pdf, 9 MiB - Infoscience - EPFL
pdf, 9 MiB - Infoscience - EPFL
pdf, 9 MiB - Infoscience - EPFL
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192 APPENDIX B. A PAIR OF PARTICLES IN A CUO 4 CLUSTER<br />
a) b b)<br />
y<br />
=χ<br />
b y<br />
=χ<br />
b -x<br />
=-χ<br />
b -x<br />
=-χ<br />
a=0 a=0<br />
b x<br />
=-χ<br />
b x<br />
=χ<br />
b -y<br />
=χ<br />
b -y<br />
=-χ<br />
Figure B.1: The two patterns of current that can be obtained when a =0is<br />
imposed, and t dp = −1, t pp < 0. The arrows indicate the current orientations.<br />
Additional complex parameters λ ij and α must be used to obtain the conservation<br />
of the current inside one triangle plaquette. For example, when a = π the current<br />
arrows on the horizontal and vertical links are reversed, and the current is then<br />
circulating around the triangle plaquettes. However, when a = π the hopping<br />
energy on these bonds is also changing sign, and therefore the energetic cost for<br />
having circulating currents is of the order of t d−px .<br />
And:<br />
〈j px−py 〉 =2t px−py (R 2 sin(b y − b x )+R 2 sin(c y − c x )<br />
+ ∑ )<br />
(L xj L yj sin(l yj − l xj ))<br />
j<br />
(B.5)<br />
Moreover, the kinetic energy on these links is obtained by replacing the sin<br />
by cos in the expressions (B.4) and (B.5). Besides, we assume for simplicity<br />
that t d−px = −1, and we consider that all the oxygen-oxygen bonds have also<br />
negative transfer integrals (t px−py<br />
< 0). This choice of the sign is equivalent to<br />
the hybridization obtained in the physical three-band Hubbard model for the<br />
cuprates. In Fig. B.1 we show two simple examples of current patterns that are<br />
obtained for two choices of the parameters b. Noteworthy is the fact that the<br />
currents cannot be oriented as a rotational flow around a triangle plaquette when<br />
t d−px < 0andt px−py < 0. When t px−py<br />
> 0 the current is a true rotational flow and<br />
the conservation of the current is easily obtained inside each triangle plaquette.<br />
By imposing the conservation of the current in all the triangle plaquettes, we get<br />
when the phases b = ±χ are chosen according to the pattern (a) of Fig. B.1:<br />
sin(χ − a)<br />
sin(−2χ) =2t pp R<br />
(B.6)<br />
t dp R a