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## The representation theory of SU(2)

Today we will give four proofs of the classification of the (finite-dimensional complex continuous) irreducible representations of $\text{SU}(2)$ (which you’ll recall we assumed way back in this previous post). As a first step, it turns out that the finite-dimensional representation theory of compact groups looks a lot like the finite-dimensional representation theory of finite groups, and this will be a major boon to three of the proofs. The last proof will instead proceed by classifying irreducible representations of the Lie algebra $\mathfrak{su}(2)$.

At the end of the post we’ll briefly describe what we can conclude from all this about electrons orbiting a hydrogen atom.

Generalities

Below, “representation” means “finite-dimensional complex continuous representation.”

Let $G$ be a compact group and suppose that, one way or another, we have found a (left-invariant) Haar measure on $G$, normalized to have total measure $1$. Such measures exist for all compact groups but are non-trivial to construct in general; in the case of $\text{SU}(2) \cong S^3$ the Haar measure can be straightforwardly described as the measure on $S^3$ inherited from Lebesgue measure on $\mathbb{R}^4$ divided by the volume of $S^3$:

$\displaystyle \int_{\text{SU}(2)} f(g) \, dg = \frac{1}{2\pi^2} \int_{x^2+y^2+z^2+w^2 = 1} f(x, y, z, w) \, dV$.

It follows that given a representation of $G$, we can define the averaging operator

$\displaystyle V \ni v \mapsto \int_G \rho(g) v \, dg \in V$.

As in the case of finite groups, the averaging operator is a projection from $V$ onto its invariant subspace. It follows that we can average an inner product on $V$ so that it is $G$-invariant, hence WLOG we talk about unitary representations only and Maschke’s theorem holds: representations are completely reducible. Schur’s theorem holds in this setting with exactly the same proof as usual.

Also as for finite groups, taking the trace of a representation defines its character $\chi$, a continuous class function $G \to \mathbb{C}$. Taking characters is additive under direct sum, multiplicative under tensor product, and conjugate under taking duals. Moreover, given two representations $\rho_1, \rho_2$ on vector spaces $V_1, V_2$ we can define the inner hom $V_1 \Rightarrow V_2$ (what I’ve denoted by something like $\mathbf{hom}(V_1, V_2)$ in previous posts, but I think this notation is less confusing), which explicitly is the space of linear transformations $V_1 \to V_2$ with action given by

$\displaystyle \rho(g)(f(v)) = \rho_2(g) f(\rho_1(g)^{-1} v)$

and which abstractly is determined by the adjunction

$\displaystyle \text{Hom}(A \otimes B, C) \cong \text{Hom}(A, B \Rightarrow C)$.

In particular, $\text{Hom}(V_1, V_2) \cong \text{Hom}(1, V_1 \Rightarrow V_2)$, so the invariant subspace of $V_1 \Rightarrow V_2$ can be identified with the space of $G$-morphisms $V_1 \to V_2$. On the other hand, $V_1 \Rightarrow V_2$ is isomorphic to $V_1^{\ast} \otimes V_2$, hence if $\chi_1, \chi_2$ denote the corresponding characters, $V_1 \Rightarrow V_2$ has character $\overline{\chi_1} \chi_2$. Since the trace of the averaging operator gives the dimension of its image (the invariant subspace of a representation), it follows that

$\displaystyle \dim \text{Hom}(V_1, V_2) = \int_G \overline{\chi_1(g)} \chi_2(g) \, dg$

and combining this result with Schur’s lemma, the orthogonality relations follow exactly as for finite groups. In particular, a representation of $G$ is uniquely determined up to isomorphism by its character.

The irreducible representations of $\text{SU}(2)$

$\text{SU}(2)$ has an obvious $2$-dimensional irreducible representation which we’ll denote by $V$. To specify the character of any representation of $\text{SU}(2)$, it suffices to specify its restriction to the maximal torus of elements of the form $g = \left[ \begin{array}{cc} z & 0 \\ 0 & z^{-1} \end{array} \right], |z| = 1$ since every element is conjugate to an element of the torus, and the character of $V$ is then $\chi_V(g) = \chi_1(g) = z + z^{-1}$.

