Radiating Dipoles in Quantum Mechanics

Radiating Dipoles in Quantum Mechanics Chapter 14 P. J. Grandinetti Chem. 4300 Oct 27, 2017 P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 1 / 26

P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 2 / 26

Electric dipole moment vector operator Electric dipole moment vector operator for collection of charges is N μ = q k r k=1 Expectation value for electric dipole moment vector is μ(t) = V Ψ ( r, t) μψ( r, t)dτ Let s focus on single bound charged QM particle. ] μ = q e r = q e [ x ex + ŷ e y + ẑ e z with time dependent energy eigenstates given by Ψ n ( r, t) = ψ n ( r)e ie nt ħ P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 3 / 26

Time dependence of electric dipole moment Energy Eigenstate Starting with ) μ(t) = Ψ V n ( r, t) μψ n ( r, t)dτ = Ψ V n ( q ( r, t) e r Ψ n ( r, t)dτ Inserting energy eigenstate, Ψ n ( r, t) = ψ n ( r)e ie nt ħ, gives ( ) μ(t) = ψ V n ( r)eie nt ħ q e r ψ n ( r)e ient ħ dτ Time dependent exponential terms cancel out leaving us with ) μ(t) = ψ V n ( q ( r) e r ψ n ( r)dτ No time dependence!! No bound charged quantum particle in energy eigenstate can radiate away energy as light or at least it appears that way Good news for Rutherford s atomic model. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 4 / 26

But then how does a bound charged quantum particle in an excited energy eigenstate radiate light and fall to lower energy eigenstate? P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 5 / 26

Time dependence of electric dipole moment - A Transition During transition wave function must change from Ψ m to Ψ n During transition wave function must be linear combination of Ψ m and Ψ n Ψ( r, t) = a m (t)ψ m ( r, t) + a n (t)ψ n ( r, t) Before transition we have a m (t = 0) = 1 and a n (t = 0) = 0 After transition a m (t = 0) = 0 and a n (t = 0) = 1 P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 6 / 26

Time dependence of electric dipole moment - A Transition To maintain normalization during transition we require a m (t) 2 + a n (t) 2 = 1. Electric dipole moment expectation value for Ψ( r, t) is μ(t) = V Ψ ( r, t) μψ( r, t)dτ = V [ a m (t)ψ m ( r, t) + a n (t)ψ n ( r, t)] μ [ a m (t)ψ m ( r, t) + a n (t)ψ n ( r, t) ] dτ μ(t) = a m (t)a m(t) V Ψ m ( r, t) μψ m ( r, t)dτ + a m (t)a n(t) V Ψ m ( r, t) μψ n ( r, t)dτ + a n (t)a m(t) V Ψ n ( r, t) μψ m ( r, t)dτ + a n (t)a n(t) V Ψ n ( r, t) μψ n ( r, t)dτ P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 7 / 26

Time dependence of electric dipole moment - A Transition μ(t) = a m (t)a m (t) Ψ m ( r, t) μψ m ( r, t)dτ +a m (t)a n (t) Ψ V m ( r, t) μψ n ( r, t)dτ V Time Independent + a n (t)a m (t) V Ψ n ( r, t) μψ m ( r, t)dτ + a n (t)a n (t) V Ψ n ( r, t) μψ n ( r, t)dτ Time Independent 1st and 4th terms still have slower time dependence due to a n (t) and a m (t) but this electric dipole variation will not lead to appreciable energy radiation. Drop these terms and focus on faster oscillating 2nd and 3rd terms μ(t) = a m (t)a n (t) V Ψ m ( r, t) μψ n ( r, t)dτ + a n (t)a m (t) V Ψ n ( r, t) μψ m ( r, t)dτ Two integrals are complex conjugates of each other. Since μ(t) must be real we simplify to { } μ(t) = R a m (t)a n (t) Ψ m ( r, t) μψ n ( r, t)dτ V P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 8 / 26

