The Heisenberg Hamiltonian representing the “exchange
interaction” between two spins (**S**_{1} and **S**_{2})
is given by H = -2f(R)**S**_{1}∙**S**_{2},
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** = **
S**_{1 }
+** S**_{2}.

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:

**S**_{1}∙**S**_{2}= ½(S^{2}– S_{1}^{2}– S_{2}^{2}). The eigenstates of**S**_{1}∙**S**_{2}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**S**_{1}∙**S**_{2}|0, 0> = -¾ħ^{2}.

For the triplet states we have**S**_{1}∙**S**_{2}|1, M_{s}> = ¼ħ^{2}.

The eigenstates of H are |0, 0> with eigenvalue (3/2)f(R)ħ^{2}and |1, 1>, |1, 0>, |1, -1> with eigenvalue -½f(R)ħ^{2}(3-fold degenerate).

Let **S**_{i}, i =
1, 2 denote the spin vectors of two spin-½ particles. The interaction is given by

H = U_{0} (**S**_{1}·**S**_{2} − 3 S_{1z}S_{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**S**_{1}·**S**_{2}are the singlet and the triplet states.

{|S,S_{z}> = |0,0>, |1,1>, |1,0>, |1,-1>}

**S**_{1}·**S**_{2}|S,S_{z}> = ½(S^{2}– S_{1}^{2}– S_{2}^{2}) |S,S_{z}> = ½(S(S+1) – 3/2)ħ^{2}|S,S_{z}>.

The singlet and the triplet states are also eigenfunctions of S_{1z}S_{2z}.

S_{1z}S_{2z}|1,1> = S_{1z}S_{2z}|++> = (ħ^{2}/4)|1,1>.

S_{1z}S_{2z}|1,0> = ½^{½}S_{1z}S_{2z}(|+-> + |-+>) = -(ħ^{2}/4)|1,0>.

S_{1z}S_{2z}|1,-1> = S_{1z}S_{2z}|--> = (ħ^{2}/4)|1,-1>

S_{1z}S_{2z}|0,0> = ½^{½}S_{1z}S_{2z}(|+-> - |-+>) = -(ħ^{2}/4)|0,0>The eigenstates of H are the the singlet and the triplet states. In that basis H is diagonal.

H |1,1> = U_{0}(1-3) (ħ^{2}/4)|1,1> = -2U_{0}(ħ^{2}/4)|1,1>

H |1,0> = U_{0}(1+3) (ħ^{2}/4)|1,0> = 4U_{0}(ħ^{2}/4)|1,0>

H |1,-1> = U_{0}(1-3) (ħ^{2}/4|1,-1> = -2U_{0}(ħ^{2}/4)|1,1>

H |0,0> = U_{0}(-3+3) (ħ^{2}/4|0,0> = 0

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:

Sz 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.

Consider the state space E = E_{1}E_{2}
of two non-identical spin ½ particles spanned by the basis vectors {|++>,
|+->, |-+>, |-->}. Use what you know about the common eigenvectors of S^{2}
and S_{z}, and find the common eigenvectors of S^{2}
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 S^{2}and S_{z}. We are to change representation from the eigenbasis of S^{2}and S_{z}to the eigenbasis of S^{2}and S_{x}. -
Details of the calculation:

The common eigenfunctions of S^{2}and S_{z}are

|11> = |++>,

|10> = ½^{½}(|+-> + |-+>),

|1-1> =|-->,

|00> = ½^{½}(|+-> - |-+>).From symmetry we therefore have that the common eigenfunctions of S

^{2}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}(|+-> - |-+>).