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and the

Since we cannot say exactly where an is, the Bohr picture of the atom, with in neat orbits, cannot be correct.

Quantum theory describes electron probability distributions:

MNW-L2 : Schrödinger Equation and Quantum Numbers

Potential for the hydrogen atom:

The -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)

= r sinθ cosφ y = r sinθ sinφ z = r cosθ Rewrite all derivatives in (r,θ,φ), gives Schrödinger equation; θ θ θ 2 ⎛ ∂ ∂ ⎞ 2 ⎛ 1 ∂ ∂θ 1 ∂2 ⎞ − h r 2 Ψ − h ⎜ sin + ⎟Ψ +V (r)Ψ = EΨ ⎜ ⎟ ⎜ 2 θ 2 ⎟ 2m ⎝ ∂r ∂r ⎠ 2m ⎝ sin ∂ ∂ sin ∂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): θ QM φ 1 ⎡ d dR 2mr2 ⎤ O Y , () r 2 + ()E −V ()r R = − θφ θ φ = λ ⎢ 2 ⎥ () R ⎣dr dr h ⎦ Y , Separation of variables MNW-L2 Where does the quantisation in QM come from ?

θ φ θ 1 ⎡ d φdR 2mr2 θ ⎤ Radial equation r 2 + ()E −θV ()rθR = λ ⎢ 2 ⎥ R ⎣dr dr h ⎦θ θ θ ⎛ 1 ∂ ∂ θ φ 1 φ∂2 ⎞ φ − ⎜ sin + ⎟Y(), OQM Y(), ⎜ sin ∂ ∂ 2 2 ⎟ Angular equation θφ ⎝ sin ∂ ⎠ − ()= () = λ Y , φ θ Y , θ θ ∂2Y ∂ ∂Yθ λ − = sin sin + sin2 θY ∂ 2 ∂ ∂ φ Once more separation ofθ variables: Y(θ,φ)= Θ(θ )Φ(φ) θ θ 1 ∂2Φ 1 ⎛ ∂ ∂Θθ λ ⎞ Derive: − = ⎜sin sin + sin2 θΘ⎟ = m2 (again arbitrary constant) Φ ∂ 2 Θ ⎝ ∂ ∂ ⎠

Simplest of the three: the azimuthal angle; φ ∂2Φφ() + m2Φ()φ = 0 ∂ 2 MNW-L2 Where does the quantisation in QM come from ?

A differential equation with a boundary condition φ ∂2Φφ() + m2Φ()φ = 0 and Φ()φ + 2π = Φ(φ) ∂ 2

Solutions: φ π Φ(φ)= eimφ Boundary condition; Φ( + 2 )= 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) θ θ θ λ angular 1 ∂ ∂Θθ ⎛ m2 ⎞ sin + ⎜ − ⎟Θ = 0 part coordinate ⎜ 2 ⎟ sin ∂ ∂ ⎝ 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

θ θ θ ⎡ 1 ∂ ∂θ 1 ∂2 ⎤ L2 = h sin + Operators: ⎢ 2 θ 2 ⎥ i ⎣sin ∂ ∂ sin ∂φ ⎦ r ∂ L = (L , L , L ) L = h x y z z i ∂φ

There is a classθ φ of functions that are simultaneous eigenfunctions

2 2 θ L Y ()θ,φ = m Y (θ,φ) L Ylm (), = l(l +1)h Ylm ( ,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 θ 2 Y = iφ ∫ Ylm ( ,φ) dΩ =1 00 Y11 = − πsinθe 4π 8 θ Ω φ Y * Y dΩ = δ δ 3 π ∫ lm l'm' ll' mm' Y10 = − πcosθ θ Ω 4 φ π 3 −iφ Y1,−1 = πsinθe () (l )()()θ MNW-L2 8 PopΥlm , = Υ − , + = − Υlm ,φ The radial part: finding the

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 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 (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 n gives the total energy. 2. The orbital quantum number l gives the angular momentum; l can take on integer values from 0 to n -1.

3. The , 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 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 , 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 θ −iEnt / h Ψnlm ()r,t = Rnl r Υ ()(lm ,φ e ρ )

For numerical use: ρ

+1 u (r) 2Z ()n − −1 ! ⎛ 2Z ⎞l ⎛ 2Zρ ⎞ R = nl u ()= l e−Zρ / n L2l+1 nl ()⎜ ⎟ n−l−1⎜ ⎟ r na0 2n n +1 ! ⎝ n ⎠ ⎝ n ⎠ πε 2 2 ρ / r = 2Z / na a = 4 0h / μe MNW-L2 Quantum analog of electromagnetic radiation

Classical electric dipole radiation Transition dipole moment

Classical oscillator Quantum jump ψ 2 2 I ∝ e&rr& * rψ rad Irad ∝ ∫ 1 er 2dτ

The atom does not radiate when it is in a ! Theμ atom has no dipole moment ψ *rψ ii = ∫ 1 r 1dτ = 0

μ 2 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 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

Prr = −rr Rule about theθ Y functions θ φ φ lm (x, y, z) → (−πx,−y,−z) θ φ PY ( , ) = (−)lY (θ,φ) ()(r, , → r, − , +π )lm lm

MNW-L2 Hydrogen Atom: Schrödinger Equation and Quantum Numbers

“Allowed” transitions between energy levels occur between states whose value of l differ by one:

Other, “forbidden,” transitions also occur but with much lower probability.

“selection rules, related to

MNW-L2 Selection rules in Hydrogen atom

Intensity of spectral lines given by μ * r r fi = ∫ Ψ f μΨi = Ψ f − er Ψi

1) Quantum number n no restrictions

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