introductory concepts: symmetry

http://aimpro.ncl.ac.uk MMG Skills Lecture Series

Introductory concepts:Symmetry

Jon Goss

MMG Skills Lecture Series2

Outline

Atomic orbitals to molecular orbitalsPoint group vs space groupPoint groups and diatomic MOsCorrelation tables“Little groups” and k-point samplingDirect productsDipole selection rules for optical transitionsVibrational mode selection rules


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.


Symmetry operations

What are the possible symmetry operations of a molecule?


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)


Symmetry operations

What additional operations are possible in a crystal?


Symmetry operations: Space groups

TranslationScrew (translation and rotation)Glide (translation and reflection)

We’re focusing on point groups in this lecture.


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?



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?



IdentityInversionRotation about the bond through any angleRotation by π about any axis perpendicular to the bond, passing through the mid-point of the bondReflection through any plane containing the bondRotation about the bond through any angle, followed by reflection in the plane perpendicular to the bond-axis containing the mid-point



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 symmetryWe gain more information about the wave-functions from the character table:



It’s not so hard



These are the symmetry operations



These are the “irreducible representations” (IRep)All aspects of the physical object (wave functions, normal modes etc) must be characterised by one of these



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



These are the “characters”The character under the identity operation tells you about the degeneracy of the IRep



These are the linear generating functionsE.g. anything which is linear in z corresponds to an A1u (IRep)

This gives information for dipoles (e.g. infrared-activity)



These are the quadratic generating functionsAs with the linear functions, but corresponding to quadratic functions, telling us about second order functions including polarisability (Raman)



For a wave functions of H2, we can determine the IReps by applying the symmetry operations to the function and determining the parityLook at inversion first.



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!



We’ll now look at another operation – which one might be most useful?



Let’s look at C2



Remember, Ψg=(1sa+1sb) and Ψu=(1sa-1sb)

C2 Ψg=Ψg

C2 Ψu=-Ψu

Therefore Ψg corresponds to A1g many-body IRep

Ψu corresponds to A1u many-body IRep

Hurrah! A one-electron picture:

1sa 1sb

a1g

a1u


From atom to molecule: HF

What symmetry operations are lost relative to H2?



Like all hetero-diatomic molecules, HF has C∞v symmetry

The electronic structure of HF is more complicated than that of H2 as there are more electrons involved

H 1s

F 2sF 2p

F 1s



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.



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?


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.


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.

Td C3v

EJT

The ideal MOs can beobtained in same way that those of H2 and HFwere (LCAO).

t2


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!

Td C3v

e

a1


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 geometryAt other points, the wave-vector of the electron in general acts as distortionThe 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!


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?


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 combine IReps together (i.e. what is the IRep of a two functions for which we know the IReps?)


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 ∫Ψ0vΨ1dVWe 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?



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?



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) = 0A2: 1x(1x1x1) + 2x(1x1x1) + 3x(1x-1x-1) = 6E: 1x(1x1x2) + 2x(1x1x-1) + 3x(1x-1x0) = 0



You divide each sum by the order of the group (the number of symmetry operations)

A1: 0/6=0A2: 6/6=1E: 0/6=0



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



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?



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+A2xEWe only need to see if A2xE contains A1It is easily shown that A2xE=E, so it doesn’t.A1 to A2 dipolar transitions are completely forbidden.



What about A1 to E?



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.



Are dipole forbidden transitions ever seen in reality?


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.


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…



Apply the C2 operation

Then apply σv(xz) (the plane of the molecule)

Note the symmetry of the molecule is never lowered.



Again, apply the C2 operationThen 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.


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?



What is the IRep of this mode?



It turns out that you need three varieties to form a degenerate group.The symmetry operations map them into on-another, or to linear combinations of them.These may be tricky to characterise.See if you can show that these form a t2 manifold.In IR-spectroscopy, it is this triplet of modes that are the high-frequency modes actually detected.


Local mode replica.

The final part of this final part is the idea that vibrational modes may couple to an electronic transition, or convert a dipole-forbidden electronic transition into an allowed transition.We adapt the previous selection rule by adding the local mode symmetry to the product:

∫Ψ0χ0vΨ1χ1dVWe assume (without any loss of generality) that the vibrational ground state is totally symmetric.We need the IRep product of the two electronic states, the dipole operator and the vibrational mode.


Direct products: Mode assisted electronic transitions

What about a A1 to A2 transition in C3v with coupled to a vibrational mode with A2 symmetry?We simply take the product:

A1 x A2 x (A1+E) x A2 = A2 x (A1+E) x A2 = (A2 + E)xA2 = A1+EHurrah: allowed


Final summary

From this introduction, you have seen some important ideas:

Point group symmetryThe assignment of irreducible representations to electronic and vibrational wave functionsThe correlation of IReps

Jahn-TellerLittle groups in k-space

The application of selection rules for spectroscopy

introductory concepts: symmetry

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originfrom atom

hardfrom atom

h2from atom

irepfrom atom

thesefrom atom

possible symmetry operations

symmetry operations