Time dependence of electric dipole moment - A Transition { } μ(t) = R a m (t)a n(t) Ψ V m ( r, t) μψ n ( r, t)dτ Inserting stationary state wave function, Ψ n ( r, t) = ψ n ( r)e E n t ħ, gives { [ ] } μ(t) = R a m (t)a n(t) ψ m ( r) μψ n ( r)dτ e i(e m E n )t ħ V μ mn ω mn = (E m E n ) ħ is angular frequency of emitted light and μ mn is transition dipole moment peak magnitude of dipole oscillation μ mn = V ψ m ( r) μψ n ( r)dτ Finally, write oscillating electric dipole moment vector as μ(t) = R { a m (t)a n(t) μ mn e iω mnt } P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 9 / 26

Transition dipole moment In summary, the superposition state Ψ( r, t) = a m (t)ψ m ( r, t) + a n (t)ψ n ( r, t) has an oscillating electric dipole moment vector given by μ(t) = R { a m (t)a n(t) μ mn e iω mnt } where ω mn = (E m E n ) ħ is angular frequency of emitted light a m (t)a n(t) gives information on time scale of transition. μ mn is transition dipole moment amplitude of dipole oscillation μ mn = V ψ m ( r) μψ n ( r)dτ These integrals give transition selection rules for various spectroscopies ( ) μx mn = ψ m μ ( ) x(t)ψ n dτ, μy mn = ψ V m μ ( ) y(t)ψ n dτ, μz mn = ψ V m μ z(t)ψ n dτ V where μ 2 mn = ( μx ) mn 2 ( ) + μy mn 2 ( ) + μz P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 10 / 26 mn 2

Okay, so... Wave function in superposition of energy eigenstates can have oscillating electric dipole moment which will emit light until system is entirely in lower energy state. But how does atom in higher energy eigenstate get into this superposition of initial and final eigenstates in first place? Problem is that we ve been talking about light as both classical E&M wave and as photons, but in equations we treat light as classical E&M wave. And as we just saw, treating light classically gives no explanation for how superposition gets formed. To explain spontaneous emission we need quantum field theory, which for light is called quantum electrodynamics (QED). Beyond scope of course to give a full QED treatment. Instead, examine Einstein s approach to absorption and stimulated emission of light and see what he teaches us about spontaneous emission. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 11 / 26

Light Absorption and Emission Potential energy of μ of quantum system interacting with ( r) is V = μ ( r) Shining light on system subjects it to time dependent ( r, t) ( r, t) = 0 ( r) cos ωt and adds a new term to the system s potential energy V(t) = μ 0 ( r) cos ωt When time dependent perturbation is present old stationary states (before the light was turned on) are no longer stationary states. Effect of electric field on a molecule or atom with no permanent electric dipole moment is to induce an electric dipole moment. If electric field is oscillating we expect induced electric dipole to oscillate. If light wavelength is long compared to system size (atom or molecule) we can ignore r dependence of and assume system is in spatially uniform (t) oscillating in time. Holds for atoms and molecules until x-ray wavelengths and shorter. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 12 / 26

Light Absorption and Emission To describe time dependence of electric dipole moment write wave function as linear combination of stationary state eigenfunction when V(t) is absent. Ψ( r, t) = n a m (t)ψ m ( r, t) m=1 Need to determine time dependence of a m (t) coefficients. As initial condition take a n (0) 2 = 1 and a m n (0) 2 = 0. Putting Ĥ(t) = Ĥ 0 + V(t) and Ψ( r, t) into time dependent Schrödinger Eq Ψ( r, t) ĤΨ( r, t) = iħ t and skipping many steps we eventually get da m (t) dt i ħ V Ψ m ( r, t) V(t)Ψ n ( r, t)dτ P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 13 / 26

Transition moment integrals Using our expression for V(t) and Ψ n = ψ n e ient ħ we get da m (t) i [ ] ψ dt ħ μ(t)ψ m n dτ e iωmnt 0 cos ωt V transition moment integral Setting a n (0) 2 = 1 and a m (0) 2 = 0 for n m (after many steps) we find a 2 ρ(ν m(t) = mn ) μ mn 2 t 6ε 0 ħ 2 ρ(ν mn ) is radiation density at ν mn = ω mn 2π. Rate at which n m transition occurs is R = a m (t) 2 t = ρ(ν mn) μ mn 2 6ε 0 ħ 2. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 14 / 26

