* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Physical Chemistry II
Lattice Boltzmann methods wikipedia , lookup
Wheeler's delayed choice experiment wikipedia , lookup
Coupled cluster wikipedia , lookup
EPR paradox wikipedia , lookup
Quantum state wikipedia , lookup
Bremsstrahlung wikipedia , lookup
X-ray fluorescence wikipedia , lookup
Renormalization group wikipedia , lookup
Dirac equation wikipedia , lookup
Copenhagen interpretation wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
Coherent states wikipedia , lookup
Identical particles wikipedia , lookup
Renormalization wikipedia , lookup
Canonical quantization wikipedia , lookup
X-ray photoelectron spectroscopy wikipedia , lookup
Atomic orbital wikipedia , lookup
Elementary particle wikipedia , lookup
Electron configuration wikipedia , lookup
Schrödinger equation wikipedia , lookup
Path integral formulation wikipedia , lookup
Tight binding wikipedia , lookup
Quantum electrodynamics wikipedia , lookup
Molecular Hamiltonian wikipedia , lookup
Probability amplitude wikipedia , lookup
Wave function wikipedia , lookup
Double-slit experiment wikipedia , lookup
Bohr–Einstein debates wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Hydrogen atom wikipedia , lookup
Atomic theory wikipedia , lookup
Particle in a box wikipedia , lookup
Wave–particle duality wikipedia , lookup
Matter wave wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
CHEM 32: Physical Chemistry II Andrew Rosen May 11, 2013 Contents 0 Math Review 0.1 Complex Numbers 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 The Dawn of the Quantum Theory 3 3 1.1 Blackbody Radiation Could Not Be Explained by Classical Physics . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Planck Used a Quantum Hypothesis to Derive the Blackbody Radiation Law . . . . . . . . . . . . . . . . . . 3 1.3 Einstein Explained the Photoelectric Eect with a Quantum Hypothesis . . . . . . . . . . . . . . . . . . . . . 4 1.4 The Hydrogen Atomic Spectrum Consists of Several Series of Lines . . . . . . . . . . . . . . . . . . . . . . . . 4 1.5 The Rydberg Formula Accounts for All the Lines in the Hydrogen Atomic Spectrum . . . . . . . . . . . . . . 4 1.6 Louis de Broglie Postulated That Matter Has Wavelike Properties . . . . . . . . . . . . . . . . . . . . . . . . 5 1.7 de Broglie Waves Are Observed Experimentally . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.8 The Bohr Theory of the Hydrogen Atom Can Be Used to Derive the Rydberg Formula . . . . . . . . . . . . . 5 1.9 The Heisenberg Uncertainty Principle States That the Position and the Momentum of a Particle Cannot be Specied Simultaneously with Unlimited Precision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 The Wave-Particle Duality 6 7 2.1 Wave vs. Particle Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Probability of an Event . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3 Alternative Expressions for the Heisenberg Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . 8 3 The Schrödinger Equation and a Particle In a Box 3.1 The Hamiltonian Operator 3.2 The Schrödinger Wave Equation 3.3 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 8 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 3.2.1 Time Dependent SWE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 3.2.2 Derivation of Temporal and Spatial Parts 9 3.2.3 Time Independent SWE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3.2.4 Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Particle In a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Wave Functions Must Be Normalized . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.4.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.4.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.4.3 Normalize the Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3.5 The Average Momentum of a Particle in a Box is Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3.6 Higher-Ordered Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.7 3.8 Two-Dimensional Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.7.1 Derivation using Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.7.