Quantum Mechanics

Question 1

Classical Mechanics

Answer 1

E=1/2mv² + V = P²/2m + V
EK - due to momentum (P)
V - due to position (x)
Need to know starting position, P and V to predict the subsequent trajectory exactly.
Newtonian mechanics - any x and p (or E) is possible.

Question 2

Rotations of bodies

Answer 2

p->J/L (angular momentum)
m-> I (moment of inertia)
v -> ω (angular velocity)
F -> T (torque)
Any rotational E/p possible, if suitable torque is applied

Question 3

Vibrations of bodies

Answer 3

F=-kx
All E possible
Internal motion

Question 4

Light

Answer 4

Classical view - light is a wave (reflection, refraction etc)
Modern view - wave-particle duality
EMR-light has different colours associated with a Δλ.
Continuous, all E and λ possible.

Question 5

Blackbody radiation

Answer 5

Material that converts heat to light. Cavity with exit pin-hole for light to escape, emitted by hot body. Increase T results in increase in total output, shifting the maxima to a lower λ.
ρ = total E in region of EM field / V
- ρ is energy density

Question 6

Wein’s law

Answer 6

Assumptions about absorption/emission of EMR.
ρ=c1/λ⁵.1/e(c2/λT), adjusting c1 and c2 to get best fit.
-really good at short λ
-not so good at long λ

Question 7

Rayleigh Jean Law

Answer 7

Treat EM field as a lot of oscillators. Calculate the number with each E, predict a curve proportional to λ-⁴
Full and detailed thermodynamic treatment.
ρ=8πkT/λ⁴
-good at long λ
-terrible at short λ
Average of each oscillator is kT, regardless of frequency. Each wave mode has the same energy.
UV catastrophe - emissions due to ∞ at 0λ. High E’s even at low T (no darkness!). All bodies lose E fast - does not reflect reality.

Question 8

Planck Distribution

Answer 8

Empirical modification of Wein’s law.
ρ=c1/λ⁵.1/e(c2/λT)-1
Essentially a perfect fit at all λ.
Each mode has a discrete amount of E (QM). Addition of the -1 term, so cannot have continuous values (as previously allowed)

Question 9

Quantum Assumptions

Answer 9

Wall of blackbody radiator contains harmonic oscillator (HO), each with characteristic frequency. E of each HO quantised: E=nhv, where n=0,1,2... HOs emit and absorb E only in discrete amounts: E=hv. 
Assumptions:
c1=8πch
c2=ch/k(B)
ρ=8πch/λ⁵.1/e(ch/λKT)-1

Question 10

Photoelectric Effect

Answer 10

Light is a particle.

When light is shone on some metals, electrons are emitted. EK of e-s remains the same. No e-s emitted for v

Question 11

e- diffraction

Answer 11

e-s diffracted from metallic Ni. Gives a diffraction pattern exactly in accord with de Broglie’s λ. Cannot think of small objects in terms of trajectory (x and p) - Heisenberg uncertainty. Rather, a probability of being in any particular place, described by the wavefunction.

Question 12

Wavefunctions Ψ

Answer 12

Mathematical function describing the probability distribution of a particle (e-). Ψ is a function of position and time.
Deal with time-independent systems -> only position (x,y,z)

Question 13

Postulate 1 - Ψ

Answer 13

In 1D Ψ, Ψ(x), allows us to determine translational motion.
Ψ contains all the dynamical information about the system. Values have no physical meaning. Get information by manipulating Ψ to return real values relating to physical properties (E, x, p…).
Operator form of SWE: H^Ψ=EΨ

Question 14

Postulate 2 - Born Interpretation

Answer 14

Wavefunction has value Ψ at some point r. Probability of finding the particle in a really small volume at r∝|Ψ²|dτ
|Ψ²| gives real, positive values (relating to where something is)

Question 15

Postulate 3 - acceptable Ψ’s

Answer 15

• continuous
• single values
• continuous first derivative
• square integrate the function (not infinite)

Question 16

Operators

Answer 16

Something you do to a function to get real information out. Different operators represent different properties (E, p, x…)

Question 17

Postulate 4 - observables

Answer 17

Represented by operators (built from x or p operators).
-position operator (x^) is x multiplied by something (operator for position along x axis). Multiply Ψ by x.
-momentum operator (p^) = ℏ/i . d/dx
Satisfy commutation relation (relation between conjugate quantities - mutually exclusive). x^ and p^ in x-direction of a particle in 1D: x^1p^ = xP-Px
-communtator of reduced Planck’s constant and i: x^p^=iℏ

