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Journal of Applied Sciences Research, 5(6): 645-647, 2009
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Exact Ground State Energy of the Hubbard Model in One Dimension
1
1
E.O. Aiyohuyin 2A. Ekpekpo
Department of Physics University of Benin, Benin City, Nigeria.
2
Department of Physics Delta State University Abraka, Nigeria
Abstract: Different solutions of the one dimensional Hubbard model are computed. The energy of the
ground state is evaluated under different restrictions. The different restrictions considered are;
i) Arbitrary filling at u = 4 and in the thermodynamic limit [3 ].
ii) Half filling and small ratios
Half filling for
and
in the thermodynamic limit
[4 ]
A variational calculation [6 ]. These results are compared to the exact diagnoalization of the Hubbard
Hamiltonian of 2 electrons on 2 sites [5 ]. The results all agree except that of [3 ]. This disagreement is
believed to be due to the weak interactions assumed in [3 ] leading to an expression of the ground state
energy which is reminiscent of that for independent electrons.
Key woords:
INTRODUCTION
that
Applying the B ethe Ansatz equations it was found
[1 ]
the energies could be written as
The Hubbard model is simple. It models the
electron-electron interaction in a lattice [2 ]. The Hubbard
hamiltonian is written as;
To find the wave vectors k j requires solving the set
of coupled nonlinear Bethe ansatz equations. [1 ]
where
creates an electron with spin ó at site i,
t is the electron hopping energy.
annihilates an
electron with spin ó at the site j. U is the coulomb
repulsion energy. The sum are over the L lattice sites,
with the sum in the kinetic energy term being restricted
to nearest neighbours. The one-dimensional chain has
cyclic boundaries thus each site has two nearest
neighbours.
It is assumed that the number of particles N is
such that
.
and
where
I j is an integer if N 9 is even and half-odd integer if N 9
is odd J a is an integer if
is odd or it is
an half-odd integer if
is even.
The ë’s are a set of ordered, unequal real numbers
The number of particles of each spin N ó is conserved
and
j runs from 1 to N In the u = 4 limit, the wave
vectors k j take the form [3 ]
It is assumed that
Corresponding Author: E.O. Aiyohuyin, Department of Physics University of Benin, Benin City, Nigeria.
645
J. Appl. Sci. Res., 5(6): 645-647, 2009
from [4]
where N 9 is even (odd)
where
when
In the infinite one dimensional system and for half
filling [4 ], the ground state energy for small ratios
the ground state energy
= - 0.3465, for u = 16, t = 1,N = 2
The exact computation
the result
is
and in the opposite limit of
is:
[3 ]
or diagonalization gives
And using u = 16, t = 1
E = - 0.2462
From the variational approach [6 ] the following
result is obtained E = - 0.2462 where we have taken u
= 16, t = 1, N = 2
Conclusions: The results of the ground state energy of
the two particles in the one dimensional Hubbard
model shows that the use of the equation
where N is the number of particles
For 2 electrons on 2 sites
2, the following three cases:
[5 ]
gives for N = 2,L =
is too severe an
approximation. It is difficult to agree with the use of
i)
ii)
as the energy of the I –
iii)
Using the equation
D Hubbard model as well as [3 ] the test carried out for
2 particles failed. The results of [4 ], [6 ] and [5 ] are in
agreement and so are much better. The conclusions
reached are also corroborated in the work [7 ] of Nerd
wisdom. The apparent weakness of the result based on
[3 ]
is the assumption that the energy of the electrons in
the one dimensional Hubbard model is like that of
independent electrons. This approximation is too
severe.
[3 ]
where
N even,
REFERENCES
for cases (i) and (ii)
Similarly, for case (iii)
1.
2.
Also when
[3 ]
3.
4.
5.
646
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238.
Hodge, W .B., N.A. Holzwarth and W .C. Kerr
Exact ground state of the Hubbard model at u =4
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J. Appl. Sci. Res., 5(6): 645-647, 2009
6.
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Hubbard model MPhil Thesis University of Benin,
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