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Higher-Ordered Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1 4 Postulates and Tunneling 14 4.1 Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.2 The Potential Step - Example of Tunneling 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 The Harmonic Oscillator - Model for Vibration 16 5.1 Hooke's Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 5.2 Quantum-Mechanical Oscillator and Hermite Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 6 The Rigid Rotor - Model for Rotation 6.1 16 Energy Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Molecular Spectroscopy 16 17 7.1 Vibration-Rotation Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 7.2 Spacing of Lines and Overtones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 8 The Hydrogen Atom 18 9 Partition Functions 18 2 0 Math Review 0.1 Complex Numbers • For an equation z = x + iy , • The asterisk symbol (e.g. the operator z∗) Re(z) = x and Im(z) = y denotes that the complex conjugate is taken |Z|2 = Z · Z ∗ • The modulus of a number is the same as the absolute value |Z| = p |Z|2 • For conversion between imaginary and Cartesian coordinates, x = r cos θ • To convert a standard complex number of form 1 are used : a + bi into a complex number in the form of reiθ , the following equations p r = a2 + b2 θ = tan−1 • To convert a complex number of the form reiθ y = r sin θ b a into one of the form a + bi, the following equations, Euler's formula is used: reiθ = r (cos θ + i sin θ) 1 The Dawn of the Quantum Theory 1.1 Blackbody Radiation Could Not Be Explained by Classical Physics • Wavelength and frequency are related by, λν = c • A blackbody is an ideal body that can absorb and emit all frequencies of light, known as blackbody radiation • The Rayleigh-Jeans law states that the radiant energy density between an innitesimal change in frequency is2 , dρ 8πν 2 = 3 kB T dν c • From this equation came the ultraviolet catastrophe, which suggested that, at high frequencies, the radiant energy density diverges as ν2 1.2 Planck Used a Quantum Hypothesis to Derive the Blackbody Radiation Law • The Planck distribution law for blackbody radiation solves the ultraviolet catastrophe by xing the average energy per oscillator using quantized values • The energy for oscillator number n 3 was derived as , E = nhν 1 When computing the Arc tangent, be careful since the calculator will simplify the result and multiple θ values can have the same Arc tangent 2 k ≡ 1.380658 × 10−23 J · K −1 B 3 h ≡ 6.6260755 × 10−34 J · S 3 • The quantized blackbody formula is thus, 1 dρ 8πhν 3 = dν c3 ehν/kB T − 1 • The Wien displacement law states, 2.90 × 10−3 T λmax = m = hc 4.965kB T 1.3 Einstein Explained the Photoelectric Eect with a Quantum Hypothesis • The photoelectric eect is the ejection of electrons from the surface of radiated metal • Unlike theory at the time, experiment showed that the kinetic energy of the ejected electrons is independent of the intensity of the incident radiation Additionally, there is a threshold frequency (ν0 ) characteristic for each metal where no e− for ν < ν0 regardless of intensity (contrasts classical mechanics) ∗ ∗ ∗ Above Above Above ν0 , ν0 , ν0 , the number of electrons is proportional to intensity the kinetic energy of electrons is independent of intensity the kinetic energy of the ejected electrons depends on ν − ν0 • As such4 , Ee− = hν − hν0 • The work function is dened as φ ≡ hν0 • Therefore, Ee− = hν − φ From this, the negated y-intercept of an electron energy vs. frequency plot will yield the work function Also, Planck's constant can be derived from the slope of the line 1.4 The Hydrogen Atomic Spectrum Consists of Several Series of Lines • The characteristic emission spectrum of an atom is called its line spectra Reciprocal wavelength is known as a wavenumber • The Balmer formula for the hydrogen spectra is, ν̃ = 1 ν = = 109680 λ c 1 1 − 2 22 n cm−1 (n = 3, 4, ...) 