Quantum Mechanics

Question 1

wavefunction

Answer 1

the state of a system is fully described by a mathematical function Ψ,

Question 2

Probability interpretation of Ψ (Born Interpretation)

Answer 2

consider a particle which can move only in the x direction

this particle is described by a wavefunction Ψ(x) - wavefunction only depends on x

Question 3

assume that the wavefunction is normalized

Answer 3

the probability of finding the
particle somewhere along the x direction is 1

the wavefunction itself has no physical meaning
- it may, at any given point in space, be positive or negative, real or complex

Question 4

observable

Answer 4

a measurable property such as bond length, dipole moment, kinetic energy

every observable B is represented by an operator
- all operators can be built from the operators for position and momentum.

Question 5

operator

Answer 5

Question 6

the Hamiltonian

Answer 6

The role of the operator is to operate on a wavefunction to yield information associated with the observable that the operator represents.

Question 7

eigenvalue equation

Answer 7

Question 8

The Schrödinger equation - for a particle moving in the x direction

Answer 8

Question 9

an exact wavefunction

Answer 9

the eigenvalue E is an energy

Question 10

orthogonal wavefunctions

Answer 10

Any two non-degenerate solutions (i.e. solutions of different energy) are orthogonal

Question 11

orthonormal wavefunctions

Answer 11

Question 12

expectation values

Answer 12

The expectation value of an operator B for a wavefunction Ψ is denoted < B > and is defined as:

• dτ tells us that the integration is being performed over all space.

Question 13

when a system is described by a wavefunction Ψ, the average value of the observable B in a series of measurements is equal to the expectation value of the corresponding operator Bˆ

when Ψ is an eigenfunction of Bˆ, determination of B always yields one result, b.
When Ψ is not an eigenfunction of Bˆ, a single measurement of B yields a single outcome which is one of the eigenvalues of Bˆ, and a large number of measurements will yield an average of the eigenvalues of Bˆ

Answer 13

Question 14

the variation principle

Answer 14

for any trial wavefunction Ψtrial , the expectation value of the energy can never
be less than the true ground state energy E0

The expectation value of the energy is an average of the true energies of the system E0 , E1 , E2 …., and this can never be less than E0

Question 15

harmonic oscillator

Answer 15

harmonic oscillations occur when a system experiences a restoring force proportional to the displacement from equilibrium, e.g. pendulums, vibrating springs.

consider a one-dimensional harmonic oscillator:

k , the constant of proportionality, is called the force constant.

Question 16

solution for the harmonic oscillator
- determine value of α required for wavefunction Ψ0 to be eigenfunctions of Hamiltonion

Answer 16

we seek a wavefunction such that the total energy E is constant as it must be an eigenvalue to be an exact solution.

Question 17

solution for the harmonic oscillator
- determine the eigenvalues of H operator acting on Ψ0

Answer 17

Question 18

wavefunctions associated with harmonic oscillator

Answer 18

the energies have even spacing, ΔE = ℏω

• If the force constant k increases, ω increases and hence so does ΔE
• This will also be the case if the mass decreases

Question 19

zero-point energy

Answer 19

the eigenvalue of the ground state E0 ≠ 0.

Question 20

harmonic oscillator - applications in spectroscopy

Answer 20

potential between two atoms in a diatomic molecule can be represented schematically by the plot of V(x) against x

well is very steep for small x due to the large replusion between the nuclei

potential tends to zero at large x as the bond weakens then breaks

harmonic oscillator potential, shown by V^(2)(x) on the plot, is a good approximation to V(x) around the equilibrium (most stable) internuclear separation xe, but clearly not away from the equilibrium region

Question 21

Morse potential

Answer 21

simple harmonic oscillator is used as a first approximation, and is most accurate for the ground state energy (v = 0), the zero-point energy

Morse potential more realistic for diatomic molecules compared to simple harmonic oscillator

term containing χe in Ev equation: reduces energy from the harmonic oscillator value and becomes increasingly important as v becomes large
Thus the vibrational energy level separation is not constant as for the harmonic oscillator, but converges as v gets larger. The molecule dissociates as v →∞

Question 22

particle on a ring

Answer 22

a particle of mass, m moving at a constant velocity, v around a circle of radius, r in the xy plane

The potential energy is constant - the Hamiltonian for this motion is just the kinetic energy part:

the particle is confined to move at a fixed radius - therefore use polar coordinates.

r is a constant, any term containing a derivative of r will vanish

Question 23

the Shrodinger equation for particle on a ring - substituting the moment of inertia, I of the particle about the z axis

Answer 23

Question 24

Normalisation constant for particle on a ring

Answer 24

N^2(2π) = 1

N = 1/ √2π

Question 25

angular momentum for particle on a ring

Answer 25

The z component of the angular momentum of a particle moving at constant velocity in a circle, Lz , can have only certain values which are multiples of ℏ The angular momentum is quantized.

