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Introductory concepts: Symmetry

Introductory concepts: Symmetry. Jon Goss. Outline. Atomic orbitals to molecular orbitals Point group vs space group Point groups and diatomic MOs Correlation tables “Little groups” and k -point sampling Direct products Dipole selection rules for optical transitions

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Introductory concepts: Symmetry

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  1. Introductory concepts:Symmetry Jon Goss MMG Skills Lecture Series

  2. Outline • Atomic orbitals to molecular orbitals • Point group vs space group • Point groups and diatomic MOs • Correlation tables • “Little groups” and k-point sampling • Direct products • Dipole selection rules for optical transitions • Vibrational mode selection rules MMG Skills Lecture Series

  3. Introduction • From concepts of atomic orbital theory, we already have some understanding of the s, p, d,… • These orbitals are all spherically symmetric • E.g. s, or (px, py, pz) • For collections of atoms, to a first order approximation we can construct molecular orbitals which are linear combinations of atomic orbitals (LCAO approximation). • This approach can be very informative. • It is useful to be able to determine and even predict the overall symmetry of the MOs. MMG Skills Lecture Series

  4. Symmetry operations • What are the possible symmetry operations of a molecule? MMG Skills Lecture Series

  5. Symmetry operations: Point groups • Reflection (σh,σv, σd) • Rotation (Cn) • Inversion (i) • Improper rotation, which are combinations of rotation and perpendicular reflection (Sn) • Also there is an identity operation (E) MMG Skills Lecture Series

  6. Symmetry operations • What additional operations are possible in a crystal? MMG Skills Lecture Series

  7. Symmetry operations: Space groups • Translation • Screw (translation and rotation) • Glide (translation and reflection) • We’re focusing on point groups in this lecture. MMG Skills Lecture Series

  8. From atom to molecule: H2 • As the most simple example, we’ll look at H2: • When the two H atoms are separated sufficiently far that we can treat them as atoms, the electrons on each can be considered as a spherically symmetric 1s state. • As they move toward each other to form a bond, the two electrons can be modelled as forming linear combinations: • Ψg=(1sa+1sb) and Ψu=(1sa-1sb) • Which is the lower in energy, and why? MMG Skills Lecture Series

  9. From atom to molecule: H2 • To move on, we need to have a nomenclature for the symmetry of the molecule and those of the wave functions. • First, what are the symmetry operations of H2? MMG Skills Lecture Series

  10. From atom to molecule: H2 • Identity • Inversion • Rotation about the bond through any angle • Rotation by π about any axis perpendicular to the bond, passing through the mid-point of the bond • Reflection through any plane containing the bond • Rotation about the bond through any angle, followed by reflection in the plane perpendicular to the bond-axis containing the mid-point MMG Skills Lecture Series

  11. From atom to molecule: H2 • It turns out (through consultation with a good symmetry book, or http://www.staff.ncl.ac.uk/j.p.goss/symmetry/) that the point group is “D∞h” • All homo-diatomic molecules (e.g. O2 and N2) have this symmetry • We gain more information about the wave-functions from the character table: MMG Skills Lecture Series

  12. From atom to molecule: H2 Yikes! MMG Skills Lecture Series

  13. From atom to molecule: H2 • It’s not so hard MMG Skills Lecture Series

  14. From atom to molecule: H2 • These are the symmetry operations MMG Skills Lecture Series

  15. From atom to molecule: H2 • These are the “irreducible representations” (IRep) • All aspects of the physical object (wave functions, normal modes etc) must be characterised by one of these MMG Skills Lecture Series

  16. From atom to molecule: H2 • These are the “characters” • These are the traces of the representative transformation matrices, but we often use the values without explicit use of their origin MMG Skills Lecture Series

  17. From atom to molecule: H2 • These are the “characters” • The character under the identity operation tells you about the degeneracy of the IRep MMG Skills Lecture Series

  18. From atom to molecule: H2 • These are the linear generating functions • E.g. anything which is linear in z corresponds to an A1u(IRep) • This gives information for dipoles (e.g. infrared-activity) MMG Skills Lecture Series

  19. From atom to molecule: H2 • These are the quadratic generating functions • As with the linear functions, but corresponding to quadratic functions, telling us about second order functions including polarisability (Raman) MMG Skills Lecture Series

  20. From atom to molecule: H2 • For a wave functions of H2, we can determine the IReps by applying the symmetry operations to the function and determining the parity • Look at inversion first. MMG Skills Lecture Series

