Degenerate PT. ψ φ λψ. When two zeroth order states are degenerate (or near degenerate), cannot use simple PT.

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1 Degenerate PT When two zeroth order states are degenerate (or near degenerate), cannot use smple PT. Degenerate PT desgned to deal wth such cases Suppose the energy level of nterest s r fold degenerate H 0, E 0,,,,..., r 0 φ r 0, cψ 0, (0) () 0, + 0, ψ φ λψ E E + λe Hψ Eψ H (0) () 0 0, φ E φ 0 (0) 0, 0 0, ( ) (0) ψ 0, H E ψ E H φ () () () (0) 0 0, 0, 0, 0 λ λ

2 ψ a ψ + a ψ () 0, 0, n n ' deg zeroth order states sum over other zeroth order states Insert nto prevous equaton and multply l l by <0,k 0k { () ()} 0, c E Sk Hk () () H k E 0 S k 0 Exactly the same result we would get from a varatonal treatment usng the r degenerate zeroth order levels.

3 Varatonal Method ε ψ ψ tral tral H ψ ψ tral tral If ψ tral has one or more parameters α, β,... ε E 0 true energy ε ε 0, 0,... α β to mnmze the energy As a specal case ψ cψ lnear varatonal problem tral ε 0 c c ( H ε S ) 0 H ε S 0 k k k k In some cases, S δ n x n matrx n values of ε, n vectors k k

4 Suppose H depends on some parameter P Hellmann Feynman Theorem de H dp P Suppose ψ s normalzed for all P EP ( ) ψ( P) HP ( ) ψ( P) de ψ ψ H H ψ + ψ H + ψ ψ dp P P P ψ ψ H E ψ + E ψ + ψ ψ P P P d H E ψ ψ + ψ ψ dp P Suppose de dp H ψ ψ P 0 H H H + Px x P x But there s a catch. We have assumed ψ s the exact wavefuncton If ψ s approxmate, there are errors due to Hellmann Feynman Theorem

5 tme dep. PT H H + H t (0) () () e.g., (0) () H H + H cos t ω Ψ HΨ t Now consder the specal case of a two level system ψ ψ E Ψ () t ψ e E Ψ () t ψ e E E (0) t/ (0) t / In the absence of the tme dep dep. perturbaton Ψ Ψ + Ψ () t a () t () t a () t () t (0) () H + H () t aψ + a Ψ a Ψ + a Ψ t

6 () a (0) a (0) H () t a a Ψ + Ψ Ψ + Ψ t t ah e + ah ( t) e () E t/ () E t/ a e + a e + E t / E t / multply on left by < (0) a H ( t ) e + a H ( t ) e ae () () E t/ () () E t/ E t/ (0) ah () te + ah () te ae () E t/ () E t/ E t/ ah () t + ah () t e a, ω E E () () ωt

7 f the dagonal elements are 0 ah () te a () ωt ah () t e a () + ωt a 0 a f no perturbaton Ψ ( t) a (0) ψ (0) e + a (0) ψ (0) e E t / E t / system s frozen at ntal composton Consder a constant perturbaton H () t V () H () t V () *

8 ωt a a Ve a av e + * ω t a t ave ω a av e ω * + t ωt d a 0 Ve a ωt dt a * V e 0 a Ω Ω ( ), 4 t t ωt/ a Ae + Be e Ω ω + V Can solve by fndng the egenvalues and egenvectors that correspond to the matrx f at t 0, a 0, a ωt a () t cosω t+ snωt e ω Ω t/ () V sn t a t Ω Ω / [ t] e ω

9 Pt () a P () t a ( ω ) 4 V () sn + 4 P t V t ω + 4 V / Rab formula f the two states are degenerate P () t sn V t System smply oscllates between the two states f ω >> 4 V P t V ω () sn ωt << small probablty that the perturbaton wll drve system nto > Skp sectons 6.3, 6.4, 6.5, 6.6

10 Ensten transton probabltes Stmulated absorpton Wf Bfρ rad( Ef) Stmulated emsson W rate ( ) W B E f fρrad f B f B f μ f 0 6ε W spont f Af NW N W W f A + f Bfρrad( Ef) N f En / kt e N 3 8π υ f A f B The two equatons 3 f c Inconsstent f f f Sorted out by Ensten need to consder spontaneous emsson

11 Lfetme + energy Ψψe Et/ for an egenstate Ψ ψ, Independent of t But f a state decays n tme, ts energy s not precse Γ t Γ Er t/ Ert / t/τ Ert / Ψ ψe e ψe e ψe τ / Γ exponentally decayng state complex energy (the energy has a wdth) Et / t/ E ' t/ e τ g E e de ge ( ') ( ') ' /πτ ( E E' ) + ( /τ ) τδ E The shorter the lfetme, the greater the wdth uncertanty lke prncple