Question 26

Particle on a sphere – rotation in 3 dimensions

Answer 26

particle of mass, m which is free to move anywhere on the surface of a sphere, at a fixed distance, r from an origin the potential energy is constant and can be neglected

Question 27

the Shrodinger equation for particle on a sphere

Answer 27

Question 28

solutions for particle on a sphere

Answer 28

energy is independent of ml and hence for every value of l there are (2l+1) states with the same energy (degenerate)

Question 29

the first nine spherical harmonics - ignoring their normalisation constants

Answer 29

Question 30

degenerate wavefunctions

Answer 30

any linear combination of them will have the exact same energy, and be a solution to the corresponding Schrödinger equation.

Question 31

rigid rotor – the rotation of diatomic molecules

Answer 31

consider a diatomic molecule AB - The atoms rotate about the centre of mass - the motion of the masses about the centre of mass is mathematically equivalent to the rotation of a single particle of reduced mass about a fixed point distance between the fixed point and the particle is equal to the bond length r - treat the rotation of a diatomic molecule as if it were the motion of a particle of mass μ on the surface of a sphere

Question 32

energies of the wavefunction for particles on a sphere

Answer 32

In rotational spectroscopy we normally use J as the symbol for the quantum number

Question 33

the hydrogen atom

Answer 33

consider general case of hydrogenic atoms - so solutions can be extended to He+ and Li2+ Coulomb’s law: gives the potential energy of the electron in the electrostatic field generated by the nucleus

Question 34

atomic units

Answer 34

QM generally uses atomic units in SI units a0 = 5.29177208x10-11 m.

Question 35

The Hamiltonian for H atom

Answer 35

the potential V is independent of the angles θ and φ the part of the Hamiltonian that depends on these angles will be the same as for the rigid rotor. The kinetic energy operator only differs in that r can now vary (r is no longer constant)

Question 36

The Schrodinger equation for H atom

Answer 36

expression for the energy is in atomic units - atomic unit of energy is also called a hartree, (1 Eh = 4.3597438 x 10-18 J = 27.211383 eV). the energies depend only on the quantum number n - This is true only for the H atom, or any other atomic ion with only one electron

Question 37

The Hamiltonian for the He atom

Answer 37

assume that the repulsion between the two electrons can be mathematically ignored and somehow absorbed into an effective potential (ignore 1/r12 term) are thus assuming that the electrons move independently of one another - assign each electron its own hydrogenic wavefunction

Question 38

wavefunction for He

Answer 38

Question 39

electron spin

Answer 39

Question 40

Pauli exclusion principle

Answer 40

No two electrons in an atom may have the same set of the four quantum numbers n, l, ml and ms consider the effect on Ψ(1,2) of interchanging the two electrons, i.e. Ψ(1,2) → Ψ(2,1) electrons are indistinguishable and hence this process cannot affect the physical properties of the system the probability distribution Ψ*Ψ must remain unchanged. - For this to be true: either Ψ(1,2) = Ψ(2,1) or Ψ(1,2) = -Ψ(2,1) the second condition is the correct restriction for electrons - therefore the more fundamental form of the Pauli exclusion principle: the total wavefunction of a system must change sign when any two electrons are interchanged - the total wavefunction must be antisymmetric

Question 41

the ground state of the He atoms

Answer 41

α - spin up β - spin down electrons are indistinguishable - cannot say for sure that electron 1 has α spin and electron 2 has β spin (or vice versa) when electrons have opposite spins, there must be equal probabilities of α(1)β(2) and α(2)β(1) - which we can ensure by taking linear combinations of these spin arrangements combine spin wavefunctions with spatial wavefunction for He atom The spatial wavefunction is symmetric with respect to the interchange of the two electrons, and hence the spin wavefunction must be antisymmetric. Only the fourth spin function is antisymmetric fourth wavefunction satisfies Pauli exclusion principle

Question 42

Slater's determinant - the ground state of the He atoms - written in determinantal form

Answer 42

Each term in the determinant has a hydrogenic spatial orbital multiplied by a spin function, and is known as a spin-orbital

Question 43

excited states of the He atom

Answer 43

He electron configuration: 1s2 Excited states of He: promote electron from the 1s orbital to the 2s to yield the configuration 1s12s1

Question 44

energy of singlet and triplet states of He atom

Answer 44

there is a difference in energy between the singlet and triplet states of 2K where the exchange integral, K > 0, and hence the triplet state is the more stable

Question 45

Born-Oppenheimer approximation

Answer 45

the total wavefunction for a molecule can be approximated as: Ψmol = Ψnuc Ψelec where Ψelec is determined from an electronic Hamiltonian (in atomic units) i and j refer to the electrons, k and l to the nuclei Thus, r_ik is the distance from electron i to nucleus k with charge Z_k R_kl is the fixed internuclear distance.

