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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? $\endgroup$ Feb 12, 2017 at 3:08

1 Answer 1

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I'm not entirely clear on the transformation you mentioned

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

however I post below my solution to the problem. This question is quite old, but I hope that someone will find it useful.


The transformation of chemical potential-like terms is rather straightforward. For now, let's focus on the hopping part, denoted as $H_t$

$$ H_t = t \sum_{\langle i, j\rangle} a^\dagger_i b_j + b_j^\dagger a_i. $$

Next, I will perform a step that transforms the neighbor summation assuming a one-dimensional lattice with a lattice vector $\mathbf{d}$. Please note that this step can be applied to different lattice types, such as 2D, 3D, triangle, etc. The final energy dispersion may vary depending on the specific case.

$$ H_t = t \sum_{\langle i, j\rangle} a^\dagger_i b_j + b_j^\dagger a_i = t \sum_i \left(a_{\mathbf R_i +\mathbf d}^\dagger b_{\mathbf R_i} + a_{\mathbf R_i -\mathbf d}^\dagger b_{\mathbf R_i}\right) + \text{h.c.} $$

By using Fourier transform, we obtain:

\begin{align} H_t &= t \sum_i \Bigg[ \left(\frac 1{\sqrt{L}}\sum_{\mathbf k} e^{-i\mathbf k (\mathbf R_i+\mathbf d)} a^\dagger_{\mathbf k}\right) \left(\frac 1{\sqrt{L}}\sum_{\mathbf k'} e^{+i\mathbf k' \mathbf R_i} b_{\mathbf k'}\right) +\\ &\left(\frac 1{\sqrt{L}}\sum_{\mathbf k} e^{-i\mathbf k (\mathbf R_i-\mathbf d)} a^\dagger_{\mathbf k}\right) \left(\frac 1{\sqrt{L}}\sum_{\mathbf k'} e^{+i\mathbf k' \mathbf R_i} b_{\mathbf k'}\right) \Bigg] + \text{h.c.} =\\ &= t \sum_{\mathbf k}\sum_{\mathbf k'}\left[ \left(e^{-i\mathbf k \mathbf d} + e^{+i\mathbf k \mathbf d}\right) a^\dagger_{\mathbf k} b_{\mathbf k' }\frac1L\sum_i e^{-i(\mathbf k - \mathbf k')\mathbf R_i} \right] + \text{h.c.} \end{align}

By utilizing the property $\delta_{\mathbf k\mathbf k'} = \frac1L\sum_i e^{-i(\mathbf k - \mathbf k')\mathbf R_i}$ and $\sum_i x_i \delta_{ij} = x_j$, we can simplify further

$$ H_t = \sum_{\mathbf k} 2t \cos(\mathbf k\mathbf d) \left(a_{\mathbf k}^\dagger b_{\mathbf k} + b_{\mathbf k}^\dagger a_{\mathbf k}\right). $$

Now, let's return to the Hamiltonian (including the potentials $\mu_{a,b}$)

$$ H = \sum_{\mathbf k}\left[ 2t \cos(\mathbf k\mathbf d) \left(a_{\mathbf k}^\dagger b_{\mathbf k} + b_{\mathbf k}^\dagger a_{\mathbf k}\right) + \mu_a a^\dagger_{\mathbf k} a_{\mathbf k} + \mu_b b^\dagger_{\mathbf k} b_{\mathbf k} \right]. $$

The Hamiltonian can be diagonalized by showing that it can be written in the following form

$$ H = \sum_{\mathbf k} \left(\begin{matrix} a_{\mathbf k}^\dagger & b_{\mathbf k}^\dagger \end{matrix}\right) % \left(\begin{matrix} \mu_a & 2t\cos(\mathbf k \mathbf d)\\ 2t\cos(\mathbf k \mathbf d) & \mu_b \end{matrix}\right) % \left(\begin{matrix} a_{\mathbf k}\\ b_{\mathbf k} \end{matrix}\right) $$

The last step is to find dispersion relation. It can be done by diagonalizing the matrix $ \left(\begin{matrix} \mu_a & 2t\cos(\mathbf k \mathbf d)\\ 2t\cos(\mathbf k \mathbf d) & \mu_b \end{matrix}\right) $, which leads to the equation:

$$ \det\left(\begin{matrix} \mu_a-\varepsilon_{\mathbf k} & 2t\cos(\mathbf k \mathbf d)\\ 2t\cos(\mathbf k \mathbf d) & \mu_b-\varepsilon_{\mathbf k} \end{matrix}\right) = 0. $$

The equation has two solutions for $\varepsilon_{\mathbf k}$

$$\boxed{ \varepsilon_{\mathbf k}^\pm = \frac{\mu_a+\mu_b}2 \pm \sqrt{\left(\frac{\mu_a-\mu_b}2\right)^2 + 4t^2\cos^2(\mathbf k\mathbf d)} .} $$

Finally, diagonalized Hamiltonian has the form

$$ \boxed{ H = \sum_{\mathbf k} \left(\begin{matrix} a_{\mathbf k}^\dagger & b_{\mathbf k}^\dagger \end{matrix}\right) U^\dagger % \left(\begin{matrix} \varepsilon_{\mathbf k}^+ & 0\\ 0 & \varepsilon_{\mathbf k}^-\\ \end{matrix}\right) % U \left(\begin{matrix} a_{\mathbf k}\\ b_{\mathbf k} \end{matrix}\right) = \sum_{\mathbf k} \left( \varepsilon_{\mathbf k}^+ c^\dagger_{\mathbf k} c_{\mathbf k}+ \varepsilon_{\mathbf k}^- d^\dagger_{\mathbf k} d_{\mathbf k} \right), } $$

where $U$ is unitary transformation (made of eigenvectors of $\left(\begin{matrix} \mu_a & 2t\cos(\mathbf k \mathbf d)\\ 2t\cos(\mathbf k \mathbf d) & \mu_b \end{matrix}\right)$, and $ \left(\begin{matrix} c_{\mathbf k}\\ d_{\mathbf k} \end{matrix}\right)= U \left(\begin{matrix} a_{\mathbf k}\\ b_{\mathbf k} \end{matrix}\right) $.

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