 Solution

ID: past-exam-of-the-quantum-condensed-matter-field-theory-course-of-the-university-of-cambridge/2016/paper-1/1/e/solution

Write

u_{k} = cosh θ_{k}

v_{k} = sinh θ_{k}

, taking real even

θ_{k}

. The transformation gives

[b_{k}, b_{q}^{†}] = (u_{k} u_{q} - v_{k} v_{q}) δ_{k q} = δ_{k q}, [b_{k}, b_{q}] = (u_{k} v_{q} - v_{k} u_{q}) δ_{k, - q} = 0.

The second equality uses

u_{- k} = u_{k}

and

v_{- k} = v_{k}

, while the first uses

u_{k}^{2} - v_{k}^{2} = 1

. The creation-creation commutator vanishes by conjugation. Therefore the $b$ modes are canonical bosons. In the paired basis,

(b _{- k}^{†} b _{k}) = (u_{k} v_{k} v_{k} u_{k}) (a _{- k}^{†} a _{k}) .

Its determinant is one and its inverse reverses the signs of the off-diagonal entries. Hence

a_{k} = u_{k} b_{k} - v_{k} b_{- k}^{†}

. At self-inverse momenta

k = 0, π

this is the corresponding one-mode squeezing transformation, still canonical for finite

θ_{k}

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