From $V$ we can construct some additional representations: we have $V \otimes V \cong S^2(V) \oplus \Lambda^2(V)$ where the former has dimension $3$ and the latter has dimension $1$. It is not hard to see that the abelianization of $\text{SU}(2)$ is trivial, so $\Lambda^2(V)$ is trivial, hence $V$ is self-dual and quaternionic. It follows that $S^2(V)$ cannot have a one-dimensional summand, hence is an irreducible representation of dimension $3$. If $g \in \text{SU}(2)$ has eigenvalues $z, z^{-1}$, then its action on $S^2(V)$ has eigenvalues $z^2, 1, z^{-2}$, so the character of this representation is given by

$\displaystyle \chi_{S^2(V)}(g) = \chi_2(g) = z^2 + 1 + z^{-2}$.

Note that the double cover $\text{SU}(2) \to \text{SO}(3)$ gives an irreducible $3$-dimensional real representation of $\text{SU}(2)$ which extends to a $3$-dimensional complex representation. In this representation $g$ acts by rotation by $2 \theta$, where $z = e^{i \theta}$, and it follows that the character of this representation agrees with $S^2(V)$, hence that the two are isomorphic.

What can we say about higher-dimensional representations? Well, given $V$, a standard strategy for constructing more representations is to apply various functors to $V$. Here it will be productive to consider the symmetric powers $S^n(V)$, getting the hint from $S^2(V)$. Since $\dim V = 2$, these representations have dimension $n+1$. Note that $S^0(V)$ is the trivial representation by definition. Recall that if $M : V \to V$ is a diagonalizable linear operator with eigenvalues $\lambda_1, ... \lambda_k$, then $S^n(M) : S^n(V) \to S^n(V)$ has eigenvalues the products of all unordered $n$-tuples of eigenvalues of $M$ (and this is easily proven by exhibiting the corresponding eigenvectors). It follows that the characters of the representations $S^n(V)$ are given by

$\displaystyle \chi_{S^n(V)}(g) = \chi_n(g) = z^n + z^{n-2} + ... + z^{-n+2} + z^{-n} = \frac{z^{n+1} - z^{-n-1}}{z - z^{-1}}$.

If $z = e^{i \theta}$, the above is therefore equal to $\frac{\sin (n+1) \theta}{\sin \theta} = U_n(\cos \theta)$ where $U_n$ is the $n^{th}$ Chebyshev polynomial of the second kind.

Proposition: All of the representations $S^n(V)$ are irreducible.

Proof. Since $V$ has real character, it is self-dual, so we can identify $S^n(V)$ with the space of homogeneous polynomials of degree $n$ on $V$. If $V$ is equipped with an invariant inner product, then letting $x, y$ denotes an orthonormal basis for $V^{\ast}$, $\text{SU}(2)$ acts on the symmetric algebra

$\displaystyle S(V^{\ast}) = \bigoplus_{n \ge 0} S^n(V^{\ast})$

of polynomial functions on $V$ as follows:

$\displaystyle \left[ \begin{array}{cc} \alpha & \beta \\ - \overline{\beta} & \overline{\alpha} \end{array} \right] f(x, y) = f(\alpha x + \beta y, -\overline{\beta} x + \overline{\alpha} y)$.

(The matrix given is the matrix of an element of $\text{SU}(2)$ with respect to an orthonormal basis $e_1, e_2$ such that $x, y$ is the corresponding dual basis.) Let $f \in S^n(V^{\ast})$ be a nonzero homogeneous polynomial of degree $n$. A minimal invariant subspace $W$ containing $f$ must also contain $f(\alpha x, \overline{\alpha} y)$ for every $|\alpha| = 1$, and by taking appropriate linear combinations of these polynomials, $W$ contains every monomial in $f$. Thus we may assume WLOG that $f$ is a single monomial $x^k y^{n-k}$. But for sufficiently small $\beta$ the action of $\text{SU}(2)$ sends $f$ to a polynomial all of whose coefficients are nonzero, hence $W$ contains every monomial in $S^n(V^{\ast})$, so $W = S^n(V^{\ast})$ as desired.

