**Problem 1:**

The quantum numbers l_{1} and l_{2} 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 |l_{1},l_{2};L,M_{L}>)
where L represents the quantum number of the total orbital momentum.

Concepts: Addition of angular momentum | |

Reasoning: We are supposed to add two angular momenta l _{1} and l_{2}
and find the possible values for L and M. | |

Details of the calculation: l _{1} = 1 and l_{2} =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: |l _{1},l_{2};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 **s**_{1} and **s**_{2}
interact through the scaled Hamiltonian H =** s**_{1}∙**s**_{2}.
Let s(s + 1) be the eigenvalue of s^{2}, where **s **=** s**_{1}
+ **s**_{2} 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 E_{0} and the energy of the highest
state E_{max}.

(c) Now suppose a magnetic field is applied and the new Hamiltonian is H',
where

H' = H - b(s_{1z} + s_{2z}). What is the residual symmetry of
the new Hamiltonian H', and what are the associated "good quantum numbers"?

Solution:

Concepts: Addition of angular momentum, common eigenbasis for J _{1}^{2}, J_{2}^{2},
J^{2}, J_{z}. | |

Reasoning: Consider two angular momentum operators, J _{1}^{2} and J_{2}^{2},
operating in two different vector spaces E_{1} and E_{2}.Let J = J_{1} + J_{2}. The operators J_{1}^{2},
J_{2}^{2}, J^{2}, and J_{z} all commute and a
common eigenbasis {|j_{1},j_{2};j,m>} exists.Here we have s = s_{1} + s_{2} and {|s_{1},s_{2};s,m_{s}>}
is a common eigenbasis of s_{1}^{2}, s_{2}^{2},
s^{2}, and s_{z}.s_{1}∙s_{2 }= ½(s^{2} - s_{1}^{2}
- s_{2}^{2}), s_{1z} + s_{2z} = s_{z},
the vectors {|s_{1},s_{2};s,m_{s}>} are also
eigenvectors of s_{1}∙s_{2 }and s_{1z} +
s_{2z}. | |

Details of the calculation: (a) s _{1}^{2}, s_{2}^{2}, s^{2}, and s_{z}
are a set of commuting operators. H = s_{1}∙s_{2}
commutes with s^{2} and s_{z}.[Similarly {s _{1}^{2}, s_{2}^{2}, s^{2},
s_{x}} and {s_{1}^{2}, s_{2}^{2}, s^{2},
s_{y}} are sets of commuting operators and H commutes with s_{x}
and s_{y}.]Therefore [H, s] = 0, H commutes with all components of
s, the
total angular momentum is a conserved quantity.(b) H|s _{1},s_{2};s,m_{s}> = ½ħ^{2}[s(s+1)
- s_{1}(s_{1}+1) - s_{2}(s_{2}+1)]|s_{1},s_{2};s,m_{s}>,s _{max} = s_{1} + s_{2}, s_{min} = |s_{1}
- s_{2}|.E _{0} = ½ħ^{2}[s_{min}(s_{min}+1) - s_{1}(s_{1}+1)
- s_{2}(s_{2}+1)],E _{max} = ½ħ^{2}[s_{max}(s_{max}+1) - s_{1}(s_{1}+1)
- s_{2}(s_{2}+1)].(c) H' commutes with s ^{2}, and s_{z}, but not with s_{x}
and s_{y}; s and m_{s} remain good quantum numbers. |

**Problem 3:**

Consider two distinguishable spin ½^{ }systems that can be repeatedly
prepared in the state

|ψ> = 2^{-½}(|++> + |-->) = 2^{-½}(|+>^{(1)}|+>^{(2)}
+ |->^{(1)}|->^{(2)}).

