Schrodinger Equation - Quantum Mechanics - Solved Past Exam, Exams of Quantum Mechanics

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Solutions to PC3130 AY0708 Paper 1(a) Time-independent Schrodinger equation: )()()( 2 2 xExxV m p ψψ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + By spherical symmetry, we write )()( rVxV = , ),()(),,()( φθφθψψ mlYrRrx == Substitute 2 2 22 r lpp r += into the Schrodinger equation, ),()(),()()( 22 2 22 φθφθ ml m l r YrERYrRrV mr l m p =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ++ Since r rri pr ∂ ∂ = 1 , r rr r rr r rr pr 2 2 222 111 ∂ ∂ −= ∂ ∂ ∂ ∂ −= and ml m l YllYl )1( 22 += , we have )()()( 2 )1( 2 )()(1)(1 2 )1(1 2 ),()(),()()( 2 )1(1 2 2 2 2 22 2 2 2 22 2 2 2 22 rErRrrRrV mr ll rm rERrrR r rV rmr ll rrm YrERYrRrV mr llr rrm m l m l =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + ∂ ∂ − =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + ∂ ∂ − =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +++ ∂ ∂− φθφθ Let )()( rrRry = , we have 0)())((2)1( )()()( 2 )1( 2 222 2 2 2 2 22 =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −+ + − ∂ ∂ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + ∂ ∂ − ryrVEm r ll r rEyryrV mr ll rm We have reduced the Schrodinger equation to 1D radial equation. (shown) 1(b) 0)( 4 1)1( 22 2 =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +− + − ∂ ∂ ρ ρρρ ynll Consider boundary conditions at 0→ρ and ∞→ρ : When 0→ρ , 0)()1( 22 2 =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − ∂ ∂ ρ ρρ yll ⇒ 1)( +≈ ly ρρ When ∞→ρ , 0)( 4 1 2 2 =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − ∂ ∂ ρ ρ y ⇒ 2/)( ρρ −≈ ey Let )(ρV to represent the behaviour of )(ρy between 0 and ∞ , 2/1 )()( ρρρρ −+=∴ eVy l (shown) 2/12/12/ )( 2 1)(')()1()(' ρρρ ρρρρρρρ −+−+− −++= eVeVeVly lll +−+++= −−−− 2/2/2/1 )( 2 1)(')1()()1()(" ρρρ ρρρρρρρ eVeVleVlly lll 2/12/12/ 2/12/12/ )( 4 3)(' 2 1)()1( 2 1 )(' 2 1)(")(')1( ρρρ ρρρ ρρρρρρ ρρρρρρ −+−+− −+−+− +−+− −+++ eVeVeVl eVeVeVl lll lll 2/12/2 )(")(')22()(4 1)1()1()(" ρρ ρρρρρρ ρρ ρ −+− +−++⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + − + = eVeVlyllly ll Substitute into the given differential equation, we get 0)1(')22(" =−−+−++ VlnVlV ρρ When 1−= nl , we get 0')2(" =−+ VnV ρρ Clearly, =)(ρV constant is a solution, we thus have 2/1const)( ρρρ −+⋅= ey l , 0 2 2 na rr == κρ 1(c) ∫∫= ππ ψφθθ 2 0 2 0 2 )(sin)( xddrrP nlmn ∫∫ −−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = π ρπ φθρφθθ 2 0 222 3 0 0 2 ),( )!2( 12sin ml n Ye nna ddr Since 0 2 na r =ρ , ρ π ρ ρρθθ −− =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ∫ ennaennadrP nn n 2 0 0 2 0 )!2( 14 )!2( 12sin)( )2( )!2( 14 212 0 ρρ ρρ −−− −= een nnadr dP nnn When 0= dr dPn , n na rn 222 0 =⇒=ρ 0 2anr =∴ (shown) 2(a) Hamiltonian of 3D harmonic oscillator: (Refer to lecture notes for more details) 22 2 2 3 2 2 2 1 2 2 3 2 2 2 1 2 1 2 )( 2 1 222 rm m pxxxm m p m p m pH ωω +=+++++= The time-independent Schrodinger equation for 3D