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I am reading some papers about quantum chemistry and in one I found a notation for excited states of a molecule that I don't understand at all.
In the picture below you can see the potential curves of biphenyl as a function of its torsion angle. In the description of the picture, the author speaks of states $1^1A$ and $1^1B_1$. What does he mean by these states?
As has already been mentioned in comments, those letters and numbers for each state correspond to their symmetry, specifically the symmetry of the (purely electronic) wavefunction. For more information regarding symmetry notation, see here, and see here for more discussion about how symmetry can be applied to molecular orbitals.
It might not be clear from the second question, but because the symmetry of the overall N-electron wavefunction results from the direct product of all occupied 1-electron wavefunctions (at least in Hartree-Fock), exciting electrons into molecular orbitals with different symmetries will lead to excited N-electron wavefunctions (excited states) with different symmetries.
To consider the specific case of biphenyl,
the $\tau = 0^{\circ}$ structure, corresponding to the leftmost values on the x-axis, will look like the 2D structure just above where both rings lie in the same plane. As an exercise left to the reader, you can figure out that the point group of the planar form of biphenyl is $D_{\mathrm{2h}}$. The point group $D_{\mathrm{2h}}$ has the following character table:
$$\begin{array}{c|cccccccc|cc} \hline D_\mathrm{2h} & E & C_2(z) & C_2(y) & C_2(x) & i & \sigma(xy) & \sigma(xz) & \sigma(yz) & & \\ \hline \mathrm{A_g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2,y^2,z^2 \\ \mathrm{B_{1g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & R_z & xy \\ \mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & R_y & xz \\ \mathrm{B_{3g}} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & R_x & yz \\ \mathrm{A_u} & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \\ \mathrm{B_{1u}} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\ \mathrm{B_{2u}} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & y & \\ \mathrm{B_{3u}} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \\ \hline \end{array}$$
The three-dimensional symmetry of any object can always be broken down into one or more irreducible representations (irreps) belonging to the object's point group; each irrep corresponds to a row in the point group's character table. The symmetry table is the leftmost column in each irrep row. To enumerate them,
You can see that these labels are almost but not quite the same as those in the figure. A minor difference is italicized versus regular/upright text; this is a matter of typographic choice and there is no physical or chemical difference between them. There are two significant differences:
The labels in the figure are prefixed with a regular number and a superscript number. The regular number helps uniquely defines the state ordering: $1^1B_{1}$ is the first state with $^1B_{1}$ symmetry, and $7^1B_{1}$ is (hypothetically) the seventh state with $^1B_{1}$ symmetry. Without the superscript number, each symmetry label only specifies the spatial symmetry, and with the superscript defined as $2S + 1$, the spin multiplicity, each label can also specify the spin symmetry. $1^1B_{1}$ is the first singlet ($S = 0$) state with $B_{1}$ symmetry, and $1^3B_{1}$ is (hypothetically) first triplet ($S = 1$) state with $B_{1}$ symmetry. See this excellent answer for more information, in particular the figure.
The labels in the figure do not have u/g in the subscript, because for most of the biphenyl structures, where $\tau \neq 0^{\circ}$, the molecular point group is no longer $D_{\mathrm{2h}}$, but one of lower symmetry. When the rings are perfectly perpendicular to each other ($\tau = 90^{\circ}$ and $\tau = 270^{\circ}$), the molecular point group will be $D_{\mathrm{2d}}$, like hydrogen peroxide, and every other intermediate value will be even lower in symmetry: $D_{\mathrm{2}}$, similar to skewed ferrocene (rather than staggered or eclipsed). To see how the irrep and its symmetry operations from a point group of higher symmetry map to those in a point group of lower symmetry, one needs a descent table. Here, most notably, all skewed structures are missing the inversion operation that would necessitate the gerade/ungerade subscript. Because most of the structures are of lower $D_{2}$ symmetry, it is easiest to represent all structures with $D_{2}$ point group labels.
For completeness,
$$\begin{array}{c|ccccc|cc} \hline D_\mathrm{2d} & E & 2S_4 & C_2 & 2C_2' & 2\sigma_\mathrm{d} & & \\ \hline \mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\ \mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\ \mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\ \mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & z & xy & \\ \mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) \\ \hline \end{array}$$
$$\begin{array}{c|cccc|cc} \hline D_2 & E & C_2(z) & C_2(y) & C_2(x) & & \\ \hline \mathrm{A} & 1 & 1 & 1 & 1 & & x^2, y^2, z^2 \\ \mathrm{B_1} & 1 & 1 & -1 & -1 & z, R_z & xy \\ \mathrm{B_2} & 1 & -1 & 1 & -1 & y, R_y & xz \\ \mathrm{B_3} & 1 & -1 & -1 & 1 & x, R_x & yz \\ \hline \end{array}$$
See how there are fewer symmetry operations (middle set of columns) in $D_{2}$ than in $D_{\mathrm{2d}}$ than in $D_{\mathrm{2h}}$.
