Madhusudhan Raman

“I started out on burgundy,
but soon hit the harder stuff.”

The Lie algebra $ \mathfrak{su}(2) $ is insufficiently generic, and it is profitable at this stage to ask: to what extent does the oscillator formalism generalise to $ \mathfrak{su}(d) $? In order to suggest a natural generalisation to $ \mathfrak{su}(d) $ it is helpful to rewrite the results of the previous section (for $ d=2 $) as follows. First, define the bilinears \begin{equation} \label{eq:Eij-defn} E _ {ij} = a ^{\dagger }_ {i} a _ {j} \ , \end{equation} whose algebra is easily derived from the canonical commutation relations of creation and annihilation operators as \begin{equation} \label{eq:uN-algebra} \left[ E _ {ij}, E _ {k \ell } \right] = \delta _ {jk}E _ {i \ell } - \delta _ {i \ell } E _ {kj} \ . \end{equation} It is easy to check that for the case of $\mathfrak{su}(2)$ \begin{equation} \label{eq:su2-Eij-pauli} K _ {a} = \frac{1}{2} \, E _ {ij} \left( \sigma _ {a} \right)_ {ij} \ , \end{equation} where we have an implicit sum over the repeated oscillator indices.

In this avatar, it is now clear how to extend the discussion of the previous section to higher $ d $. The algebra in \eqref{eq:uN-algebra} is the Lie algebra for $ \mathfrak{u}(d) $, which differs from $ \mathfrak{su}(d) $ by an additional tracelessness condition on the generators. Therefore, increasing the number of oscillators from $ d=2 $ to (say) $ d=3 $ should give us the unitary irreducible representations of $ \text{su}(d) $. Consequently, we start with the Hamiltonian of an isotropic three-dimensional oscillator with unit angular frequency:

\[\begin{align} H &=\sum_ {i=1}^{3} E _ {ii} + \frac{3}{2} \ , \\ &=\sum_ {i=1}^{3} N _ {i} + \frac{3}{2} = N + \frac{3}{2} \ . \label{eq:3d-hamiltonian} \end{align}\]

where $i \in \left\lbrace 1,2,3 \right\rbrace$ and $ N $ is the total number operator. The degeneracy is easily counted to be $\frac{1}{2}(N+1)(N+2)$, and there are six operators that move one within a degenerate subspace which are of the form $ E _ {ij} $ with $ i \neq j $.

As in the previous section, we will need to augment this list of six operators in order to derive a closed algebra. Notice that from \eqref{eq:uN-algebra}, \begin{equation} \left[ E _ {ij}, E _ {ji} \right] = E _ {ii} - E _ {jj} = N _ {i} - N _ {j} \ . \end{equation} This would seem to suggest there are three additional operators we need to add, each of the form $ N _ {i} - N _ {j} $ for $ i \neq j $, but notice that these operators are not linearly independent. Therefore, we just need to add two, giving a total of eight operators that generate the $ \mathfrak{su}(3) $ Lie algebra.

This is to be expected, since any element of $ \text{SU}(d) $ can be written as $ e^{i a F} $ where $ F $ is a $ d \times d $ traceless Hermitian matrix. Now, a $ d \times d $ Hermitian matrix has $ d ^{2} $ independent components, but tracelessness supplies one condition, leaving $ d ^{2}-1 $ independent parameters, which for $ d = 3 $ once again gives eight independent generators. These are conventionally taken to be proportional to the Gell-Mann matrices $ t _ {a} = \frac{1}{2} \lambda _ {a} $. You can look up the Gell-Mann matrices here.

These matrices are normalised so that $\operatorname{Tr}\left(t_ {a} t_ {b}\right)=\frac{1}{2} \delta_ {a b}$, and their algebra is \begin{equation} \label{eq:su3-algebra} \left[t_ {a}, t_ {b}\right]=i f_ {a b c} t_ {c} \ , \end{equation} where the $f_ {a b c}$ are called structure constants. This is the $ \mathfrak{su}(3) $ Lie algebra.

Analogously to \eqref{eq:su2-Eij-pauli}, we consider the operators \begin{equation} T_ {a} = E _ {ij} \left( t_ {a} \right)_ {ij} \ . \end{equation} In the above equation, the indices $ \left\lbrace i, j, \cdots \right\rbrace \in \left\lbrace 1, 2 , 3 \right\rbrace $ and $ \left\lbrace a, b, c, \cdots \right\rbrace \in \left\lbrace 1, \cdots , 8 \right\rbrace $. Now, the $ T ^{a} $ are Hermitian, courtesy of the Hermiticity of the Gell-Mann matrices. From \eqref{eq:uN-algebra} and \eqref{eq:su3-algebra}, one can easily verify that the $ T _ {a} $ satisfy the $ \mathfrak{su}(3) $ Lie algebra. Finally, each of the $ T _ {a} $ commute with $ N _ {i} = E _ {ii} $, and therefore from \eqref{eq:3d-hamiltonian} that the Hamiltonian commutes with each of the $ T _ {a} $, i.e. the $ T _ {a} $ will only move states within a degenerate subspace. We can conclude at this point that the degenerate states \begin{equation} \label{eq:oscillator-su3-states} \left[\prod _ {i=1} ^{3} \frac{1}{\sqrt{N _ {i}!} } \left( a _ {i}^{\dagger } \right)^{N _ {i}} \right] \left\vert 0 \right\rangle \ , \end{equation} with fixed $ N = \sum_ {i=1}^{3} N _ {i} $ of the three-dimensional isotropic harmonic oscillator form unitary irreducible representations of $ \text{SU}(3) $ of dimensions \begin{equation} \label{eq:oscillator-su3-dimension} \frac{1}{2} \left( N+1 \right) \left( N+2 \right) \ . \end{equation}

These conclusions are easily generalised to $ \text{SU}(d) $. In much the same way as we have outlined here, we first construct operators of the form $T_ {a}= E _ {ij} \left( t _ {a} \right)_ {ij} $ where the $t^{a}$ are $ d ^{2}-1 $ traceless $d \times d$ Hermitian matrices, and $a_ {i}^{\dagger}$ and $a_ {i}$ are the creation and annihilation operators of a $d$-dimensional isotropic harmonic oscillator, whose degenerate subspaces form unitary irreducible representations of $ \text{SU}(d) $.

Categories:

Updated: