BIOC331 In Class

1/23 - Quantum Mechanics Introduction

• QM can be defined as a consequence of the duality of light & matter

  • Light can behave like a wave and a particle

  • Matter is both a particle and a wave

• Waves

  • Oscillates in space and time

  • Properties:

    • peak-to-peak distance AKA wavelength (λ)(meters)

    • may travel at a specific speed (m/s)

      • all light waves travel at the speed of light in a vacuum (C = 2.998×10^8 m/s)

    • frequency (ν)(cycles/s, s^-1, Hz)

      • how many peaks pass by a point per second

    • λν = C

      • λ and v are inversely related

    • Wave number (𝜈̃)

      • v/C

Red light has a wavelength of about 650nm. What is the frequency of red light?

• λν = C → ν =C/λ

• ν = 2.998×10^8 m/s / 6.5×10^-7 m = 4.6×10^14 s^-1

• Light as a particle

  • Photoelectric effect

• Ejects e- off metal surface using energy from light

• Observed:

  • When light below a certain threshold frequency was shone on the metal, no electrons were ejected, no matter the intensity/brightness of the light (Brighter light = higher amplitude)

  • Once light has a frequency above the threshold, e- are ejected

    • Increasing the frequency further will eject e- with higher kinetic energy → move faster

    • Increasing the amplitude/brightness/intensity, more e- are ejected but with the same kinetic energy as before

  • Waves → if amplitude increases, energy must increase

  • Particles → if frequency increases, energy increases per particle

    • If brightness increases, number of particles increases

• KE= 1/2mv²

• y=mv+b

• Slope of the line is a constant, h / Planck’s constant = 6.626×10^-34 J*s

• 1/2mv²=hv → energy per photon depends on the frequency, leftover energy goes into increasing the KE of the e-

• λν = C

• Ephoton= hv

• Ephoton = hC/λ

A photon produced by an X-ray machine has an energy of 4.70×10^-16J. What is the frequency of the photon?

• Ephoton= hv

• 4.70×10^-16 J= hv

• v = 7.09×10^17 s-1

Would the frequency of violet light be higher or lower than the previous frequency?

• Lower

Would the wavelength of a radiowave be longer or shorter than the previous wavelength?

• Longer

• Light as a particle and a wave

  • KE=1/2mv²

  • λdB = h/m*v= h/P, where P = momentum

Estimate the de Broglie wavelength of a baseball thrown by an average Red Sox pitcher.

• λdB = h/m*v

• λdB = 6.626×10^-34 J*s/(0.1kg)(10m/s) = 6.626×10^-34 m

  • Where the baseball is 0.1kg and travels at 10m/s

• Can be ignored, because the value is so much smaller than any length scale of matter that is relevant

Estimate the de Broglie wavelength of an electron traveling at 0.1% the speed of light.

• λdB = h/m*v

• λdB = 6.626×10^-34 J*s/(9.11×10^-31 kg)(2.998×10^5m/s) = 2.44×10^-9 m / 24Å

  • Cannot be ignored, because it is spread out over molecular length scales

1/27 - Quantum Mechanics p2

The minimum frequency of light that can cause the photoelectric effect in cesium is 4.60×10^14 Hz. If light with a wavelength of 4.40×10^-7 strikes a cesium surface, calculate the dB wavelength of the emitted photoelectrons.

KE=Ephoton-minimum frequency

1/2mv²=h(C)/λ - hv0

1/2(9.11×10^-3kg)v² = (6.626×10^-34 Js)(2.998×10^8 m/s)/4.40×10^-7) - (6.626×10^-34Js)(4.60×10^14)

v=567,000 m/s

λdB= h/mv

λdB= (6.626×10^-34 Js)/(9.11×10-31)(567,000 m/s)

λdB= 1.28nm or 1.28×10^-9 m

Quantum Mechanics

• λdB for an e- was big enough to be significant over molecular length scales → cannot be ignored like it can for larger length scales

• e- and other small mass must be treated as waves

• Quantum mechanics is a branch of physics that focuses on accounting for the wave-like nature of matter

• Quantized → can only exist in discrete quantities, cannot take on all values, stairs/ladder

• Continuous → can exist in any quantity, can take on any value, ramp/elevator

Why does treating matter like a wave lead to quantized states?

