In 1D heisenberg chain hamilltonian.

1. “Quantum Phase Transition”
In the 1D Heisenberg Model
RITTWIK CHATTERJEE
West Bengal State University
Under the guidance of Dr.Sankhasubhra Nag

2. The Heisenberg Chain Hamiltonian
The Heisenberg Chain Hamiltonian is
H =
N
X
i=1
(BSi
z + J~
Si
.~
Si+1
)
So considering the 2-spin case we can write,
H =
2
X
i=1
(BSi
z + J~
Si
.~
Si+1
)
Now if the external magnetic field is zero the Hamiltonian is
reduced to
H =
N
X
i=1
J~
Si
.~
Si+1
J > 0 if the sample is antiferromagnetic and if J < 0 the sample is
ferromagnetic.

3. Two spin Hamiltonian and Pauli spin matrices:
(in the unit of ~ = 1),Si = σi /2 ~
σi = (σi
x , σi
y , σi
z) in which σi
x,y,z
are the Pauli matrices. So
H = B(Sz
1 + Sz
2 ) + J~
S1
· ~
S2
=
J
4
(σx
1 · σx
2 + σy
1 · σy
2 + σz
1 · σz
2) +
B
2
(σz
1 + σz
2)
Here
σx =
0 1
1 0
!
and
σy =
0 −i
i 0
!
and
σz =
1 0
0 −1
!

4. Eigenstates and Eigenvalues of the Hamiltonian
Eigenstates of Hamiltonian: Evidently the triplet states with
s = 1 are,
|↑↑i
1
√
2
(|↑↓i + |↓↑i)
|↓↓i





triplet states
And for s = 0 the singlet state is,
1
√
2
(|↑↓i − |↓↑i)
o
singlet state.
Eigenvalues of S2 operator:
S2|↑↑i = 2~2|↑↑i
S2 1
√
2
(|↑↓i + |↓↑i) = 2~2 1
√
2
(|↑↓i + |↓↑i)
S2|↓↓i = 2~2|↓↓i





All three states have
the same energy i.e. the states are degenerate. And the energy is
2~2.
S2 1
√
2
(|↑↓i + |↓↑i) = 0
o
The energy eigenvalue is zero for
singlet state.

5. Eigenstates and Eigenvalues of the Hamiltonian(continued)
Eigenvalues of ~
S1. ~
S2 operator:
~
S1. ~
S2|↑↑i = ~2
4 |↑↑i
~
S1. ~
S2
1
√
2
(|↑↓i + |↓↑i) = ~2
4
1
√
2
(|↑↓i + |↓↑i)
~
S1. ~
S2|↓↓i = ~2
4 |↓↓i







All three states
have the same energy i.e. the states are degenerate. And the
energy is ~2
4 .
~
S1. ~
S2
1
√
2
(|↑↓i + |↓↑i) = −3~2
4
1
√
2
(|↑↓i + |↓↑i)
o
The energy
eigenvalue is −3~2
4 for singlet state.
Eigenvalues for the perturbation term if the magnetic field
term is non zero: If the ~
B 6= 0 then the perturbation hamiltonian
is, ∆H = B(S1z + S2z)
B(S1z + S2z)B(S1z + S2z)|↑↑i = 2B~|↑↑i
B(S1z + S2z)(|↑↓i + |↓↑i) = 0
B(S1z + S2z)|↓↓i = −2B~|↓↓i





for s=1
B(S1z + S2z) 1
√
2
(|↑↓i − |↓↑i) = 0
o
for s=0.

6. Entanglement for pure states:
To measure the entanglement, starting from the density
matrix ρ, we first need to calculate the spin flipped density
matrix e
ρ = (σy ⊗ σy )ρ∗(σy ⊗ σy ).
And then we have to calculate the time reversed matrix
R = ρe
ρ. So now the concurrence is written as ,
Ci = max(
p
λ1 −
p
λ2 −
p
λ3 −
p
λ4, 0) = 1
Here λ1, λ2, λ3, λ4 are the eigenvalues of the corresponding density
matrix.

7. Quantum Phase Transition:
Figure: Determining the critical value of magnetic field Bc when the
quantum phase transition occurs:

8. Entanglement for mixed states:
At constant temperature a particle lies at a mixture of states.At a
pure state |ψi the density matrix is:
ρ = |ψihψ|
For mixed state the density matrix is:
ρ =
X
i
e−βEi
Z
ρ̂
. Hence at finite temperatures the system is in a mixed state,
ρ =
1
Z
(|ψ1ihψ1|e−β(4gB+J/4)
+ |ψ2ihψ2|e−βJ/4
+ |ψ3ihψ3|e−β(−4gB+J/4)
+
Where β = 1
kBT and
Z = Tr(ρ) = e−βJ/4 + eβ3J/4 + e−β(−4gB+J/4) + e−β(4gB+J/4).

9. Entanglement for mixed states(continued):
Here the formula of the concurrence C is given by:
C = 0
for e8J/KT ≤ 3
C =
e8J/KT − 3
1 + e−2B/KT + e2B/KT + e8J/KT
for e8J/KT > 3 Hence we can plot the concurrence as a function of
the magnetic field B for a certain value of temperature and J.

10. Plots of Concurrence:
(a) Concurrence C with B at kT= 0.1
and J = 1.
(b) Concurrence C with B and kT for
J = 1

11. The END:
Thank You!