 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2018/1/c/solution

Use

c_{j} = (γ_{j}^{A} + i γ_{j}^{B}) /2

and

c_{j}^{†} = (γ_{j}^{A} - i γ_{j}^{B}) /2

. Moving operators from different modes past one another introduces a minus sign. Direct expansion gives

c_{1}^{†} c_{2} + c_{2}^{†} c_{1} c_{1}^{†} c_{2}^{†} + c_{2} c_{1} = \frac{i}{2} (γ_{1}^{A} γ_{2}^{B} - γ_{1}^{B} γ_{2}^{A}), = - \frac{i}{2} (γ_{1}^{A} γ_{2}^{B} + γ_{1}^{B} γ_{2}^{A}) .

Thus, for real hopping and pairing parameters,

H = \frac{i}{2} [(Δ - t) γ_{1}^{A} γ_{2}^{B} + (t + Δ) γ_{1}^{B} γ_{2}^{A}] .

Δ = t \neq = 0

this becomes

H = i t γ_{1}^{B} γ_{2}^{A} .

Use

τ

for time to distinguish it from the hopping amplitude

t

. Each unused Majorana operator anticommutes with both factors in

H

, so commuting it through their product gives two minus signs. Hence

[H, γ_{1}^{A}] = [H, γ_{2}^{B}] = 0

d O / d τ = (i /ℏ) [H, O]

therefore gives

γ_{1}^{A} (τ) = γ_{1}^{A} (0), γ_{2}^{B} (τ) = γ_{2}^{B} (0) .

These are the two uncoupled Majorana zero modes.

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