P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 15 / 26

Light Absorption and Emission In 1916, before quantum theory, Einstein used Planck s distribution ( ) ρ(ν) = 8πν2 hν c 3 e hν kbt 1 0 to examine how atom interacts with light inside cavity full of radiation. For Light Absorption he reasoned that atom s rate of light absorption, R n m, depends on light frequency, ν mn, that excites N n atoms from level n to m, and is proportional to light intensity shining on atom R n m = N n B nm ρ(ν mn ) B nm is Einstein s proportionality constant for light absorption. For Light Emission, when atom drops from the m to n level, Einstein proposed two processes: Spontaneous Emission and Stimulated Emission Absorption Spontaneous Emission Stimulated Emission P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 16 / 26

Light Emission - Spontaneous and Stimulated Spontaneous Emission When no light is in cavity, atom spontaneously radiates away energy at rate proportional only to number of atoms in mth level, R spont m n = N ma mn A mn is Einsteins proportional constant for spontaneous emission. Einstein knew oscillating dipoles radiate and he assumed same for atom. Remember, in 1916 he didn t know about stationary states and Schrödinger Eq. incorrectly predicting atoms do not spontaneously radiate. Stimulated Emission From E&M Einstein guessed that emission was also generated by external oscillating electric fields light in cavity and that stimulated emission rate is proportional to light intensity and number of atoms in mth level. R stimul m n = N m B mn ρ(ν mn ) B mn is Einsteins proportional constant for stimulated emission. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 17 / 26

Light Absorption and Emission Taking 3 processes together and assuming that absorption and emission rates are equal at equilibrium we obtain R n m = R spont m n + Rstimul m n N n B nm ρ(ν mn ) = N m [A mn + B mn ρ(ν mn )] Einstein knew from Boltzmann s statistical mechanics that n and m populations at equilibrium depends on temperature N m N n = e (E m E n ) k B T = e ħω k B T We can rearrange the rate expression and substitute for N m N n to get and then get A mn + B mn ρ(ν mn ) = N n N m B mn ρ(ν mn ) = e ħω k B T B nm ρ(ν mn ) A mn ρ(ν mn ) = B nm e ħω kbt B mn P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 18 / 26

Light Absorption and Emission When Einstein compared to Planck s distribution A mn ρ(ν mn ) = B nm e ħω kbt B mn ρ(ν) = 8πν2 c 3 0 ( hν ) e hν kbt 1 he realized that B nm = B mn, i.e., stimulated emission and absorption rates must be equal to agree with Planck. This leads to ρ(ν mn ) = A ( ) mn 1 B nm e ħω kbt 1 He found relationship between spontaneous and stimulated emission rates A mn B nm = 8πν2 mn c 3 0 hν mn = 8πhν3 mn c 3 0 Amazing Einstein got this far without full quantum theory and QED. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 19 / 26

Light Absorption and Emission Earlier we found that the stimulated emission rate was R = a m(t) 2 t = ρ(ν mn) μ mn 2 6ε 0 ħ 2 = B mn ρ(ν mn ) setting this equal to Einstein s rate for stimulated emission gives B mn = μ mn 2 6ε 0 ħ 2 from which we can calculate the spontaneous emission rate A mn = 8πhν3 mn c 3 0 μ mn 2 6ε 0 ħ 2 QED tells us that quantized electromagnetic field has zero point energy. It is these vacuum fluctuations that stimulate charge oscillations that lead to spontaneous emission process. For H atom, spontaneous emission rate from 1st excited state to ground state is 10 8 /s in agreement with what A mn expression above would give. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 20 / 26

Transition Selection Rules In all spectroscopies you find that transition rate between certain levels will be nearly zero. This is because corresponding transition moment integral is zero. For electric dipole transitions we found that transition rate depends on μ mn = V ψ m ( r) μψ n ( r)dτ ψ m and ψ n are stationary states in absence of electric field. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 21 / 26