Question 18

Postulate 5 - Heisenberg

Answer 18

Impossible to specify simultaneously both p and x of particle. Can relate any pair of observables with operators that do not commutate.
ΔpΔx>ℏ/x
px-xp=h/2πi
If x=0, Δp=infinity - complete uncertainity in momentum
If p=0, Δx=infinity - complete uncertainity in position

Question 19

Time independent SWE

Answer 19

Interpretation of de Broglie (wave-particle duality)
-ℏ²/2m . d²Ψ/dx² + V(x)Ψ = EΨ
Gives real values
Solutions: Ψ=coskx + isinkx
-k can take any value, thus E should be able to take any value. But acceptable Ψ restricts E (quantised)

Question 20

K(B)

Answer 20

Boltzmann Constant

Question 21

H^

Answer 21

Hamiltonian Operator

Question 22

τ (tau)

Answer 22

Really small volume

dτ=dxdydz

Question 23

|Ψ²

Answer 23

Probability density

Question 24

i

Answer 24

Imaginary number = root(-1)

Question 25

ℏ

Answer 25

Reduced Planck’s constant = h/2π (Js/rad)

Question 26

Simple model systems

Answer 26

Translational (movement in 1D), rotational (motion about an axis) and vibrational (internal motion).

Question 27

Translational motion

Answer 27

Description of motion in 1D

Particle in a box and at a barrier

Question 28

Particle in a box

Answer 28

Particle of mass m can lie anywhere along a line of length L, creating a box where you can find the e- (V=∞ at walls and V=0 at base). Probability of being where V=∞ is zero, Ψ=0 at x=0 and x=L.
-ℏ²/2m . d²Ψ/dx² = EΨ as V=0
General solutions: Ψ=Ae(ikx) + Be(-ikx)
E=ℏk²/2m describes Ψ and E of a free particle.
As the particle is confined to a region, the acceptable Ψ must satisfy boundary conditions.
Ψ=(A+B)coskx + (A-B)isinkx = Csin(kx) + Dcos(kx)
x=0, Ψ(0)=0.
-D=0, thus, Ψ = Csin(kx)
x=L, Ψ(L)=Csin(kx)=0.
-if C=0, then Ψ=0 for all values of x, conflicting with the Born Interpretation (particle is somewhere)
-sin(kL)=0 and kL=πn, n=1,2,3…
-Ψ(x)=Csin kπx/L = Csin nπx/L
E=n²h²/8mL²
C=root(L/2), thus, Ψ(x)=root(2/L)sin(nπx/L) for 0

Question 29

Vibrating string

Answer 29

Classical analogy: only certain wavelengths possible as the ends are fixed.
Ψ = sin 2πx/λ, λ restricted to 2L/n, n=1,2,3…
-quantisation forced by boundary conditions

Question 30

Energy levels - translational motion

Answer 30

E levels increase as n² increases, separation increases with n.
E(n)=n²h²/8mL², n=1,2,3…

Question 31

Zero-point energy (ZPE)

Answer 31

n cannot = 0, so E cannot = 0.
n=1, E(1)=h²/8mL² = ZPE (lowest E a particle can possess).
-must have non-zero kinetic energy (Heisenberg)
-Ψ=0 at walls, but smooth, continuous and non-zero everywhere else.

Question 32

Correspondence Principle

Answer 32

The behaviour of QM systems becomes more classical in the limit of large qns.
E(n+1)-E(n)=(2n+1)h²/8mL²
E will increase with decreasing m and L (L has a greater effect due to ²).
-As m tends to infinity, E tends to 0. Thus, E level separation becomes inseparable and is no longer quantised
Implications - atoms/molecules in beakers treated as if their translational motion is not quantised, impacting spectroscopy and material properties.

Question 33

Applications - predict dye colour

Answer 33

To be coloured, dyes must absorb in the visible region of the EM spectrum. Interaction of light with e-s. Absorption of light/photons (change of quantised E). Object absorbing certain λ or v of white light, reflects complementary colour. E=hv=hc/λ
π e-s in p-orbitals (conjugated systems - alternating single and double bounds).
Terminantor atoms - denote the end of the box, disrupt the flow of e-s along molecule (N)
1) box length: 1.39Å x #bond lengths
2)#e-s (lp and π) - for multiple e-s, assume no interaction between e-s apart from Pauli Exclusion Principle. Populate using Aufbau’s rule.
3) ΔE=(2n+1)h²/8mL² (should be E-19 J or kgm²/s²). Convert to kJ/mol by multiply by N(A)/1000.
4) λ = hc/E. Absorbs this λ and reflects complementary colour
5) v = E/h

Question 34

Particle at a Barrier

Answer 34

Tunnelling - particle penetrates through the walls to classically forbidden regions. Walls thin, enough E for e- to pass through.
E=ℏ²k²/2m
kℏ²=root(2mE)
V->infinity at barrier, but if it doesn’t, E A’e(ikx).
Particle has a probability of being beyond the barrier, and within the barrier. Ψ must be continuous and dΨ/dx must be continuous entering and leaving the barrier.
Tunnelling consequence of wave character behaviour of matter (ex: radio waves pass through walls). Matter can pass through walls on the quantum level. Probability decreases with increasing thickness of walls and m of particle (e->proton).