It follows that the characters $\chi_n$ are orthogonal and satisfy $\langle \chi_n, \chi_n \rangle = 1$. This can also be checked with an explicit formula for integrating a class function on $\text{SU}(2)$, but we will not need to do this.

The representations $S^n(V)$ all have real characters, so are all self-dual. A computation of the characters of both sides, or an explicit argument with bases, shows that

$\displaystyle S^n(V) \otimes V \cong S^{n+1}(V) \oplus S^{n-1}(V)$.

It follows by induction that any irreducible subrepresentation of $V^{\otimes k}$ is isomorphic to $S^n(V)$ for some $n$. Since they are all self-dual, these are in some sense all of the representations of $\text{SU}(2)$ obtainable from $V$ via universal methods.

Theorem: Every irreducible representation of $\text{SU}(2)$ is isomorphic to $S^n(V)$ for some $n \ge 0$.

Proof 1

Our first proof is based on the following important observation: since the character of a representation of $\text{SU}(2)$ is determined by its restriction to any maximal torus $S^1$, the restriction functor $\text{Rep}(\text{SU}(2)) \to \text{Rep}(S^1)$ is essentially injective. So we should try to understand the second category in order to understand what kind of characters are possible.

Theorem: Regard $S^1$ as the unit complex numbers $\{ z \in \mathbb{C} : |z| = 1 \}$. Then every irreducible representation of $S^1$ is isomorphic to the representation $z \mapsto z^n$ for some $n \in \mathbb{Z}$.

This result is closely related to the existence of Fourier series; see Pontryagin duality for a general discussion.

Proof. Since $S^1$ is abelian it is true as for finite groups that any irreducible representation has dimension $1$ (since any eigenvector of a non-identity element spans an invariant subspace). Thus an irreducible representation is just a continuous homomorphism $S^1 \to C^{\ast}$. By compactness the image of this map is contained in $S^1$, so an irreducible representation is just a continuous homomorphism $f : S^1 \to S^1$. By path lifting, $f$ lifts to a continuous homomorphism $\mathbb{R} \to \mathbb{R}$, which must therefore be of the form $x \mapsto rx$ for some real $r$. This gives

$\displaystyle f(e^{i \theta}) = e^{ir \theta}, r \in \mathbb{R}$

which is a homomorphism if and only if $r \in \mathbb{Z}$, and the conclusion follows.

Any irreducible representation $W$ of $\text{SU}(2)$ decomposes into the direct sum of finitely many irreducible representations of a given maximal torus $S^1 \subset \text{SU}(2)$. Suppose the representation $z \mapsto z^n$ occurs $a_n$ times. Then the character of $W$ must be given by

$\displaystyle \chi_W(g) = \sum a_n z^n$

where $g$ has eigenvalues $z, z^{-1}$. That is, it must be a Laurent polynomial in $z$ with integer coefficients. Moreover, since we can swap the order of the eigenvalues, it follows that $\chi_W$ is invariant under the substitution $z \mapsto z^{-1}$, so it must be a symmetric Laurent polynomial in $z$:

$\displaystyle \chi_W(g) = \sum_{n \ge 0} a_n (z^n + z^{-n})$.

However, it is not hard to see that the characters $\chi_n$ of the symmetric powers $S^n(V)$ give a $\mathbb{Z}$-basis of the symmetric Laurent polynomials in $z$, so it follows that there exist $c_n \in \mathbb{Z}$ such that

$\displaystyle \chi_W = \sum_{n \ge 0} c_n \chi_n$.

In particular, $\langle \chi_W, \chi_n \rangle \neq 0$ for some $n$, hence $W \cong S^n(V)$ for some $n$ as desired.