• Waves, when constrained, can only take on certain discrete states

Heisenberg Uncertainty Principle

• A quantum-mechanical object does not simultaneously have a well-defined position (as it does when acting like a particle) and a well-defined momentum (as it does when acting like a wave)

• σx σp ≥ h/4π

  • σx = uncertainty in position

  • σp = uncertainty in momentum

Schrodinger Equation

• A QM object is described by a wave, represented by a function called a wave function

• ψ aka “psi” gives you the amplitude of an object as a function of space

• |ψ|² gives you the probability that, if you were to measure the particle’s location, you would find it at a given location

How do we get ψ for a particular object?

Solve Schrodinger Equation

→ for a one dimension, time-independent Schrodinger Equation:

• -ℏ²/2m d²/dx² ψ(x) + V(x)ψ(x) = Eψ(x)

  • ℏ = h/2π

  • m = mass of object

  • d²/dx² = second derivative

  • ψ(x) = wave function

  • V(x) = potential energy of the system: mathematically describes the constraints and environment of the object

    • high V(x) value = unstable, unhappy

    • low V(x) value = stable, happy

  • E = energy

When we start to solve, we only know m and V(x)

• plug those in to start

• Goal: solve for ψ(x) and associated Es (energies)

  • 2 unknowns, one equation → differential equation, infinite # of discrete solutions

• Shorthand notation for Schrödinger

  • Ĥ ψ(x) = E ψ(x)

    • Ĥ aka “Hamiltonian Operator” = -ℏ²/2m d²/dx² + V(x)

    • Operator is a set of symbols that acts on what comes afterwards (only to the right)

      • Ex: Â = d/dx

        f(x) = 5x

        What is Âf(x)?

        = d/dx(5x) = 5

      • Ex: Â = d²/dx²

        f(x) = e^5x

        What is Âf(x)?

        = d²/dx²(e^5x) = 25e^5x

      • Ex: Â = d/dx+5x

        f(x) = 2x²

        What is Âf(x)?

        = d/dx+5x(2x²) = 5x(2x²) = 4x + 10x³

  • Goal: Find ψ(x) such that when operated on by Ĥ, we get back ψ*E

    • Ĥ ψ(x) = E ψ(x)

      • ψ(x) = eigenfunctions

      • E = eigenvalue

      • Ĥ = operator

        • d/dx(e^3x) = 3(e^3x) → 3 is eigenvalue and (e^3x) is eigenfunction

1/29 Schrodinger Equation + PIB

Eigenfunctions of d/dx

d/dx (e^3x) = 3(e^3x)

3 = eigenvalue

e^3x = eigenfunction

d/dx = operator

Generic equation: d/dx e^kx = k(e^kx)

Schrödinger equation

• You can only solve Schrödinger exactly for a handful of V(x) values → only so many environments

• Can:

  • Solve an “exact” model exactly (with exact math) ← hard

  • Solve an approximate model exactly

  • Solve an “exact” model approximately

  • Solve an approximate model approximately

Particle in a (1-dimensional) box

• aka particle confined to a line segment

  FIGURE 3

• What is V(x)?

  • Cannot exist outside of box

    • V(x) outside the box = ∞ → no amount of energy will allow the particle to escape

  • Only exists inside the box → no preference for location, consistent energy in all areas of the box

    • V(x) = 0 → can be any real value but 0 is convenient

  FIGURE 4

General Solution:

d/dx sin k x = k cos k x

d/dx(k cos k)=k² sin k x

let k² = 2mE/ħ²

• General solution: A sin k x + B cos k x

• Verify: d²/dx² [Asinkx+Bcoskx]

  • = d/dx[kAcoskx - kBsinkx]

  • =-k²Asinkx-k²Bcoskx]

  • =-k²[Asinkx+Bcoskx]

Impose Boundary Conditions

Wave function has to become zero at points on box x=0 and x=L

Conditions:

• ψ(x) =. Asinkx+Bcoskx

• ψ(0) = 0

  • ψ(0)= Asink(0) + Bcosk(0) = 0

  • B = 0 → removes cos term

• ψ(L) = 0

  • ψ(0)= Asink(L) = 0

  • kL = 0, π , 2π, 3π → nπ, where n=integer

    • n=quantum number

    • kL=nπ → k = nπ/L

    • previous k value → sqrt(2mE/ħ) = nπ/L

      • E = n²π²ħ²/2mL²

PIB Solutions:

• ψ(n) = A sin (nπx/L), En = n²π²ħ²/2mL² ← general solutions, A sin (nπx/L) is Asin(kx) substituted

