Energy Shifts & Probabilities in Quantum Mechanics with Perturbation Theory, Study notes of Physics

Download Energy Shifts & Probabilities in Quantum Mechanics with Perturbation Theory and more Study notes Physics in PDF only on Docsity! 1 Perturbation Theory: Time Independent and Time Dependent Let’s go back to a 2-state system and use the general perturbation analysis. Time independent perturbation 0 0 0 mH Em mψ ψ= m 1,2= H Eψ ψ= Note the same 0H Hamiltonian, but 0H H H'= + ; H' perturbation≡ 0 1 1ψ ≡ 0 2 2ψ ≡ Use 0 01 1 2 2 1 2a a a 1 a 2ψ ψ ψ= + = + ( ) ( )1 2a H E 1 a H E 2 0− + − = 1 1 2 2a 1 H 1 a E 1 1 a 1 H 2 a 1 2 0E− + − = Because 1 and 2 are orthogonal: 1 11 1 2 12a H a E a H 0− + = And, similarly: 1 21 2 22 2a H a H a E 0+ − = Solve for 1a and 2a : 11 12 21 22 H E H 0 H H E − = − determinant General Solution: ( ) ( ){ }1/ 2211 22 11 22 12 211 1E H H H H 4H H2 2± = + ± − + ; Exact case for a 2-state system If mmH 0′ = , then ( ) ( ){ }1/ 22 21 2 1 21 1E E E E E 42 2 ε± = + ± − + 2 2 12 21 12H H Hε ′ ′ ′= = The stronger the perturbation, the larger the energy shift of the two levels. ( 2ε is larger) Also, the smaller 1 2E E− , the larger the perturbation. The closer 1 2E E− , the larger the effect of the perturbation. Note when 1 2E E= (degenerate), E E 2ε+ −− = 2 Now assume ( )22 1 2E Eε << − small perturbation Expand E± : ( ) ( )1 2 1 2 2 1 1 2E E E E E 1 . . . 2 2 E ε ± ⎡ ⎤= + ± − + +⎢ ⎥∆⎣ ⎦ So, to 2nd order: 2 1E E E ε + ≅ − ∆ 2 2E E E ε − ≅ + ∆ The shift in energy = 2 E ε ∆ ; this shows the size of pert. is important and E∆ is important. We can solve for the perturbed wave functions: If we write 0 0 1 2 0 0 1 2 cos sin sin cos ψ ψ ζ ψ ζ ψ ψ ζ ψ ζ + − ⎧ = +⎪ ⎨ = −⎪⎩ Then we have, ( ) 12 1 2 2 Htan 2 E Eζ ′ = − Say, 1 2E E= ; then 4 πζ = ( )tan 2ζ = ∞ Then we can write, ( )0 01 212ψ ψ ψ+ = + ( ) 0 0 1 2 1 2 ψ ψ ψ− = − So, each state is a 50% mixture of the two original states. If 2 1E E− is large, then 12 -2 Htan 2 2 Eζ ζ ′ ≅ = ∆ and sinζ ζ∼ , cos 1ζ ≈ 120 01 2 H E ψ ψ ψ+ ′ ≈ − ∆ 120 02 1 H E ψ ψ ψ− ≈ + ∆ ***Note the mixing of the two states. We have already calculated the general case in the previous lecture. 5 ***APPLICATION*** Let’s apply the closure approximation to a real – but simple – case. Ground states of a H-atom; perturbation static electric field ˆH zEe′ = E = electric field ẑ = direction of the E-field Note: ( )0 H 0 0′ = , by symmetry. Note: There is no second order Hamiltonian; so, 0 H 0 0′′ ≡ 2 20 0 0 n 0 0 E E E En n z n z e ≠ < >< >′′ = −∑ Remember 0 0 0z< >= and apply closure 2 2 2 2E 0 0e zε = < > 2 2 2 0 E 0 r 0 E 3 E e′′ = ∆ because 2 210 0 0 0 3 z r< >= < > 2 2 2 2r x y z= + + How to find approximate solutions & energies of complex molecular systems: The Rayleigh ratio: * trial trial * trial trial H E d d ψ ψ τ ψ ψ τ = ∫ ∫ Variation theorem says that 0E E> for any trialψ .We used this already. As homework you will prove it. Expand in orthonormal functions: ( )iψ trial ci iψ ψ=∑ , , , , c c i H c c E i j i j ij i j i j i j i j ij i j i j j H c c i j c c S = = ∑ ∑ ∑ ∑ i H Hi jj dψ ψ τ≡ ∫ ; find the minimum value of E 6 We want the minimum value of E. 2 , , c c c c Hc H c H E0 c c c c c j kj i ik i j ijj kj i ik j i ij i k i j ij i j i j ij i j S S S S ⎛ ⎞ ++ ⎜ ⎟ ∂ ⎝ ⎠= = − ∂ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ∑ ∑ ∑∑ ∑ ∑ ∑ Using the expression for E (see previous page), ( ) ( ) , , c H E c H E E0 c c c c c j kj kj i ik ik j i k i j ij i j ij i j i j S S S S − − ∂ = = + ∂ ∑ ∑ ∑ ∑ The numerators must be zero. ( )c H E 0i ik ik i S− =∑ Secular Equation for every k These linear equations have a solution provided that det H E 0ik ikS− = We solve this determinant for all the values for E that solve the resulting polynomial equation. Then for each value of E we solve for the ci values, to give us trial ci iψ ψ=∑ . Note, the variation principle leads to an upper bound to the energy of a system. (1) The energy may be a good approximation; however, other properties, such as the dipole moment, may not be good approximations. (2) Methods for finding lower bounds of E true E sandwiched between E and En . 