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

Use

ℏ = 1

throughout. Let

N_{a} = a^{†} a

N_{b} = b^{†} b

and

N = N_{a} + N_{b}

. Operators belonging to different boson species commute. The proposed Schwinger boson representation gives

S^{+} S^{-} S^{-} S^{+} = a^{†} b b^{†} a = N_{a} (N_{b} + 1), = b^{†} a a^{†} b = N_{b} (N_{a} + 1) .

[S^{+}, S^{-}] = N_{a} - N_{b} = 2 S^{z} .

[N_{a}, a^{†}] = a^{†}

[N_{a}, a] = - a

and their

b

counterparts give

[S^{z}, S^{+}] [S^{z}, S^{-}] = \frac{1}{2} ([N_{a}, a^{†}] b - a^{†} [N_{b}, b]) = a^{†} b = S^{+}, = - b^{†} a = - S^{-} .

[N, S^{\pm}] = [N, S^{z}] = 0

: each fixed-

N

sector is invariant. To identify its spin, calculate the Casimir operator:

S^{2} = (S^{z})^{2} + \frac{1}{2} (S^{+} S^{-} + S^{-} S^{+}) = \frac{1}{4} (N_{a} - N_{b})^{2} + N_{a} N_{b} + \frac{1}{2} (N_{a} + N_{b}) = \frac{N}{2} (\frac{N}{2} + 1) .

s

therefore requires

N = 2 s

, and its normalized Fock basis correspondence is

∣ s, m ⟩ ⟷ ∣ s + m, s - m ⟩ = \frac{( a ^{†} ) ^{s + m} ( b ^{†} ) ^{s - m}}{( s + m )! ( s - m )!} ∣0, 0 ⟩ .

Conversely,

s = (n_{a} + n_{b}) /2

and

m = (n_{a} - n_{b}) /2

. The nonnegative integer occupation numbers give precisely

m = - s, - s + 1, \dots, s

, and there are

2 s + 1

such states. As a check, the spin raising operator acts as

a^{†} b ∣ s + m, s - m ⟩ = (s + m + 1) (s - m) ∣ s + m + 1, s - m - 1 ⟩,

s (s + 1) - m (m + 1)

(a_{n}^{†} a_{n} + b_{n}^{†} b_{n}) ∣ ψ ⟩ = 2 s ∣ ψ ⟩ at every site n .

The full two-species bosonic Fock space also contains other local irreducible spin representations; fixing only the total boson number does not remove them.

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