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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

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