Harmonic Oscillator Selection Rules Consider transitions of quantum harmonic oscillator. Imagine oscillating dipole moment of charged particle in harmonic oscillator potential or vibration of diatomic molecule with electric dipole moment. This is 1D problem so we write electric dipole moment operator of harmonic oscillator in series expansion about its value at equilibrium μ( r) = μ(r e ) + dμ(r e ) ( r r dr e ) + 1 2 d 2 μ(r e ) ( r r dr 2 e ) 2 + 1st term in expansion, μ(r e ), is permanent electric dipole moment of harmonic oscillator associated with oscillator at rest. 2nd term describes linear variation in electric dipole moment with changing r. We will ignore 3rd and higher-order terms in expansion. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 22 / 26

Harmonic Oscillator Plug 1st two terms of expansion into transition moment integral ( μ nm = ψ m (r) μ(r e ) + dμ(r ) e) ( r r dr e ) ψ n (r)dr, ( ) = μ(r e ) ψ m (r)ψ n (r)dr + dμ(re ) ψ dr m (r)( r r e )ψ n (r)dr ( μ nm = ψ m (r) μ(r e ) + dμ(r ) e) ( r r dr e ) ψ n (r)dr, = μ(r 0 ( ) e ) ψ m (r)ψ dμ(re ) n(r)dr + ψ dr m (r)( r r e)ψ n (r)dr Since m n we know that 1st integral is zero as stationary state wave functions are orthogonal leaving us with ( ) dμ(re ) μ mn = ψ dr m (r)( r r e )ψ n (r)dr In quantum harmonic oscillator it is convenient to transform into coordinate P. J. Grandinetti ξ(chem. using4300) x = rradiating r and Dipoles ξ = in Quantum αx to obtain Mechanics Oct 27, 2017 23 / 26

Harmonic Oscillator With harmonic oscillator wave function, χ n (ξ), we obtain ( ) dμ(re ) μ mn = A dr m A n H m ξh n e ξ2 dξ Using recursive relation, ξh n = 1 H 2 n+1 + nh n 1, we obtain ( ) [ dμ(re ) 1 μ mn = A dr m A n H 2 m H n+1 e ξ2 dξ + n To simplify expression we rearrange A m A n H m (ξ)h n (ξ)e ξ2 dξ = δ m,n to ] H m H n 1 e ξ2 dξ H m (ξ)h n (ξ)e ξ2 dξ = δ m,n A m A n Substitute into expression for μ nm gives ( ) [ dμ(re ) 1 A μ mn = n δ dr 2 A m,n+1 + n A ] n δ n+1 A m,n 1 n 1 P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 24 / 26

Harmonic Oscillator Recalling 1 A n 2 n n! π 1 2 we finally obtain transition dipole moment for harmonic oscillator ( ) dμ(re ) [ ] n + 1 n μ mn = dr 2 δ m,n+1 + 2 δ m,n 1 For absorption, m = n + 1, transition is n n + 1 and μ mn 2 gives R n n+1 = ρ(ν mn ) μ mn 2 6ε 0 ħ 2 = ρ(ν mn ) ( ) dμ(re ) 2 n + 1 6ε 0 ħ 2 dr 2 For emission, m = n 1, transition is n n 1 and μ mn 2 gives R n n 1 = ρ(ν mn ) μ mn 2 6ε 0 ħ 2 = ρ(ν mn ) ( ) dμ(re ) 2 n 6ε 0 ħ 2 dr 2 P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 25 / 26

Harmonic Oscillator Selection rule for harmonic oscillator is Δn = ±1. Also, for allowed transitions ( dμ(r e ) dr ) must be non-zero. For allowed transition it is not important whether a molecule has permanent dipole moment but rather that dipole moment of molecule varies as molecule vibrates. In later lectures we will examine transition selection rules for other types of quantized motion, such as quantized rigid rotor and orbital motion of electrons in atoms and molecules. P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 26 / 26