This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Because I*I=e, we showed in L2 that the inversion symmetry of H2 leads to the result that every eigenstate of H2 is either g or u
For multielectron systems, the inversion symmetry inverts all electron coordinates simultaneously (xk,yk,zk)è (-xk,-yk,-zk) for k=1,..N But I*I takes (xk,yk,zk)è (+xk,+yk,+zk) for k=1,..N That is I*I=e the identity operator
Permutational symmetry from L2 For a two-electron system the Hamiltonian is invariant (unchanged) upon transposition of the electrons (changing both spatial and spin coordinates simultaneously) Tspace-spin H(1,2) = H(2,1) = H(1,2) But the Tspace-spin * Tspace-spin = e (identity) Thus for every eigenstate of the Hamiltonian we obtain either Ψs(1,2) = +1 Ψs(1,2) Ψa(1,2) = -1 Ψa(1,2) But our Hamiltonian does not depend on spin. Hence Ψ(1,2) = Φ(1,2)χ(1,2) Also H(1,2) is separately unchanged by transposing either just the spatial coordinates or the spin coordinates Thus either Φ(2,1) = +Φ(1,2) or Φ(2,1) = -Φ(1,2) and either χ(2,1) = +χ(1,2) or χ(2,1) = -χ(1,2)
Permutational symmetry, summary Our Hamiltonian for H2, H(1,2) =h(1) + h(2) + 1/r12 + 1/R Does not involve spin This it is invariant under 3 kinds of permutations Space only: ρ1 ⎜⎝ ρ2
Spin only: σ1 ⎜⎝ σ2 Space and spin simultaneously: (ρ1,σ1) (ρ2,σ2) Since doing any of these interchanges twice leads to the identity, we know from previous arguments that
Ψ(2,1) = ± Ψ(1,2) symmetry for transposing spin and space coord Φ(2,1) = ± Φ(1,2) symmetry for transposing space coord Χ(2,1) = ± Χ(1,2) symmetry for transposing spin coord
In this course we are concerned with the solutions of the Schrodinger equation, HΨ=EΨ, but we do not actually want to solve this equation. Instead we want to extract the maximum information about the solutions without solving it. Symmetry provides a powerful tool for doing this. Some transformation R1 is called a symmetry transformation if it has the property that R1 (HΨ)=H(R1Ψ) The set of all possible symmetries transformations of H are collected into what is called a Group.
1). Closure: If R1,R2 ε G (both are symmetry transformations) then R2 R1 is also a symmetry transformation, R2 R1 ε G 2. Identity. The do-nothing operator or identity, R1 = e ε G is clearly is a symmetry transformation
3. Associativity. If (R1R2)R3 =R1(R2R3).
4. Inverse. If R1 ε G then the inverse, (R1)-1 ε G ,where the inverse is defined as (R1)-1R1
= e.
For the case of the inversion symmetry, the group is {e, I} Since I*I = e we see that (I)-1 = I
For the case of the transposition symmetry, the group is {e, τ} Since τ * τ = e we see that (τ)-1 = I
The degenerate eigenfunctions of H form a representation
If HΨ=EΨ then H(R1Ψ)= E(R1Ψ) for all symmetry transformations of H.
Thus the transformations amount the n denegerate functions, {S=(RiΨ), where Ri Ψi ε G} lead to a set of matrices that multiply in the same way at the group operators. The Mathematicians say that these functions form a basis for a representation of G. Of course the functions in S may not all be different, so that this representation can be reduced. The mathematicians went on to show that one could derive a set of irreducible representations that give all possible symmetries for the H. reorientations from which one can construct any possible.
For the inversion and transposition groups, all representations are nondegenate, just symmetric and antisymmetric
For an atom any rotations about any axis passing through the nucleus is a symmetry transformation. This leads to the group denoted as SO(3) by the mathematicians [O(3) indicates 3 three-dimensional real space, S because the inversion is not included).
The irreducible representations of O(3) are labeled as
Consider the stereographic projection of the points on the surface of a sphere onto a plane, where positive x are circles and negative x are squares.
Start with a general point, denoted as e and follow where it goes on various symmetry operations. This make relations between the symmetry elements transparent. e.g. C2zσxz= σyz
Consider now the singly occupied Npy orbital C2z changes the sign but, σxz does not. Thus Npx transforms as b1 and The total wavefunction transforms as B1. Include the S= 1/2 , leads to The 2B1 state
Soon we will consider the triplet state of CH2 in which one of the 2s nonbonding electrons (denoted as σ to indicate symmetric with respect to the plane of the molecule) is excited to the 2px orbital (denoted as π to indicate antisymmetric with respect to the plane)
Since we know that the two CH bonds are invariant under all symmetry operations, from now on we will write the wavefunction as A{[(CHL)2(CHR)2](Cσα)1(Cπα)1}
Here σ is invariant (a1) while π transforms as b1. Since both s and p are unpaired the ground state is triplet or S=1
We will consider a system such as NH3, with three equal bond lengths. Here we will take the z axis as the symmetry axis and will have one H in the xz plane.
The other two NH bonds will be denoted as b and c.
The E symmetry (irreducible representation) is of degree 2, which means that if φpx is an eigenfunction of the Hamiltonian, the so is φpy and they are degenerate.
