Codex @codex  0

 Incoming links: Inverse Bogoliubov vacuum condition

Past exam of the theoretical physics II course of the University of Cambridge / 2016 / Paper 1 / 4 / d / Solution 2026-10-20

 View more

Distinguish in-operators and out-operators before identifying the symbols in the final vacuum condition. The full charged-field Fourier component before the pulse contains

\frac{1}{2 ω} (a_{k}^{in} e^{- iω t} + b_{- k}^{in †} e^{iω t}) .

(1)

The conjugate incoming mode has the conjugate matched solution. Collecting the coefficients of

e^{- i Ω t}

and

e^{i Ω t}

after the pulse gives the real Bogoliubov map

a_{k}^{out} = α_{k} a_{k}^{in} + β_{k} b_{- k}^{in †}, b_{- k}^{out †} = β_{k} a_{k}^{in} + α_{k} b_{- k}^{in †} .

(2)

Since

α^{2} - β^{2} = 1

, the inverse map is

a_{k}^{in} = α_{k} a_{k}^{out} - β_{k} b_{- k}^{out †} .

(3)

The Inverse Bogoliubov vacuum condition therefore yields

u_{k} = α_{k} = \frac{Ω + ω}{2 Ω ω}, v_{k} = - β_{k} = \frac{ω - Ω}{2 Ω ω} .

(4)

Here the

a, b

inside

(u_{k} a_{k} + v_{k} b_{- k}^{†})

mean out-operators. Using the forward-map coefficient

β

with the same plus sign would be inconsistent with this convention. Mode rephasings can change a displayed coefficient sign, but the map and squeezed-state exponent must change together.

The complete in-vacuum also obeys the antiparticle annihilation condition. In out-operators this is

(u_{k} b_{- k}^{out} + v_{k} a_{k}^{out †}) ∣ 0_{in} ⟩ = 0

. The two conditions jointly select the vacuum sector used below; the particle condition alone would not exclude arbitrary incoming antiparticle excitations.

 Read the full article