1.5 The Rydberg Formula Accounts for All the Lines in the Hydrogen Atomic Spectrum • The more general Rydberg formula states5 , ν̃ = RH 1 1 − 2 n21 n2 cm−1 (n2 > n1 ) 4 φ is the work function of a metal, which is analogous to an ionization energy of an isolated atom. It typically has units of electron volts (eV) where 1 eV = 1.602 × 10−19 J 5 The Rydberg constant, R , is equal to 109677.581 cm−1 H 4 1.6 Louis de Broglie Postulated That Matter Has Wavelike Properties • Relating related the momentum of a photon to wavelength as follows, λ= h h = p mv 1.7 de Broglie Waves Are Observed Experimentally • X-ray diraction occurs because the interatomic spacings in the crystal are about the same as the wavelength of the x-rays The distance between rings has to do with the interatomic spacing • Electron microscopes are devices that used electrons, specically their wave-like properties, to observe matter that a light microscope simply cannot 1.8 The Bohr Theory of the Hydrogen Atom Can Be Used to Derive the Rydberg Formula I have included the following derivation. It should not be memorized, but it's helpful to understand it. I have also tried to make this equation relatively general by including Z µ. It still only works for atoms with one electron. Z = 1. Also, by rearranging constants it is possible to have a0 are typically easier to use on a calculator since the other and mentioning However, for the most frequently used hydrogen Bohr model, alternate yet equal versions of the equations. The ones with equations raise very small constants to higher powers (exception: TI-nspire) : 1. Coulomb's law relates the radius of orbit (r ), charge on the electron (e), and the force holding the electron in a circular 6 orbit (f ) , f= qnucleus · qelectron 4π0 r2 (a) Equating this to the centripetal force of the electron orbit (this assumes that the electron circles the nucleus in a stationary, circular orbit) yields the following, where f= Z is the proton number: Ze2 me v 2 = 2 4π0 r r 2. Classical physics assumes that the electron emits electromagnetic radiation, loses energy, and should collapse into the nucleus (a) The assumption that the de Broglie waves of the orbiting electron must be in phase and that the electron had stationary orbits solved this issue. This quantum approach is given by, 2πr = nλ (b) Since λ= h h = , p mv 7 substituting this into equation 2a yields the angular momentum , mvr = (c) Plugging equation 2b into the equation 1 and solving for r nh 2π yields the following equation. Physically, it means the electron has a radii of orbit that must satisfy: r= 4π0 n2 ~2 me e2 Z 6 is the permitivity of free space and has a value of 8.85419 × 10−12 C2 · N−1 ·m−2 0 7~ ≡ h 2π 5 i. Solving for v instead yields, e2 4π0 n~ v= ii. Setting n=1 Z=1 and gives the Bohr radius for the hydrogen atom, a0 = 4π0 ~2 = 5.2917725 × 10−11 m m e e2 iii. Using the Bohr radius is helpful since rn = a0 n2 3. The potential energy of the electron in the Bohr model separated from the positive nucleus by a distance r is, 2 V (r) = − Ze 4π0 r 4. Consequently, E = KE + V (r) = 1 Ze2 me v 2 − 2 4π0 r 5. Then, a little algebra is used: me v 2 (a) Rearrange the equation from step 1a to solve for (b) Then substitute equation 2c into 5a by substituting for 4 En = − and substitute it into equation 4 r. The simplied result is, me e Z e2 Z 2 = − 820 h2 n2 8π0 a0 n2 2 i. The negative sign indicates that the energies are less when the proton and electron are innitely separated ii. n=1 n is the ground state energy and higher values are excited states 6. Finally, for hydrogen or a hydrogen-like ion with only one electron (and using the eective mass µ in place of me ) the change in energy can be found as, m e e4 Z 2 ∆E = 820 h2 (a) Using a0 1 1 − 2 2 n1 n2 = hν = hc λ instead gives: e2 Z 2 ∆E = 8π0 a0 1 1 − 2 2 n1 n2 = hν = hc λ • The Lyman series occurs when excited electrons return to ground-state, and the Balmer series occurs when excited electrons fall back to n=2 7. For accuracy, the eective mass, µ, can be substituted for mass of the electron: µ= (a) Note that a0 8. Optional: Since ν̃ with me is simply me where m1 will be the mass of the nucleus, and m2 is the m1 m2 m1 + m2 is not a perfect constant since it does not use the reduced mass λ−1 , one can