I am trying to understand exactly what IRREPs are and, in order to know more about it, I started reading the chem.libretext course on molecular symmetry which up to this point seemed fairly comprehensive and understandable to me. However, I blocked right at the moment the irreducible representations were introduced in the context of this course.

What I understand:

  • Symmetry operations that can be applied to molecules form point groups, with "point" being there because there is always a point in the molecule that is not modified by the symmetry operations, as opposition to "space" groups in crystals where translations become important
  • Symmetry operations can be represented as matrices that act on vectors of basis functions, leading to the calculation of linear combinations of the basis functions
  • A set of matrices that represent symmetry operations is called a "representation"
  • There is a process of reducing representations into various "pieces" called irreducible representations which involves decomposing the rotations matrices into a set of smaller matrices which when "directly-summed" (with this odd ⨁ symbol) reconstitute the original matrices.

But then, at This page, the author suddenly, out of the blue, introduces that it is possible to somehow replace the basis functions used as reference with what are called "symmetry adapted linear combinations" (SALCs) of basis functions. Suddenly, she specifies the actual SALC, but I don't understand how she gets there, and the SALC that are depicted seem to be completely random, they do not make any sense to my eye.

SALC for symmetry group of ammonia

Why are these specific basis functions SALC for the symmetry group of ammonia? This is completely puzzling to me. Especially $s_{2}'$ and $s_{3}'$ do not make any sense to my intuition.

How do we get these SALCs allowing us to decompose the transformation matrices the most?

Does this have anything to do with "block diagonalization"?

As an annex question, what exactly is an IRREP? Can we have a molecular feeling of what an IRREP is? Can we have a visual depiction of an IRREP? Everytime I try to google this question, I end up with abstract character tables and I get confused by the last column of these character tables with all the x, y, z which seem to be out of place here.

Thanks in advance!

  • $\begingroup$ You may find this corresponding question on Math.SE useful. $\endgroup$ Sep 10, 2021 at 19:59

1 Answer 1


The answer to this is very deep, and is rooted in mathematics, not physics. I don't think any chemistry textbook seriously explains what an irreducible representation "really" is, precisely because it's almost impossible to teach it to the average chemist. To properly understand this, I'd suggest reading about group theory, and then representation theory if you're interested.

However, let's see if I can try to provide a brief-ish overview. As will become very clear, covering this in full is not feasible in a Stack Exchange answer. Also, you will likely need some linear algebra background to understand this. I simply can't explain everything from the ground up, so apologies to any readers left behind.

We start from the definition of a group. This is a set of objects $G$, equipped with a binary operation $\cdot$, which obeys the four group axioms, namely:

  • closure (for all $g, h \in G$, $g\cdot h \in G$)
  • associativity (for all $g, h, k \in G$, $(g \cdot h) \cdot k = g \cdot (h \cdot k)$)
  • identity (there exists some element $e \in G$ such that for all $g \in G$, $e\cdot g = g \cdot e = g$)
  • inverse (for all $g \in G$, there exists some element $g^{-1} \in G$ such that $g\cdot g^{-1} = g^{-1} \cdot g = e$)

In molecular symmetry, the objects $g$ are symmetry operations. That is, the point group consists of all symmetry operations which preserve the molecular geometry. These elements form the top row of the character table.*

Often in (basic) group theory we are concerned with the elements of the group $g$, i.e. the symmetry operations. However, in chemistry we are really more interested in the molecules or the orbitals, which are not elements of the group! These are actually objects which the group elements act on.

Mathematically speaking, these objects are vectors which inhabit a vector space $V$. For example, in ammonia, we might have a vector space which consists of the three hydrogen 1s orbitals, as well as the 2s and 2p orbitals on nitrogen. It is left as an exercise to the reader to convince themselves that the group elements (i.e. symmetry operations) convert an element of the vector space (e.g. a hydrogen 1s) to another element of the same vector space (e.g. a rotation converts it to the 1s orbital of a different hydrogen).