This set of degenerate functions would be denoted as {ex,ey} and said to belong to the E irreducible representation.
The characters in this table are used to analyze the symmetries, but we will not make use of this until much later in the course.
The character table for C3v
Thus an atom in a P state, say C(3P) at a site with C3v symmetry, would generally split into 2 levels, {3Px and 3Py} of 3E symmetry and 3Pz of 3A1 symmetry.
Consider first the effect of σxz. This leaves the NHx bond pair invariant but it interchanges the NHb and NHc bond pairs. Since the interchange two pairs of electrons the wavefunction does not change sign. Also the σLP orbital is invariant.
We will write the wavefunction for NH3 as
where we combined the 3 Valence bond wavefunctions in 3 pair functions and we denote what started as the 2s pair as σLP
Next consider the C3 symmetry operator. It does not change σLP. It moves the NHx bond pair into the NHb pair, moves the NHb pair into the NHc pair and moves the NHc pair into the NHx pair. A cyclic permutation on three electrons can be written as (135) = (13)(35). For example. φ(1)φ(3)φ(5) è φ(3)φ(5)φ(1) (say this as e1 is replaced by e3 is replaced by e5 is replace by 1) This is the same as φ(1)φ(3)φ(5) è φ(1)φ(5)φ(3) è φ(3)φ(5)φ(1) The point is that this is equivalent to two transpostions. Hence by the PP, the wavefunction will not change sign. Since C3 does this cyclic permutation on 6 electrons,eg (135)(246)= (13)(35)(24)(46), we still get no sign change.
Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, ρ = sqrt(x2+y2), α, z (axis along z) Since the wavefunction has period of 2π in α, the α dependence of any wavefunction can be expanded as a Fourier series,
φ(ρ,α,z)=f(ρ,z){a0 + Σ m=1m=∞ [am cos mα + bm sin mα]
Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, ρ = sqrt(x2+y2), α, z (axis along z) Since the wavefunction has period of 2π in α, the α dependence of any wavefunction can be expanded as a Fourier series,
φ(ρ,α,z)=f(ρ,z){a0 + Σ m=1m=∞ [am cos mα + bm sin mα]
Clearly the kinetic energy will increase with m, so that for the same f(ρ,z), we expect m=0 lowest, then m=±1, then m=±2, m=±3, etc
Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, ρ = sqrt(x2+y2), α, z (axis along z) Since the wavefunction has period of 2π in α, the α dependence of any wavefunction can be expanded as a Fourier series,
φ(ρ,α,z)=f(ρ,z){a0 + Σ m=1m=∞ [am cos mα + bm sin mα]
Clearly the kinetic energy will increase with m, so that for the same f(ρ,z), we expect m=0 lowest, then m=±1, then m=±2, m=±3, etc Also if we rotate the molecule about the z axis by some angle β, the states with the same m get recombined [cos m(α+β)] = (cos mα)(cosmβ) – (sin mα)(sin mβ) [sin m(α+β)] = (sin mα)(cosmβ) + (cos mα)(sin mβ)
O=C=O z x
Which means that the wavefunctions with the same m are degenerate
The symmetry operators are: Rz(α): counterclockwise rotation by an angle α about the z axis σxz: reflection in the xz plane (this takes +α into –α) σ’ = Rz(α) σxz Rz(-α); reflection in a plane rotated by an angle α from the xz plane (there are an infinite number of these)
The symmetry operators are: Rz(α): counterclockwise rotation by an angle α about the z axis σxz: reflection in the xz plane (this takes +α into –α) σ’ = Rz(α) σxz Rz(-α); reflection in a plane rotated by an angle α from the xz plane (there are an infinite number of these) e: einheit (unity) This group is denoted as C∞v,The character table (symmetries) are
Application to FH The ground state wavefunction of HF is
A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](αβ-βα)}
In C∞v symmetry, the bond pair is σ (m=0),while the px and py form a set of π orbitals (m=+1 and m=-1). Consider the case of up spin for both πx and πy Ψ(1,2) = A{φxαφyα}=(φxφy- φyφx) αα
Rotating by an angle γ about the z axis leads to φa = cosγ φx + sinγ φy and φb = cosγ φy - sinγ φx This leads to (φaφb- φbφa) = [(cosγ)2 +(sinγ)2] }=(φxφy- φyφx)
Continuing with FH Thus the (px)2(py)2 part of the HF wavefunction
A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](αβ-βα)}
Since both αα and ββ transform like Σ; the total wavefunction transforms as Σ
The symmetry table, demands that we also consider the symmetry with respect to reflection in the xz plane. Here px is unchanged while py changes sign. Since there are two electrons in py the wavefunction is invariant. Thus the ground state of FH has 1Σ+ symmetry
A{(πxα)(πyα)} transforms like Σ. thus we need examine only the transformations of the downspin orbital. But this transforms like π.
Another way of describing this is to note that A{(πx)2(πy)2} transforms like Σ and hence one hole in a (π)4 shell, (π)3 transforms the same way as a single electron, (π)1
Ground state has 1A1 symmetry. For optimum bonding, the pz orbital should point at the Hz while the px orbital should point at the Hx. Thus the bond angle should be 90º.
As NH2 (103.2º) and OH2 (104.5º), we expect CH2 to have bond angle of ~ 102º