• n=1, A sin (πx/L), En = π²ħ²/2mL²

FIGURE 5

FIGURE 6

1/30 - PIB solution + spectroscopy model

FIG 7

Recall that ψ² tells you the probability of finding a particle at x. Since particles have to be found somewhere, the total sum of probabilities must =1

Normalization condition : ∫|ψ(x) |² dx =1

for PIB:

∫[0,L] A²sin² (nπx/L) dx = 1

• A² ∫[0,L] sin² (nπx/L) dx = 1

• cos2x=1-2sin²x → sin²x = ½ - cos2x/2

• A² ∫[0,L] [1/2-1/2(cos(2nπx/L)))]

• A² [1/2x - L/4nπ sin(2nπx/L)] ∫[0,L]L = 1

• A² [L/2 - 0]-[0-0]]

• A² [L/2] = 1

• A=sqrt (2/L)

Eigenstate → physical interpretation

Eigenfunction → mathematical representation

Draw the n=1 eigenstate for a quantum mechanical particle confined to move along a line segment

FIG 8

• Where are you most likely to find the particle? How do you know?

  • L/2, the highest value on the graph

• How much energy would I need to add to the system to move the particle into the n=2 state?

  • FIG 9

  • h²/8mL²

• Now suppose that the particle is in the n=2 eigenstate. At what points are you most likely to find the particle?

  • L/4, 3L/4

• What is the probability of finding the particle right in the middle of the box now?

  • The middle is a node, so = 0

• Does the particle have a well-defined postion? Does it have a well-defined energy?

  • The particle does not have a well-defined position. It does have a well-defined energy

• What does it mean to say that a wave function is normalized? Why must a wave function be normalized?

  • Probability of finding particle needs to = 1.

What would happen to the energy spacings for the particle in a box eigenstates if:

• The box were made wider?

  • The energy spacing would be smaller → L is in denominator, if L is big enough, the energy spacing become continuous → less constraint

• The mass of the particle were increased?

  • The energy spacing would be smaller → mass is in denominator

Expectation (or average) value

FIG 10

All PIB article states are liquid serum

Expectation value (for what when) <x> average position

By symmetry , <x>, /s

1,2,3,4,5,6

= 1/6(1)+2\1/6(2) + (1/6)3…..

Weighted die (50% of a 1, 10% of each 2-6

• 1/2×1 + 1/10×2 + 1/10 × 3….

Weighted average

sum of values i Pi (i)

<x>= F |psi(x)|² ] ∫ ψ(x)*x*ψ(x) dx

∫0 to L xsin² (n(pi)x/L) dx

Calculate the expectation value of the position <x> and momentum <p> for the n=1 eigenstate of a PIB: Do your answers make sense? Would <x> and <p> be for any other eigenstate of PIB?

<x> = F |psi(x)|² ] ∫ ψ(x)*x*ψ(x) dx

<p> = 0 for all PIB eigenstates

• F psi(x)*phat*psi(x)dx = F sqrt(2/L sin (n(pi)x/L) (-ihbar*d/dx) sqrt(2/L(npix)) = 0

PIB as a (coarse) model for real things

1. e- in conjugated pi systems (ie benzene)

2. E- within metals (conduction) (not discussed in this class)

In conjugated pi systems

• delocalized/spread out (like a wave) across molecule

• Why is the model coarse/approximate?

  • Ignores nuclear attraction

  • e- can actually leave the molecule IRL

  • Ignores all the other e-

Delta E for n=3 orbital = (3² - 2²)*h²/8mL²

delta E = h * frequency

n=1 ground state

n=2 HOMO

n=3 LUMO

2/3 - Spectroscopy + Wave Function

• As a model for spectroscopy

• HOMO n=2

• LUMO n=3

• IMAGE 12

• Energy = (3² - 2²) h²/8mL²

• Less constrained = more classical E/ wavelength

Wave Functions

• The four time-independent postulates:

  1. All information about a quantum mechanical system is contained in the wave function

     • non-deterministic behavior → particle movement is inherently unpredictable (also don’t know position for position/velocity)

       

  2. To every observable in classical mechanics, there corresponds a linear operator in quantum mechanics

     • Observable → a property or characteristic

       • 

     • Linear if:

       • Satisfies the distributive property aka: Â(f1(x)+f2(x)) = Âf1(x) = Âf2(x)

       • Can remove constant and operator aka: Â(cf(x)) = cÂ(f(x))

