### Two spin ½ particles

#### Problem:

The Heisenberg Hamiltonian representing the "exchange interaction" between two spins (S1 and S2) is given by H = -2f(R)S1S2, 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:
S1S2 = ½(S2 - S12 - S22).  The eigenstates of S1S2 are the singlet state and the triplet states, {|S, Ms>}, S = 0, 1.
In that basis H is diagonal.
For the singlet state we have S1S2 |0, 0> = -¾ħ2.
For the triplet states we have S1S2 |1, Ms> = 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 Si, i = 1, 2 denote the spin vectors of two spin-½ particles. The interaction is given by
H = U0 (S1·S2 − 3 S1zS2z).
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,Sz> = |0,0>, |1,1>, |1,0>, |1,-1>}
S1·S2|S,Sz> = ½(S2 - S12 - S22) |S,Sz> = ½(S(S+1) - 3/2)ħ2|S,Sz>.
The singlet and the triplet states are also eigenfunctions of S1zS2z.
S1zS2z |1,1> = S1zS2z |++> = (ħ2/4)|1,1>.
S1zS2z |1,0> = ½½ S1zS2z (|+-> + |-+>) = -(ħ2/4)|1,0>.
S1zS2z |1,-1> = S1zS2z |--> = (ħ2/4)|1,-1>
S1zS2z |0,0> = ½½ S1zS2z (|+-> - |-+>) = -(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> = -2U02/4)|1,1>
H |1,0> = U0(1+3) (ħ2/4)|1,0> = 4U02/4)|1,0>
H |1,-1> = U0(1-3) (ħ2/4|1,-1> = -2U02/4)|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)> = ½½(|+>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 Siz 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 Sz, and find the common eigenvectors of S2 and Sx.  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 S1z and S2z and
{|S,Sz>} = {|1,1>,|1,0>,|1,-1>,|0,0>},
the common eigenbasis of S2 and Sz.  We are to change representation from the eigenbasis of S2 and Sz to the eigenbasis of S2 and Sx.
• Details of the calculation:
The common eigenfunctions of S2 and Sz are
|11> = |++>,
|10> = ½½(|+-> + |-+>),
|1-1> =|-->,
|00> = ½½(|+-> - |-+>).

From symmetry we therefore have that the common eigenfunctions of S2 and Sx are
|11>x = |++>x,
|10>x = ½½(|+->x + |-+>x),
|1-1>x = |-->x,
|00>x = ½½(|+->x + |-+>x).

We have
|++>x = ½½(|+> + |->)½½(|+> + |->) = ½(|++> + |+-> + |-+> + |-->),
|-->x = ½½(|+> - |->)½½(|+> - |->) = ½(|++> - |+-> - |-+> + |-->),
|+->x = ½½(|+> + |->)½½(|+> - |->) = ½(|++> - |+-> + |-+> - |-->),
|-+>x = ½½(|+> - |->)½½(|+> + |->) = ½(|++> + |+-> - |-+> - |-->).

Therefore
|11>x = ½(|++> + |+-> + |-+> + |-->),
|10>x = ½½(|++> - |-->),
|1-1>x = ½(|++> - |+-> - |-+> + |-->),
|00>x = ½½(|+-> - |-+>).