1000 Solved Problems in Modern Physics

164 3 Quantum Mechanics – II The first term on the RHS is zero at both the limits. − ∫ ∂^2 ψ∗ ∂x^2 xψdx=− ∫ dψ∗ dx ( ψ+x dψ dx ) ...

3.3 Solutions 165 Integrating by parts twice d dt <Px>=− ∫ ψ∗ [ ∂ ∂x (Vψ)−V ∂ψ ∂x ] dτ =− ∫ ψ∗ ∂V ∂x ψdτ=< −∂V ∂x > ...

166 3 Quantum Mechanics – II Following the same procedure g(φ) sinθ d dθ sinθ df(θ) dθ + f(θ) sin^2 θ d^2 g(φ) dφ^2 +λf(θ)g(φ)= ...

3.3 Solutions 167 force” is supplied by the potential energy, and hence adds to theV(r) which appears in (13) for the radial mot ...

168 3 Quantum Mechanics – II ( 1 sinθ ) ∂ ∂θ ( sinθ df dθ ) +λf(θ)=0(2) ( d dθ ) = ( d dμ ) · ( dμ dθ ) =−sinθ d dμ Writingf(θ)= ...

3.3 Solutions 169 Or d^2 ψ(x) dx^2 + ( 2 mE ^2 ) ψ(x)=0(2) Writing α^2 = 2 mE ^2 (3) Equation (2) becomes d^2 ψ dx^2 +α^2 ψ=0( ...

170 3 Quantum Mechanics – II ∫a 0 ψn∗(x)ψn(x)dx= 1 A^2 ∫α 0 sin^2 (nπx a ) dx= 1 ( A^2 2 )( x−cos ( 2 nπx a ))∣ ∣ ∣ a 0 =A^2 a= ...

3.3 Solutions 171 Fig. 3.7Deuteron wave function and energy whereμis the reduced mass=M/ 2 ,M,being neutron of proton mass. With ...

172 3 Quantum Mechanics – II The solutions are u 1 (r)=Asinkr+Bcoskr;r<R (11) u 2 (r)=Ce−γr+Deγr;r>R (12) Boundary conditi ...

3.3 Solutions 173 3.21 u=Ce−kr ∫∞ 0 |u| (^2) dr=c 2 ∫∞ 0 e − 2 kr=c^2 2 k=^1 C= √ 2 k The probability that the neutron – proton ...

174 3 Quantum Mechanics – II The ground state corresponds ton=1 and the first excited state ton= 2 ,m= 8 meandL=1nm= 106 fm. Put ...

3.3 Solutions 175 d^2 ψ dx^2 + ( 2 mE ^2 ) ψ= 0 d^2 ψ dx^2 +α^2 ψ=0(6) withα^2 = 2 mE ^2 (7) ψ 2 =Csin Odd αx+Dcos even αx (8) ...

176 3 Quantum Mechanics – II Fig. 3.9η−ξcurves for class I solutions. For explanation see the text Note that from (15) and (2),α ...

3.3 Solutions 177 Fig. 3.10η−ξcurves for class II solutions. For explanation see the text (After Leonard I. Schiff, Quantum mech ...

178 3 Quantum Mechanics – II ∫R 0 |ψ 1 |^2 dτ+ ∫∞ R |ψ 2 |^2 dτ= 1 ∫R 0 u^21. 4 πr^2 dr/r^2 + ∫∞ R u^224 πr^2 dr/r^2 = 1 A^2 ∫ R ...

3.3 Solutions 179 α^2 a^2 = 2 mEa^2 ^2 = n^2 π^2 4 E= n^2 π^2 ^2 8 ma^2 (neven) ThusE= n^2 π^2 ^2 8 ma^2 ,n= 1 , 2 , 3 ... 3. ...

180 3 Quantum Mechanics – II Region 3: (x>a)V= 0 Solution:ψ 3 =Dexp(ik 1 x) (b) Boundary conditions: ψ 1 (0)=ψ 2 (0)→ 1 +A=B+ ...

3.3 Solutions 181 Using (11) in (9) and notingk^21 +k^22 =^2 mV 2 b k^22 = 2 mVb ^2 andk^21 k^22 = ( 2 m ^2 ) 2 E(Vb−E) we fi ...

182 3 Quantum Mechanics – II <E>=<ψ|H|ψ> =<(C 1 ψ 1 +C 2 ψ 2 )|H|(C 1 ψ 1 +C 2 ψ 2 )> =<C 1 ψ 1 +C 2 ψ 2 |C ...

3.3 Solutions 183 G= 2  (2m) 1 2 ∫b a ( zZe^2 r −E ) 1 / 2 dr wherez= 2 Now at distancebwhere the alpha energy with kinetic ene ...