ψ 2 ψ 3 h 2 d 2 2m dx 0 x > a V (x) = V0 x a x < a = B sin kx or B'cos kx a x a = Ce κx x > a k =, κ = h Since the solution has to be odd, ψ 2

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1 .0 Proble Set 3 Solution.Solution The energy eigenvalue proble for the given syste is: Ĥψ = Eψ h d Ĥ = + V (x) dx 0 x > a V (x) = V0 x a For bounded particle, the solution is: ψ = Ae κx x < a ψ = B sin kx or B'cos kx a x a ψ 3 = Ce κx x > a ( E + V0 ) E k =, κ = h h Since the solution has to be odd, ψ = B sin kx Also we know, ψ (a) = ψ 3 (a), if ψ 3 (x) is decaying, there has to be at least a quarter of wave fro 0 to a within the well. This eans λ / 4 a. Just being bounded eans λ / 4 = a and E = 0. π π V0 π h This gives us k = = = V 0 = λ 4a h 8a.Solution iboff 5. (a) Proof [ Â + Bˆ, Ĉ] = [ Â, Ĉ] + [Bˆ, Ĉ] HS = [ Â + Bˆ, Ĉ] = ( Â + Bˆ)Ĉ Ĉ( Â + Bˆ) = Â Ĉ + BˆĈ ĈÂ ĈBˆ RHS = [ Â, Ĉ] + [Bˆ, Ĉ] = Â Ĉ ĈÂ + BˆĈ ĈBˆ = HS (b) Proof [ AB, ˆ ˆ C] ˆ = A[B,C] ˆ ˆ ˆ + [ A, ˆ C]B ˆ ˆ HS = ABC ˆ ˆ ˆ CAB ˆ ˆ ˆ RHS = Â(BˆĈ ĈBˆ) + ( Â Ĉ ĈÂ)Bˆ = Â BˆĈ Â ĈBˆ + Â ĈBˆ ĈÂ Bˆ = Â BˆĈ ĈÂ Bˆ = HS iboff 5.9 (ϕ,ϕ 3 ) = {sin kx,exp(ikx)} Since we know ϕ 3 = exp(ikx) is a coon eigenfunction of Ĥ and pˆ, we d like to keep it as one of the two coon eigenfunctions coposed of (ϕ,ϕ 3 ). If we denote it as ϕ a, thenϕ a = a ϕ + a 3 ϕ 3 = ϕ 3, which eans a = 0 and a 3 =. () Also we know ϕ b = exp( ikx) is a coon eigenfunction of Ĥ and pˆ and is linearly independent of ϕ a = exp(ikx), we hope we could construct ϕ b as a linear cobination of (ϕ,ϕ 3 ). ϕ b = b ϕ + b 3 ϕ 3 = b sin kx + b 3 exp(ikx) = b sin kx + b 3 (cos kx + i sin kx)

2 .0 Proble Set 3 Solution ϕ b = b 3 cos kx + (b + ib 3 )sin kx = exp( ikx) = cos kx i sin kx b 3 =, b = i () Su () and (), we get: ϕ a = 0 ϕ + ϕ 3 = exp(ikx) ϕ b = i ϕ + ϕ 3 = exp( ikx) ϕ and ϕ 3 are a pair of linearly independent coon eigenfunctions of Ĥ and pˆ. iboff 5.5 Use uncertainty principle: If [ Â, Bˆ] = C 0, then A B < C > (a) x E [xˆ, Ê]ψ = xˆêψ Ê xˆψ h d xˆ = x, Ê = dx + V (x) h d ψ h d [xˆ, Ê]ψ = x[ dx + V (x)ψ ] [ dx + V (x)][xψ ] h d ψ h d = x + xv (x)ψ + [xψ ] xv (x)ψ dx dx h d ψ h d dψ h d ψ h dψ h d ψ h dψ = x x + dx dx [ψ + x dx ] = x dx dx dx dx [xˆ, Ê] = h d dx d pˆ x = ih [xˆ, Ê] = i h pˆ x dx h Fro uncertainty principle, we get x E < i pˆ x > Thus x E h < pˆ x > (b) p x E d h d pˆ x = ih dx, Ê = dx + V (x) [ pˆ x, Ê]ψ = pˆ x Êψ Ê pˆ x ψ

