Institute for Theoretical Physics, Faculty of Physics and Astronomy, UU.
Made available in electronic form by the TBC of A−Eskwadraat In 2005/2006, the course SFT was given by Prof. Dr. Stoof.
Statistical Field Theory Midterm take home (SFT) 29 November 2005
Question 1. Hubbard-Stratonovich transformation
Consider an interacting gas N spinless fermions on a one-dimensional line with length L. The action for this gas at temperature T = 1/kBβ and chemical potential µ is given by
S[φ∗, φ] = Z ~β
0
dτ Z
dxφ∗(x, τ )
~
∂
∂τ − ~2 2m
∂2
∂x2 − µ
φ(x, τ ) + (1)
1 2
Z ~β 0
dτ Z
dx Z
dx0φ∗(x, τ )φ∗(x0, τ )V (x − x0)φ(x0, τ )φ(x, τ ) (2)
where m is the mass of the fermions and X(x−x0) the interaction potential. We want to perform a Hubbard-Stratonovich transformation to the field n(x, x0, τ ) that on average is equal to the density matrix of the Fermi gas, so hn(x, x0, τ )i = hφ∗(x, τ )φ(x0, τ )i. To still be able to calculate the exact fermionic Green’s function G(x, τ ; x0, τ0), even after the Hubbard-Stratonovich transformation and integrating out the fermionic fields, we add also the current terms
S[φ∗, φ, J ] = −~
Z ~β 0
dτ Z
dx(φ∗(x, τ )J (x, τ ) + J∗(x, τ )φ(x, τ ))
to the action. In the following we denote the fermionic Green’s function with the Fock-like selfen- ergy ~Σ(x, τ ; x0, τ0) = δ(τ − τ0)X(x − x0)n(x0, x, τ ) by G(x, τ ; x0, τ0; n).
a) Perform the usual Hubbard-Stratonovich transformation that decouples the interaction be- tween the fermions and integrate out the fermions. Determine the effective action Sef f[n; J, J∗] ≡ Sef f[n] + Sef f[J, J∗]. Show that in particular that
Sef f[J, J∗] = ~ Z ~β
0
dτ Z
dx Z ~β
0
dτ0 Z Z
dx0J∗(x, τ )G(x, τ ; x0, τ0; n)J (x0, τ0).
b) Prove now the the exact fermionic Green’s function can be obtained from G(x, τ ; x0τ0) = hG(x, τ ; x0, τ0; n)i
where the average in the left hand side denotes the averega over the field n(x, x0, τ ) using the effective Sef f[n].
It is possible to obtain also all the correlation functions of the product φ∗(x, τ )φ(x, τ ) in the above manner, but in this case it is easier to perform the calculation differently. Instead of adding the current term S[φ∗, φ; J, J∗], we now add the source term
S[φ∗φ; K] = −~
Z ~β 0
dτ Z
dx Z
dx0φ∗(x, τ )φ(x0, τ )Kx0, x, τ ) to the action S[φ∗, φ]. We can assume that K∗(x0x, x0, τ ) = K(x, x0, τ ).
c) Perform a Hubbard-Stratonovich transformation that simultaneously decouples the interac- tion between the fermions and also removes the above source term from the fermionc part of the action. Integrate out the fermions and calculate the effective action Sef f[n; K].
d) Prove from the last result that
hφ∗(x, τ )φ(x0, τ )i = hn(x, x0, τ )i, and
hφ∗(x, τ )φ(x0, τ )φ∗(x00, τ0)φ(x000, τ0)i = hn(x, x0, τ )n(x00, x000, τ0)i
− ~V−1(x0, x, x00, x000)δ(τ − τ0),
with V (x, x0, x00, x000) ≡ V (x − x0)δ(x − x00)δ(x0− x000). If you have not calculated the effective action Sef f[n; K] in part c), deduce from the above relations how it should look like.
