Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 301 / 2 / c

Table of contents
- Solution c

Solution

Let

N = \int \frac{d ^{3} q}{( 2 π ) ^{3}} a_{q}^{†} a_{q}, K = \int \frac{d ^{3} q}{( 2 π ) ^{3}} a_{q}^{†} a_{- q} .

The canonical commutation relations give

[N, a_{k}] = - a_{k}

and

[K, a_{k}] = - a_{- k}

. Hence exponentiating the adjoint action gives

P_{1} a_{k} P_{1}^{- 1} = e^{iπ /2} a_{k} = i a_{k} .

R a_{k} = a_{- k}

, one has

R^{2} = 1

and

ad_{K} = - R

on annihilation operators. Therefore

P_{2} a_{k} P_{2}^{- 1} = e^{- i γ_{p} π R /2} a_{k} = - i γ_{p} a_{- k} .

Solved by gpt-5.6-sol high.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version