3 .0 Proble Set 3 Solution [ pˆ x, Ê]ψ = ih d [ h d ψ h d + V (x)ψ ] [ + V (x)][ ih dψ ] dx dx dx dx h 3 d 3 ψ dv (x) dψ h d 3 ψ dψ = i ih ψ ih V (x) i + ihv (x) dx 3 dx dx dx 3 dx = ih dv (x) ψ dx p x E < ih dv (x) > = h < dv (x) > dx dx (c) x T h d xˆ = x, Tˆ = dx [xˆ,tˆ]ψ = xˆtˆψ Tˆ xˆψ h x d ψ h d h x d ψ h d dψ = dx + dx (xψ ) = dx + dx (ψ + x dx ) h x d ψ h dψ h dψ h x d ψ h dψ = = dx dx dx dx dx [xˆ,tˆ] = i h pˆ x h h x T < i pˆ x > = < pˆ x > (d) p x T d h d pˆ x = ih dx, Tˆ = dx [ pˆ x,tˆ]ψ = pˆ x Tˆψ Tˆ pˆ x ψ d h d ψ h d dψ h 3 d 3 ψ h 3 d 3 ψ = ih dx ( dx ) + dx ( ih dx ) = i dx 3 i dx 3 = 0 p x T = 0 3.Solution The D Schrodinger equation is: ih ψ (x, t) = Ĥψ (x, t) (3.) t Separate the variables of the solution as ψ (x,t) = C n (t)ϕ n (x) (3.) Plug (3.) into (3.), we get: ih ϕ n (x) dc n (t) = C n (t)ĥϕ n (x) (3.3) n dt n Insert Ĥϕ n (x) = E n ϕ n (x) into (3.3), we get: n

4 .0 Proble Set 3 Solution ih ϕ n (x) dc n (t) = C n (t)e n ϕ n (x) n dt n Match every ter, ih dc n (t) = En C n (t) (3.4) dt Solve (3.4) C n (t) = C n (0)e iϖ n t, ϖ n = E n / h (3.5) Replace C n (t) in (3.) with (3.5) ψ (x, t) = C n (0)ϕ n (x)e iϖ nt, ϖ n = E n / h (3.6) n Use initial condition, ψ (x,0) = a ϕ (x) + a 3 ϕ 3 (x), we get: C (0) = a, C 3 (0) = a 3, C i (0) = 0 for i,3 (3.7) Insert (3.7) into (3.6): ψ (x, t) = a e iϖ t ϕ (x) + a 3 e iϖ 3t ϕ 3 (x) (3.8) Fro proble set, we know ϕ n (x) = πn sin kn x, k n =, put into (3.8) ψ(x,t) = a e iϖ t sink x + a 3 e iϖ3 t sink 3 x k = π, ϖ = E = hπ, k = 3π 3, ϖ 3 = E 3 = 9hπ h h The probability of finding the particle between x = and x = + x for sall x is: P(x) x= / x = ψ (x, t) x= / x P(x) x= / x = a e iϖ t sin k x + a 3 e iϖ 3 t sin k 3 x x x= / π 3π sin k x x= / = sin( ) =, sin k 3 x x= / = sin( ) = P(x = / ) x = x a e iϖ t a 3 e iϖ 3t If we write a and a 3 as a = a e iθ and a 3 = a 3 e iθ 3 P(x = / ) x = x a e i (ϖ θ )t a 3 e i(ϖ 3 θ 3 )t = x ( a + a 3 a a 3 cos[(ϖ θ ) (ϖ 3 θ 3 )]) Probabilities of energy easureent: a) P ( E ) = a exp( iω t ) = a b) P(E ) = 0 c) P ( E 3 ) = a 3 exp( iω 3 t ) = a 3