derive the Rydberg Constant by solving for 4 R∞ = 4 ṽ . Thus, 2 µe me e e ≈ 2 3 = 2 3 80 ch 80 ch 8π0 a0 1.9 The Heisenberg Uncertainty Principle States That the Position and the Momentum of a Particle Cannot be Specied Simultaneously with Unlimited Precision • The act of locating the electron leads to a change in its momentum • As such, the Heisenberg Uncertainty Principle states, ∆x∆p ≥ 6 ~ 2 2 The Wave-Particle Duality 2.1 Wave vs. Particle Facts • In the double slit experiment, light exhibited diraction - a wave property • Einstein's photoelectric eect proved that light has particle-like properties • de Broglie was the rst to predict that matter can behave as a wave according to λ= h h = p mv Note: The de Broglie wavelength does NOT work in for anything but light waves ν= c λ since c is the speed of light! Do not use this equation • Davisson and Germer's experiment proved the de Broglie wavelength for electron diraction • When h is much smaller when compared to other variables, the quantum limit approaches the classical limit • For normal (90◦ ) incidence (d is the spacing of scattering centers), ∆l = d sin θ • Each photons needs to feel both slits simultaneously so it can interfere with itself (diuse and large) However, the photon is still detected as a tiny localized spot on lm (photon is small) • Wave nature of light and particles is interpreted as a statistical prediction • The measurement process can and does aect the outcome of the event 2.2 Probability of an Event The recipe for predicting what happens: • The probability of an event happening is given by the square of the absolute magnitude of a probability amplitude (ψ ∗ ) P = ψ ∗ ψ = |ψ|2 • When an event can occur in two or more ways, then ψ for the event is the sum of the ψi values for each path considered separately P = |ψ1 + ψ2 |2 • If we perform an experiment that can identify which way the event occurred, then the probability is the sum of probabilities for each alternative P = |ψ1 |2 + |ψ2 |2 • The measurement process forces the event to have a single outcome If many events are averaged, you get a probability distribution • Wave Particle Duality states: The Photoelectric Eect had 1 path, and light behaved as a particle The Double Slit Experiment had 2 paths, and light was wave-like • A plane wave has the form eikx where k= 2π λ • A standing wave has the form e−iωt since it becomes a number at any given t 7 2.3 Alternative Expressions for the Heisenberg Uncertainty Principle • Three forms of the Heisenberg Uncertainty Principle are: 3 ∆x∆px ≥ ~ 2 ∆E∆t ≥ ~ 2 ∆ω∆t ≥ ~ 2 The Schrödinger Equation and a Particle In a Box 3.1 The Hamiltonian Operator • The mathematical operators8 used in this course are all linear (order matters though). This means that an operator follows the following rule: Ω̂ Ω̂ (c · f ) = c Ω̂ (f ) • The commutation relationship states that, x̂P̂ − P̂ x̂ = i~ • The Hamiltonian operator is dened as follows: Ĥ = T̂ + V̂ • Kinetic Energy is described as, T = p2 1 mv 2 = 2 2m • The corresponding quantum mechanical operato T̂ = −~2 2 ∇ 2m To not restrict the discussion to one-dimension, ∇2 ≡ ∂2 ∂2 ∂2 + + ∂x2 ∂y 2 ∂z 2 • Finally, the Hamiltonian operator can be expressed as follow: Ĥ = − ~2 2 − ∇ + V (→ r , t) 2m 3.2 The Schrödinger Wave Equation 3.2.1 Time Dependent SWE • The time-dependent Schrödinger Wave Equation (SWE) states, i~ − ∂Ψ (→ r , t) − = ĤΨ (→ r , t) ∂t The solution to a SWE is called a wave function and is the probability amplitude that encodes all information about the system it describes • As an example, here is the time-dependent SWE with variables substituted for the operators in one dimension, i~ ∂Ψ (x, t) ~2 ∂ 2 Ψ(x, t) =− + V (x, t)Ψ(x, t) ∂t 2m ∂x2 8 All operators have a carat above the letter 8 (1D Time Dependent SWE) 3.2.2 Derivation of Temporal and Spatial Parts If we assume − − Ψ (→ r , t) = ψ (→ r ) θ (t): 1. Therefore, i~ ∂ − − [ψ (→ r ) θ (t)] = Ĥ [ψ (→ r ) θ (t)] ∂t ∂ θ (t) − − i~ψ (→ r) = θ (t) Ĥψ (→ r) ∂t 2 2 i~ ∂θ (t) 1 −~ ∇ → − → − → − → − ψ ( r ) θ (t) yields: = → ψ( r ) + V ( r )ψ( r ) θ (t) ∂t 2m ψ(− r) 2. Using properties of partial derivatives, 3. Dividing