In general, maximal tori are very important in the classification and representation theory of compact Lie groups. The Lie-algebraic analogue of a maximal torus is a Cartan subalgebra, which plays a corresponding role in the classification and representation theory of semisimple Lie algebras.

Proof 2

Our second and third proofs are both motivated by the standard result that if $G$ is a finite group and $V$ a faithful representation of $G$, then every irreducible representation of $G$ occurs in $V^{\otimes n}$ for some $n$. This result can be proven in many ways, some of which don’t generalize to compact groups, and in fact the result is not true as stated for compact groups: the example of $G = S^1$ and $V$ the representation $z \mapsto z$ shows that we need to at least consider $V^{\otimes n} (V^{\ast})^{\otimes m}$ for integers $m, n$. This result is true (see for example this MO question), but Proof 2 will not be enough to prove it in general. Nevertheless, it works in the special case of $\text{SU}(2)$.

First, note that $\chi_1(g) = \chi_1(h)$ if and only if they have the same eigenvalues (which are determined by their real part), hence if and only if they are conjugate in $\text{SU}(2)$. So letting $\theta$ denote the eigenvalues $e^{i \theta}, e^{-i \theta}$ of $g$, the space of conjugacy classes of $\text{SU}(2)$ is naturally identified with $\{ \theta : 0 \le \theta \le \frac{\pi}{2} \}$, and $\chi_1$ separates points on this space. It follows by the complex Stone-Weierstrass theorem that the smallest subalgebra of the algebra of class functions containing $\chi_1$ and closed under conjugation is dense in the space of continuous class functions with the uniform norm. But since $V$ is self-dual, this algebra is spanned by the characters of the representations $V^{\otimes k}$, which are all direct sums of the representations $S^n(V)$.

It follows that given an irreducible representation $W$, we can find a sequence of class functions which are linear combinations of the characters $\chi_n$ converging uniformly to $\chi_W$. It again follows that $\langle \chi_W, \chi_n \rangle \neq 0$ for some $n$, and again we are done.

This proof is not enough to give the general result about faithful representations because $\chi_V$ does not necessarily separate conjugacy classes in general. However, it is still possible to use the Stone-Weierstrass theorem to prove the general result, and this is the subject of the next proof.

Proof 3

Our third proof relies on a companion result to the orthogonality relations for characters. For us, a matrix coefficient of a compact group $G$ is a continuous function $G \to \mathbb{C}$ of the form $\langle w, \rho(g) v \rangle_V$ for a unitary representation $\rho : G \to \text{U}(V)$ and vectors $v, w \in V$.

Theorem (orthogonality for matrix coefficients): Let $f_1 = \langle w_1, \rho_1(g) v_1 \rangle_{V_1}, f_2 = \langle w_2, \rho_2(g) v_2 \rangle_{V_2}$ be matrix coefficients of a compact Lie group $G$, where $\rho_1, \rho_2$ are irreducible unitary representations on $V_1, V_2$ with characters $\chi_1, \chi_2$. If $V_1, V_2$ are non-isomorphic, then

$\displaystyle \langle f_1, f_2 \rangle_{L^2(G)} = \int_G \overline{f_1(g)} f_2(g) \, dg = 0$.

(The full statement of the orthogonality relations includes an expression for the inner product of matrix coefficients from the same irreducible representation, but there is a constant in it that I can’t figure out, and in any case we won’t need it.)

Proof. Our convention is that inner products are conjugate-linear in the first variable. Fix $v_1$ and $v_2$. We can write

$\displaystyle \langle f_1, f_2 \rangle_{L^2(G)} = \int_G \langle \rho_1(g) v_1, w_1 \rangle_{V_1} \langle w_2, \rho_2(g) v_2 \rangle_{V_2}$.

For fixed $g$, the integrand gives a bilinear map $V_1 \times V_2^{\ast} \to \mathbb{C}$ (since it is linear in $w_1$ but conjugate-linear in $w_2$); moreover, the natural action of $g$ by precomposition gives a representation isomorphic to $V_1^{\ast} \otimes V_2$ on the space of such bilinear maps, and the above integral computes precisely the averaging operator on this representation. If $V_1$ is not isomorphic to $V_2$, the corresponding tensor product does not have any copies of the irreducible representation, so the averaging operator sends any vector to zero and the conclusion follows.