         

  3. In any measurement of the observable associated with operator Â, the only values that will ever be measured are eigenvalues of the operator Â

     • ÂΨa = aΨa → where a is the eigenvalue

     • Example: Suppose I have a PIB. The only values I can get for the energy (E) are 1h²/8mL² ; 4h²/8mL² ; 9h²/8mL² ; 16h²/8mL² → these are the only allowed quantized energies

       • Suppose the particle is in an eigenstate of Â. (Â = Ĥ in this example)

         • 

           • Ψ(x) matches the Ψn = sqrt(2/L sin (nπx/L))

           • What happens if I measure E? → I will get E2 with 100% certainty. The particle has a well-defined energy. The particle matches an existing defined eigenstate.

       • Suppose the particle is not in an eigenstate of Â. (Â = Ĥ in this example)

         • 

           • Ψ(x) does not match the Ψn = sqrt(2/L sin (nπx/L))

           • ÂΨ(x) =/= EΨ(x)

           • The particle does not have a well-defined energy. It does not match an existing eigenstate.

           • Must still get an eigenvalue from a measurement, but unsure which one (E=2? 4?)

             • Suppose I get E2 → after the measurement, Ψ(x) “collapses” and becomes Ψ(2) → Copenhagen interpretation

             • 

             • Now energy is defined. Remeasurements will always result in Ψ(2)

2/5 - 3rd/4th postulate of QM

• Any function can be written as a weighted sum of eigenstates

IMAGE 22

aka Fourrier series

Probability of measuring En = Cn² → aka (weight)²

• the higher the weight, the better the probability of measuring that energy

A weighted sum of states is called a “superposition of eigenstates” → it has both energy levels at the same time

Example

• Does the particle above have a well-defined energy?

  • No, you cannot pull out an eigenvalue so the energy is not well-defined

    IMAGE 23/24

ImG 25 → Can swtich between defining position and energy, but not both

IMG 26

Postulate 4:

• For any observable with operator Â, the average value (or expectation value) of the observable for a wavefunction Ψ(x) is:

IMG 27

Orthogonality

• “independent” or “have no overlap”

  • For vectors, “perpendicular” is used

  • Orthogonality is the same concept, but applied more generally

IMG 28

• For any given operator in quantum mechanics, all its eigenfunction are mutually orthogonal

  • Functions do not contain any other functions within them, they are not dependent on them

2/6 - Orthogonality + New Model: Vibrations

• <E> = integral (psi* ˆH psi dx)

  New model: Vibrations of a chemical bond

2/10 - Vibrations model

• V(r ) → V(x) = 1/2 kx²

  • X=0 → at equillibrium bond length

  • X>0 stretched bond

  • X<0 compressed bond

2/12 - Spectroscopy

2/18 - Vibrational-Rotational (Microwave) Spectroscopy

Rigid rotor- model for rotation of molecules

slower than vibrations

can be excited → moves faster

2/19 - Rigid Rotor + Rotational-Vibrational IR Spec

2/20 - Hydrogen Atoms

• So far:

  • PIB (exact) (models electrons)

  • Harmonic oscillator (exact) (models nuclei)

  • rigid rotor (exact) (models whole molecules)

  • hydrogen atom (models electron)

• Hydrogen:

  • 1 proton (nucleus) +1

  • 1 electron -1

• Assume proton is fixed at the origin

• Electrostatic attraction b/t proton and electron → distance b/t is ‘r’

  • V( r ) = (-e²)/(4πε0r)← coulomb’s law - potential energy b/t two things is V= k*q1*q2/r1,2

  • Proton charge : 1.6 ×10^-19 C = +e

  • Electron charge: -e

  • K=1/(4πε0)

    • ε0 = permittivity of free space

H^ for hydrogen

• -(hbar)²/2(mass of electron) * [ 1/r² (partial derivative) (r² Partial derivative) + (1/r² sin (theta)) (partial derivative of theta) sin theta (partial derivative of theta)) +(1/r² sin (theta) second partial derivative of phi)] + (-e²/4πε0r) psi (r, theta, phi) = E psi r, theta, phi

• Laplacian + V( r ) = E(psi)(r, theta, phi)

  • 3D problem → 3 quantum numbers (psi n,l,m(l))

    • =Rnl(r ) U(superscrpt ml) (subscript m) (theta, phi) ← radial function, spherical harmonic

  • Can be solved exactly

2/24 - Polyelectronic Atoms