Homework Set 3

From phys814
Jump to navigationJump to search

Problem 1: Action on fermionic creation and annihilation operators on Fock states

Using the following convention

|11111000=c^5c^4c^3c^2c^1|vacuum

where |vacuum>=|0000:

(a) Evaluate c^3c^6c^4c^6c^3|111110000.

(b) Write |1101100100 in terms of excitations about the "filled Fermi sea" |1111100000. Interpret your answer in terms of electron and hole excitations.

(c) Find Φ|N^|Φ, where |Φ=A|100+B|111000 and N^=ic^ic^i is the operator of the total particle number.

Problem 2: Density-density correlation function for noninteracting electrons

Compute the desity-density correlation function:

Dσσ(𝐫𝐫)=GS|Ψ^σ(𝐫)Ψ^σ(𝐫)Ψ^σ(𝐫)Ψ^σ(𝐫)|GS

for a gas of noninteracting electrons in the Fermi sea ground state |GS. Since Fock space is orthonormal, to get a non-zero result you should consider only those cases where particles are created and annihilated by the four operators in exactly the same states. Consider both cases with |𝐤i|<𝐤F and |𝐤i|>𝐤F. Express your result in terms of pair-correlation function discussed in the class.

Problem 3: Hubbard model for a triangular molecule

Consider a triangular molecule made of identical atoms, each of which has an s-type valence orbital. The simplest model describing this system is the corresponding Hubbard model with hopping t and on-site Coulomb repulsion U.

(a) Assuming that there are two spin-up electrons in the system, find their ground-state energy.

(b) In the absence of magnetic field, the eigenstates of two electrons can be classified either as being a singlet (total spin S=0) or triplet (total spin S=1). In (a) you found the triplet ground-state energy (for the component Sz=+1, but you can check that the Sz=0,1 triplet solutions also have precisely the same groundstate energy). Find the singlet ground-state energy if U. Obviously infinite repulsion prevents the two electrons from ever being on the same site.

(c) Find the singlet ground-state energy for a finite U, i.e. when the two electrons can be at the same site.

Problem 4: BSC superconductor in Zeeman field

Consider a thin film superconductor whose thickness is small enough to allow the penetration of the magnetic field into the interior, while any coupling of magnetic field to orbital degrees of freedom is neglected. Its BCS Hamiltonian in the presence of a magnetic field 𝐁=B𝐞z, coupling only to the spin, can then be written as:

H^=𝐤,σ(ε𝐤μσμBB)c^𝐤σc^𝐤σ+𝐤,𝐤V𝐤,𝐤c^𝐤c^𝐤c^𝐤c^𝐤

Here V𝐤,𝐤=V0/Ω (Ω is volume) for 𝐤,𝐤 within a shell of energy of width ωD on either side of the Fermi surfaace and zero otherwise; μB is the Bohr magneton; and σ=+1 for spin and σ=1 for spin.

(a) Show that the expectation value of H^ in the BCS variational ansatz for ground state wavefunction

|Φ=𝐤(u𝐤+v𝐤c^𝐤c^𝐤)|0

has the same energy as for B=0. Thus, minimization of energy with respect to the pair occupation amplitude gives the same result as at B=0.

(b) For small B, ignore the interaction term and calculate the lowering of the ground state energy of free electron gas caused by the Zeeman coupling, to second order in B.

(c) Combine (a) and (b) to obtain the reduction in the condensation energy of the superconductor caused by the Zeeman coupling. At what value of the magnetic field is superconductivity destroyed by the Zeeman coupling?

(d) Calculate the minimum excitation energy gap of the BCS superconductor as a function of the field at zero temperature T=0. When does the gap collapse? Compare this result with the one obtained in (c). This is the so-called "Pauli limiting field" also known as "Chandrasekhar-Clogston limit".