Question 35

Applications of particle at a barrier - Scanning Tunnelling Microscopy (STM)

Answer 35

STM - dependence of e- tunnelling on thickness of region between point and surface. Needle scans across surface of conducting solid and e-s tunnelled across space.
Constant current mode - fixed distance between surface and stylus -> map
Quantum biology - does tunnelling occur during mutations of DNA? - Life and cancer research.

Question 36

c=normalising constant

Answer 36

c=root(L/2)

Question 37

Vibrational motion

Answer 37

Two models, Harmonic oscillator (HO) and Anharmonic oscillator (AO)

Question 38

Harmonic Oscillator (HO)

Answer 38

Hooke’s law: F=-kx
K is force constant. Large K = narrow potential well = stiff spring (more F required to displace spring)
V=1/2kx² - parabolic shape. No displacement means particle is at eqm. Particle tends towards infinity, no sharp boundary (unlike particle in a box)
SWE: -ℏ²/2m . (1/2x²)Ψ = EΨ
-particle trapped by infinite walls, results in quantisation and boundary conditions. Ψ and Ψ² don’t equal 0 at any point (except infinity).
-particle can be anywhere, regardless of E (even chemically forbidden zone -> tunnelling).
E levels: E=(v+1/2)ℏω, where v=0,1,2… and ω=root(k/m)
E spacings are even (ΔE=ℏω). For large objects ℏω is negligible.
ZPE: lowest E level at 1/2ℏω (E cannot be 0 - always in motion and position restricted). Walls rise to infinity (curved).
X-H: heavy atom (x) is stationary and light atom (H) moves, vibrating as a simple harmonic oscillator. Bonds act like springs.

Question 39

HO for diatomics

Answer 39

Atoms with similar masses move relative to each other.
Push atoms together (e- repel, increase E) and pull apart (until bond breaks) when vibrating with respect to r(eq).
HO: models bottom of potential well well, but not at higher E.
Parabolic V: close to r(eq) can approx V by parabola (V=1/2kx²), x=r-r(eq). k is a measure of curvature of V close to eqm extension of bond. Steep walls = stiff bond = large k.
Frequency of oscillator depends on the two masses
-homonuclear (F2): m1=m2
-heteronuclear (NaCl): m1 doesn’t equal m2
SWE: -ℏ²/2m(eff) . (1/2x²)Ψ = EΨ
E=(v+1/2)ℏω, v=0,1,2… and ω=root(k/m(eff))
Curvature of walls leads to some Ψ extending outside of the parabola (tunnelling). ΔE=ℏω, so E level spacings are even.
Tunnelling - probability of being in a forbidden region, independent of k and m.
ex: inversion of NH3. Doesn’t go over E barrier of planar form (not superimposable). Absorption ~900cm-1 (barrier ~2000cm-1), corresponding to v0->v1 (lowest vibrational state).

Question 40

Effective mass (M(eff))

Answer 40

Measure of mass moved during vibration

m(eff)=m1m2/(m1+m2)

Question 41

Reduced mass

Answer 41

Emerges from separation of relative internal and overall translational motion. Meff and u are not the same overall (except have same values for diatomics).

Question 42

Vibrational term (G~)

Answer 42

Expresses E of vibrational states in terms of wavenumber (cm-1). E=hcG~(v)
G~(v)=(v+1/2)v~
v=ω/2π and v~=ω/2πc

Question 43

IR spectroscopy

Answer 43

Frequency of E level transition (v~10E14 Hz). Use IR to look at vibrations of molecules.
Molecular fingerprint
-k proportional to bond strength
-bond strength depends on molecular environment
-selection rule: for a vibration to be observed, the dipole must change (heterogeneous, asymmetric or bent)

Question 44

Anharmonic Oscillators (AO)

Answer 44

AOs can break bonds to make new bonds (limitation of HO - based on a parabola). HO describes the small changes in bond length well (parabola), but is wrong for increased E/Ψ.
At higher vibrational excitations, movement of atoms allows molecules to explore regions of V curve where parabolic approx. is poor. F no longer proportional to x.
Curve less confined than parabola. E levels less widely spaced at high excitations (weaker bond) - Morse potential

