Question

1- An infinite potential box containing a particle of mass m is Perturbed as shown below. i. Find the 1st order correction to the G.S. Energy of the particle. (10 points) 0.5 -0.5. E°(n=1) ii- Find the 1st order correction (the 1st non-zero term) to the G.S. Wave Function of the particle. (15 points). iii- Find the 2nd order correction to the G.S. Energy of the particle. (15 points)

Answer 1

Saution ci pertesbation - 0.5 on 0.5 order correction in ground state - 0.5 Ground state wave function of a infinite potential is ht - sim (nominal) V L L Here 1=1, WINKHUIZZUEEN nal Ground state - Yar sin (MTA) a energy correction first order is • Ground state sinction) (2015) Sinbox) . <YolHoly = Eol = in2tu) dan = (-0.5) 2 = (-0.5) 42) ' sinkera) con E". -3.0 lsinkrimi dhe do t'e-costunu) dhe = +1'on !' cost way un] - [62) - sta sillo TE = -2 - = -0.5 ground state energy This correction in the is the first order

1 to the ground state energy of the (1) first order correction partiale, is 'n mah (0) (od - Em " Y c) = Grosand state wave fumetion = Yn ? Y Y = excited state wave function, Y") a sinton) & (0) = a sin (24) - <ul '|th Va ) sinprix) (0.5) tz sinbou) du - - 0.5 X2 sin arx) sin(tn) dan sin A sinB = 1 [las (A-B) = cos(A+B) Jusing this = <!um 1'= -0.5*2 [$cosmhu da - 'cas a touche = <42 1*p1 827 =toist [ 1 sin mn]! - 1 2 3 4 [sin ert)! ] o - o n. [ <Am ) ' 19m) 1 = 0 min Elo Em

elli) second onder correction to the ground state pression is given by correction energy expression Second order 2 K for 1 hp lens 1? Ap= -0.5=-6 = perturbed Hamiltonian. ET- Emos . MIE - Em Y = Y = Le groung state = a sincron) excited state = r2 sin carth) second excited State= V2 sin (374) <&ml Hp1 8) = < 4 'Hp 1440 from equation & UDIWAN <lm I Molde) s < Y3 IHp 1T1) = 0 J sinara sinkone) de <931 Hp/41) = 2 -0.582 = -0.5x2 { s ces anndan d sinaran an ] OS X 2 2 = 0 correction in energy is 3ds in ground State.