Question 46

Linear Combination of Atomic Orbitals (LCAO)

Answer 46

Question 47

Linear Combinations of Molecular Orbitals

Answer 47

same approach for LCAO is taken for molecular wavefunctions. the electronic wavefunction for a molecule with n electrons is written as the product of one-electron wavefunctions: this is not actually a product wavefunction, but the leading diagonal of a Slater determinant there will be as many molecular orbitals Ψi as there are atomic orbitals φk in the basis set.

Question 48

secular equations

Answer 48

consider a diatomic molecule with - atomic orbital φ1 on atom 1 - atomic orbital φ2 on atom 2 separated by a fixed distance R Consider also a trial wavefunction for the molecular orbitals (MOs): Ψtrial = c1φ1 + c2φ2 HOW DO WE DETERMINE c1 AND c2

Question 49

secular equations in matrix form

Answer 49

secular equations have non-trivial solutions only when the secular determinant equals zero:

Question 50

Two-orbital systems: Identical orbitals and zero overlap

Answer 50

φ1 and φ2 are identical - the combination of any two identical atomic orbitals on adjacent atoms of the same type (e.g. in homonuclear diatomics or the π orbitals of ethene, where φ1 and φ2 are each 2pz orbitals on the C atoms) φ1 and φ2 are assumed real - we can therefore replace the Coulomb resonance integrals with α and β respectively

Question 51

molecular orbital diagram formed from two identical atomic orbitals when we assume that there is no overlap i.e. S_12 = 0

Answer 51

the bonding orbital is stabilised by as much as the antibonding orbital is destabilised. Coulomb integral, α - the energy an electron would have in a molecule if it occupied either AO φ1 or φ2 - more negative then the energy electron would have if it occupied φ1 or φ2 in the isolated atoms - due to Coulomb attraction between electron and second nucleus Resonance Integral, - energy of interaction between φ1 and φ2 in the molecule - normally negative - typically about five times smaller than α for adjacent, identical orbitals

Question 52

normalise coefficients

Answer 52

Question 53

Two-orbital systems: Identical orbitals and non-zero overlap

Answer 53

anti-bonding orbital is destabilised by more than the bonding orbital is stabilised He2 is not stable - two electrons in both the bonding and anti-bonding orbital - giving an energy of 2E+ + 2E- which is less stable than the isolated atoms at all separations.

Question 54

Two-orbital systems: Different orbitals and zero overlap

Answer 54

in heteronuclear diatomics, φ1 and φ2 are not the same

Question 55

Two-orbital systems: Different orbitals and zero overlap - when φ1 and φ2 have very different energies

Answer 55

α1 and α2 will be very different from one another and δ^2 >> β^2 atomic orbitals will mix together strongly to form molecular orbitals only if Iα1 - α2I ≈ IβI If two atomic orbitals have very different energies (and hence α1 and α2 are very different) then they will not mix together strongly.

Question 56

Huckel Theory

Answer 56

In conjugated organic molecules such as ethene or benzene, the π orbitals usually have higher (i.e. less negative) energies than the σ orbitals - are the most important in chemical reactions and spectroscopy. Huckel theory: an approach to the electronic structure of conjugated molecules - ignores the σ orbitals (only using them to define the molecular geometry) - determines the π MOs for systems with n conjugated carbon atoms

Question 57

general approach to Huckel theory

Answer 57

Question 58

approximations in Huckel theory

Answer 58

Question 59

Huckel theory - secular equations written in matrix notation

Answer 59

Question 60

Huckel theory - setting up the matrix

Answer 60

α - E along the leading diagonal β along the off-diagonals when atom j is next to atom k zero along the off-diagonals when atom j is not next to atom k

Question 61

Huckel theory - coefficients are normalised

Answer 61

Question 62

Huckel theory - total energy of molecule

Answer 62

There are n electrons for an uncharged molecule with n C atoms contributing to the π system and these can now be put in pairs into the MOs. ni (the occupation number of the ith MO) takes the values 0, 1 or 2 (it is the number of electrons)

Question 63

Huckel theory example - the allyl radical

Answer 63

there are three 2p orbitals, one on each carbon atom, that can contribute to the π MOs - π MOs are perpendicular to the molecular plane there are three atomic orbitals therefore, there must be three π MOs secular determinant must equal zero for non-trivial solutions

Question 64

Huckel theory example - the allyl radical: for each E value, the MO coefficients are found by substituting E back into the secular equations

Answer 64

Question 65

Huckel theory example - the allyl radical: MO diagram

Answer 65

Question 66

electron population on atom k (the π electron density)

Answer 66

since each atom k has a residual nuclear charge of +1. These sort of charge data are useful for determining the most likely position in a molecule for attack by an electrophile or nucleophile

Question 67

π bond order between atoms j and k

Answer 67

Answer 68