# Rotational barrier of biphenyl in ground and excited electronic states

I'm studying how the energy of biphenyl varies with its dihedral angle. I know how to describe the dependence quantitatively, but still, I don't know how to explain why the rotational barrier of biphenyl in its excited state is distinctly higher than in its ground state. This is illustrated in the plot below, where you can see that the excited states have a much greater increase in $V(\tau)$ at $\tau = 90^\circ$.

Additionally, I've obtained the HOMO and LUMO of biphenyl at $\tau = 90^\circ$. Is it possible to explain this jump in energy using them?