Vector spaces are defined by a few more axioms, which can be easily found in any reference. Each group element $g_i \in G$ is therefore a map: it acts on a vector $v \in V$ and produces another vector $w \in V$. Of course, this map is different for every group element. So each group element $g_i$ is a different map $\rho_i: V \to V$. The act of "going from group elements $g_i$ to vector mappings $\rho_i$" is called a group homomorphism, or a representation. $V$ is called the representation space. There is a condition on this, namely that all of the $\rho_i$'s are linear: $$\begin{align} \rho_i(v_1 + v_2) &= \rho_i(v_1) + \rho_i(v_2) \\ \rho_i(kv_1) &= k\rho_i(v_1) \end{align}$$ where $v_1, v_2 \in V$ and $k$ is some complex number.†

A different representation simply means a different set of $\rho_i$'s, which may well act on a different vector space $V'$.

If our vector space has finite dimensions, let's say $n$ dimensions, then each vector $v \in V$ can be associated with a $n \times 1$ (complex-valued) column vector. The exact numbers in the column vector depends on the choice of basis. (Again, I can't explain all of this here; this is basic linear algebra.) Then, the fact that the $\rho_i$'s are linear maps on $V$, means that the $\rho_i$'s are actually $n \times n$ matrices. Again, the matrix elements will depend on the basis chosen for $V$. Often you will read that these matrices are elements of the general linear group $GL(n; \mathbb{C})$. Maths textbooks don't like to choose an explicit basis for $V$, but it's necessary for our purposes here.

Let's take stock of where we are. We've explained how group elements may be represented by $n \times n$ complex matrices (assuming a finite-dimensional representation space). These matrix elements $\rho_{i,pq}$ depend on the basis chosen.

$$g_i \to \rho_i = \begin{pmatrix} \rho_{i,11} & \rho_{i,12} & \cdots & \rho_{i,1n} \\ \rho_{i,21} & \rho_{i,22} & \cdots & \rho_{i,2n} \\ \vdots & \vdots & \ddots & \vdots \\ \rho_{i,n1} & \rho_{i,n2} & \cdots & \rho_{1,nn} \\ \end{pmatrix}$$

The act of decomposing a representation refers to finding a basis of $V$ which make all $\rho_i$'s have (the same) block diagonal form. Mathematicians don't like to speak about bases of $V$, and so this is often framed in terms of "subspaces" and "quotient spaces"; you may see that in books. But we will stick with our explicit matrices and bases.

Let's talk about a three-dimensional basis for now; so the $\rho_i$'s are $3 \times 3$ matrices. (Essentially, $V = \mathbb{C}^3$.) And let's suppose that we find a basis in which

$$g_i \to \rho_i = \begin{pmatrix} \rho_{i,11} & 0 & 0 \\ 0 & \rho_{i,22} & \rho_{i,23} \\ 0 & \rho_{i,32} & \rho_{i,33} \\ \end{pmatrix}$$

and that every matrix $\rho_i$, for all group elements $g_i$, has the same block-diagonal form in this basis. Essentially, what a block-diagonal matrix does is it disentangles different basis vectors from one another. If the matrix $\rho_i$ acts on the vector $v_1 = (a, 0, 0)^T$, then we can guarantee that the resulting vector $\rho_i v_1$ is of the form $(a', 0, 0)^T$ (and furthermore $a' = \rho_{i,11}a$). Likewise, if we take $\rho_i$ and act on some other vector $v_2 = (0, b, c)^T$, then the resulting vector $\rho_i v_2$ is $(0, b', c')^T$ where the first element is still $0$.

This means that we have essentially obtained two different representations from this.

  • The first representation is defined by the vector space $V_1 = \{(a, 0, 0)^T\}$ for some $a \in \mathbb{C}$. We can get rid of the two zero entries such that $V_1$ is basically a 1-dimensional space.