Observer 1 measure the X^{(1)} = σ_{x}^{(1)} or Y^{(1)}
= σ_{y}^{(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:

Concepts: The state space of two spin-½ particles | |

Reasoning: The Pauli operators operating on different systems all commute. Different Pauli operators operating on the same system anticommute. σ _{x}^{(i)}σ_{y}^{(i)} = -σ_{y}^{(i)}σ_{y}^{(i)},
σ_{x}^{(i)}σ_{y}^{(j)} = σ_{y}^{(j)}σ_{x}^{(i)}.
| |

Details 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 σ _{x} are |±>_{x} = (1/√2)(|+>
± |->),and the eigenvectors of σ _{y} are |±>_{y} = (1/√2)(|+> ±
i|->).Therefore in terms of the eigenbasis of σ _{x}|±> = (1/√2)(|+> _{x} ± |->_{x})and in terms of the eigenbasis of σ _{x}|+> = (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 |+>|+> + |->|-> = ½[(|+> _{x} + |->_{x})(|+>_{y} +
|->_{y}) - i(|+>_{x} - |->_{x})(|+>_{y} - |->_{y})]= ½[|+> _{x} (|+>_{y} + |->_{y} - i|+>_{y}
+ i|->_{y}) + |->_{x} (|+>_{y} + |->_{y}
+ i|+>_{y}
- i|->_{y})]The given |ψ> therefore can be written as ½(e ^{-iπ/4}|+>_{x}|+>_{y }+ e^{iπ/4}|->_{x}|->_{y})
+ ½(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> = ½(e ^{-iπ/4}|+>_{y}|+>_{x
}+ e^{-iπ/4}|->_{y}|->_{x}) + ½(e^{π/4}|+>_{y}|->_{x
}+ e^{iπ/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(-r^{2}/α^{2})
with α real. If L_{z} and L^{2} are
measured, what are the probabilities of finding 0 and 2ħ^{2}?

Solution:

Concepts: The spherical harmonics, the postulates of quantum mechanics | |

Reasoning: 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) = Σ _{i}|<a^{i}|ψ>|^{2}. | |

Details of the calculation: Express the wave function in terms of spherical harmonics. x + y + z = r sinθ cosφ + r sinθ sinφ + r
cosθ. ψ(x,y,z) = N*rexp(-r |

**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 k_{B}T >> Δ.

Solution:

Concepts: Angular momentum and magnetic dipole moment, the energy of a magnetic dipole in an external magnetic field, Boltzmann statistics | |

Reasoning: 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. | |

Details of the calculation: (a) The magnetic moment μ of a
molecule is proportional to S, μ
= -(γμ_{B}/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 isE = -μ _{z}B = γ(μ_{B}/ħ)BS_{z}
= γμ_{B}Bm, 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 E _{0}(S = 0) = 0, E_{1}(S = 1, m = 1) = Δ + γμ_{B}B,
E_{2}(S = 1, m = 0) = Δ, E_{3}(S = 1, m = -1) = Δ - γμ_{B}B.
The ration N _{n}/N_{0 }of the number of molecules in the
excited state with E_{n} to the number of molecules in the ground
state with E_{0} is N _{n}/N_{0} = exp(-E_{n}/(k_{B}T)).<μ _{z}> = (N_{0}μ_{0z} + N_{1}μ_{1z}
+ N_{2}μ_{2z} + N_{3}μ_{3z})/(N_{0}
+ N_{1} + N_{2} + N_{3}) = (N_{1}μ_{1z}
+ N_{3}μ_{3z})/(N_{0} + N_{1} + N_{2}
+ N_{3})= (-N _{0}exp(-(Δ + γμ_{B}B)/(k_{B}T))*γμ_{B}
+ N_{0}exp(-(Δ - γμ_{B}B)/(k_{B}T))*γμ_{B})/(N_{0}
+ N_{0}exp(-(Δ + γμ_{B}B)/(k_{B}T)) + N_{0}exp(-Δ/(k_{B}T)
+ N_{0}exp(-(Δ - γμ_{B}B)/(k_{B}T)))= -γμ _{B*}(exp(-(Δ + γμ_{B}B)/(k_{B}T)) - exp(-(Δ -
γμ_{B}B)/(k_{B}T)))/(1 + exp(-(Δ + γμ_{B}B)/(k_{B}T))
+ exp(-Δ/(k_{B}T) + exp(-(Δ - γμ_{B}B)/(k_{B}T)))= average magnetic moment <μ _{z}
> per molecule in the presence of a magnetic field B.(b) Let k _{B}T >> Δ >>
2γμ_{B}B.Expand e ^{y} = 1 + y in the numerator and e^{y} = 1 in the
denominator.Then <μ _{z}> ~ ½B(γμ_{B})^{2}/(k_{B}T).χ = N d<μ _{
z} >/dB = ½N(γμ_{B})^{2}/(k_{B}T). |