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I am working on a problem in introductory solid-state physics and have a question. I feel hesitant to ask my professor because I think it might be a simple question. In the problem, an electron can move on a lattice. Sites $a$ and $c$ have an on-site energy of $E_0$, while sites $b$ have zero on-site energy. The Hamiltonian is provided as:

$H = E_0 \sum_{n}(|a_n\rangle\langle a_n| + |c_n\rangle\langle c_n|) + \sum_{n}(g_{a}(|a_n\rangle\langle b_n|+ |b_n\rangle\langle a_n|) + g_{c}(|c_n\rangle\langle b_n|+|b_n\rangle \langle c_n|) - \gamma (|b_n\rangle \langle b_{n+1}| +|b_{n+1}\rangle \langle b_{n}|))$

Lattice

This is how I am solving it: I have realized that the system has two symmetries: discrete translational symmetry with operator $T$ and reflection symmetry with operator $P$ about the axis passing through the central row containing sites $c$. The basis is ${\vert a_n\rangle, \vert b_n\rangle, \vert c_n\rangle}$, and the dimension of the Hilbert space is $3N$, where $N$ is the number of unit cells.

I see that $\vert c_{n}\rangle$ is an eigenstate of $P$ with an eigenvalue of +1, while the other two eigenstates $\vert \Psi_{n}^{\pm}\rangle$ can be written as: \begin{align} \vert \Psi_{n}^{\pm}\rangle &= \frac{1}{\sqrt{2}} (\vert a_n\rangle \pm \vert b_n\rangle)\end{align}

Now, we can diagonalize $H$ in each $N$-dimensional block. To calculate the matrix elements, it is useful to evaluate the action of $H$ on the three types of basis states. However, I am uncertain about my computation for $H |c_n\rangle$: \begin{align} H\vert a_n \rangle &= E_0 \vert a_n \rangle + g_a \vert b_n \rangle \\ H\vert c_n \rangle &= E_0 \vert c_n \rangle + g_c \vert b_n \rangle \\ H\vert b_n \rangle &= g_a \vert a_n \rangle + g_c \vert c_n \rangle - \gamma (\vert b_{n-1} \rangle + \vert b_{n+1} \rangle) \end{align} Is this computation correct under Periodic Boundary Conditions and $N$ very large?

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  • $\begingroup$ While symmetries play important role in crystallography and facilitate computations, I think here using them would complicate your life. It is like using symmetry for solving a free particle SE - one gets sine and cosine solutions, but one would have easier life using plane waves. $\endgroup$
    – Roger V.
    Apr 25, 2023 at 9:03
  • $\begingroup$ I understand, I will try that way. However here my doubt is more about the way I make the computation using the Dirac notiation, I don't know if I find it confusing because of symmetry or because I am missing something $\endgroup$
    – mlp
    Apr 25, 2023 at 9:08
  • $\begingroup$ I think the point of the exercise is to move from $3N$ basis functions to $3-by-3$ eigenvalue equation, using the Bloch theorem. You can thing of Dirac notation as basis vectors: $|n\rangle \rightarrow \mathbf{e_n}$, $\langle n| \rightarrow \mathbf{e}_n^T$, $\langle n|m\rangle=\delta_{n,m}\rightarrow\mathbf{e}_n^T\cdot\mathbf{e}_m=\delta_{n,m}$. $\endgroup$
    – Roger V.
    Apr 25, 2023 at 9:20
  • $\begingroup$ Im quite sure your reflection symmetries part confused $c_n$ and $b_n$. That was quite confusing to read. $\endgroup$ Apr 25, 2023 at 9:34
  • $\begingroup$ @naturallyInconsistent I actually changed the wordings from the exercise so that's why $\endgroup$
    – mlp
    Apr 25, 2023 at 14:51

1 Answer 1

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A similar but simpler model in second quantization formalism is considered, for example, in Christophe Mora's lecture, section 3.2, example 2. Also, see Altland & Simons Condensed Matter Field Theory, second edition, chapter 2, Tight-binding systems.

The tool to diagonalize such Hamiltonians is the Fourier transform: it always works when the translational symmetry is present. For your problem, new plane wave basis vectors $|a_p\rangle, |b_p\rangle, |c_p\rangle$ should be defined as follows (as Eq. 2.22 in Altland & Simons): \begin{equation} \begin{gathered} |a_n\rangle = \frac{1}{\sqrt{N}}\sum_p e^{ipn} |a_p\rangle,\\ |b_n\rangle = \frac{1}{\sqrt{N}}\sum_p e^{ipn} |b_p\rangle,\\ |c_n\rangle = \frac{1}{\sqrt{N}}\sum_p e^{ipn} |c_p\rangle, \end{gathered} \end{equation} where $p \in \left\{0, \frac{2\pi}{N}, \frac{4\pi}{N}, \dots, \frac{2\pi(N-1)}{N}\right\}$. Assuming periodic boundary conditions, the Hamiltonian splits in new basis into $N$ uncoupled $3\times 3$ blocks for each $p$. Then, each block can be diagonalized analytically.

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