Question 45

Morse Potential

Answer 45

V=hcD(e){1-e(-a(r-r(e)))}²
-a=root(m(eff)ω²/2hcD~(e)) - width of potential
-r - internuclear separation
-D~(o) - dissociation energy from v=0 to highest occupied v
-D~(e) - dissociation energy from r(eq) to highest occupied v
D(e)-D(o) = ZPE
Near minimum vibration of v with displacement resembles a parabola. Allows for dissociation at large displacements.
Permitted E levels: G~(v)=(v+1/2)v~-(v+1/2)²X(e)v~
-X(e) = a²ℏ/2m(eff)ω = anharmonicity constant
-G~(v) becomes smaller, reaching v(max) at G~=0.
Δv=+/-1, so ΔG~(v+1/2)=G~(v+1)-G~(v) -> converge
Dissociation limit - point where v=v(max). The separation between E levels vanishes and ΔG(v+1/2)=0.
v(max)=(v~/2v~X(e))-1
D(o) - when several vibrational transitions are detectable, can determine D(o) of bond. Sum of successive intervals from v=0 to v(max).
Birge-Sponer Plot - plot ΔG~(v+1/2) against v+1/2. X-axis intercept will give v(max). Inaccessible part of spectrum estimated by linear extrapolation. Integrate to get D(o) - often overestimated as realistically not linear.

Question 46

HO vs MO

Answer 46

HO
+gives bond strength; apply to polynuclear systems
-doesn’t give D(o) and doesn’t fit experimental data well
MO
+gives bond strengths and D(o) and fits experimental data well
-cannot be applied to polyatomic molecules

Question 47

Particle on a sphere

Answer 47

Ψ must satisfy 2 cyclic boundary conditions, resulting in 2 quantum number’s for state of angular momentum. θ and ϕ. Ψ must match over the poles (θ) and round the equator (ϕ)
SWE: -ℏ²/2m . ∇² = EΨ
-∇² (laplacian) = d²/dx²+d²/dy²+d²/dz²
-Once object is set in motion in a vacuum, no E required to keep it spinning, so V=0.
Rotational qn’s: 2 orbits -> 2 qn’s
L=angular momentum qn, L=0,1,2..
m(l) = component of orbital angular momentum about the x-axis, +L to -L.
E particle: restricted to certain values due to boundary conditions, continuous Ψ around sphere. Independent of m(l).
E=L(L+1)ℏ²/2I, where I=mr²
Classical model - centre of mass depends on m1/m2
-homogenic diatomic - m1=m2
-heterogenic diatomic - m1r1=m2r2 and r=r1+r2

Question 48

Rigid Rotor

Answer 48

Fixed radius (rotation doesn’t result in stretching of the bond). Thus, a body does not distort under the stress of rotation, due to negligible centrifugal force.
Simplification, use μ to transform problem to that of a single mass rotating around the origin.
μ=m1m2/m1+m2 x 1.66o54E-27
I depends on masses of atoms present and molecular geometries. 3 I defined about 3 perpendicular axes: I(c)>I(b)>I(a)
Linear rotor - E quantised.
E(J)=J(J+1)ℏ²/2I, J=0,1,2…
E level ladder: E expressed in terms of a rotational constant (B~)
B~=ℏ/4πcI or hcB~=ℏ/2I
E(J)=hcB~J(J+1)=nBJ(J+1)
-E level spacings get further apart with increasing J.
-B expressed as frequency = B~c
E rotational state: rotational term F~(J)
F~(J)=B~J(J+1)=E(J)/hc
Frequency: E=hb=hv

Question 49

Rotational-Vibrational Spectrum

Answer 49

Heteronuclear diatomics in the gas phase.
Rotational and vibrational motion is not discrete and combine to give special spectral information. As the molecule vibrates, the bond length changes. Rotational E changes with bond length (longer bond = slower spin).
Bond spectrum - high resolution vibrational spectrum has bands. n peaks = 2 isotopes (more abundant = bigger)
Selection rule: ΔJ=+/-1 and sometimes ΔJ=0 during a vibrational transition.
Spectral branches - each vibrational E level has a series of rotational E levels associated with it. Appearance of S~.
S~=G~(v)+F~(J) = (v+1/2)v~+B~J(J+1)
Vibrational transition from V+1

Question 50

P Branch

Answer 50

ΔJ=-1

Transition: v+1

Question 51

Q Branch

Answer 51

ΔJ=0

Transition: v+1

Question 52

R Branch

Answer 52

ΔJ=+1

Transition: v+1