Quantum Mechanics (1) · Problem 14

Identical Particles

Physics — Quantum Mechanics · group-theory-rotation-group-cubic-symmetry

Problem

This problem concerns the representation theory of rotation groups and crystal field splitting. (b) We will need the character table of this group. Since all elements in the same class have the same character, we pick a convenient element in each class by considering rotations about the

z

-axis,

R = (\alpha, z)

(means rotate by angle

\alpha

about the

z

-axis). Thus:

\hat{P}_{(\alpha,z)}Y_{\ell m} = e^{-im\alpha}Y_{\ell m}

(which you should convince yourself of). Find the character ``table'' of the rotation group, that is, find

\chi^{\ell}(\alpha)

, the character of representation

D^\ell

for the class of rotations through angle

\alpha

. If you find an expression for the character in the form of a sum, do the sum, expressing your answer in as simple a form as you can. (c) At last we are ready to put our atom into a potential with cubic symmetry. Now the symmetry of the free Hamiltonian is broken, and we are left with the discrete symmetry of the cube. The symmetry group of proper rotations of the cube is a group of order 24 with 5 classes. Call this group ``

O

''. Construct the character table for

O

. (d) Consider in particular how the

f

-level (

l = 3

) of the free atom may split when it is placed in the ``cubic potential''. The seven eigenfunctions which transform according to the irreducible representation

D^3

of the full group will most likely not transform according to an irreducible representation of

O

. On the other hand, since the operations of

O

are certainly operations of

D

, the eigenfunctions will generate some representation of

O

. Determine the coefficients in the decomposition:

D^3 = a_1 O^1 \oplus a_2 O^2 \oplus a_3 O^3 \oplus a_4 O^4 \oplus a_5 O^5,

where

O^i

are the irreducible representations of

O

. Hence, show how the degeneracy of the 7-fold level may be reduced by the cubic potential. Give the degeneracies of the final levels. Note that we cannot say anything here about the magnitude of any splittings (which could ``accidentally'' turn out to be zero!), or even about the ordering of the resulting levels -- that depends on the details of the potential, not just its symmetry.

Intuitive explanation

The key insight is that character theory lets you decompose any representation into irreducibles by taking inner products: the multiplicity of irrep $i$ in a reducible representation is $a_i = \frac{1}{|G|}\sum_{\text{classes}} \chi_{\text{reducible}}(C) \chi_i^*(C) n_C$ , where $n_C$ is the class size. For part (b), recognize that rotating spherical harmonics by angle $\alpha$ about $z$ picks up phase factors $e^{-im\alpha}$ for each $m$ -value; summing these phases over all $m$ from $-\ell$ to $\ell$ gives the character. For parts (c) and (d), you'll identify the five conjugacy classes of the cubic rotation group $O$ (identity, face diagonals, edges, face centers, body diagonals), build its character table using similar rotation arguments, then apply the decomposition formula to see how the seven $f$ -level states split into smaller irreducible blocks under cubic symmetry.

Intuitive explanation

Related theory

Gauge Symmetry Yang Mills Theory

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