Consider a generalized 1D tight-binding model (without spin) with the following Hamiltonian \begin{equation} \mathcal{H}\left(\{\chi_{r,r+1}\}\right)=\sum_{r}(\chi_{r,r+1}c^\dagger_rc_{r+1}+h.c.)~, \end{equation} and suppose that the complex number $\chi_{r,r+1}$ is a variational parameter. The $r$ index runs over a lattice sites of a 1D chain (you can think of this as a periodic boundary condition problem with $N$ lattice sites).

Q: Given an arbitrary set of $\chi_{r,r+1}$, is there a systematic procedure to find the ground-state of the model (numerically or analytically) ? I'm not looking for a detailed answer with the calculation here. Perhaps just a good reference or starting point. For instance, in the case where $\chi_{r,r+1}=\chi$ $\forall r$, the system can be diagonalized by a simple lattice Fourier transform.

  • $\begingroup$ Is this a sensible question to ask? Take any $\chi_{r,r+1}$, solve it. Then if the ground state is negative, just rescale $\chi \to \lambda \chi$ and if the ground state is positive, rescale $\chi \to -\lambda \chi$. As $\lambda \to \infty$ the ground state energy $\to -\infty$. Which is erm, minimum. What are you looking for? $\endgroup$
    – nervxxx
    Jan 11 '14 at 19:43
  • $\begingroup$ Of course, when I say "minimize" this is in units of this $\chi$. Otherwise the question would not make sense as you have noted. To be more specific, take a random set of $\chi_{r,r+1}$... Now, how do you solve this, i.e. how do you find the spectrum ? $\endgroup$ Jan 11 '14 at 19:59
  • $\begingroup$ Are the particles fermions or bosons? What is the system size? How many particles are there? $\endgroup$ Jan 11 '14 at 21:23
  • $\begingroup$ These are fermions on say a $N$-site lattice. However the number of fermion depends on the choice of $\chi_{r,r+1}$. $\endgroup$ Jan 11 '14 at 22:10
  • $\begingroup$ I am not sure I understand. Why the number of fermions depends on the choice of $\chi_{r,r+1}$? Are you working in the grand-canonical ensemble? $\endgroup$ Jan 11 '14 at 22:19

OK, it is probably a bad idea to exchange in comments. Let me expand what I said in the comments.

If my understanding is correct, the OP wants to know, as the first step toward solving the whole problem, the ground state energy of the many-body Hamiltonian $\mathcal{H}$ defined by $$ \mathcal{H} = \sum_{r,s}H_{rs}c^\dagger_r c_{s}, $$ for a given set of parameters $\{ H_{rs}\}$. Here $c^\dagger_{r}$ and $c_{r}$ are standard fermion creation and annihilation operators. The subscripts $r,s$ run over all lattice sites from 1 to $N$. The Hermiticity requires that $$ H_{rs} = H^\ast_{sr}. $$ In other words, the $N\times N$ matrix $H$, whose $(r,s)$ entry is defined to be $H_{rs}$, must be Hermitian. In some literature, $H$ is known as the "first-quantized Hamiltonian". Note that the above $\mathcal{H}$ takes a slightly more general form than the one described by OP.

The first step is to diagonalize $\mathcal{H}$. To this end, we introduce a new set of fermion operators: $$ c_{r} = \sum_{m}V_{rm}f_{m};\quad{}c^\dagger_{r}=\sum_{m}V^\ast_{rm}f^\dagger_{m}. $$ We demand that the new fermion operators obey the standard fermion algebra. It can be seen that this is amount to demand $$ \sum_{m}V_{rm}V^\ast_{sm}=\delta_{rs}, $$ or equivalently $VV^\dagger=1_N$, i.e. $V$ is a unitary $N\times N$ matrix.

Substituting the above in, we find $\mathcal{H}$ written in terms of new fermion operators, $$ \mathcal{H} = \sum_{r,s,m,n}V^\ast_{rm}V_{sn}H_{rs}f^\dagger_m f_n = \sum_{m,n}(V^\dagger HV)_{mn}f^\dagger_m f_n. $$ Since $H$ is Hermitian, we can always find a unitary $V$ so that $H$ is diagonalized: $$ V^\dagger HV = \Lambda. $$ Here $\Lambda = \textrm{diag}(\lambda_1,\lambda_2\cdots,\lambda_N)$. $\lambda_i\in\mathbb{R}$ are eigenvalues of $H$. Thus, $$ \mathcal{H} = \sum_{m}\lambda_m f^\dagger_m f_m. $$ This is the desired diagonalized form of $\mathcal{H}$.

The second step is to find the ground state energy of $\mathcal{H}$. We see that all eigenstates of $\mathcal{H}$ are labeld by the occupation numbers $f^\dagger_mf_m$. It is easy to see that the ground state of $\mathcal{H}$ is constructed by filling up all modes with negative energy. In other words, in the ground state, $$ f^\dagger_m f_m=\left\{\begin{array}{cc} 1 & \lambda_m<0\\ 0 & \lambda_m>0 \end{array} \right. . $$ There will be degeneracy if some $\lambda_m = 0$. Then, the ground state energy is $$ E_{G}=\sum_{m,\lambda_m<0}\lambda_m. $$

  • $\begingroup$ Great ! I had not thought of this general kind of BG transformation. $\endgroup$ Jan 14 '14 at 13:58

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