# Hilbert space decomposition into irreps

I'm currently following a course in representation theory for physicists, and I'm rather confused about irreps, and how they relate to states in Hilbert spaces.

First what I think I know:

If a representation $D: G \rightarrow L(V)$ of a group $G$ is reducible, then that means there exists a proper subspace of $V$, such that that for all $g$ in $G$ the action of $D(g)$ on any vector in the subspace is still in that subspace: that subspace is invariant under transformations induced by $D$.

Irreducible means not reducible: my interpretation is that an irrep is a representation restricted to its invariant subspace. In other words, an irrep $R$ only works the subspace that it leaves invariant. Is this a correct view?

Now, my confusion is the following:

Say we have a system invariant under the symmetries of a group $G$. If this group is finite then any rep $D$ of $G$ can be fully decomposed into irreps $R_i$. We could write any $D(g)$ as the following block diagonal matrix:

$D(g) = \left( \begin{array}{cccc} R_1(g) & & & \\ & R_2(g) & & \\ & & \ddots & \\ & & & R_n(g) \end{array} \right)$

I suppose the basis of this matrix is formed by vectors in the respective subspaces that are left invariant by $R_i(g), \forall g \in G$, but here is where I'm not clear on the meaning of it all. How does such a matrix transform states in the Hilbert space, when Hilbert space is infinite dimensional, and this rep $D$ isn't?

I've found a book that gives an example of parity symmetry, using $Z_2 = \{ e,p \}$.

The Hilbert space of any parity invariant system can be decomposed into states that behave like irreducible representations.

So we can choose a basis of Hilbert space consisting of such states, which I suppose would be the basis of the matrix $D(g)$ above? Then the Hilbert space is the union of all these invariant subspaces? In the case of parity there exist two irreps: the trivial one (symmetric) and the one that maps $p$ to $-1$ (anti-symmetric). I suppose this is also a choice of basis, but in this basis $D(g)$ is $2$-dimensional, so I don't understand how this could possibly work on the entire Hilbert space.

I apologize for the barrage of questions, but I honestly can't see the forest for the trees anymore.

• The representation should be $D:G\to \mathcal{L}(V)$, i.e. the group is mapped to linear (bounded) operators of some Hilbert space, not to the vectors itself. The vectors correspond to states and the group should act on them. This seems to be some confusion, because in this sense, a representation on an infinite dimensional Hilbert space always consists of infinite matrices, no matter what. – Martin Sep 29 '14 at 16:07
• Your definitions are a bit confused. A representation $G \ni g \mapsto R(g) : V \to V$ is reducible is there is a proper subspace $V_0 \subset V$ which is invariant under the action of $R$ (proper means $V_0 \neq \{0\}$ and $V_0 \neq V$). In other words $R(g)V_0 \subset V_0$ for all $g\in G$. A representation is irreducible if there are not invariant proper subspaces $V_0\subset V$. Above the $R(g)$ are linear operators on $V$ and it does not matter if they are bounded or not in the definitions I stated. Finally the invariant subspaces may be finite or infinite dimensional. – Valter Moretti Sep 29 '14 at 16:07
• Martin: You're right, my notation is rather sloppy; I'll fix it. I'm not sure I understand in what sense a representation of an infinite dimensional Hilbert space consists of infinite matrices, though. How would this become apparent in the example of parity, for instance? @ValterMoretti: Indeed, I see that didn't write out the definitions as unambiguously as I would have hoped. I agree with your definitions, and I'll try to fix the OP. – Timsey Sep 29 '14 at 16:52

Your understanding of reducible and irreducible representations is a little bit muddled. Let me try to clarify this a bit:

• A reducible representation $D:G\to \text{GL}(V)$ is one that has a nontrivial invariant subspace $W$. That is, there exists a nonzero $W<V$ such that for all $g\in G$ and all $w\in W$, the action $D(g)w\in W$ remains in the subspace.

• By contrast, an irreducible representation is one where no such subspace exists. That is, for any nonzero proper subspace $W$, there exist a $g\in G$ and a $w\in W$ such that $D(g)w \notin W$.

After that, the main source of your confusion, I think, is the fact that the invariant subspaces to not need to be finite-dimensional. This is why formulations of the type $$D(g) = \left( \begin{array}{cccc} R_1(g) & & & \\ & R_2(g) & & \\ & & \ddots & \\ & & & R_n(g) \end{array} \right)$$ can be rather misleading. It is indeed possible to construct finite direct sums of vectors and of operators which are infinite-dimensional, and to represent them graphically using matrices; it's a little bit involved but I think it will help clarify the issue.

