Quantum Mechanics and the hydrogen atom Since we cannot say exactly where an electron is, the Bohr picture of the atom, with electrons in neat orbits, cannot be correct. Quantum theory describes electron probability distributions: MNW-L2 Hydrogen Atom: Schrödinger Equation and Quantum Numbers Potential energy for the hydrogen atom: The time-independent Schrödinger equation in three dimensions is then: Equation 39-1 goes here. MNW-L2 Where does the quantisation in QM come from ? The atomic problem is spherical so rewrite the equation in (r,θ,φ) x = r sinθ cosφ y = r sinθ sinφ z = r cosθ Rewrite all derivatives in (r,θ,φ), gives Schrödinger equation; 2 ⎛ ∂ ∂ ⎞ 2 ⎛ 1 ∂ ∂ 1 ∂2 ⎞ − h r 2 Ψ − h ⎜ sinθ + ⎟Ψ +V (r)Ψ = EΨ ⎜ ⎟ ⎜ 2 2 ⎟ 2m ⎝ ∂r ∂r ⎠ 2m ⎝ sinθ ∂θ ∂θ sin θ ∂φ ⎠ This is a partial differential equation, with 3 coordinates (derivatives); Use again the method of separation of variables: Ψ()r,θ,φ = R(r)Y(θ,φ) Bring r-dependence to left and angular dependence to right (divide by Ψ): QM 1 ⎡ d dR 2mr2 ⎤ O Y()θ,φ r 2 + E −V r R = − θφ = λ ⎢ 2 ()() ⎥ R ⎣dr dr h ⎦ Y()θ,φ Separation of variables MNW-L2 Where does the quantisation in QM come from ? 1 ⎡ d dR 2mr2 ⎤ Radial equation r 2 + E −V r R = λ ⎢ 2 ()() ⎥ R ⎣dr dr h ⎦ ⎛ 1 ∂ ∂ 1 ∂2 ⎞ QM − ⎜ sinθ + ⎟Y()θ,φ O Y()θ,φ ⎜ sinθ ∂θ ∂θ sin2 θ ∂φ 2 ⎟ Angular equation − θφ = ⎝ ⎠ = λ Y()θ,φ Y ()θ,φ ∂2Y ∂ ∂Y − = sinθ sinθ + λ sin2 θY ∂φ 2 ∂θ ∂θ Once more separation of variables: Y(θ,φ) = Θ(θ )Φ(φ) 1 ∂2Φ 1 ⎛ ∂ ∂Θ ⎞ Derive: − = ⎜sinθ sinθ + λ sin2 θΘ⎟ = m2 (again arbitrary constant) Φ ∂φ 2 Θ ⎝ ∂θ ∂θ ⎠ Simplest of the three: the azimuthal angle; ∂2Φ()φ + m2Φ()φ = 0 ∂φ 2 MNW-L2 Where does the quantisation in QM come from ? A differential equation with a boundary condition ∂2Φ()φ + m2Φ()φ = 0 and Φ()φ + 2π = Φ(φ) ∂φ 2 Solutions: Φ(φ) = eimφ Boundary condition; Φ(φ + 2π ) = eim(φ+2π ) = Φ()φ = eimφ e2πim =1 m is a positive or negative integer this is a quantisation condition General: differential equation plus a boundary condition gives a quantisation MNW-L2 Where does the quantisation in QM come from ? with integer m imφ First coordinate Φ(φ) = e (positive and negative) angular 1 ∂ ∂Θ ⎛ m2 ⎞ sinθ + ⎜λ − ⎟Θ = 0 part Second coordinate ⎜ 2 ⎟ sinθ ∂θ ∂θ ⎝ sin θ ⎠ angular Results in λ = +1 with = 0,1,2, momentum l l(l ) l K and m = −l,−l +1,K,l −1,l 1 ⎡ d dR 2mr2 ⎤ r 2 + E −V r R = +1 Third coordinate ⎢ 2 ()()() ⎥ l l R ⎣dr dr h ⎦ Differential equation Results in quantisation of energy radial part 2 Z 2 Z 2 ⎛ e2 ⎞ m E = − R = − ⎜ ⎟ e n 2 ∞ 2 ⎜ ⎟ 2 n n ⎝ 4πε0 ⎠ 2h with integer n (n>0) MNW-L2 Angular wave functions ⎡ 1 ∂ ∂ 1 ∂2 ⎤ L2 = h sinθ + Operators: ⎢ 2 2 ⎥ Angular momentum i ⎣sinθ ∂θ ∂θ sin θ ∂φ ⎦ r ∂ L = (L , L , L ) L = h x y z z i ∂φ There is a class of functions that are simultaneous eigenfunctions 2 2 L Y θ,φ = m Y θ,φ L Ylm ()θ,φ = l(l +1)h Ylm (θ,φ) z lm () h lm ( ) with l = 0,1,2,K and m = −l,−l +1,K,l −1,l Spherical harmonics (Bolfuncties) Ylm (θ,φ) Vector space of solutions 1 3 2 Y = iφ ∫ Ylm (θ,φ) dΩ =1 00 Y11 = − sinθe 4π 8π Ω * 3 ∫YlmYl'm'dΩ = δll'δmm' Y = − cosθ Ω 10 4π Parity 3 −iφ Y1,−1 = sinθe P , , l , MNW-L2 8π opΥlm ()θ φ = Υ (π −θ φ +π )()()= − Υlm θ φ The radial part: finding the ground state 2 1 ⎡ d 2 dR 2mr ⎤ r + E −V r R = λ 2 2 ⎢ 2 ()() ⎥ Ze R ⎣dr dr h ⎦ Prefactor for 1/r: h − = 0 ma 4πε0 Find a solution for l = 0, m = 0 Solution for the length scale paramater 2 ⎛ 2 ⎞ Ze2 − h ⎜ R"+ R'⎟ − R = ER 2m ⎝ r ⎠ 4πε r 4πε 2 0 a = 0h 0 2 Bohr radius Physical intuition; no density for r → ∞ Ze m trial: R()r = Ae−r / a A −r / a R R'= − e = − Solutions for the energy a a A −r / a R 2 R"= e = 2 ⎛ e2 ⎞ m 2 2 E = − h = −Z 2⎜ ⎟ e a a ⎜ ⎟ 2 2ma ⎝ 4πε0 ⎠ 2h 2 ⎛ 1 2 ⎞ Ze2 − h ⎜ − ⎟ − = E 2m 2 ar 4πε r ⎝ a ⎠ 0 Ground state in the Bohr model (n=1) must hold for all values of r MNW-L2 Hydrogen Atom: Schrödinger Equation and Quantum Numbers There are four different quantum numbers needed to specify the state of an electron in an atom. 1. The principal quantum number n gives the total energy. 