1000 Solved Problems in Modern Physics

204 3 Quantum Mechanics – II Classically,E=ka 2 2 = ( n+^12 ) ω(quantum mechanically)≈nω(n→∞) Thereforea^2 =^2 nkω= ( 2 n k ...

3.3 Solutions 205 P(x)= 1 π √ a^2 −x^2 (5) 3.57 Schrodinger’s equation in one dimension is ( − ^2 2 m ) d^2 ψ dx^2 +V(x)ψ=Eψ (1 ...

206 3 Quantum Mechanics – II = Nn^2 α^3 ∫∞ −∞ Hn^2 (ξ)ξ^2 e−ξ 2 dξ = ( α √ π 2 nn! )( 1 α^3 ) (2n+1) 2 2 nn! √ π = 1 α^2 ( n+ 1 ...

3.3 Solutions 207 c ( 1 λj+ 1 − 1 λj ) =c×(20.556 cm−^1 )= ^2 I 0 Moment of inertia I 0 =  4 π^2 c× 20. 556 = 6. 63 × 10 −^2 ...

208 3 Quantum Mechanics – II It is seen from the last column of the table that the degeneracy D is given by the sum of natural n ...

3.3 Solutions 209 Fig. 3.24ψ 1 (x)forSHO Substitute Bxexp(−x^2 / 2 a)=ψ 1 anda= ( mω ) 1 / 2 and rearrange to get − ( ^2 2 m ) ...

210 3 Quantum Mechanics – II ∇^2 = d^2 dr^2 + ( 2 r ) d dr (2) Inserting (2) in (1) and performing the integration we get <T& ...

3.3 Solutions 211 u^2210 dτ= ( 1 8 πa^30 )[∫∞ 0 y^4 e−y ( dy 4 a^20 ) a^50 ][ cos^3 θ 3 ] 1 − 1 (2π) = ( 1 24 ) ×4!= 1 Similarly ...

212 3 Quantum Mechanics – II Writing sin^2 θ= 1 −cos^2 θand simplifying we getu∗u=^29 A^23 e−^2 xx^4 which is independent of bot ...

3.3 Solutions 213 Fig. 3.25Probability distribution of electron in n=2 orbit of H-atom hν= 0 .088 MeV λ= 8.^12418 × 104 nm= 0 .0 ...

214 3 Quantum Mechanics – II =− ∂ ∂b ( k b^2 +k^2 ) = 2 kb (b^2 +k^2 )^2 Therefore the integral in (4) is evaluated as ( 2 k a 0 ...

3.3 Solutions 215 3.3.6 Angular Momentum ................................ 3.77 L= ∣ ∣ ∣ ∣ ∣ ∣ ijk xyz px pypz ∣ ∣ ∣ ∣ ∣ ∣ =i(ypz ...

216 3 Quantum Mechanics – II ∴(Sp+Sn)·(Sp+Sn)=S·S Sp^2 +Sn^2 +2Sp·Sn=S^2 = 0 (^1) / 2 (1/ 2 +1)+ (^1) / 2 (1/ 2 +1)+2Sp·Sn= 0 Or ...

3.3 Solutions 217 x=rsinθcosφ y=rsinθsinφ (2) z=rcosθ r^2 =x^2 +y^2 +z^2 (3) tan^2 θ= x^2 +y^2 z^2 (4) tanφ= y x (5) Differentia ...

218 3 Quantum Mechanics – II NowLzg(φ)=−i∂φ∂g=mg(φ) Thus the z-component of angular momentum is quantized with eigen value. 3 ...

3.3 Solutions 219 Apart from the factor 1/r^2 , the angular part is seen to be ∂^2 ψ ∂θ^2 +cotθ ∂ψ ∂θ + 1 sin^2 ∂^2 ψ ∂φ^2 3.84 ...

220 3 Quantum Mechanics – II (Jx) 12 =< 1 |Jx| 2 >= 〈 1 2 , 1 2 |Jx| 1 2 ,− 1 2 〉 = 1 2 [( 1 2 − 1 2 + 1 )( 1 2 + 1 2 )] 1 ...

3.3 Solutions 221 =[(1−0)(1+ 0 +1)]^1 /^2 | 1 >= √ 2 | 1 > J+ ∣ ∣ ∣ 3 >=[[1−(−1)](1− 1 +1)] (^12)  ∣ ∣ ∣ 2 >= √ 2 ...

222 3 Quantum Mechanics – II Clearly the statesψ ( 3 2 , 3 2 ) andψ ( 3 2 ,− 3 2 ) can be formed in only one way ψ ( 3 2 , 3 2 ) ...

3.3 Solutions 223 Thus ψ ( 1 2 , 1 2 ) = √ 2 3 φ(1,1)φ ( 1 2 ,− 1 2 ) − √ 1 3 φ(1,0)φ ( 1 2 , 1 2 ) (11) Similarly ψ ( 1 2 ,− 1 ...