Using

B^{2} = 0

and

B B^{†} = 1 - B^{†} B

H^{2} = B^{†} B B^{†} B = B^{†} (1 - B^{†} B) B = H .

Thus $H$ is a projection and its eigenvalues are $0$ and $1$ .

ii

Solution

The state

B ∣1 ⟩

has norm

∥ B ∣1 ⟩ ∥^{2} = ⟨ 1∣ B^{†} B ∣1 ⟩ = 1

and satisfies

H B ∣1 ⟩ = B^{†} B^{2} ∣1 ⟩ = 0

. Hence it is the normalized ground state, denoted

∣0 ⟩

. Also

∥ B^{†} ∣1 ⟩ ∥^{2} = ⟨ 1∣ B B^{†} ∣1 ⟩ = ⟨ 1∣ (1 - H) ∣1 ⟩ = 0,

B ∣1 ⟩ = ∣0 ⟩, B^{†} ∣1 ⟩ = 0.

In the ordered basis

(∣0 ⟩, ∣1 ⟩)

B = (0010), B^{†} = (0100), H = (0001) .

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