Two spin ½ particles

Problem:

The Heisenberg Hamiltonian representing the "exchange interaction" between two spins (S₁ and S₂) is given by H = -2f(R)S₁∙S₂, where f(R) is the so-called exchange coupling constant and R is the spatial separation between the two spins.  Find the eigenstates and eigenvalues of the Heisenberg Hamiltonian describing the exchange interaction between two electrons.
HINT: The total spin operator is S = S₁ + S₂.

Solution:

• Concepts:
  The state space of two spin-½ particles
• Reasoning:
  We have to diagonalize the matrix of H in the state space of two spin ½ particles
• Details of the calculation:
  S₁∙S₂ = ½(S² - S₁² - S₂²).  The eigenstates of S₁∙S₂ are the singlet state and the triplet states, {|S, M_s>}, S = 0, 1.
  In that basis H is diagonal.
  For the singlet state we have S₁∙S₂ |0, 0> = -¾ħ².
  For the triplet states we have S₁∙S₂ |1, M_s> = ¼ħ².
  The eigenstates of H are |0, 0> with eigenvalue (3/2)f(R)ħ² and |1, 1>, |1, 0>, |1, -1> with eigenvalue -½f(R)ħ² (3-fold degenerate).

Problem:

Let S_i, i = 1, 2 denote the spin vectors of two spin-½ particles. The interaction is given by
H = U₀ (S₁·S₂ − 3 S_1zS_2z).
Find the energy eigenstates and eigenvalues.

Solution:

• Concepts:
  The state space of two spin-½ particles
• Reasoning:
  We have to diagonalize the matrix of H in the state space of two spin ½ particles
• Details of the calculation:
  The eigenstates of S₁·S₂ are the singlet and the triplet states.
  {|S,S_z> = |0,0>, |1,1>, |1,0>, |1,-1>}
  S₁·S₂|S,S_z> = ½(S² - S₁² - S₂²) |S,S_z> = ½(S(S+1) - 3/2)ħ²|S,S_z>.
  The singlet and the triplet states are also eigenfunctions of S_1zS_2z.
  S_1zS_2z |1,1> = S_1zS_2z |++> = (ħ²/4)|1,1>.
  S_1zS_2z |1,0> = ½^½ S_1zS_2z (|+-> + |-+>) = -(ħ²/4)|1,0>.
  S_1zS_2z |1,-1> = S_1zS_2z |--> = (ħ²/4)|1,-1>
  S_1zS_2z |0,0> = ½^½ S_1zS_2z (|+-> - |-+>) = -(ħ²/4)|0,0>

  The eigenstates of H are the the singlet and the triplet states.  In that basis H is diagonal.
  H |1,1> = U₀(1-3) (ħ²/4)|1,1> = -2U₀(ħ²/4)|1,1>
  H |1,0> = U₀(1+3) (ħ²/4)|1,0> = 4U₀(ħ²/4)|1,0>
  H |1,-1> = U₀(1-3) (ħ²/4|1,-1> = -2U₀(ħ²/4)|1,1>
  H |0,0> = U₀(-3+3) (ħ²/4|0,0> = 0

Problem:

Two spin ½ particles, 1 and 2, are in the singlet state shown below. Here |+> and |-> refer to spin up and spin down with respect to the z-axis.
|ψ(1,2)> = ½^½(|+>₁|->₂ - |->₁|+>₂).
(a)  What is the probability of measuring the spin of particle 1 along an axis u(θ,φ), and finding the particle in the state |+>_u?
Assume the spin of particle 1 is measured along the z-axis and found to be ħ/2.
(b)  A simultaneous measurement of the spin of particle 2 along the z-axis would yield which results with which probabilities?
(c)  A simultaneous measurement of the spin of particle 2 along the x-axis would yield which results with which probabilities?

Solution:

• Concepts:
  Entangled particles
• Reasoning:
  S_z and S_iz are incompatible observables for this two-particle system.
• Details of the calculation:
  (a)  The probability of measuring the spin of particle 1 along an axis u(θ,φ), and finding the particle in the state |+>_u is ½.
  (b)  A simultaneous measurement of the spin of particle 2 along the z-axis would yield which -ħ/2 with probability 1.
  (c)  A simultaneous measurement of the spin of particle 2 along the z-axis would yield which -ħ/2 or ħ/2, each with probability 0.5.

Problem:

Consider the state space E = E₁E₂ of two non-identical spin ½ particles spanned by the basis vectors {|++>, |+->, |-+>, |-->}.  Use what you know about the common eigenvectors of S² and S_z, and find the common eigenvectors of S² and S_x.  Express these eigenvectors in terms of the basis vectors {|++>, |+->, |-+>, |-->}.

Solution:

• Concepts:
  The state space of two spin ½ particles, change of representation
• Reasoning:
  Two eigenbases for the state space of two spin ½ particles are 
  {|++>, |+->, |-+>, |-->},
  the common eigenbasis of S_1z and S_2z and 
  {|S,S_z>} = {|1,1>,|1,0>,|1,-1>,|0,0>},
  the common eigenbasis of S² and S_z.  We are to change representation from the eigenbasis of S² and S_z to the eigenbasis of S² and S_x.

• Details of the calculation:
  The common eigenfunctions of S² and S_z are
  |11> = |++>,
  |10> = ½^½(|+-> + |-+>),
  |1-1> =|-->,
  |00> = ½^½(|+-> - |-+>).

  From symmetry we therefore have that the common eigenfunctions of S² and S_x are
  |11>_x = |++>_x,
  |10>_x = ½^½(|+->_x + |-+>_x),
  |1-1>_x = |-->_x,
  |00>_x = ½^½(|+->_x + |-+>_x).

  We have
  |++>_x = ½^½(|+> + |->)½^½(|+> + |->) = ½(|++> + |+-> + |-+> + |-->),
  |-->_x = ½^½(|+> - |->)½^½(|+> - |->) = ½(|++> - |+-> - |-+> + |-->),
  |+->_x = ½^½(|+> + |->)½^½(|+> - |->) = ½(|++> - |+-> + |-+> - |-->),
  |-+>_x = ½^½(|+> - |->)½^½(|+> + |->) = ½(|++> + |+-> - |-+> - |-->).

  Therefore
  |11>_x = ½(|++> + |+-> + |-+> + |-->),
  |10>_x = ½^½(|++> - |-->),
  |1-1>_x = ½(|++> - |+-> - |-+> + |-->),
  |00>_x = ½^½(|+-> - |-+>).