standard model - Why more than one Higgs?

Since by introducing one Higgs Boson we can give a mass to the leptons and gauge bosons of the weak interaction:

Why should we consider more than one Higgs (doublet) once we go beyond the standard model?

Answer

After some research on the net, I found the following answer (with the framework of the supersymmetric model of two Higgs doublets.) Please feel free to comment if I'm wrong and give some additional information.

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In the Minimal Supersymmetry Standard Model one needs two Higgs doublets to break the symmetry to $SU(2)_L \times U(1)_\text{em}$. There are two reasons:

1. In the SM the Higgs doublet has hypercharge $Y=1/2$ and one needs to introduce an adjoint doublet $\tilde{H}=i\sigma_2H^*$, which has hypercharge $Y=-1/2$, to give the upper component of the left-handed quark doublets a mass.

However, in a supersymmetric model, the conjugate of a (super)field is not allowed in the Lagrangian. Therefore, one cannot use the $\tilde{H}$-'trick' and we are forced to introduce a second doublet to give the upper-components of the fermion-doublet a mass.

2. The other reason is related to chiral anomalies related to triangular fermionic loops. For the anomalies to vanish (we don't want the theory to be renormalisable) there turns out to be a simple criterion: $$Tr(Y_f)= Tr(Q_f)=0. $$ (Since $Q=I^3_W+Y$ and the generators of a unitary group are always traceless: $Tr(I_w^3)$.) This is indeed the case in the SM: $$\sum_f Q_f= (-1)+3 \times (2/3)+3\times (-1/3)=0. $$ But in supersymmetry we get an additional charged spin-$1/2$ particle, the higssino, associated to the Higgs and the sum will no longer add up to zero. However, by introducing a second doublet with opposite hypercharge we again get a nice cancelation.