harmonic oscillator is: ),,(),,( 2 1 2 22 2 φθψφθψω rErrm m p =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + We can transform the above equation into 1D radial equation, just as the case for hydrogen atom. By considering the boundary conditions, we can obtain the radial wavefunction as a product of different functions. /1)( 2/2 2 w Ji w JieR w εε ε +==− Since )( 2ε−= wRR , [ ] /1,1 22 2 wvu JiJJ ε ε +=+ [ ] wvu JiJJ =, (shown) 4(a) Consider the Hamiltonian of a pair of particle and its anti-particle, e.g. electron and positron. This Hamiltonian commutes with the permutation operator, but the states need not be symmetrized or antisymmetrized. 4(b)(i) Identical bosons: Ground state: )sin()sin(2),( 212111 b x b x b xx ππψ = Energy of ground state, ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 22 11 2 2 mb E π 1st excited state: ⎥⎦ ⎤ ⎢⎣ ⎡ += )sin()2sin()2sin()sin(2),( 21212112 b x b x b x b x b xx ππππψ Energy of 1st excited state, ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 22 12 2 5 mb E π 2nd excited state: )2sin()2sin(2),( 212122 b x b x b xx ππψ = Energy of 2nd excited state, ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 22 22 2 8 mb E π 4(b)(ii) Identical fermions: Ground state: ⎥⎦ ⎤ ⎢⎣ ⎡ −= )sin()2sin()2sin()sin(2),( 21212112 b x b x b x b x b xx ππππψ Energy of ground state, ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 22 12 2 5 mb E π 1st excited state: ⎥⎦ ⎤ ⎢⎣ ⎡ −= )sin()3sin()3sin()sin(2),( 21212113 b x b x b x b x b xx ππππψ Energy of 1st excited state, ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 22 13 2 10 mb E π 2nd excited state: ⎥⎦ ⎤ ⎢⎣ ⎡ −= )2sin()3sin()3sin()2sin(2),( 21212123 b x b x b x b x b xx ππππψ Energy of 2nd excited state, ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 22 23 2 13 mb E π Ground state wavefunction of 3 identical fermions: ⎢ ⎣ ⎡ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= b x b x b x b x b x b x b 321321 3123 sin 3sin2sin3sin2sinsin 3 2 ππππππψ ⎥ ⎦ ⎤ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ b x b x b x b x b x b x b x b x b x b x b x b x 321321 321321 sin2sin3sin3sinsin2sin 2sin3sinsin2sinsin3sin ππππππ ππππππ 4(c) 0,00,0 )2()1()2()1( baba SSSS = 2 1, 2 1 2 1, 2 1 2 1 2 1, 2 1 2 1, 2 1 2 1 2 1, 2 1 2 1, 2 1 2 1, 2 1 2 1, 2 1 2 1 )2()1()2()1( )2()1( −−+−−= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−−= baba ba SSSS SS Taking )1(3 )1( SSa = , let b̂ be the vector on 31xx plane, θθ sincos )2( 1 )2( 3 )2( SSSb += 2 1, 2 1)sincos( 2 1, 2 1 2 1 2 1, 2 1)sincos( 2 1, 2 1 2 1 )2( 1 )2( 3 )1( 3 )2( 1 )2( 3 )1( 3 )2()1( −+−+ −+−= θθ θθ SSS SSSSS ba θ θθθθ cos 4 2 1, 2 1 2 1, 2 1sin 4 cos 42 1, 2 1)sincos( 2 1, 2 1 2 22 )2( 1 )2( 3 )1( 3 −= −+−=−+− SSS θ θθθθ cos 4 2 1, 2 1 2 1, 2 1sin 4 cos 42 1, 2 1)sincos( 2 1, 2 1 2 22 )2( 1 )2( 3 )1( 3 −= −−−−=−+− SSS θθθ cos 4 cos 42 1cos 42 1 222)2()1( −=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −+⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=ba SS (shown)