# Download Applied quantum mechanics: Solutions manual by Levi A.F.J. PDF

By Levi A.F.J.

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**Additional resources for Applied quantum mechanics: Solutions manual**

**Sample text**

There is now only one term on the right-hand-side and we can write ih 4 i ωm nt d a m ( t ) = a n ( t ) W mn e dt Integration from the time when the perturbation is applied at t = 0 gives t′ = t iω mn t ′ 1 a m ( t ) = ----- ∫ W mn e d t′ ih t ′ = 0 (b) The transition probability from state | n〉 to state |m〉 is P nm = a m ( t ) . Using our solution in part (a) we have 2 1 P nm = -----2 h t′ = t ∫ 〈m |V ( x, t ) |n〉 e iω mn t ′ 2 t′ = t 2 V = -----20 h d t′ t′ = 0 ∫ 〈m |x 3 | n〉 e ( iωm n –1 ⁄ τ )t ′ 2 d t′ t′ = 0 In the problem we have initial state |n = 0〉 and we consider the long time limit, t → ∞ , this allows us to write t′ = ∞ 2 2 V0 2 3 ( imω –1 ⁄ τ )t ′ e d t′ P nm = -----2 〈m | x |0〉 ∫ h t′ = 0 where, for the harmonic oscillator, the non-zero positive integer m multiplied by the frequency ω is related to the difference in energy eigenvalue by mω = ( E m – E 0 ) ⁄ h .

2π h (b) Current is assumed proportional to applied voltage, electron velocity v ( E ) , transmission coefficient T ( E ) , and density of states. EF + e V I = e ∫ E F + eV v ( E )T ( E )D 2 ( E ) d E = e EF ∫ EF 2mE m - --h-- --------- ---------- m h 2 - T ( E ) 2 πh 2 dE If we assume T ( E ) = T ( E F ) = 1 , then 2m I = e ------------2 2πh EF + e V ∫ EF 2m 2 3 ⁄ 2 EF + eV 2m 3 ⁄ 2 eV 3 ⁄ 2 E d E = ------------2 ⋅ - [ E ] EF = ------------2 E F 1 + ------ – 1 EF 2π h 3 3πh When E F « eV , conductance per electron spin I 2m G = --- = ------------2 V 3 πh V For the other situation when EF » e V we use the binomial expansion 3 eV n eV 3 ⁄ 2 ( 1 + x ) = 1 + nx + … so that 1 + ------ = 1 + -- ------ .

2π h (b) Current is assumed proportional to applied voltage, electron velocity v ( E ) , transmission coefficient T ( E ) , and density of states. EF + e V I = e ∫ E F + eV v ( E )T ( E )D 2 ( E ) d E = e EF ∫ EF 2mE m - --h-- --------- ---------- m h 2 - T ( E ) 2 πh 2 dE If we assume T ( E ) = T ( E F ) = 1 , then 2m I = e ------------2 2πh EF + e V ∫ EF 2m 2 3 ⁄ 2 EF + eV 2m 3 ⁄ 2 eV 3 ⁄ 2 E d E = ------------2 ⋅ - [ E ] EF = ------------2 E F 1 + ------ – 1 EF 2π h 3 3πh When E F « eV , conductance per electron spin I 2m G = --- = ------------2 V 3 πh V For the other situation when EF » e V we use the binomial expansion 3 eV n eV 3 ⁄ 2 ( 1 + x ) = 1 + nx + … so that 1 + ------ = 1 + -- ------ .