A Hard-Core Boson Model in K space

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The discussion centers on the hardcore boson model expressed through a Hamiltonian that incorporates nearest and next-nearest neighbor interactions. The model enforces that two bosons cannot occupy the same site, leading to specific constraints in momentum space. Key questions arise regarding the correctness of the derivation, the validity of the bosonic commutation relations in momentum space, and the appropriate range for momentum values. Additionally, the impact of particle filling on the model and the block diagonalization of momentum sectors are explored, particularly in relation to parity blocks. The inquiry seeks clarity on these complex concepts to enhance understanding and computational implementation.
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I am asking about how to convert the Hard-Core boson Model into K-space.
Hello,

I am interested in the following model:

$$
H = \sum_{<i,j>} -t (c_i c_j^{\dagger} + \text{H.C.}) + U (n_i n_j) + \sum_{<<i,j>>} -t' (c_i c_j^{\dagger} + \text{H.C.}) + U' (n_i n_j)
$$

where \( <i,j> \) indicates nearest neighbors, and \( <<i,j>> \) indicates next-nearest neighbors interactions. \( c_i \) and its conjugate are the annihilation and creation operators, respectively.

The hardcore boson model also establishes that
$$
c_i^{\dagger} c_i^{\dagger} = c_i c_i = 0
$$
when acting on a state, so a particle may only inhabit one site at a time.

If we establish that there are \( L \) sites on a periodic lattice (so \( i+L = i \)), then this Hamiltonian can be rewritten:

$$
H = \sum_{i} -t (c_i c_{i+1}^{\dagger} + \text{H.C.}) + U (n_i n_{i+1}) - t' (c_i c_{i+2}^{\dagger} + \text{H.C.}) + U' (n_i n_{i+2})
$$

This Hamiltonian can be block diagonalized into different sectors of momenta, \( K \). For this reason, I'd like to express this Hamiltonian in terms of momentum states.

Using
$$
\omega = \frac{2 \pi}{L}
$$
and
$$
c_j = \sum_{k} c_k e^{-i \omega k j}
$$
we can derive the following two expressions:

$$
\sum_{j} n_j n_{j+x} = \sum_{j} \sum_{k, k', q, q'} c^{\dagger}_{k'} c^{\dagger}_{q'} c_k c_q e^{i (k'+q'-k-q) j \omega} e^{i (q'-q) x \omega}
$$

which simplifies to
$$
\sum_{j} n_j n_{j+x} = \sum_{k, k', q, q'} \delta(k'+q'-k-q) e^{i (q'-q) x \omega} c^{\dagger}_{k'} c^{\dagger}_{q'} c_k c_q
$$

The diagonal elements are more simple:

$$
\sum_j c^{\dagger}_{j+x} c_j + \text{H.C.} = \sum_k 2 \cos(k \omega x) n_k
$$

Now these terms can be substituted into the Hamiltonian for \( x = 1 \), \( x = 2 \).

My questions are as follows:

1. Was this derivation correct?

2. In momentum space, does
$$
c_k^{\dagger} c_k^{\dagger} = c_k c_k = 0
$$
hold?

3. Should values of $$ K $$ vary from $$( (-L/2, -L/2+1, \ldots, L/2-1, L/2) $$ or from $$ (0, 1, \ldots, L-1) $$? Is there a difference?

4. I'm interested in this model when there is an appreciable amount of filling. Say $$N = L/3 $$. I have been attempting to program a script that computes the matrix elements of this Hamiltonian for an arbitrary $$L $$ and $$ N $$. I first find all available basis states in momentum space, then I separate them into total momentum sectors. Say three particles with momenta $$ (1, 2, 5) $$ are in the same sector as $$ (1, 3, 4) $$ as they sum to the same number $$ K = 8$$. The action of the $$ n_j n_{j+x}$$ term in momentum space is to couple these states together (as well as any other states that are reachable via a momentum-conserving interaction). But let's say $$ L = 6 $$, then these two states are of $$K \% L $$, correct? It would be correct to say that these are of momentum $$ K = 2 $$, because it must lie in the first Brillouin zone?

5. Are these $$ K $$ sectors block diagonalizable themselves? Some papers I've seen make reference to parity blocks, but none define what this would look like. I can see, for instance, very clearly that $$ (1, 3, 4)$$ and $$(1, 2, 5)$$ are in the same block of total momentum $ K $, but how can I see if they are in the same parity block?

Any help or guidance would be greatly appreciated. I have tried a lot of academic papers and Google-searching but I have not been able to feel assured in my understanding yet.
 
Last edited:
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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