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I am working on a problem based on Problem 1.6 in Gerald D. Mahan - Many Particle Physics, 3rd edition.

The problem:

Consider a tight-binding solid which has alternating atoms of type A and B. The Hamiltonian is given $$H = t\sum_{\langle i, j \rangle} (a_i^\dagger b_j + b_j^\dagger a_i) + \sum_i(\mu_a a_i^\dagger a_i + \mu_b b_i^\dagger b_i)$$

where $t$, $\mu_a$ and $\mu_b$ are constants, and $a_i^\dagger$ ($b_i^\dagger$) represents the electron creation operator at site $i$ on atom $A$ ($B$). $\langle i, j \rangle$ denotes nearest neighbors. Set the distance between $A$ and $B$ to be $\mathbf{d}$.

(a) Write the Hamiltonian above in momentum space, $\mathbf{k}$.

(b) Diagonalize the Hamiltonian obtained in (a) and find the exact energy eigenvalues.

My attempt:

(a) I believe I have this part correct:

Using:

$$\sum_{\langle i, j \rangle} \rightarrow \sum_i \sum_\mathbf{d}$$

and

$$ a_i = \frac{1}{\sqrt{N}}\sum_\mathbf{k} e^{i\mathbf{k} \cdot \mathbf{R_i}} a_\mathbf{k}$$

etc., where $\mathbf{R}_i$ is the position of site $i$, and $N$ is the number of sites, I obtain:

$$ H = t \sum_\mathbf{d} \sum_\mathbf{k}(e^{i\mathbf{k} \cdot \mathbf{d}} a^\dagger_\mathbf{k} b_\mathbf{k} + e^{-i\mathbf{k} \cdot \mathbf{d}} b^\dagger_\mathbf{k} a_\mathbf{k}) + \sum_\mathbf{k}(\mu_a a_\mathbf{k}^\dagger a_\mathbf{k} + \mu_b b_\mathbf{k}^\dagger b_\mathbf{k})$$

(b) I am not sure how to diagonalize the Hamiltonian. I suspect that one can define new creation/annihilation operators in terms of the old ones, which leave the Hamiltonian in diagonal form.

Any ideas or references to similar problems would be appreciated.

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  • $\begingroup$ Can you think of each term in the sum over k as a two by two matrix, which you can then diagonalize? $\endgroup$ – user2309840 Feb 12 '17 at 3:08

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