Quantum Phase Transition
- 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