Assignment 8, solutions

Problem 1:

The quantum numbers l1 and l2 of the orbital momenta of particle A and particle B are 1 and 2, respectively.  Find the 15 possible kets' in the coupled representation (notation |l1,l2;L,ML>) where L represents the quantum number of the total orbital momentum.

Solution:
bulletConcepts:
Addition of angular momentum
bulletReasoning:
We are supposed to add two angular momenta l1 and l2  and find the possible values for L and M.
bulletDetails of the calculation:
l1 = 1 and l2 =2.  The possible values for L are L = 3, 2, 1.
L = 1:  Possible values M are 1, 0, -1.
Corresponding kets: |1,2; 1,1>, |1,2; 1,0>, |1,2; 1,-1>.
(notation: |l1,l2;L,m>)
L = 2:  Possible values M are 2, 1, 0, -1, -2.
Corresponding kets: |1,2; 2,2>, |1,2; 2,1>, |1,2; 2,0>, |1,2; 2,-1>, |1,2; 2,-2>.
L = 3:  Possible values M are 3, 2, 1, 0, -1, -2 -3.
Corresponding kets: |1,2; 3,3>, |1,2; 3,2>, |1,2; 3,1>, |1,2; 3,0>, |1,2; 3,-1>, 
|1,2; 3,-2>, |1,2; 3,-3>.
We have listed 15 kets. 

Problem 2:

A pair of magnetic ions with individual spins s1 and s2 interact through the scaled Hamiltonian H = s1s2.  Let s(s + 1) be the eigenvalue of s2, where s = s1 + s2 is the total angular momentum.  Note ħ = 1.
(a)  Show that the total angular momentum is a conserved quantity.
(b)  Find the ground state energy E0 and the energy of the highest state Emax.
(c)  Now suppose a magnetic field is applied and the new Hamiltonian is H', where
H' = H - b(s1z + s2z).  What is the residual symmetry of the new Hamiltonian H', and what are the associated "good quantum numbers"?

Solution:

bulletConcepts:
Addition of angular momentum, common eigenbasis for J12, J22, J2, Jz.
bulletReasoning:
Consider two angular momentum operators, J12 and J22, operating in two different vector spaces E1 and E2.
Let J = J1 + J2.  The operators  J12, J22, J2, and Jz all commute and a common eigenbasis {|j1,j2;j,m>} exists.
Here we have s = s1 + s2 and {|s1,s2;s,ms>} is a common eigenbasis of s12, s22, s2, and sz.
s1s2 = (s2 - s12 - s22),  s1z + s2z = sz,  the vectors {|s1,s2;s,ms>} are also eigenvectors of s1s2 and  s1z + s2z.
bulletDetails of the calculation:
(a)  s12, s22, s2, and sz are a set of commuting operators.  H = s1s2 commutes with s2 and sz.
[Similarly {s12, s22, s2, sx} and {s12, s22, s2, sy} are sets of commuting operators and H commutes with sx and sy.]
Therefore [H, s] = 0, H commutes with all components of s, the total angular momentum is a conserved quantity.
(b)  H|s1,s2;s,ms> = ħ2[s(s+1) - s1(s1+1) - s2(s2+1)]|s1,s2;s,ms>,
smax = s1 + s2, smin = |s1 - s2|.
E0 = ħ2[smin(smin+1) - s1(s1+1) - s2(s2+1)],
Emax = ħ2[smax(smax+1) - s1(s1+1) - s2(s2+1)].
(c)  H' commutes with s2, and sz, but not with sx and sy;  s and ms remain good quantum numbers.

Problem 3:

Consider two distinguishable spin systems that can be repeatedly prepared in the state
|ψ> = 2-(|++> + |-->) = 2-(|+>(1)|+>(2) + |->(1)|->(2)).
Observer 1 measure the X(1) = σx(1) or Y(1) = σy(1), i.e. the x- or y-component of the spin in units of ħ/2, of system 1, and observer 2 measures X(2) or Y(2) of system 2.
(a)  Show that the product operators X(1)Y(2) and Y(1)X(2) commute and therefore form a complete set of commuting operators for the system.
(b)  What are the possible results of measuring X(1), X(2), Y(1), or Y(2) for the given |ψ>?
(c)  What are the possible outcomes of measuring X(1)Y(2) or Y(1)X(2)?

