1000 Solved Problems in Modern Physics

184 3 Quantum Mechanics – II Fig. 3.13 For (a) the inside and outside wave functions are as in the deuteron Problem 3.19. For (b ...

3.3 Solutions 185 3.37 (a)un= ( 2 L ) 1 / 2 sin (nπx L ) <x>= ∫ L 0 u∗nxundx= ( 2 L )∫L 0 xsin^2 (nπx a ) dx = L 2 + ( L 4 ...

186 3 Quantum Mechanics – II H= ( − ^2 2 m ) ∇^2 +a [ x^2 +y^2 +z^2 − 5 6 x^2 ] = ( − ^2 2 m ) ∇^2 +a [ x^2 6 +y^2 +z^2 ] (2) ...

3.3 Solutions 187 3.39 (a) [( − ^2 2 m ) ∇^2 +V ] ψ(x,y,z)=Eψ(x,y,z)(1) PutV= 0 ( − ^2 2 m )[ ∂^2 ∂x^2 + ∂^2 ∂y^2 + ∂^2 ∂z^2 ] ...

188 3 Quantum Mechanics – II Going back to (3) 1 Y ∂^2 Y ∂y^2 + 1 Z ∂^2 Z ∂z^2 =−α^2 1 Z ∂^2 Z ∂z^2 =− 1 Y ∂^2 Y ∂y^2 −α^2 (6) E ...

3.3 Solutions 189 None of the numbersnx,ny,ornzcan be zero, otherwiseψ(x,y,z) itself will vanish. For an infinitely deep potenti ...

190 3 Quantum Mechanics – II ψ 2 =Cexp(ik 2 x)(5) where k^22 = 2 m(E−U 0 ) ^2 (6) It represents the transmitted wave to the rig ...

3.3 Solutions 191 Dividing (10) by (9) gives ik(A−B) A+B =−α (11) Diagrams forψat aroundx= 0 3.42k 1 = ( 2 mE ^2 ) 1 / 2 ;k 2 = ...

192 3 Quantum Mechanics – II This is a direct result of the fact that the current density is constant for a steady state. Thus|A ...

3.3 Solutions 193 3.44 (a) Fig. 3.18Penetration of a rectangular barrier (b) Region 1,x< 0 d^2 ψ dx^2 +k^2 ψ= 0 withk^2 =^2 m ...

194 3 Quantum Mechanics – II Therefore 2αL= 2 × 8. 8748 × 109 × 0. 3 × 10 −^9 = 5. 3249 T= 16 ( 2 5 )( 1 − 2 5 ) e−^5.^3249 = 0. ...

3.3 Solutions 195 When the boundary conditions are imposed, β= n 2 πy a ψy=Gsin (n 2 πz a ) Thusψ(x,y)=ψxψy=Ksin (n 1 πx a ) sin ...

196 3 Quantum Mechanics – II or kR=nπ→k= nπ R (4) Complete unnormalized solution is u(r)=Asin (nπr R ) (5) The normalization con ...

3.3 Solutions 197 Fig. 3.19Class I and Class II wave functions 3.49 The analysis for the reflection and transmission of stream o ...

198 3 Quantum Mechanics – II ever the conditionk 2 a=nπis satisfied, the width of the maxima becomes broader as the electron’s i ...

3.3 Solutions 199 Whenm =0, (2) can be written as [ ∂^2 ∂r^2 + ( 2 r ∂ ∂r − m^2 c^2 ^2 )] φ(r)= 0 or 1 r^2 ∂ ∂r ( r^2 ∂φ ∂r ) = ...

200 3 Quantum Mechanics – II 3.52u 0 = [ √α π ] e−ξ (^2) / 2 H 0 (ξ);ξ=αx P= 1 − ∫a −a |u 0 |^2 dx= 1 − 2 ∫a 0 (α/ √ π)e−ξ (^2) ...

3.3 Solutions 201 The boundary condition thatu/rbe finite atr=0 demands thatb=0. Thus,ψis proportional torl. The probability tha ...

202 3 Quantum Mechanics – II Br^2 =x (8) Substitute (8) in (7) and simplify, to obtain xd^2 v dx^2 + dv dx (C−x)−Av=0(9) which i ...

3.3 Solutions 203 Similarly<Δpx>^2 =<p^2 > (^1) / 2 mω (^2) <x (^2) >· ( 1 2 m ) <px>^2 ≤ 1 4 ( ω 2 ) 2 ...