7 Time Dependent Perturbation Before going to the general case we will look at just two states. 0H H H (t)′= + H (t) 2H cos( t)ω′ ′= basis functionsnψ ⇒ ( )H i tψ ψ ∂ = ∂ ; 0H En n nψ ψ= ( )-i E /( ) n tn nt eψ ψ= ( )0H n ni tψ ψ ∂ = ∂ Assume the state can be expressed as: 1 1 2 2( ) a ( ) ( ) a ( ) ( )t t t t tψ ψ ψ= + We want to calculate 2a ( )n t ; probability of being in state n at time t. Note both a ( )n t and ( )n tψ are functions of time. ( )H i t ψ ψ∂= ∂ 1 2 1 1 1 2 2 2 a aH a ad di t dt t dt ψ ψ ψ ψ ψ∂ ∂⎛ ⎞= + + +⎜ ⎟∂ ∂⎝ ⎠ [ ]0 1 1 2 2H H H a aψ ψ ψ′⎡ ⎤= + +⎣ ⎦ a an n d dt ≡ ; So, [ ]1 1 2 2 1 1 2 2a H (t) a H (t) a aiψ ψ ψ ψ′ ′+ = + 1- E / 1 1 i teψ = ; 2 . . .ψ = ( )1 2 1 2- E / - E / - E / - E /1 2 1 2a H (t) 1 a H (t) 2 a 1 a 2i t i t i t i te e i e e′ ′+ = + Multiply by 1 . . . & integrate ( )1 2 1- E / - E / - E /1 11 2 12 1a H (t) a H (t) ai t i t i te e i e′ ′+ = 10 *1. Note the exact solution of a 2-state system requires the solution of a 2nd order differential equation. The exact solution of an n-state system would require the solution of an nth order differential equation. *2. The exact solution is only possible with a trivially simple perturbation (e.g. constant in time). *Use the system of the “variation of constant” developed by Dirac . Many Level System: Variation of Constants: - E /0( ) ni tn nt eψ ψ= 0H n ni t ψ ψ∂= ∂ - E /0( ) a ( ) ( ) a ( ) ni tn n n n n n t t t t eψ ψ ψ= =∑ ∑ H i t ψψ ∂= ∂ Do the same as for the 2-level case. 0H a ( )H ( ) ( ) a ( )H ( ) ( )n n n n n n t t t t t tψ ψ ψ′= +∑ ∑ H a ( ) a( ) ( )nn n n n i t i i t t t t ψψψ ψ∂∂= = + ∂ ∂∑ ∑ a ( )H ( ) ( ) a( ) ( )n n n n n t t t i t tψ ψ′ =∑ ∑ 0nn ψ= Now include the t-dependence: - E / - E /a ( )H ( ) a ( )n ni t i tn n n n t t n e i t n e′ =∑ ∑ Now multiply by k on the left and integrate. . . - E / - E /a ( ) H ( ) an ki t i tn k n t k t n e i e′ =∑ Define: E Ekn k nω = − knH Hk n′ ′= 11 Then, kn 1a ( ) a ( )H ( ) kni tk n n t t t e i ω′= ∑ This equation is exact kn0 1a ( ) a (0) a ( )H ( ) kn t i t k k n n t t t e i ω′− = ∑∫ The last equation is a problem; we would need to know all a ( )n t ? Assume the initial state i is the only one with a high probability. Assume a ( ) 0n t ≈ except for i. 0 1a ( ) a ( )H ( ) fi t i t f i fit t t ei ω′≅ ∫ **Note that a ( ) 1i t ≈ , so 0 1a ( ) H ( ) fi t i t f fit t ei ω′≅ ∫ NO INTERNEDIATE STATES Now let’s look at the case where the perturbation is turned on slowly. H (t)′ = ( )-kt 0 H 1 e ⎧⎪ ⎨ ′ −⎪⎩ 0 0 t t < ≥ Use the equation, 0 1a ( ) H ( ) fi t i t f fit t ei ω′= ∫ ( )-kt 0 H a ( ) 1 fi t i tfi f t e e dti ω ′ = −∫ ( ) ( ) -H 1 1a ( ) k fifi k i ti t fi f fi fi e et i i i ωω ω ω −⎡ ⎤′ − − ⎢ ⎥= + −⎢ ⎥⎣ ⎦ Now assume 1/t k>> and 2 2k fiω<< , so the kte− term 0→ . Then we can calculate 2 a ( )f t∼ 12 22 2 2a ( ) Hf fi fit ω′= Compare the 1st order correction to the time independent perturbed wavefunction. 15 110fi sω −≈ , 3 110k s−≈ So for the usual frequency perturbation this approximation holds. The molecule just slowly (adiabatically) changes the nature of the wave function. Note: the higher the frequency fiω , the slower the transition to the new state. fi f iE Eω = −