identify the basis?

Answer 1

The proofs of properties of reducible representations don't depend on the matrices being block-diagonal in any basis; you just need one. The fact that one basis with that property exists already tells you something about the representation $D$, and you can use that basis to prove things about $D$.

Or to put it in a more invariant way: a representation is reducible if it's possible to write the space on which it acts as a (nontrivial) direct sum of subspaces, such that each subspace is invariant under the action of all the $D(g)$. This is a basis-independent property of $D$, and for mathematicians that's more than enough: they don't need to speak about block-diagonal matrices, they can prove everything just from there. But if you actually want the matrices, it's better to choose a basis adapted to those subspaces.

It's a bit like orthogonal matrices. An orthogonal linear transformation (that is, one that satisfies $(Tv, Tw) = (v,w)$) won't have an orthogonal matrix in any basis; only in an orthonormal basis. This doesn't mean that the concept of an orthogonal matrix is useless; it means that orthonormal bases are better (or at least, better adapted to the transformation at hand).

Answer 2

"Reducible" and "irreducible" can be defined in a basis-independent way.

Think of the matrices $D(g)$ in the representation as acting on a vector space $V$ of column-matrices. The representation is irreducible if $V$ does not have any non-trivial subspace that is self-contained under the action of all of the matrices $D(g)$. Otherwise, the representation is reducible.

("Non-trivial subspace" means a subspace that is smaller than $V$ itself but containing more than just the zero-vector.)

Loosely speaking, within a given reducible representation, different irreps correspond to vectors that can't be mixed with each other by any $D(g)$. (I'm referring to the vectors themselves here, not to their components in any basis.) That's why irreps are important.