∣ n_{1} n_{2} ⟩

n_{a} \in {0, 1}

, are exact eigenvectors, with

H_{tot} ∣ n_{1} n_{2} ⟩ = (E_{1} n_{1} + E_{2} n_{2}) ∣ n_{1} n_{2} ⟩ .

Thus the energies attached to

∣00 ⟩, ∣01 ⟩, ∣10 ⟩, ∣11 ⟩

are respectively

0, E_{2}, E_{1}, E_{1} + E_{2} .

ii

Solution

In the ordered basis

(∣00 ⟩, ∣01 ⟩, ∣10 ⟩, ∣11 ⟩)

B_{1} = 0000000010000100, B_{2} = 0000100000000010,

B_{1}^{†}

and

B_{2}^{†}

are their transposes, and

H_{tot} = diag (0, E_{2}, E_{1}, E_{1} + E_{2}) .

For

H_{0} = E_{2} B_{2}^{†} B_{2}

0

and

E_{2}

are each two-dimensional because

n_{1}

E_{1} B_{1}^{†} B_{1}

within each space. It is already diagonal in the product basis, with eigenvalues

0, E_{1}

. Therefore the first-order eigenvectors remain

∣00 ⟩, ∣10 ⟩, ∣01 ⟩, ∣11 ⟩,

and their energies are

0, E_{1}, E_{2}, E_{2} + E_{1}

, respectively. These are the exact answers, so all higher corrections vanish.

There are no discussions about this article yet.

There are currently no matching articles.