Solution:

bulletConcepts:
The state space of two spin- particles
bulletReasoning:
The Pauli operators operating on different systems all commute.  Different Pauli operators operating on the same system anticommute.
σx(i)σy(i) = -σy(i)σy(i),  σx(i)σy(j) = σy(j)σx(i).
bulletDetails of the calculation:
(a)  X(1)Y(2) Y(1)X(2) - Y(1)X(2) X(1)Y(2) = Y(2)X(2) X(1)Y(1) - X(2)Y(2)Y(1)X(1)
= Y(2)X(2) X(1)Y(1) - Y(2)X(2) X(1)Y(1) = 0.
X(1)Y(2) and Y(1)X(2) commute.  We have a system with two degrees of freedom, therefore the operators form a complete set of commuting operators for the system.

(b) The possible results of each measurement X(1), X(2), Y(1), or Y(2) are 1.

(c)  The eigenvectors of σx are |>x = (1/√2)(|+> |->),
and the eigenvectors of σy are |>y = (1/√2)(|+> i|->).
Therefore in terms of the eigenbasis of σx
|> = (1/√2)(|+>x |->x)
and  in terms of the eigenbasis of σx
|+> = (1/√2)(|+>y + |->y), |-> = (-i/√2)(|+>y -|->y).

The eigenvalues of X(1)Y(2) are 1.  The eigenvalues are degenerate.
|+>x|+>y and |->x|->y and all linear combinations of those two states are eigenstates with eigenvalue +1,
|+>x|->y and |->x|+>y and all linear combinations of those two states are eigenstates with eigenvalue -1.
We can write
|+>|+> + |->|-> = [(|+>x + |->x)(|+>y + |->y) - i(|+>x - |->x)(|+>y - |->y)]
= [|+>x (|+>y + |->y - i|+>y + i|->y) + |->x (|+>y + |->y + i|+>y - i|->y)]
The given |ψ> therefore can be written as
(e-iπ/4|+>x|+>y + eiπ/4|->x|->y) + (e-π/4|+>x|->y + e-iπ/4|->x|+>y).
It is a linear combination of eigenstates of X(1)Y(2).
Possible outcomes of a measurement of X(1)Y(2) are 1, each with 50% probability.
Similarly we show |ψ> =  (e-iπ/4|+>y|+>x + e-iπ/4|->y|->x) + (eπ/4|+>y|->x + eiπ/4|->y|+>x).
It is a linear combination of eigenstates of Y(1)X(2).
Possible outcomes of a of Y(1)X(2) measurement are 1, each with 50% probability.

Problem 4:

A system has a wave function ψ(x,y,z) = N*(x + y + z)*exp(-r22) with α real.  If Lz and L2 are measured, what are the probabilities of finding 0 and 2ħ2?

Solution:

bulletConcepts:
The spherical harmonics, the postulates of quantum mechanics
bulletReasoning:
If we want to find the probability of measuring the eigenvalue a of an observable A we must express the wave function of the system as a linear combination of eigenfunctions of A.  Then  P(a) = Σi|<ai|ψ>|2.
bulletDetails of the calculation:
Express the wave function in terms of spherical harmonics.

x + y + z = r sinθ cosφ + r sinθ sinφ + r cosθ.
cosφ + sinφ = (e+ e-iφ) - i(e- e-iφ)
= (e(1 - i) + e-iφ(1 + i)) = 2-(ee-iπ/4 + e-iφeiπ/4).
x + y + z = r sinθ 2-(ee-iπ/4 + e-iφeiπ/4) + r cosθ
= r (4π/3)(-e-iπ/4Y11(θ,φ) + eiπ/4Y1-1(θ,φ) + Y10(θ,φ)).