through by (a) We have separated time dependence from spatial dependence (b) To have a solution, each side must be equal to a constant, and that constant is energy i~ ∂θ(t) ∂θ(t) = E → i~ = Eθ (t) θ (t) ∂t ∂t 2 −~ 2 → − − − ∇ + V ( r ) ψ (→ r ) = Eψ (→ r) 2m 4. The temporal piece is: 5. The spatial piece is: 6. Using separation of variables on −iE ∂θ (t) = dt θ (t) ~ yields θ (t) = Ae−(iE/~)t 3.2.3 Time Independent SWE • The Time-Independent SWE states, − − Eψ (→ r ) = Ĥψ(→ r) • An example of this in one dimension is given as, Eψ(x) = − ~2 d2 ψ(x) + V (x)ψ(x) 2m dx2 (1D Time Independent SWE) 3.2.4 Postulates 1. Quantum mechanical systems are described by a wavefunction, (a) All information about the system is encoded in (b) Born Interpretation states time i. − Ψ (→ r , t) − Ψ (→ r , t) is a probability amplitude for the presence of the particle at position t − |Ψ (→ r , t) |2 = Probability (c) Additionally, using ψ Density as spatial-dependent and τ is the generic volume ˆ ∞ ψ ∗ ψ dτ ≡ 1 −∞ (d) − Ψ (→ r , t) dτ = sin θdθdφr2 dr is used for spherical coordinates 9 element, → − r and 3.3 Particle In a Box • Above is a diagram of the theoretical particle in a box. Outside of the domain, • Inside the box Ĥ = − ~2 d2 2m dx2 and outside the box Ĥ = − V (x) = ∞ ~2 d 2 +∞ 2m dx2 • The particle in a box problem is a good model for electrons in a conned geometry, such as molecules with conjugated π systems and quantum dots/wells ψ(x) = 0 • Since it is impossible for the particle to be outside of the box, Since ψ(x) outside of the domain is a position, it must be a continuous function • The general solution of the 1D Time Independent SWE where V (x) = 0 is ψ(x) = A cos (kx) + B sin (kx) √ k≡ • ψ(0) = 0 because ψ=0 This implies that • Since for x≤0 A=0 and because ψ 2mE ~ is continuous cos(0) = 1. It does not provide any information for A = 0, ψ(L) = B sin (kx) = 0 • B=0 is rejected because it yields ψ(x) = 0 for all x, which isn't true Alternatively, we can state the following to make sin be zero, k= Therefore, En = h2 n2 8mL2 The energy of transition for an electron is, ∆En = nπ L h2 n2f − n2i 2 8me L The corresponding wave function is, ψn (x) = B sin 10 nπx L B 0≤x≤L 3.4 Wave Functions Must Be Normalized • The wave function must be normalized such that the sum of all probabilities is 1 ψnorm = N ψ ∞ ˆ ∗ −1/2 ψ ψ dτ N= −∞ ˆ • It is also true that ∗ ψm ψn dτ = δm,n This relationship holds: δm,n = 1 m = n, when and δm,n = 0 (orthonormal) when For spherical coordinates and functions that are spherically symmetrical: ˆ ∞ ˆ π ˆ ˆ 2π ∞ f (r)4πr2 dr f (r)r2 sin θdrdθdφ = 0 0 0 0 • The ground state energy (lowest energy state, m 6= n E1 ) This is known as the zero-point energy and is is always greater than zero h2 8mL2 E1 = 3.4.1 Example 1 Normalize 1. If 2 ψ = e−ax 2 ψ = e−ax 2. Then, : for instance, it can be normalized by doing N = q´ ∞ −∞ = ψ ∗ ψdx 3. Finally, 1 ψnorm = 2a π 1/4 2a π ´∞ ψ ∗ ψ dx = −∞ ´∞ 2 e−2ax dx = −∞ r π 2a 1/4 2 e−ax • For even functions that have symmetry, ψ(x) = ψ(−x) • For odd functions that have antisymmetry ψ(−x) = −ψ(x) • The following rules apply: odd × even = odd, even × even = even, odd × odd = even • The integrals of odd functions are zero 3.4.2 Example 2 Normalize ψ0 = Ae−x 2 1. Set up the integral: 2. Factoring out constants yields: 3. ψ0 2 ψ1 = Bxe−ax ´ ∗ ´∞ 2 2 ψ0 ψ1 dτ = −∞ A∗ e−x Bxe−x dx and A∗ B = ´∞ −∞ 2 xe−2x dx is an even function due to its symmetry (the Gaussian curve) and 4. Therefore, the integral is 0 since x is odd and e−2x 2 ψ1 is an odd function due to its antisymmetry is even and is already normalized 11 3.4.3 Normalize the Particle in a Box 1. Set up the integral: ´∞ ψ ∗ ψdx = −∞ 2. Simplify the integral: ´L 0 B ∗ B sin2 ´0 0∗ 0dx + −∞ nπx ´L 0 B ∗ sin nπx L B sin nπx L dx + ´∞ L 0∗ 0dx dx L (a) There is a nifty integral trick that is as such: For ´H L nπx sin2 H = H − L ´ nπ 2 H − L nπ H − L sin z dz = = 0 nπ nπ 2 2 i. This also works for cosine (b) Therefore, the integral is equal to 3. As such, N =B= p 1 L/2 r = (B ∗ B) L 2 2 L (a) This can also be found because the wavefunction must be equal to one along the domain, so (B ∗ B) L =1 