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The off-diagonal term of the 1-D Bose-Hubbard model is : $K=\sum_l a_l^\dagger a_{l+1}+a_{l+1}^\dagger a_l$.

We can diagonalize it by introducing the Fourier transformation:$a_l=\frac {1}{\sqrt{N}}\sum_k e^{ikl}a_k$.

Then the Hamiltonian can be written as: $K=2\sum_k a_k^\dagger a_k \cos k$, which means the eigenenergies are $E_k=2\cos k$.

However, I have a lot of questions about this transformation:

  1. Suppose the largest occupation number in each site is n-1, then the original occupation representaion will have $n^N$ states, written as : $\{|n_1,n_2,\dots,n_i,\dots,n_N \rangle \}$,$n_i=0,1,\dots,n-1$. So we can know that the dimension of matrix of this Hamiltonian is $n^N\times n^N$ and it will have $n^N$ eigen energies after diagonalization. Why does the dimension of the Hamiltonain becomes N after Fourier transformation since the summation of k takes form 1 to N.

  2. What's the tranformation between the original states and the states after Fourier transformation? We know that the original representation is the occupation number representation in real space and the latter representation is the occupation number representation in momentum space. What's the largest occupation number in momentum space and what's the relation between the states in these two representations?

I cannot find any explanations in textbooks. There are only the Fourier transformation of the operators when introducing the diagonalization of Hamitonians in textbooks without any explanations about the transform of the states. Maybe you can give me some references if you think my question is too basic and can be found in some quantum mechanics books.

Thanks for your answers and sorry for the grammatical mistakes I may make.

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  • $\begingroup$ well dimension of single particle hamiltonian is $N$ but many particle hamiltonian is $H\otimes H...$ so in many particle case you need to represent you creation and anhiliation operators as matrix form than the matrix rep of you hamiltonian will have correct dim. $\endgroup$
    – physshyp
    Oct 22, 2020 at 13:03
  • $\begingroup$ also in momentum space the largest filled state will be the state with lowest energy, actually only that state will be filled rest will be empty. $\endgroup$
    – physshyp
    Oct 22, 2020 at 13:07

3 Answers 3

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$E_k$ are the single particle energies. The full spectrum of the many body Hamiltonian consists of all possible combinations of single particle energies $$ E = \sum_k n_k E_k $$ where $n_k$ is the number of particles in the single particle energy eigenstate $k$. The possible combinations of $n_k$s gives you back your $n^N$ states.

For the second part of your question remember that in Fock space we usually choose to represent states in terms of creation operators, so we can write a single particle momentum state as \begin{align} |k\rangle &= a_k^\dagger |0\rangle\\ &= \frac{1}{\sqrt{N}}\sum_l e^{ikl}a_l^\dagger |0\rangle\\ &= \frac{1}{\sqrt{N}}\sum_l e^{ikl}|l\rangle\;. \end{align} The generalization to many-particle states is obvious but cumbersome. We can use a similar approach to find the maximum occupation of a momentum state. The momentum state number operator is \begin{align} a_k^\dagger a_k = \frac{1}{N}\sum_{l,l'}e^{ik(l-l')}a_l^\dagger a_{l'} \end{align} The total number of particles is then \begin{align} \sum_k a_k^\dagger a_k &= \frac{1}{N}\sum_{k,l,l'}e^{ik(l-l')}a_l^\dagger a_{l'}\\ &=\sum_{l,l'}\delta_{l,l'}a_l^\dagger a_{l'}\\ &=\sum_{l}a_l^\dagger a_{l} \end{align} that is, it is unchanged. Since there are the same number of momentum states as position states and there is no reason for the maximum occupation to be different for different momenta, the maximum possible occupation of a state in momentum states is the same as the maximum for position states.

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Let's do this for fermions, so $n=2$. If you have $N$ sites, the original fermionic Fock space (Hilbert space) has dimnsion $2^N$. After using Fourier for diagonalizing you have $a_{k}$'s etc, and there are $N$ distinct allowed values of $k=2\pi m/N$, $m=0,\ldots,N-1$. Each $k$ state can be occupied or empty, so there are still $2^N$ states.

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  1. Regarding the reduced dimensionality: when performing the Fourier transform you do what is a so-called block-diagonalization, i.e. you take advantage of some discrete symmetry of your many-body Hamiltonian (in this case translational invariance) to take it, from a $n^{N}\times n^{N}$ matrix, into blocks associated to the eigenvalues related to the discrete symmetry, which are in this case the momenta $k$ over which you perform the sum. As you have $N$ allowed values for the momentum, there are $N$ such blocks. You can think of your original $n^{N}\times n^{N}$ matrix, if written in the appropriate basis, as consisting of these $N$ $n\times n$ blocks along the diagonal.
  2. I guess the number of allowed states must remain unchanged, so having $N$ momenta implies the maximum occupation number for each of these is $n-1$ as well.
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