ψ(x,y,z) = N*rexp(-r22)(4π)  * (1/3)(-e-iπ/4Y11(θ,φ) + eiπ/4Y1-1(θ,φ) + Y10(θ,φ))
= f(r) * χ(θ,φ).
0πsinθ dθ∫0dφ |χ(θ,φ))|2 = 1,  ∫0r2 |f(r)|2 dr = 1, for properly chosen N.

χ(θ,φ) = Σl,mdlmYlm(θ,φ), dlm = 0 for l ≠ 1.
Let P(l,m) denote the probability of finding the eigenvalues l(l + 1)ħ2 and mħ .
P(l,m) = |dlm|2.  P(1,1) = 1/3,  P(1,0) = 1/3,  P(1,-1) = 1/3,  P(l,m)l≠1 = 0.
The probabilities of finding Lz = 0 is 1/3 and the probability of finding L2 = 2ħ2 is 1.

Problem 5: 

Some organic molecules have a triplet (S = 1) excited state that is located at an energy Δ above the singlet (S = 0) ground state.  Consider an ensemble of N such molecules where N is of the order of Avogadro's number
(a)  Find the average magnetic moment <μ> per molecule in the presence of a magnetic field B.  Assume Boltzmann statistics.  You may also assume that Δ is large compared to the field-induced level splittings.
(b) Show that the magnetic susceptibility χ = N d<μ>/dB is approximately independent of Δ when kBT >> Δ.

Solution:

bulletConcepts:
Angular momentum and magnetic dipole moment, the energy of a magnetic dipole in an external magnetic field, Boltzmann statistics
bulletReasoning:
The magnetic moment of a molecule is proportional to its angular momentum. Molecules with S = 0 have no magnetic moment but molecules with S = 1 have a magnetic moment μ proportional to S.  The energy of a magnetic moment in an external magnetic field is quantized.  The three possible different projections of the magnetic moment onto the direction of the magnetic field have different energies.  The relative probabilities for the different projections are found using Boltzmann statistics.  Once we have the relative probabilities, we can find the average magnetic moment.
bulletDetails of the calculation:
(a)  The magnetic moment μ of a molecule is proportional to S, μ = -(γμB/ħ)S.
The energy of a magnetic moment in a magnetic field is given by E = -μ∙B.
Let the magnetic field B point into the z-direction. Then for the molecules with S = 1 the energy of the magnetic moment is
E = -μzB = γ(μB/ħ)BSz = γμBBm, with m = -1, 0, 1.
(The aligned spin, anti-aligned magnetic moment has the highest energy.)
With the appropriate choice of the zero of the potential energy, the total energy of the molecule energy can be written as
E0(S = 0) = 0, E1(S = 1, m = 1) = Δ + γμBB, E2(S = 1, m = 0) = Δ, E3(S = 1, m = -1) = Δ - γμBB.
The ration Nn/N0 of the number of molecules in the excited state with En to the number of molecules in the ground state with E0 is
Nn/N0 = exp(-En/(kBT)).
z> = (N0μ0z + N1μ1z + N2μ2z + N3μ3z)/(N0 + N1 + N2 + N3) = (N1μ1z + N3μ3z)/(N0 + N1 + N2 + N3)
= (-N0exp(-(Δ + γμBB)/(kBT))*γμB +  N0exp(-(Δ - γμBB)/(kBT))*γμB)/(N0 + N0exp(-(Δ + γμBB)/(kBT)) + N0exp(-Δ/(kBT) + N0exp(-(Δ - γμBB)/(kBT)))
= -γμB*(exp(-(Δ + γμBB)/(kBT)) - exp(-(Δ - γμBB)/(kBT)))/(1 + exp(-(Δ + γμBB)/(kBT)) + exp(-Δ/(kBT) + exp(-(Δ - γμBB)/(kBT)))
= average magnetic moment <μz > per molecule in the presence of a magnetic field B.

(b)  Let kBT >> Δ >> 2γμBB.
Expand ey = 1 + y in the numerator and ey = 1 in the denominator.
Then <μz> ~ B(γμB)2/(kBT).
χ = N d<μ z >/dB = N(γμB)2/(kBT).