2 4. Therefore, r ψn (x) = • To nd the probability density, use nπx 2 sin L L (1D Particle in a Box) ψ∗ ψ • To nd the probability a particle is in a certain region, ˆ x2 ψ ∗ ψ dx = x1 2 L ˆ x2 sin2 nπx x1 • For a one-dimensionl box, each energy state has a corresponding L dx n-half-cycle of sine where the numbers of nodes are is n − 1, • The average momentum of a particle in a box is zero, indicating it is likely to move in either direction 3.5 The Average Momentum of a Particle in a Box is Zero ˆ 2 hp i = ψn∗ (x)ψn (x)P̂x2 dx • Evaluating this integral along the limits of the box (0 to L) yields the following where σp2 = hp2 i = n2 π 2 ~2 nπ~ ∴ σP = 2 L L 12 P̂x2 = d2 −~2 2 : dx • To prove the Heisenberg Uncertainty Principle, ~ σx σP = 2 1/2 ~ π 2 n2 > −2 3 2 This is true because the square-root can never be less than unity 3.6 Higher-Ordered Particle in a Box • For a three-dimensional particle in a box, ψnx ,ny ,nz = 8 Lx Ly Lz 1/2 sin nx πx Lx sin ny πy Ly sin nz πz Lz • As a result, Enx ,ny ,nz h2 = 8m n2y n2z n2x + + (Lx )2 (Ly )2 (Lz )2 ! 3.7 Two-Dimensional Particle in a Box 3.7.1 Derivation using Separation of Variables The process is similar for a two-dimensional box. For our case, let's assume it's a square 2D box such that 1. If ψ(x, y) = X(x)Y (y), 2. Inside the box, 3. Dividing by then V (x, y) = 0 X(x)Y (y) 6. so yields − ∂2 ∂2 X(x)Y (y) + 2 X(x)Y (y) + V (x, y) = EX(x)Y (y) ∂x2 ∂y ~2 (Y (y) + X 00 (x) + X(x)Y 00 (y)) = E (X(x)Y (y)) 2m X 00 (x) Y 00 (y) −2mE + = X(x) Y (y) ~2 Y 00 (y) −2mEy = Y (y) ~2 r r 2 2 nx πx ny πy Separation of variables then yields X(x) = sin sin and Y (y) = Lx Lx Ly Ly ! n2y h2 n2x Additionally, Enx ,ny = + 8m (Lx )2 (Ly )2 4. Solving the above yields 5. ~2 − 2m Ĥψ(x, y) = Eψ(x, y) −2mEx X 00 (x) = X(x) ~2 and 3.7.2 Properties • If the box is square (Lx = Ly ) then the energies can be written out as (1, 1), (1,2), (2, 1), (2, 2), etc., for values The energies will be degenerate for (1, 2) and (2, 1), • Free Electron Molecular Orbital Model (FEMO) Treat electron in a MO as a particle in a box 13 for instance nx and ny 3.8 Higher-Ordered Particle in a Box • For a three-dimensional particle in a box, ψnx ,ny ,nz = 8 Lx Ly Lz 1/2 sin nx πx Lx sin ny πy Ly sin nz πz Lz • As a result, Enx ,ny ,nz • Each energy level can hold 2 4 h2 = 8m n2y n2z n2x + + 2 2 (Lx ) (Ly ) (Lz )2 ! π -electrons Postulates and Tunneling 4.1 Postulates • Postulate 1: ψ completely describes a quantum-mechanical system • Postulate 2: For every observable in classical mechanics, there is a corresponding quantum mechanical linear operator (specically, Hermitian Operators) where we choose x̂ x̂ = x • Postulate 3: For an operator Â, and p̂ so that p̂ = −i~ x̂p̂ − p̂x̂ = i~ ∂ ∂x the only observables are equal to the eigenvalues of Case 1: If the system is in a state described by an eigenfunction, Case 2: ψ = c cos (kx). Here, ψ is not a function of p̂. ψn , of  When ◦ ◦ p̂ acts on an 1 ikx 1 −ikx ce + ce with 2 2 then we always measure We can also write ψ theorem. ∗  as Euler's ψ = c cos (kx): 1) Only returns e-values of p̂ for answers 2) The probability of measuring a particular e-value is given by the square of the coecient weighting the corresponding e-function in the superposition sum ◦ 3) We can only predict the average outcome of many experiments, or the possible outcomes and their probabilities for a single measurement • Postulate 4: Expectation Value = If i h Â, B̂ = ÂB̂ − B̂  = 0, • Postulate 5: Free particle where ´ ∗ ψ Âψdτ hai = hÂi = ´ ∗ ψ ψdτ then we can measure a and b without uncertainty V (x) = 0 A free particle is not bound and has no quantization 4.2 The Potential Step - Example of Tunneling 14 • When E ≥ V0 , x=0 the particle decelerates at • In quantum mechanical regions I and II, x>0 but either stops or continues on to ψI (x = 0) = ψII (x = 0) and 0 ψI0 (x = 0) = ψII (x = 0) so that the wave functions are nite and continuous • As seen in the diagram, an energy below V0 will have a decaying exponential, termed an evanescent wave, when it hits the potential step −h2 d2 ψI = EψI 2m dx2 • Region I: • Region II: −h2 d2 ψII + V0 ψII = EψII , 2m dx2 • Therefore, ψI = Aeikx + Be−ikx • The restrictions on ψ and must be that 0 0 ψII = A0 eik x + B 0 e−ik x ψ(x) h2 d2 ψII = (E − V0 )ψII 2m dx2 p √ 2m (E − V0 ) 2mE 0 where k = and k = ~ ~ which is equivalent to: − is continuous, nite, and that ψ 0 (x) V =∞ is continuous except if for all x This can give you boundary conditions to solve for the constants. However, two will be missing One more constraint can be found from the assumption that the particles originate from the −x direction as a right-going wave such that there are no particles coming in from the rightmost region (this makes the rightgoing constant of the last region zero) 2 The last constraint can be found from the ux, which is dened as Flux=v|A| N constant representing probability density when squared, N, where A is the pre-exponential is the total number of particles, and v is the velocity in that region ∗ ∗ Recall that p = mv . Therefore, v= k~ m Reected ux is found based on the leftgoing wave before the boundary (B ), and transmitted ux is found 0 based on the rightgoing wave after the boundary (A ) • To nd the transmittance, simply divide the rightgoing ux after the boundary (region II in this case) by the rightgoing ux before the boundary (region I in this case). As in, it's the amount that went through In this example, it would be This will simplify down to, vII |A0 |2 N vI |A|2 N T = T = k 0 4k 2 k (k + k 0 )2 • To nd the fraction reected, simply divide the leftgoing ux before the boundary by the rightgoing ux before the boundary In this example, it would be This will simplify to vI |B|2 vI |A|2 R= 2 ∗ R= (k − k 0 ) 2 , which is always greater than or equal to zero (k + k 0 ) This goes against classical mechanics because, even when • Additionally, there should be no particles lost; therefore, • When E < V0 , k E > V0 , there will be some particles reected R+T =1 is imaginary; therefore, the exponential terms of the wavefunction are real and not plane waves When the wave has a probability for transmittance even when • The larger total energy of the particle (lower V0 and/or higher E) E < V0 , the wave is called an evanescent wave means larger p and thus shorter amplitude • The smaller total energy of the particle means smaller p and thus longer 15 λ. Also, larger amplitude λ. Also, smaller 5 The Harmonic Oscillator - Model for Vibration 5.1 Hooke's Law • The classical harmonic oscillator is • The total energy is conserved as 1 V = kx2 , F = −kx, 2 E =V +T = 1 2 kx 2 and 1/2 k ω = 2πν = µ for a spring oscillating between −x and +x 5.2 Quantum-Mechanical Oscillator and Hermite Polynomials • The fundamental vibrational frequency is given by • Note that the value of k 1 2πc ṽ = 1/2 k µ is the same regardless of the isotope (even though µ is dierent). k is only a function of bond order ~2 ∂ 2 ψ 1 2 + kx ψ = Eψ 2m ∂x2 2 √ kµ 2 , Hn α1/2 x e−αx /2 where α = ~ 1 α 1/4 − • The SWE can be set up as follows (note that there is potential energy now): ψn (x) = Nn Hn • The wavefunction for the harmonic oscillator is given by Hermite Polynomial, and Nn is the normalization constant given by Nn = 1/2 (2n n!) is the π Hermite polynomials are either even or odd. The Hermite polynomial has the same parity as n 2 The symmetric integral of an odd function is zero, and because ψn (x) is always even, hxi must be odd. Therefore, the average displacement of a harmonic oscillator is zero for all states. This is the same for hpi n ≥ 0, • There is quantum tunneling with the harmonic oscillator, and there are n nodes, and a zero-point energy of 1 hν 2 (for a 1-D case), which is dierent from the PIB • Allowable energies are E= n+ 1 2 hν = n+ ~ω where n is the quantum number There is thus equal spacing between the energies of the vibrational states For the harmonic oscillator, ∆n = ±1 (typically confusing for me to use since it looks like 6 1 2 v is used for the harmonic oscillator energy state, but it's too ν) The Rigid Rotor - Model for Rotation 6.1 Energy Levels • To describe a rotating body, we use the moment of inertia, • The possible energies9 for a rigid rotor are It is also true that As such, EJ = I, ~2 J (J + 1), 2I such