Theorem: Let $G$ be a compact group with a faithful representation $V$. Then the span of the matrix coefficients of the representations $V^{\otimes n} (V^{\ast})^{\otimes m}$ is dense in $L^2(G)$.

Proof. The sum of any two matrix coefficients of a given representation $V$ is another matrix coefficient of the same representation $V$. Furthermore, the complex conjugate of a matrix coefficient of $V$ is a matrix coefficient of $V^{\ast}$, and the product of a matrix coefficient of $V$ and a matrix coefficient of $W$ is a matrix coefficient of $V \otimes W$. It follows that if $G$ has a faithful representation $V$, then by the complex Stone-Weierstrass theorem, the space spanned by the matrix coefficients of the representations $V^{\otimes n} (V^{\ast})^{\otimes m}$ is dense in $L^2(G)$. In particular, if $W$ is any other irreducible representation, then we can approximate its matrix coefficients with the above matrix coefficients, and we find by orthogonality that $W$ is a subrepresentation of $V^{\otimes n} (V^{\ast})^{\otimes m}$ for some $m, n$.

Proof 4

For our final proof we will provisionally assume that all representations are smooth so that we can pass from a representation $G \to \text{GL}(V)$ to the induced map on Lie algebras $\mathfrak{g} \to \mathfrak{gl}(V)$. As it turns out, all continuous homomorphisms between Lie groups are automatically smooth (although we won’t prove this), so we can assume this WLOG.

Recall that a representation of a Lie algebra $\mathfrak{g}$ is a homomorphism of Lie algebras $\mathfrak{g} \to \mathfrak{gl}(V) \cong \text{End}(V)$ for $V$ a (finite-dimensional complex) vector space. There is an obvious notion of direct sum of representations. As for groups, we say that a representation is irreducible if it has no non-trivial $\mathfrak{g}$-invariant subspaces.

Proposition: Let $G$ be a Lie group all of whose elements are contained in a one-parameter subgroup of $G$, $G \to \text{GL}(V)$ a smooth representation, and $\mathfrak{g} \to \mathfrak{gl}(V)$ the induced representation of $\mathfrak{g}$. Then $V$ is irreducible as a representation of $G$ if and only if it is irreducible as a representation of $\mathfrak{g}$.

Proof. By assumption, $V$ is irreducible if and only if it has no non-trivial subspaces invariant under all of the one-parameter subgroups of $G$. Let $e^{Dt}, D \in \mathfrak{g}$ be such a one-parameter subgroup. Let $v \in V$ be a nonzero vector. Then

$\displaystyle \lim_{t \to 0} \frac{e^{Dt} v - v}{t} = Dv$

and

$\displaystyle e^{Dt} v = \sum_{n=0}^{\infty} \frac{t^n}{n!} D^n v$.

It follows that any point in the minimal $\mathfrak{g}$-invariant subspace containing $v$ is a limit of points in the minimal $G$-invariant subspace containing $v$, and vice versa, hence that the two coincide.

In particular, we now know that the representations of $\mathfrak{su}(2)$ associated to $S^n(V)$ are all irreducible. Moreover, given a representation of $\mathfrak{su}(2)$ which is known to come from a representation of $\text{SU}(2)$, by exponentiating we can recover the action of a maximal torus, hence the character of the representation. It follows that if we show that every irreducible representation of $\mathfrak{su}(2)$ is of the form $S^n(V)$, we have also shown the same statement for $\text{SU}(2)$. (Note that we don’t need to know how to lift representations of $\mathfrak{su}(2)$ to representations of $\text{SU}(2)$ to do this.)

Recall that $\mathfrak{su}(2)$ can be explicitly presented as the imaginary quaternions under the inherited bracket

$\displaystyle [A, B] = 2C, [B, C] = 2A, [C, A] = 2B$.