5 .0 Proble Set 3 Solution As can be seen fro the result, the energy easureent probabilities don t change over tie. 4.Solution If ] = 0, p x conserves. If ] = 0, p y conserves. If ] = 0, p z conserves. Ĥ = Tˆ + Vˆ and [Tˆ, pˆ x ] = [Tˆ, pˆ y ] = [Tˆ, pˆ z ] = 0 ] = [(Tˆ + Vˆ), pˆ x ] = [Tˆ, pˆ x ] + [Vˆ, pˆ x ] = [Vˆ, pˆ x ] According to the sae logic, ] = [(Tˆ + Vˆ), pˆ y ] = [Tˆ, pˆ y ] + [Vˆ, pˆ y ] = [Vˆ, pˆ y ] ] = [(Tˆ + Vˆ), pˆ z ] = [Tˆ, pˆ z ] + [Vˆ, pˆ z ] = [Vˆ, pˆ z ] (a) V (x, y, z) = ax + by + cz 3 ]ψ = [Vˆ, pˆ x ]ψ = (ax + by + cz 3 ) ih ψ + ih aψ + (ax + by + cz 3 ) ψ = ihaψ x x ] = iha 0. The oentu does not conserve in x direction. ]ψ = [Vˆ, pˆ y ]ψ = (ax + by + cz 3 ) ih ψ + ih byψ + (ax + by + cz 3 ) ψ = ihbyψ y y ] = ihby 0. The oentu does not conserve in y direction. ]ψ = [Vˆ, pˆ z ]ψ = (ax + by + cz 3 ) ih ψ + ih 3cz ψ + (ax + by + cz 3 ) ψ = 3ihcz ψ z z ] = 3ihcz 0. The oentu does not conserve in z direction. (b) V (x, y, z) = a arctan(x / ) / ψ ]ψ = [Vˆ, pˆ x ]ψ = a arctan(x / ) ih ψ + ih aψ + a arctan(x / ) + (x / ) x x / ] = iha 0. + (x / ) The oentu does not conserve in x direction. ψ ψ ]ψ = [Vˆ, pˆ y ]ψ = a arctan(x / ) ih y + ih a arctan(x / ) y = 0 The oentu conserves in y direction.

6 .0 Proble Set 3 Solution ]y = [Vˆ, pˆ z ]y = a arctan(x / ) y ih z The oentu conserves in z direction. (c) V (x, y, z) = b exp( y / d ) + ih a arctan(x / ) ]ψ = [Vˆ, pˆ y ]ψ = b exp( y / d ) ih ψ + ih b exp( y / d ) ψ y + b exp( y / d )( )ψ d 0 y y. The oentu does not conserve in y direction. y z = 0 ]ψ = [Vˆ, pˆ x ]ψ = b exp( y / d ) ih ψ + ih b exp( y / d ) ψ = 0 x x ] = 0. The oentu does conserve in x direction. ]ψ = [Vˆ, pˆ z ]ψ = b exp( y / d ) ih ψ + ih b exp( y / d ) ψ = 0 z z The oentu conserves in z direction. 5.Solution Ĥ = Tˆ + Vˆ = h + Vˆ, Pˆψ (x, y, z) = ψ ( x, y, z) If [Ĥ, Pˆ] = 0, energy and parity operators have coon eigenfunctions. First, we can proof that [Tˆ, Pˆ] = 0. Then we check if [Vˆ, Pˆ] = 0. If it is true, Ĥ and Pˆ have coon eigenfunctions. Prove [Tˆ, Pˆ] = 0 ˆ h h h T = = T ˆ + T ˆ + T ˆ x y z x y z Pˆψ (x, y, z) =ψ ( x, y, z) = Pˆ x Pˆ y Pˆ zψ (x, y, z) If we can show [Tˆ x ] = [Tˆ y, Pˆ y ] = [Tˆ z, Pˆ z ] = 0, [Tˆ, Pˆ] = 0 is proved. Show [Tˆ x ] = 0 Since ψ (x, y, z) is a linear cobination of odd functions and even functions, if we can prove that [Tˆ x ]ψ (x) = 0, where ψ (x) = ψ ( x) and [Tˆ x ]ψ (x) = 0, where ψ (x) = ψ ( x), it is true that [Tˆ x ]ψ (x, y, z) = 0. [Tˆ x ]ψ (x) = h ψ ( x) + h Pˆ ψ (x) = h ψ (x) + h ψ (x) = 0 x x x x x