  21. From atom to molecule: H2 • Remember, Ψg=(1sa+1sb) and Ψu=(1sa-1sb) • i Ψg=Ψg • i Ψu=-Ψu • Since both functions are non-degenerate, the IReps of a and b must be Ag and Au, respectively. • However, we are yet to be precise! MMG Skills Lecture Series

  22. From atom to molecule: H2 • We’ll now look at another operation – which one might be most useful? MMG Skills Lecture Series

  23. From atom to molecule: H2 • Let’s look at C2 MMG Skills Lecture Series

  24. a1u 1sa 1sb a1g From atom to molecule: H2 • Remember, Ψg=(1sa+1sb) and Ψu=(1sa-1sb) • C2Ψg=Ψg • C2 Ψu=-Ψu • Therefore • Ψg corresponds to A1g many-body IRep • Ψu corresponds to A1u many-body IRep • Hurrah! • A one-electron picture: MMG Skills Lecture Series

  25. From atom to molecule: HF • What symmetry operations are lost relative to H2? MMG Skills Lecture Series

  26. From atom to molecule: HF • Like all hetero-diatomic molecules, HF has C∞vsymmetry • The electronic structure of HF is more complicated than that of H2 as there are more electrons involved ? H 1s F 2p F 2s F 1s MMG Skills Lecture Series

  27. From atom to molecule: HF • We now have a more complicated problem as the atomic orbitals we start with include degeneracies. • How does the loss of spherical symmetry in HF affect the 2p orbitals? (Choose the HF axis along z and consider px, py and pz.) • This is an elementary example of a crystal field splitting. MMG Skills Lecture Series

  28. From atom to molecule: HF • Let us assume that the molecule is ionic, H+F-. • The wave functions in order of increasing energy are • F(1s) • F(2s) (+ a little H(1s)) • F(2pz)+H(1s) • F(2px) • F(2py) • What are their IReps? MMG Skills Lecture Series

  29. Correlation • If you look carefully at the character tables of the H2 and HF molecule examples, you’ll see that the latter is a subset of the former. • The C∞v group is a sub-group of D∞h. • The IReps of the sub-group are all correlated with IReps in the main-group. • For example, the A1g IRep in D∞h is correlated with A1 in C∞v. • This is a very useful relationship to know about. MMG Skills Lecture Series

  30. Correlation: Jahn-Teller • For systems with orbitally degenerate many-body states, there is the potential for a reduction in the total energy by distorting the structure that removes the degeneracy. • This is the Jahn-Teller effect, and this occurs in molecules, solids and importantly for us, in point defects. • The simplest model for the J-T effect can be understood from the diagram, representing a positively charged vacancy in Si. t2 EJT Td C3v The ideal MOs can be obtained in same way that those of H2 and HF were (LCAO). MMG Skills Lecture Series

  31. Correlation: Jahn-Teller • The correlation of IReps tells us exactly what the IReps in the distorted case will be, but not their order. • There is no need to go through a derivation for the IReps, as they are completely specified! e a1 Td C3v MMG Skills Lecture Series

  32. Correlation: Little groups • Correlation also serves us in the splitting of bands in the Brillouin-zone for non-zero k. • The wave-functions at the Γ-point reflect the symmetry of the atomic geometry • At other points, the wave-vector of the electron in general acts as distortion • The symmetry of the wave-functions for a general k-point must be a sub-group of that at the Γ-point. • Therefore the splitting of degenerate band along a high-symmetry branch in reciprocal space (such as those at the valence band top of a cubic material such as diamond, silicon, GaAs,…) can be qualitatively predicted purely on symmetry grounds. • For example, along the <111> branch of a cubic material, the little groups are trigonal: triply degenerate bands are split into e and a. • Looking at such features may help you spot problems in calculations! MMG Skills Lecture Series

  33. Correlation: Little groups In the diamond band-structure along Γ-X, what do you expect to happen to the four valence bands which are a and t at the zone centre? Hint: what is happening along y and z? MMG Skills Lecture Series

  34. Direct products • In the final part of the lecture, we’ll look at another use of the IReps: determining which electronic and vibrational transitions are optically active. • To do this we need to know how to combineIReps together (i.e. what is the IRep of a two functions for which we know the IReps?) MMG Skills Lecture Series