(I’m not using the symbols $i, j, k$ because we’re about to extend scalars to $\mathbb{C}$.) Since we only care about complex representations of this real Lie algebra, we can pass to the complexification

$\displaystyle \mathfrak{g} = \mathfrak{su}(2) \otimes_{\mathbb{R}} \mathbb{C}$

(a technique which highlights the algebraic convenience of working with Lie algebras), and now that we are working over $\mathbb{C}$ we can ask for a more convenient basis of $\mathfrak{g}$. One idea is to look at the adjoint action $[A, -] : \mathfrak{g} \to \mathfrak{g}$. By inspection the eigenvectors of this linear transformation are $A, iB - C, iB + C$ with eigenvalues $0, -2i, 2i$ respectively. So one convenient choice of basis is $H = i A, X = \frac{iB - C}{\sqrt{2}}, Y = \frac{iB + C}{\sqrt{2}}$, in which the relations are given by

$\displaystyle [H, X] = 2X, [H, Y] = -2Y, [X, Y] = H$.

This basis is convenient for the following fundamental reason. Let $W$ be any representation of $\mathfrak{g}$. (We won’t distinguish between an element of $\mathfrak{g}$ and its action on $W$.) Suppose $v \in W$ is an eigenvector of $H$ of eigenvalue $\lambda$. Then

$\displaystyle (HX - XH)v = 2Xv \Leftrightarrow HXv = (\lambda + 2)Xv$

and similarly

$\displaystyle (HY - YH)v = -2Yv \Leftrightarrow HYv = (\lambda - 2)Yv$

hence $Xv$ is (zero or) an eigenvector of $H$ of eigenvalue $\lambda + 2$ and $Yv$ is (zero or) an eigenvector of $H$ of eigenvalue $\lambda - 2$! The eigenspaces spanned by these eigenvectors are known as the weight spaces of $W$, and correspond to the decomposition of $W$ under the action of a maximal torus of $\text{SU}(2)$. But unlike in the first proof above using maximal tori, we can now use elements of the Lie algebra to move between weight spaces.

The sequence of vectors $v, Xv, X^2 v, ...$ are all eigenvectors with distinct eigenvalues, so since $W$ is finite-dimensional the sequence must eventually terminate. Hence we may assume WLOG that $Xv = 0$. In this case $v$ is known as a highest weight vector. Since $[X, Y] = 0$, it follows that

$\displaystyle \text{span}(v, Yv, Y^2 v, ...)$

is a $\mathfrak{g}$-invariant subspace of $W$. If $W$ is irreducible it must therefore be all of $W$. If $\dim W = n + 1$, we conclude that

$W = \text{span}(v, Yv, ... Y^n v)$.

Hence $W$ has a basis of eigenvectors of $H$ with eigenvalues $\lambda, \lambda - 2, ... \lambda - 2n$. In particular, it follows that the trace of $H$ acting on $W$ is

$\displaystyle \sum_{i=0}^n (\lambda - 2i) = (n+1) \lambda - n(n+1)$.

But since $[X, Y] = H$ and the trace of a commutator is zero, it follows that we must have $\lambda = n$. The resulting weights are precisely the weights of $S^n(V)$, and it follows that $W \cong S^n(V)$ as desired.

Note that, unlike all of the other proofs, this one explicitly constructs the irreducible representations (at least as representations of the Lie algebra) without requiring that we knew what they were in advance.

The representation theory of $\text{SO}(3)$

An electron orbiting a hydrogen atom can be described by four quantum numbers, which are really labels for a direct sum decomposition of the corresponding Hilbert space of states. Using what we now know about the representation theory of $\text{SU}(2)$, we can explain two of these quantum numbers, although we will do so in more detail in a later post.

