I encounter a problem in understanding the eigenvectors of the Hamiltonian and the true ground state of the 1D lattice. Suppose my lattice has 5 electrons and we only consider the on-site potential of the them. Therefore, the system is described by the following Hamiltonian: \begin{equation} \hat{H} = - \mu \sum_{i=1}^{N = 5} c^{\dagger}_{i} c_{i} ~~,~~ \mu > 0 \end{equation} From our intuition, the configuration having lowest energy is that all electrons sit on each site. We can write the ground state as: \begin{equation} |\Psi_{GS } \rangle = c^{\dagger}_{1}c^{\dagger}_{2}c^{\dagger}_{3}c^{\dagger}_{4}c^{\dagger}_{5} | 0\rangle = |11111 \rangle = |1\rangle_{1} \otimes|1\rangle_{2} \otimes |1\rangle_{3} \otimes |1\rangle_{4} \otimes |1\rangle_{5} ~~,~~ E_{GS} = -5 \mu \end{equation}

Next, we try to see what information contained in the Hamiltonian by solving its energy eigenvalues and eigenstates: \begin{equation} \hat{H} = \begin{pmatrix} -\mu & 0 & 0 & 0 & 0 \\ 0 & -\mu & 0 & 0 & 0 \\ 0 & 0 & -\mu & 0 & 0 \\ 0 & 0 & 0 & - \mu & 0 \\ 0 & 0 & 0 & 0 & -\mu \end{pmatrix} \end{equation} We therefore get the energy spectrum of this Hamiltonian $E_{i} = -\mu$, which is 5-fold degenerate. However, the energy eigenvalues of the Hamiltonian is not the true ground state as we know that the true ground state should have the energy $E = -5\mu$. Therefore, I want to ask what is the physical meaning of the eigenstates and eigenvalues of $\hat{H}$? Given a Hamiltonian $\hat{H}$, how can we find the true ground state?


1 Answer 1


The matrix representation you wrote is only correct when $H$ is regarded as a single particle Hamiltonian. If you constrain your system to have exactly one electron, then the spectrum you derived makes complete sense: there are 5 different degenerate states in which you can place your electron, each with energy $-\mu$. On the other hand, the ground state you wrote using creation operators is a 5 particle state. If you consider states of any variable particle number, then clearly the dimension of your Hilbert is not 5 but $2^5$ (you can place either a particle or a hole at each of your 5 sites). You can still easily write down a matrix representation of your Hamiltonian in this larger multidimensional Hilbert space (known as a Fock space) since it will be diagonal, but it will be considerably larger than 5x5.

  • $\begingroup$ Thank you for your answer, Zack. I know understand the fault in my question. Besides, I want to ask that if I want to construct a Hamiltonian for two sites and two electrons, can I write the hamiltonian as $H = H_{onsite,1} \otimes I_{2} + I_{1} \otimes H_{onsite,2}$, where the matrix form of $H_{onsite,i} =[ [ 0,0] , [0, -mu ] ]$ ? $\endgroup$
    – Ricky Pang
    Commented Apr 24, 2021 at 14:09
  • $\begingroup$ Sure, that totally works. One way to think about it is with a Jordan-Wigner type mindset, where every fermion mode can be just as well thought of as a single qubit (ie spin-1/2) degree of freedom. A site being unoccupied corresponds to the state $|0 \rangle = | \uparrow \rangle$, while the state being occupied corresponds to $|1 \rangle = | \downarrow \rangle$. Then it's clear that the matrix form you wrote is a reasonable approach. $\endgroup$
    – Zack
    Commented Apr 24, 2021 at 15:17

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