  35. Direct products: Electronic transitions • The probability for a transition between electronic states Ψ0 and Ψ1 coupled by an electric dipole (photon) with electric field pointing along a given direction v is related to ∫Ψ0vΨ1dV • We have already seen how to determine the IReps of the wave functions, and actually, we’ve also seen how to get the IRep for the electric dipole field (the linear generating function). • It can be shown that the integral will be exactly zero if the IRep of the product is other than even parity for all symmetry operations: generally A, A1, Ag or A’. • This can be qualitatively understood by an extension of the idea that the integral between symmetric limits of an odd function is always zero. • So, how do we obtain the IRep of the product? MMG Skills Lecture Series

  36. Direct products: Electronic transitions • You’ll be happy to learn that there is a simple method to determine the products simply from the character tables. • Let’s take the example of C3v point group. • What is the direct product of A1 and A2? MMG Skills Lecture Series

  37. Direct products: Electronic transitions • You start by calculating the sum over all operations of the products of the characters with each line in turn: • A1: 1x(1x1x1) + 2x(1x1x1) + 3x(1x-1x1) = 0 • A2: 1x(1x1x1) + 2x(1x1x1) + 3x(1x-1x-1) = 6 • E: 1x(1x1x2) + 2x(1x1x-1) + 3x(1x-1x0) = 0 MMG Skills Lecture Series

  38. Direct products: Electronic transitions • You divide each sum by the order of the group (the number of symmetry operations) • A1: 0/6=0 • A2: 6/6=1 • E: 0/6=0 MMG Skills Lecture Series

  39. Direct products: Electronic transitions • The product A1 x A2 contains each IRep this many times! • A1 x A2 = 0 x A1 + 1 x A2 + 0 x E • It’s that easy  • In fact is always true that A1 x ΓX = ΓX. • Now try E x E MMG Skills Lecture Series

  40. Direct products: Electronic transitions • We now have to include all three terms, Ψ0,v, and Ψ1. • There are more terms, but the method is the same. • Is an electric dipole transition allowed between two states with A1 and A2 symmetry? MMG Skills Lecture Series

  41. Direct products: Electronic transitions • We already know that A1xA2 is A2, and we can see that the electric dipole will transform (in general) as (A1+E). • We need to see if (A2 x(A1+E)) contains A1. • The normal distributive laws apply, and the products commute: • A2(A1+E)=A2xA1+A2xE=A2+A2xE • We only need to see if A2xE contains A1 • It is easily shown that A2xE=E, so it doesn’t. • A1 to A2 dipolar transitions are completely forbidden. MMG Skills Lecture Series

  42. Direct products: Electronic transitions • What about A1 to E? MMG Skills Lecture Series

  43. Direct products: Electronic transitions • The product of interest is • (A1 x E x (A1+E) ) = (E x A1 + E x E) = E + (A1+A2+E) • Dipole allowed! • Note, that if we had polarized light along z so that the dipole only transforms as A1, the transition would not occur – only light with electric field amplitude in the x-y polarisation couples to A1-E transitions. MMG Skills Lecture Series

  44. Direct products: Electronic transitions • Are dipole forbidden transitions ever seen in reality? MMG Skills Lecture Series

  45. Infrared and Raman modes • The final section is on vibrational mode characterisation. • Vibrational modes are IR-active or Raman active depending upon symmetry. • Formally, the IR-active mode selection rule is the same as that of the dipole transitions, but now we’re talking about vibrational wave functions, not electronic ones. • Just like electronic problems, the characterisation of which modes can be seen experimentally is dependent (at least in part) upon the assignment of IReps to the modes of vibration. MMG Skills Lecture Series

  46. Infrared and Raman modes: H2O • Let us look at the example of water: • Each O-H bond can be viewed as an oscillaor. • There are two possible combinations (as with the two 1s electrons in H2): in-phase and anti-phase. • We assign the point group first: in the interests of brevity, I’ll tell you that it’s C2v • We now apply the operations to the displacement vectors… MMG Skills Lecture Series

  47. Infrared and Raman modes: H2O • Apply the C2 operation • Then apply σv(xz) (the plane of the molecule) • Note the symmetry of the molecule is never lowered. MMG Skills Lecture Series

  48. Infrared and Raman modes: H2O • Again, apply the C2 operation • Then apply σv(xz) (the plane of the molecule) • Note, in general the symmetry of the molecule is less than C2v during the anti-symmetric stretch. MMG Skills Lecture Series

  49. Infrared and Raman modes: CH4 • The breathing mode is very simple as the symmetry of the molecule is Td at all times. • A1 symmetry • Is this IR-active? • Raman active? MMG Skills Lecture Series

  50. Infrared and Raman modes: CH4 • What is the IRep of this mode? MMG Skills Lecture Series

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