7 .0 Proble Set 3 Solution [Tˆ x ]ψ (x) = h ψ ( x) h + Pˆ ψ (x) h ψ = (x) h ψ + (x) x = 0 x x x x According to the sae logic, [Tˆ y, Pˆ y ] = [Tˆ z, Pˆ z ] = 0 Thus [Tˆ, Pˆ] = 0 (a) V (x, y, z) = e / r [V (x, y, z), Pˆ]ψ = (e / r)ψ ( x, y, z) Pˆ[(e / r)ψ (x, y, z)] = (e / r)ψ ( x, y, z) (e / r)ψ ( x, y, z) = 0 Thus [Ĥ, Pˆ] = 0 The energy operator and the parity operator have coon eigenfunctions. (b) V (x, y, z) = ax + by + cz 3 [V (x, y, z), Pˆ]ψ = (ax + by + cz 3 )ψ ( x, y, z) Pˆ[(ax + by + cz 3 )ψ (x, y, z)] = (ax + by + cz 3 )ψ ( x, y, z) ( ax + by cz 3 )ψ ( x, y, z) 0 Thus [Ĥ, Pˆ] 0 The energy operator and the parity operator do not have coon eigenfunctions. (c) V (x, y, z) = axy [V (x, y, z), Pˆ]ψ = (axy)ψ ( x, y, z) Pˆ[(axy)ψ (x, y, z)] = (axy)ψ ( x, y, z) (axy)ψ ( x, y, z) = 0 Thus [Ĥ, Pˆ] = 0 The energy operator and the parity operator have coon eigenfunctions. 6.Solution iboff9.3 The possible values of an angular oentu easureent are the square roots of eigenvalues of the total angular oentu operator ˆ. They are = h l( l + ), l = 0,,,.... In another word, ˆϕ l = h l(l + )ϕ l. Fro the proble, we know = h 56, which gives us l = 7, since = h 56 = h 7 (7 + ) The possible values of an angular oentu projection easureent on x axis are the eigenvalues of ˆ x. They are x = 7h,6h,..., 6h, 7h. The angle θ is given by cosθ = x / θ = arccos( x / ). Thus θ in = arccos( x,ax / ) = arccos(7h / h 56) = 3 ibiff9.5(a) The solution of eigenvalue probles for ˆ and ˆ z are: ˆY l = h l(l + )Y l, l = 0,,,...

8 .0 Proble Set 3 Solution ˆ zy l = hy l, = l, l +,..., l This eans the possible values of easureents of are h l( l + ) and those of z are h. The given state wave function is expanded with Y l as: ψ (θ,φ,0) = 3 Y + 4 Y7 + 3 Y The possible values of and z are: (, z ) = {(,) (7,3) (7,)}. The possibilities of these easureents are: P( =, z = ) = P( = 7, z = 3) = P( = 7, z = ) = ibiff = = = ˆ, l >= h l(l + ), l >, l = 0,,,... ˆ z, l >= h, l >, = l, l +,..., l The above eigenvalue proble solutions are true for both particles, which eans: ˆ h l (l + ) ˆ z h l > l >= l > l > ˆ h l (l + ) ˆ z h The eigenvalues of the set of operators (ˆ, ˆz, ˆ, ˆ z ) corresponding to the product eigenstate l > l > are: { h, h l (l + ), h, h l (l + )}. ibiff9.34 l =, =, l =, l =, l = =,0, l = =,0, Due to coupling, = + (, ) = (,0)or(0, ) Thus, l l >= C,0,,,0 > +C 0,,0,, > Find the coefficients as: ˆ,,, >= 0 = (C,0 + C 0, ), >, > C,0 = C 0, To noralize, C,0 + C 0, = C,0 = C 0, = ±

9 .0 Proble Set 3 Solution l l >= ±,,,0 >,0,, > The possible values of z are h with probability of and 0 with probability of. ibiff9.35 l =, =, l =, l =, l = =,0, l = =,0, Due to coupling, = + (, ) = (, ) Thus, l l >=,,, > There is only one set of possible quantu nubers. The joint probability of finding the two electrons with z = z = h is one.