Consider, then, a vector space $V$ which is the direct sum of its subspaces $W_1,\ldots,W_n\leq V$. By definition, this means that for every $v\in V$ there exist unique vectors $w_j\in W_j$ such that $v=\sum_j w_j$. It is possible, in this case to represent $v$ using the notation $$v = \left( \begin{array}{c} w_1\\ \vdots \\ w_n \end{array} \right).$$ However, it is important to note that the $w_j$ are not numbers; instead, they are vectors in as-yet-unspecified vector spaces $W_j$. Moreover, these could indeed be infinite-dimensional. (Indeed, if $V$ is infinite-dimensional then at least one of the $W_j$ needs to be.)

Linear transformations $T:V\to V$ can be treated similarly. For any $w_j\in W_j$, $T(w_j)$ is in $V$ which means that it can be decomposed as $T(w_j)=w'_1+\ldots+w'_n$, with each $w'_j\in W_j$. These new vectors are unique for each $w_j$, which means that they are functions of it, and it's easy to show that the dependence is linear. This allows us to get new sub-functions $T_{kj}:W_j\to W_k$, which have the property that for every $w_j\in W_j$ $$T(w_j)=\sum_k T_{kj}(w_j).$$ This then extends, by linearity, to the action of $T$ on a general vector $v=\sum_j w_j\in V$, which is then written $$T(v)=\sum_{k,j} T_{kj}(w_j).$$ With this machinery in place, you can represent $T$ as a matrix, $$T = \left( \begin{array}{cccc} T_{11} & T_{12} & \cdots & T_{1n} \\ T_{21} & T_{22} & \cdots & T_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ T_{n1} & T_{n2} & \cdots & T_{nn} \end{array} \right).$$ The advantage of this notation is that the matrix-vector product works perfectly: $$T\, v = \left( \begin{array}{cccc} T_{11} & T_{12} & \cdots & T_{1n} \\ T_{21} & T_{22} & \cdots & T_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ T_{n1} & T_{n2} & \cdots & T_{nn} \end{array} \right)\left( \begin{array}{c} w_1\\ \vdots \\ w_n \end{array} \right).$$

So why have I gone to such lengths to define matrices? The important thing is that the submatrices need not be finite dimensional.

To bring things down to something more concrete, consider the specific case of parity on $L_2(\mathbb R)$. Here $L_2$ (dropping the $\mathbb R$) splits into an even and an odd component, $$L_2=L_2^+\oplus L_2^-,$$ which is just the statement that every function $f$ can be seen as the sum of its even and odd parts $f_+$ and $f_-$, or in column vector notation $$f=\begin{pmatrix}f_+\\f_-\end{pmatrix}.$$

Similarly, the parity operator splits into two constant parts, the identity $\mathbb I:L_2^+\to L_2^+$ on even functions, and minus the identity on odd functions, $-\mathbb I:L_2^-\to L_2^-$. In matrix notation, $$P=\begin{pmatrix}\mathbb I&0\\ 0&-\mathbb I\end{pmatrix},$$ and $$Pf=\begin{pmatrix}\mathbb I&0\\ 0&-\mathbb I\end{pmatrix} \begin{pmatrix}f_+\\f_-\end{pmatrix} =\begin{pmatrix}f_+\\-f_-\end{pmatrix}.$$ As before, the individual subrepresentations $R_j(g)=\pm\mathbb I$ are infinite-dimensional operators, and the fact that $D(g)$ is written as a matrix with finite rows and columns does not imply that it is finite-dimensional. This aspect of the discussion can get dropped from textbooks (and is never very prominent to begin with), so it's perfectly understandable to be confused about it.

I hope this helps clarify the issue but let me know if it doesn't.

• Thank you very much! If I understand correctly you're saying that an irrep $R_j(g)$ can be infinite dimensional, and thus $D(g)$ has dimensionality equal to the entire Hilbert space? Then notation like $D(p) = (1,0 ; 0,-1)$ actually refers to the 1,-1 being $\mathbb{I}$,-$\mathbb{I}$: the (-)identity in the basis of states which behave like the irreps (the quote from the book)? – Timsey Sep 29 '14 at 17:46
• Yes, exactly that. – Emilio Pisanty Sep 29 '14 at 17:51