Manipulations with Hubbard operators

In summary: Your Name]In summary, the forum user is asking about the Fourier transform of a hopping term in a lattice Hamiltonian with bosons, and why the product of two Hubbard operators does not simplify in a certain equation mentioned in a reference. The response explains that this is due to the use of a cluster perturbation theory approach, where the hopping term is treated as a perturbation to the Hamiltonian, rather than the dominant term. In such cases, the simplification of the product of two Hubbard operators in momentum space is not possible.
  • #1
Arman88
1
0
Hi, I'm starting to study how to use Hubbard operators and I cannot understand one property:
Consider the hopping terms for a lattice Hamiltonian with bosons:
$$\sum_{i,j\neq i} t_{i,j} b^\dagger_i b_j$$
when writing this term in the basis of Hubbard operators $$X^{a,b}_i =| a,i \rangle \langle b,i | $$ becomes:
$$\sum_{i,j\neq i} t_{i,j} \sum_{p} \gamma_p \gamma_q X^{p+1,p}_i X^{q,q+1}_j$$ being $$\gamma_q = \sqrt{q+1}$$
My question refers to the Fourier transform of this term. I would expect that due to the translation symmetry of the hopping, this term can be diagonalized in momentum space and therefore it should be possible to contract the two Hubbard operators to:
$$\sum_k t_k X^{p+1,p}_k X^{q,q+1}_k = \sum_k t_k X^{p+1,p+1}_k$$
but I see that for example in ref. "Cluster Perturbation Theory in Hubbard Model Exactly Taking into Account the ShortRange Magnetic Order in 2 × 2 Cluster" with DOI: 10.1134/S1063776110100146, Eq. 13 does not simplify the product of these two Hubbard operators. Could someone explain to me why?

Thank you in advance
 
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  • #2
for your help!
Thank you for your question. The reason why the product of the two Hubbard operators does not simplify in the equation mentioned in the reference is because the authors are using a cluster perturbation theory approach, where the hopping term is treated as a perturbation to the Hamiltonian. In this case, the translation symmetry of the hopping term is broken and the term cannot be diagonalized in momentum space. Instead, the authors are using a cluster basis to represent the Hamiltonian, where the hopping term is treated as a perturbation within each cluster.

In general, the simplification of the product of two Hubbard operators in momentum space is only possible when the hopping term is the dominant term in the Hamiltonian and the system is translationally invariant. In other cases, such as in the reference mentioned, the hopping term is treated as a perturbation and the simplification cannot be made.

I hope this helps clarify your question. Please let me know if you have any further inquiries.
 

1. What are Hubbard operators?

Hubbard operators are mathematical operators used in quantum mechanics to describe the behavior of interacting particles. They are named after physicist John Hubbard, who first introduced them in 1963.

2. What is the significance of Hubbard operators in quantum mechanics?

Hubbard operators are used to describe the behavior of particles in a system where interactions between the particles are important. They are essential in understanding many physical phenomena, such as superconductivity and magnetism.

3. How are Hubbard operators manipulated in quantum mechanics?

Hubbard operators can be manipulated using mathematical techniques such as commutation and anticommutation relations. These manipulations allow for the calculation of important physical quantities, such as energy and correlation functions.

4. Can Hubbard operators be used in other fields of science?

While Hubbard operators were originally developed for use in quantum mechanics, they have also been applied in other fields such as condensed matter physics, materials science, and even in the study of social networks.

5. Are there any limitations to using Hubbard operators in quantum mechanics?

One limitation of using Hubbard operators is that they are most effective in systems with a finite number of particles. They also do not take into account relativistic effects, so they are not suitable for describing high-energy particles.

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