Keeping in mind the double cover $\rho : \text{SU}(2) \to \text{SO}(3)$, any irreducible representation of $\text{SO}(3)$ pulls back to an irreducible representation of $\text{SU}(2)$. The ones we get in this way are precisely the ones in which $-I \in \text{SU}(2)$ acts trivially, and by inspection these are the irreducible representations $S^{2n}(V)$. Hence $\text{SO}(3)$ has exactly one irreducible representation of each odd dimension $2n+1$. (Of course all of the irreducible representations of $\text{SU}(2)$ are projective representations of $\text{SO}(3)$, so we should expect the even-dimensional ones to also be important in quantum mechanics.)

It follows that in any quantum mechanical system with physical $\text{SO}(3)$ symmetry, the eigenstates corresponding to a particular energy eigenvalue will organize themselves into clumps with an odd number of members. For electrons around a hydrogen atom, the corresponding clumps are referred to in chemistry as subshells and delineated by their azimuthal quantum number $\ell$, which measures orbital angular momentum. In chemistry, subshells are also traditionally referred to by letters based on what the corresponding atomic spectra looked like:

1. $s$ (sharp) refers to $S^0(V)$,
2. $p$ (principal) refers to $S^2(V)$,
3. $d$ (diffuse) refers to $S^4(V)$,
4. $f$ (fundamental) refers to $S^6(V)$,

and so forth. Experimentally, $s$ subshells hold $2$ electrons, $p$ subshells hold $6$ electrons, $d$ subshells hold $10$ electrons, $f$ subshells hold $14$ electrons, and so forth. This is off by exactly a factor of $2$ from what can be predicted purely on the basis of $\text{SO}(3)$ symmetry (where we would expect $1, 3, 5, 7$ electrons, respectively), and this is because there is a second symmetry here coming from spin. Nevertheless, the representation theory of $\text{SO}(3)$ alone is already enough to address abstractly the origin of one quantum number, and in fact by fixing a maximal torus and considering its eigenvectors we get another one, the magnetic quantum number $m$. The reason it takes values between $- \ell$ and $\ell$ is that these are the possible weights of $S^{2\ell}(V)$.

Since an argument based on $\text{SO}(3)$ symmetry is independent of the choice of Hamiltonian, we cannot use symmetry alone to figure out what the energy eigenspaces are, so we still can’t explain the principal quantum number without actually looking at the Hamiltonian, and we also haven’t explained spin. But two out of four isn’t bad!

### 8 Responses

1. […] Example. Let and let be the defining representation. The character is just the trace of regarded as a matrix, which completely determines its conjugacy class; consequently, already separates points, and to understand Haar measure on the conjugacy classes of it suffices to understand the moments of . But these are just the dimensions of the invariant subspaces of the tensor powers of . The explicit description […]

2. on March 29, 2012 at 6:04 pm | Reply Mozibur Ullah

Very nice explanation, and clearly stated. My chemistry is now pretty faint, but I do recall the electron shells having certain shapes. Can representation theory help in saying anything about them?

Also, is it fair to say that the electrons in the same orbital have lost their own ‘individuality’? That is, if I put an electron in an orbital, and then later on take one out, can I meaningfully ask whether it is the same electron I put in and get a straight answer?

• 1) Yes. The keyword here is “spherical harmonic.” 2) My understanding is that two particles of the same type do not have an individual existence in quantum mechanics. They are identical on a fundamental level.

• on March 31, 2012 at 6:33 am | Reply Mozibur Ullah

Thanks for the tip. Sure, electrons are fundamentally identical, but surely that doesn’t mean that we can’t distinguish them under certain situations. Take for example two electrons separated by some large distance. Even with the uncertainty principle at work, I think the two ‘identical in every way’ electrons can still be distinguished in space. Or am I making a silly blunder here?

• I suppose.

3. How do you like your internship? Are you studying GARCH and ARIMA models?

4. Quite impressive to me -I am not fluent in representation theory.

Thanks alot for the post.
It would be even nicer with (more) references to your sources and possible extensions of the results -which are numerous I suspect. But this has already been alot of work, thanks.

• The classical results of Lie theory extending these results can be found in just about any good text on Lie theory, e.g. Fulton and Harris or Serre.