Hard-Core Boson Model in K space

In summary, the "Hard-Core Boson Model in K space" explores the behavior of hard-core bosons, which are particles that obey Bose-Einstein statistics but cannot occupy the same quantum state. The study focuses on the K space representation, where momentum states are analyzed to understand the system's properties, including correlations and phase transitions. The model highlights the effects of interactions and confinement on bosonic systems, providing insights into phenomena like superfluidity and collective excitations in low-dimensional systems.
  • #1
partyday
3
0
TL;DR Summary
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:
Back
Top