    • The $\rho_i$'s in this representation are then just $1 \times 1$ matrices with the single element $\rho_{i,11}$.
  • The second representation is defined by the vector space $V_2 = \{(0, b, c)^T\}$ for $b, c \in \mathbb{C}$. Again, we can drop the zero entry at the beginning.

    • $\rho_i$ in this representation is then the bottom-right $2\times 2$ block of the previous matrix.
  • Note that the old $3 \times 3$ $\rho_i$ matrix can be obtained through a direct sum of the $1 \times 1$ and $2 \times 2$ matrices. This is where the direct sum comes in.

  • Alternatively, you could leave in all the zeroes that we took out; and instead of the direct sum we would just have the ordinary sum of two matrices.

These two new representations are called subrepresentations. An irreducible representation is then a subrepresentation which cannot be reduced further in this manner. It remains to make a few points:

  • The SALCs that you see are the basis vectors of $V$ which allow for this decomposition.
  • The character listed for each irrep in the character table refers to the trace of the matrix $\rho_i$. Note that each group element corresponds to a different matrix with (in general) a different trace, so the character of an irrep under each symmetry operation is (in general) not the same.
  • If an irrep is 2-dimensional (for example), then the basis in this 2-dimensional vector space can be arbitrarily chosen (the fact is that it can't be further decomposed into block-diagonal form, regardless of basis: so it doesn't matter what the basis is). So, the SALCs for 2- and higher-dimensional irreps are actually arbitrarily chosen. This is true for the $s_2'$ and $s_3'$ irreps in the question: any linear combination of $s_2'$ and $s_3'$ is a valid pair of SALCs.
  • Changing the basis within an irrep does not change the characters of the irrep, because the trace of a matrix is invariant under a change of basis.


* Note that in a character table, elements that are conjugate to one another are grouped into the same entry. Two elements $g, h \in G$ are said to be conjugate if there exists some element $t \in G$ such that $g = tht^{-1}$.

† We are dealing with complex-valued vector spaces. Most of the simple character tables don't have complex numbers. But counterexamples are easy to find: see e.g. the character table for $C_3$. http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=203&option=4

  • $\begingroup$ Thank you for this answer; it's nice to have a perspective grounded in chemistry but oriented around the math. Could you confirm if the following summary is accurate? 1. A point group is a group of isometries with at least one fixed point. 2. All molecules can be classified under a point group. 3. A representation of a group is the action of the group elements on a specific vector space V (here, of molecules). 4. A representation can be reduced by partitioning $V$ into disjoint subspaces that are left invariant by the representation. $\endgroup$ Sep 10, 2021 at 21:11
  • $\begingroup$ 5. The restriction of a representation to such a subspace is a subrepresentation. 6. A subrepresentation that cannot be further reduced is an irreducible representation (irrep). 7. The original representation can be recovered as a direct sum of irreps. 8. The set of SALCs is a basis for $V$ in which a representation is maximally block diagonal. 9. Each block corresponds to an irrep and an invariant subspace of $V$. $\endgroup$ Sep 10, 2021 at 21:11
  • $\begingroup$ Still have no idea what character tables are, but that's a question for a different time. ;) $\endgroup$ Sep 10, 2021 at 21:15
  • 1
    $\begingroup$ @a-cyclohexane-molecule (1) Excluding translations, I think. (3) Given a particular molecule, I think the orbitals form a more useful vector space than the molecule itself: because under the group operation the molecule always maps to itself, the molecule should always transform under the totally symmetric irrep. It's a valid vector space but just not particularly useful. (6) You'd probably like the formal definition, which is that an irrep is a representation that has no proper subrepresentations (i.e. no subrepresentations for which the vector subspace $V'$ is a proper subspace of $V$). $\endgroup$ Sep 10, 2021 at 22:21
  • 1
    $\begingroup$ The rest agrees with what I understand of the topic (which is not all that much beyond what I've just written here...!) The character tables just collate the traces of the matrices $\rho_i$ for each irreducible representation of the group. $\endgroup$ Sep 10, 2021 at 22:23

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