ABSTRACT - DRUM - University of Maryland

at here is the expectation value of the fermion parity operator in the ground state
subspace, namely 〈 ˆP 0 〉 = 〈iˆγ 1ˆγ 2 〉.
Suppose at t = 0 we start from the ground state |g〉 with even fermion parity
〈g| ˆP 0 |g〉 = 1 and the excited level is unoccupied, too. After the braiding at time T
the expectation value of ˆP 0 becomes
〈 ˆP 0 (T )〉 = 1− 8β2 ET
sin2
E2 2 . (6.46)
where ˆP 0 (T ) = Û † (T ) ˆP 0 Û(T ). This confirms that fermion parity is not conserved
anymore. For |β| ≪ ε, the coupling to excited state can be understood as a small
perturbation. 〈 ˆP 0 〉 only slightly deviates from the non-perturbed value. In the
opposite limit |β| ≫ ε, 〈 ˆP 0 〉 can oscillate between 1 and −1 so basically fermion
parity is no longer well-defined.
Now we can sum up the contributions from each excited state and (6.46) is
replaced by:
〈 ˆP 0 (T )〉 = 1− ∑ λ
8|β λ | 2
E 2 λ
sin 2 E λT
2 . (6.47)
The sum over the continuum states can be replaced by an integral over energy.
We assume that the couplings β λ dependes only weakly on the energy ε λ so it can
be factored out as β λ ≈ β. Then we obtain
∫ ∞
〈 ˆP 0 (T )〉 = 1−8|β| 2 ν(ε)
dε
εT ε 2 + 4|β| 2 sin2 2 . (6.48)
∆ 0
The density of states ν(ε) depends on the microscopic details of the underlying
superconducting phase.
For simplicity we take the typical BCS-type density of
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