Quantum Physics Lecture 6 Bohr model of hydrogen atom (cont.) Line spectra formula Correspondence principle Quantum Mechanics formalism General properties of waves Expectation values Free particle wavefunction 1-D Schroedinger Equation
Experimental Evidence for Bohr model From optical emission spectrum of hydrogen: Consists of line spectra (in contrast to blackbody continuum) Balmer series; fitted to formula 1 λ = R( 1 2 2 1 ) n 2 for n = 3, 4, 5... Experimental value of R = 1.097 x 10 7 m -1 In general, use two integers n i (initial) and n f (final) 1 λ = R 1 2 n 1 2 f n i for n i > n f Where n f = 1 (Lyman), = 2 (Balmer), = 3 (Paschen) etc
Connect expt. with Bohr model 1 λ = R 1 2 n 1 2 f n i E n = me 4 ( ) 8ε o 2 h 2 1 n 2 Optical emission: result of a down transition of electron from a higher energy orbit (n i ) to a lower energy orbit (n f ) Energy difference is emitted as a photon: ω = E ni E n f = me4 1 8ε 2 0 h 2 2 n 1 2 i n f = hc λ 1 λ = me4 8ε 0 2 h 3 c 1 n f 2 1 n i 2 R = me4 8ε 2 0 h 3 c
Centre-of-mass correction R = me4 8ε o 2 h 3 c =1.097x107 m 1 This value of R agreed with expt. of the time! However, later (more accurate) experiments gave R expt = 1.0967785 whereas model gave R = 1.0973731 Replace m with m* = mm/(m + M) = 0.99945(m) In centre-of-mass description of electron (m) and proton (M) Correspondence Principle The greater the quantum number. the closer Quantum Physics approaches Classical Physics!
Correspondence Principle Compare orbit frequency (f) and emitted photon frequency (ω/2π) f = v 2πr = e 2π 4πε o mr = me 4 2 3 8ε 2 o h 3 ν = ω/2π = E/h me4 1 8ε 2 o h 3 2 n 1 2 f n i Note same pre-factor! Write n i = n and n f = n - p n 3 1 n f 2 1 n i 2 = 1 ( n p) 2 1 n 2 = When n >> p, (2np - p 2 ) ~ 2np and (n - p) 2 ~ n 2 then ω/2π ~pf and letting p=1, a transition: n to (n-1) gives ω/2π ~ f 2np p 2 ( n p) 2 n 2 2 p n 3
Bohr criterion for allowed orbits The Bohr requirement for orbits can also be stated as: angular momentum is quantised in units of ħ mvr = n h 2π ( ) = n 2πr = n h mv ( ) = nλ i.e. Equivalent to fitting de Broglie wavelengths In fact, the concept of quantised angular momentum nħ is the fuller and broader criterion, with further consequences to be seen; the other is merely illustrative, not factual! Complete description of atom requires at least three different quantum numbers (see later lectures)
General properties of waves Recall 1-D wave: y = Acos(ωt - kx) This is just one possible solution of the 1-D wave equation: δ 2 y δx = 1 δ 2 y 2 v 2 δt 2 [ ( )] where i = 1 ( ) isin( ωt kx) - the general (complex) solution. In general y = Aexp i ωt kx [ ] = A cos ωt kx For waves (of existence) in QM use wavefunction ψ Recall UP and probability: y 2 is a measure of probability of finding a particle at location x In QM, ψ is in general complex, and not usually a measureable parameter (such a momentum etc.) However, ψ 2 is! So must retain full complex solution, not just real part.
General properties of waves cont. Three other required properties of wavefunction ψ (1) single-valued and continuous (2) derivative (dψ/dx) single-valued and continuous (3) normalisable: ψ 2 dx = 1 - i.e. integrated probability density over all space is unity i.e. for 1-D case require If ψ is complex, what about ψ 2? ψ 2 = ψ*ψ where ψ* is complex conjugate of ψ ψ = A + ib and ψ* = A - ib ψ 2 dx = 1 ψ*ψ = (A - ib)(a + ib) = A 2 - (ib) 2 = A 2 + B 2 (i.e. real) +
Expectation values Expectation value: the most probable value of a variable Multiply variable by probability density ( ψ 2 ) and integrate! eg. expectation value of 1-D position: x = If ψ is normalised then denominator = 1, in which case x = + and for a general variable G(x) the expectation value is G x + xψ 2 dx ( ) = G x + ( )ψ 2 dx + xψ 2 dx ψ 2 dx
Free particle wavefunction.think simple wave, rather than wavegroup ψ = Aexp[ i( ωt kx) ] ω ω = 2πν = 2π E h = E k = 2π = 2π p λ h = p ψ = Aexp i E t p x Replacing wave-notation (ω, k) with particle notation (E, p) What is the equation of motion, just as in Newton s Laws, but for quantum particle?
Free particle functions Now consider partial differential with x Re-arranged: Or: i.e. if we operate on ψ with iħδ/δx - we get the value of momentum p multiplied by ψ iħδ/δx is the momentum operator ˆp 2 Expect Kinetic Energy operator 2m = 2 2 2m x 2 General Equation: ψ = Aexp iet + ipx = Ae iet e ipx ψ x = Ae iet. ip eipx = ip ψ i x ψ = pψ Operator on ψ = value ψ ˆp i x ψ = pψ
1-D Schroedinger Equation Is the equation for energy. Consider: ψ = Aexp i E t p x ψ x = +ip ψ 2 ψ x = p2 2 ψ 2 ψ 2 p2 ψ = 2 x 2 ψ t = i E ψ Eψ = +i ψ t Compare with ordinary (non-relativistic) mechanics
1-D Schroedinger Equation developed E = KE + PE = p 2 /2m + U(x,t) multiply across by ψ Eψ = (p 2 /2m)ψ + Uψ and substitute for Eψ and p 2 ψ i ψ t = 2 2 ψ 2m x +Uψ 2 1-D Schroedinger Equation i ψ t = 2 2 ψ 2m x + 2 ψ 2 y + 2 ψ 2 z 2 +Uψ 3-D Schroedinger Equation Patently true for free particle (U=0), also found to be true for constrained particle a basic principle
Steady State Simplification When U is not a function of t, get considerable simplification: - The time-independent, or steady state Schroedinger Equation. Recall free particle wavefunction: ψ = Aexp i E t p x ψ = Aexp( i E t) exp i p x ψ = ψ exp( i E t) where ψ = Aexp i p x Fortunately, this separation of time and position dependences is also possible for all wavefunctions when U is indep. of t! substitute this form of ψ into 1D Schroedinger Equation.
Steady State Schroedinger Equation i ψ t = 2 2 ψ 2m x + Uψ 2 LHS = i t ( ψ exp ( i E t )) = E ψ exp i E t ( ) 2 RHS = 2 ψ exp i E t 2m x 2 = 2 2m exp i E t E ψ = 2 2m drop the dash & re-write as : ( ( )) + U ( ) 2 ψ x + U ψ exp i E 2 t 2 ψ x + U ψ 2 ( ψ exp( i E t )) ( ) 2 ψ x + 2m 2 2 ( E U )ψ = 0