Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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? – nervxxx Jan 11 '14 at 19:43
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 ? – VanillaSpinIce Jan 11 '14 at 19:59
Are the particles fermions or bosons? What is the system size? How many particles are there? – Isidore Seville Jan 11 '14 at 21:23
These are fermions on say a $N$-site lattice. However the number of fermion depends on the choice of $\chi_{r,r+1}$. – VanillaSpinIce Jan 11 '14 at 22:10
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? – Isidore Seville Jan 11 '14 at 22:19
up vote 1 down vote accepted

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 $$ E_{G}=\sum_{m,\lambda_m<0}\lambda_m. $$

share|cite|improve this answer
Great ! I had not thought of this general kind of BG transformation. – VanillaSpinIce Jan 14 '14 at 13:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.