that, where I= J ≥0 P (integers) ∆J = ±1 ~2 ∆EJ = (J + 1) I • A molecule must posses a permanent dipole moment to have rotational energy • Frequently, the constant B= h 8π 2 I is used such that B̃ = h 8π 2 cI 9 Note that the units for energy here are in Hz and must be multiplied by h to yield joules 16 mi ri2 = µR2 • Additionally10 , ν = 2B (J + 1), This is useful because a frequency (or wavenumber) can be read o of a vibrational spectrometer. This could then I I = µR2 , be used to calculate nd 7 ∆ν = 2B and thus for an energy transition R from using the denition of B shown above and the fact that ∆ν = 2B . Finally, with I, you can and thus you have the bond length Molecular Spectroscopy 7.1 Vibration-Rotation Interaction • Since a molecule can rotate and vibrate at same time, we need to combine the rigid rotor and SHO models to nd a better approximation of the internuclear energies • The vibrational energy of a molecule in wavenumbers is • The rotational energy of a molecule in wavenumbers if • As such, Ẽ = G(v) + F (J) and v, where ∆J = +1 it is called an R-Branch (high frequency side) If ∆J = −1 it is called a P-Branch (low frequency side) There are no lines at ṽ because the transition of ν̃obs are inaccurate, as ν̃R To derive the equations of and ν̃P , B̃ consider a ∆J = 0 as described earlier v B̃ for a v=0→1 is dependent on ∆ER = ν̃ + 2B̃ (J 00 + 1) v transition and apply ν̃R = Ẽ1,J+1 − Ẽ0,J and ν̃P = Ẽ1,J−1 − Ẽ0,J ∗ ν̃R = ν̃ + 2B̃1 + 3B̃1 − B̃0 J + B̃1 − B̃0 J 2 ∗ ν̃P = ν̃ − B̃1 + B̃0 J + B̃1 − B̃0 J 2 The vibration rotation interaction states that is forbidden is a function of Example: • The energies of each branch are F (J) = B̃J (J + 1) so is the change in energy If • Technically, the equations of J F (J) where ∆E = E 0 − E 00 where we dene • Since energy is a function of G(v) 1 ν̃ G(v) = v + 2 and v ∆E transition such that 1 B̃v = B̃e − αe v + 2 ∆EP = ν̃ − 2B̃J 00 7.2 Spacing of Lines and Overtones • The spacing of spectral lines are not even because the centrifugal force causes the bond to stretch slightly The corrected equation is • To account for overtones, F (J) = B̃J (J + 1) − D̃J 2 (J + 1) 2 1 1 G(v) = ν̃e v + − x̃e ν̃e v + , 2 2 2 where where x̃e D̃ is the centrifugal distortion constant is the anharmonicity constant • The terminology is that the 0 to 1 transition is the fundamental transition, 0 to 2 is the rst overtone, 0 to 3 is the second overtone, etc. • The selection rule for anharmonic oscillator is any integral value for • The electronic transition energy is given by ∆v with higher integral values being weaker Ẽ = T̃e + G0 (v 0 ) − G00 (v 00 ) + F 0 (J 0 ) − F 00 (J 00 ) • The Franck-Condon Principle states that electron motion does not cause appreciable nuclei motion 10 Note: This can be rewritten using ν̃ and B̃ 17 8 The Hydrogen Atom • For the hydrogen atom, ψ(r, θ, φ) = Rn,` (r)Y`m` (θ, φ) • The angular solution is based on Legendre Polynomials (m = 0) and Associated Legendre Functions (any m) The shape of the orbitals • The radial solution is based on the Laguerre Polynomials Tells how the probability density dies o • The wavefunctions of a rigid rotor are spherical harmonics given by • L̂2 Y`m` (θ, φ) = ` (` + 1) ~2 Y`m` (θ, φ) • L̂z Y`m` (θ, φ) = m` ~Y`m` (θ, φ), • Ylm 's Note that m = 0, ±1, ±2, ..., ±` 1/2 Rn,` = − so (n − ` − 1)! 2n (n + `)! Such that En = 2 na0 where ` (` + 1) ~2 `+3/2 is the eigenvalue of L̂2 with an eigenvalue of m` ~ ` = 0, 1, 2, 3, ... r` e−r/(na0 ) L2`+1 n+` 2r na0 where L is the Laguerre Polynomial and a0 = 0 h2 πµe2 e2 8π0 a0 n2 • The variation principles states that we can pick any guess at L̂z are the eigenfunctions of and that Y This means that a trial wavefunction, φ, ψ, we'll call it φ, and then ´ ∗ φ Ĥφdτ ´ = Eφ ≥ E0 φ∗ φdτ can be created. To improve the accuracy of this model the energy can be minimized by changing the parameter(s). Adding more parameters to alter can also improve the accuracy of this trial wavefunction 9 Partition Functions • For partition functions, a = g exp (−E/kT ), where g is the number of degeneracies • Furthermore, the fraction of molecules in this state is fi = ai /Q 18 where Q= P a