2. Griffith s 4.19 a. The commutator of the z-component of the angular momentum with the coordinates are: [L z. + xyp x. x xxp y.

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1 Physics Homework #8 Spring 6 Due Friday 5/7/6. Download the Mathematica notebook from the notes page of the website ( or the homework page of the website ( and answer the questions in the Mathematica file. Print out your plots to go with the answers to your questions. Each one of you should do a different set of quantum numbers in part 3.6. Answers will vary.. Griffith s 4.9 a. The commutator of the z-component of the angular momentum with the coordinates are:, x = [xp y, x = xp y x x xxp y + xyp x = x[ p y, x+ y[x = i!y, y = [xp y, y = xp y y y yxp y + yyp x = x[ p y, y+ y[y = i!x,z = [xp y,z = xp y z z zxp y + zyp x = x[ p y,z+ y[z = where we have used the results from Griffith s problem 4., where if i j [r i, p j = [ p i,r j = i!δ ij = i!. if i = j The commutator of the z-component of the angular momentum with the momentum operators are: = [xp y = [xp y [yp x = {x[ p y + [x p y {y[ p x + [y p x = + i!p y = i!p y = [xp y, py x = [xp y [yp x = {x[ p y + [x p y {y[ p x + [y p x = + i!p x = i!p x = [xp y = [xp y [yp x = {x[ p y + [x p y {y[ p x + [y p x = + = where in both parts, we ve used from homework #6, problem Griffith s 3.3, where [AB,C = A[B,C+ [A,CB

2 b. In order to do this part, we require two relationships between commutators. The first is a footnote from page 6, [A, B + C = [A, B+ [A,C and the second is from homework #6, problem Griffith s 3.3, where [AB,C = A[B,C+ [A,CB and by analogy [A, BC = [A, BC + B[A,C. Thus, =, yp z zp y =, yp z,zp y = {, yp z + y {,zp y + z = { i!xp z + { zi!p x ( ) = i! xp z zp x = i!l y where we used [A, BC = [A, BC + B[A,C again. c. The commutator of the z-component of the angular momentum and the r operator is:,r = {L z, x + y + z =, x +, y +,z,r =, xx+, yy+,zz { + {, yy + y, y + {,zz + z,z,r =, xx + x, x { + { i!xy i!yx + { +,r = i!yx + i!yx,r = The commutator of the z-component of the angular momentum and the p operator is:, p = {L z + p y + p z = + +, p = p x + p y + p z { + { p y + p y + p z + p z { + { i!p x p y i!p y p x + { +, p = p x + p x, p = i!p y p x + i!p x p y {, p = d. Since p and r commute with each component of L, L commutes with the Hamiltonian.

3 3. Griffith s 4.4 a. A rigid rotator is a rod that can rotate in 3D about its center, but the center point is fixed in space (it cannot translate.) This could model a rotating molecule. Suppose that the two masses (have mass m ) and are separated by a distance a. The magnitude of the angular momentum is! L = L = rmv = a mv = amv = ap. The Hamiltonian H = T + V = p m + = L ma = L ma. Now, H ψ = L ma ψ = E ψ = ( ). ( ) λ ma ψ =! l l + ψ. Therefore the energy ma is given as E i =! ma l l + b. The eigenfunctions of the square of the angular momentum are the spherical harmonics with degeneracy l Griffith s 4.9 a. The eigenspinors are determined (following the examples in class) as: ψ S y ψ =! i i ψ ψ = c ψ ψ i! ψ = cψ ψ = i! c ψ i! ψ = cψ i! i! c ψ = cψ c = ±! ψ = iψ So, we define two states (the eigenspinors) = c' i and = c' i with eigenvalues +! and! respectively. We need to normalize these states. The normalization for each state is given as: = = c'* ( i )c' i = c' c' = and = = c'* ( i )c' i = c' c' =. Thus the normalized eigenspinors are:

4 = i and = i. b. The general state is given by X = on each of the states is given by: P + = ψ = P = ψ = ( i a )* b ( + i a )* b The sum of the probabilities is: P = P + + P = ( a + ib) + a ib ( ) a b. The probability of a measurement = ( a + ib) = ( a ib) = a + ib ( ) + a ib ( ) a* ib * P = aa* iab * + iba * + bb * + aa * + iab * iba * + bb * P = aa * + bb * = a + b = ( ) ( ) a * + ib * c. We operate on the general state with the S y operator where S y =! 4 i i i i =! 4. Operating on the general state would return! times the general state. The probability would of course be 4 unity since the states have been normalized. 5. Griffith s 4.3 Construct S x, S y, and for a particle of spin. For a spinparticle, { so there are three eigenstates and we can label them as m s =,, X = ; X = Let s construct. ; X =.

5 ψ =!m s ψ X =! X X = X X =! X =! Let s construct S x and S y, where as usual S x = + S and S y = S i and S are the raising and lowering operators respectively. X =, where X =! s(s +) m s (m s +) X =! (+) ( +) X =! X X =! (+) ( )( +) X =! X S X =! s(s +) m s (m s ) X =! (+) ( ) X =! X S X =! (+) ( )( ) X =! X S X = =! & S =! Therefore S x and S y, are respectively S x =! S y = i!. and

6 6. Griffith s 4.33 a. The Hamiltonian for the system looks like H = γ B! S! = γ B " cos ωt ( ) = γ B " cos ( ωt ) γ B " cos ( ωt ) b. We have to solve the time-dependent Schoridinger wave equation. H ψ = i! d ψ γ B! cos ( ωt ) γ B! cos ( ωt ) iγ B cos ( ωt )ψ = dψ + + ψ + ψ = i! dψ + dψ iγ B cos ( ωt ) = dψ + ψ + = A + e ψ + iγ B ω sin ( ωt ) iγ B ψ ( t) = cos ( ωt )ψ = dψ A + e iγ B sin ( ωt ) ω A e iγ B sin ( ωt ) ω iγ B cos ( ωt ) = dψ = iγ B e sin ( ωt ) ω e iγ B ω sin ( ωt ) ψ ψ = A e iγ B sin ( ωt ) ω The normalization constant is determined from the initial conditions: ψ ( ) = X x + = = = A + A c. The probability of getting! (a spin down state) if you measure S x is given by:

7 ( ) P = ψ ( t) = iγ B e sin ( ωt ) ω e iγ B ω sin ( ωt ) P = iγ B 4 e sin ( ωt ) ω e iγ B ω ( ) = 4 isin γ B sin ( ωt ) ω sin ωt P = isin γ B sin ( ωt ) ω isin γ B sin ( ωt ) ω γ B P = sin sin ( ωt ) ω d. The minimum magnetic field needed to cause a complete spin flip is when the argument of the sine is a maximum, which occurs at π. Thus we have, π = γ B ω B = πω = πω. γ γ