2. The orbital quantum number lgives the angular momentum; l can take on integer values from 0 to n -1. 3. The magnetic quantum number, m , gives the l direction of the electron’s angular momentum, and can take on integer values from –l to +l . MNW-L2 Hydrogen Atom Wave Functions The wave function of the ground state of hydrogen has the form: The probability of finding the electron in a volume dV around a given point is then |ψ|2 dV. MNW-L2 Radial Probability Distributions Spherical shell of thickness dr, inner radius r and outer radius r+dr. Its volume is dV=4πr2dr Density: |ψ|2dV = |ψ|24πr2dr The radial probablity distribution is then: 2 2 Pr =4πr |ψ| Ground state MNW-L2 Hydrogen Atom Wave Functions This figure shows the three probability distributions for n = 2 and l = 1 (the distributions for m = +1 and m = -1 are the same), as well as the radial distribution for all n = 2 states. MNW-L2 Hydrogen “Orbitals” (electron clouds) Represents the probability to find a particle At a location r at a time t Ψ rr,t 2 () The probability density The probability distribution Max Born The Nobel Prize in Physics 1954 "for his fundamental research in quantum mechanics, especially for his statistical interpretation of the wavefunction" MNW-L2 Atomic Hydrogen Radial part Analysis of radial equation yields: Z 2 Enlm = − R∞ n2 m e4 R = e with ∞ 3 8ε0h c Wave functions: r −iEnt / h Ψnlm ()r,t = Rnl ()(r Υlm θ,φ )e For numerical use: +1 u (r) 2Z ()n − −1 ! ⎛ 2Zρ ⎞l ⎛ 2Zρ ⎞ R = nl u ρ = l e−Zρ / n L2l+1 nl () ⎜ ⎟ n−l−1⎜ ⎟ r na0 2n()n +1 ! ⎝ n ⎠ ⎝ n ⎠ 2 2 ρ / r = 2Z / na a = 4πε0h / μe MNW-L2 Quantum analog of electromagnetic radiation Classical electric dipole radiation Transition dipole moment Classical oscillator Quantum jump 2 2 I ∝ e&rr& * r rad Irad ∝ ∫ψ1 erψ 2dτ The atom does not radiate when it is in a stationary state ! The atom has no dipole moment *r μii = ∫ψ1 rψ1dτ = 0 2 μ fi Intensity of spectral lines linked B = to Einstein coefficient for absorption: if 2 6ε0h MNW-L2 Selection rules Mathematical background: function of odd parity gives 0 when integrated over space ∞ ∞ In one dimension: * with * Ψf x Ψi = ∫ Ψf xΨidx = ∫ f (x)dx f (x) = Ψf xΨi −∞ −∞ ∞ 0 ∞ 0 ∞ ∞ ∞ ∫ f (x)dx = ∫ f (x)dx + ∫ f (x)dx = ∫ f (−x)d()− x + ∫ f (x)dx = ∫ f (−x)dx + ∫ f (x)dx −∞ −∞ 0 ∞ 0 0 0 ∞ and opposite parity = 2∫ f (x)dx ≠ 0 if f (−x) = f (x) Ψi Ψf 0 = 0 if f (−x) = − f (x) Ψi and Ψf same parity Electric dipole radiation connects states of opposite parity ! MNW-L2 Selection rules depend on angular behavior of the wave functions Parity operator r All quantum mechanical wave functions r have a definite parity θ Ψ(− rr) = ±Ψ(rr) φ r Ψf r Ψi ≠ 0 − rr If Ψ f and Ψ i have opposite parity r r Y Pr = −r Rule about the lm functions (x, y, z) → (−x,−y,−z) PY (θ,φ) = (−)lY (θ,φ) ()(r,θ,φ → r,π −θ,φ +π ) lm lm MNW-L2 Hydrogen Atom: Schrödinger Equation and Quantum Numbers “Allowed” transitions between energy levels occur between states whose value of ldiffer by one: Other, “forbidden,” transitions also occur but with much lower probability. “selection rules, related to symmetry” MNW-L2 Selection rules in Hydrogen atom Intensity of spectral lines given by * r r μ fi = ∫ Ψ f μΨi = Ψ f − er Ψi 1) Quantum number n no restrictions Balmer series 2) Parity rule for l Δl = odd 3) Laporte rule for l Angular momentum rule: Lyman series r r r l f = li + 1 so Δl ≤1 From 2. and 3. Δl = ±1 MNW-L2.