ABSTRACT - DRUM - University of Maryland

From (??) we have 〈Ψ 01 | ˙Ψ 01 〉 = 〈Ψ 02 | ˙Ψ 02 〉 = 0. So M only has off-diagonal
elements. Furthermore, we can show that M +− must be a real number following the
Majorana condition Ψ ∗ = Ψ. Write Ψ = (u, u ∗ ) T , we have
∫
〈Ψ 01 | ˙Ψ 02 〉 =
d 2 r (u ∗ 1 ˙u 2 + u 1 ˙u ∗ 2), (6.32)
from which we can easily see 〈Ψ 01 | ˙Ψ 02 〉 ∈ R. The same for 〈Ψ 02 | ˙Ψ 01 〉. The integral
in M +− can be further simplified:
M +− = ωα, α = 1 2
∫
d 2 r Ψ † (r + R)Ψ(r), (6.33)
where R = R 1 − R 2 . The form of α is not important apart from the fact that
|α| ∼ e −R/ξ .
Therefore the matrix M takes the following form:
⎛ ⎞
0 α
M = ω ⎜ ⎟
⎝ ⎠ . (6.34)
α 0
Since both α and E ± are functions of R, they are time-independent. We now
have to solve essentially the textbook problem of the Schrödinger equation of a spin
1/2 in a magnetic field, the solution of which is well-known:
⎛ ⎞ ⎛
⎞ ⎛ ⎞
⎜
c + (T )
⎟
⎝ ⎠ = ⎜
cos ET − iE + sin ET iωα
sin ET
E E
c + (0)
⎟ ⎜ ⎟
⎝
⎠ ⎝ ⎠
√E , E = + 2 + ω 2 α 2 .
iωα
c − (T )
sin ET cos ET + iE +
sin ET c
E E
− (0)
(6.35)
We can translate the results into transformation of Majorana operators:
ˆγ 1 →
(
cos ET + iωα
E sin ET )
ˆγ 2 − E +
ˆγ 2 → − E +
E sin ET ˆγ 2−
E sin ET ˆγ 1
(
cos ET + iωα )
. (6.36)
E sin ET ˆγ 1
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