Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford / 2023 / 2 / b / i

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- Solution i

Solution

For the specified positive-exponent discrete Fourier transform, the inverse discrete Fourier transforms have negative exponents:

a_{n} = \frac{1}{L} q \sum e^{- i q n} A_{q}, b_{n} = \frac{1}{L} q \sum e^{- i q n} B_{q} .

Every sum runs over

q = 2 πj / L

,

j = 0, \dots, L - 1

. Indeed, orthogonality of roots of unity gives

\frac{1}{L} \sum_{q} e^{i q (r - n)} = δ_{r n},

so substituting the forward discrete Fourier transform into either proposed inverse recovers the original operator. Taking Hermitian conjugates supplies the inverse formulas for the creation operators.

Multiplying the inverse discrete Fourier transforms gives the local number operator combination

\frac{1}{2} (a_{n}^{†} a_{n} + b_{n}^{†} b_{n}) = \frac{1}{2 L} q, k \sum e^{i (q - k) n} (A_{q}^{†} A_{k} + B_{q}^{†} B_{k}) .

This local expression includes off-diagonal momentum terms. Only after summing over sites does orthogonality of roots of unity give

\frac{1}{2} \sum_{n} (a_{n}^{†} a_{n} + b_{n}^{†} b_{n}) = \frac{1}{2} \sum_{q} (A_{q}^{†} A_{q} + B_{q}^{†} B_{q}) .

In a physical fixed-

s

state the first boxed operator acts as

s

at every site, by the local spin constraint in the Schwinger boson representation.

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