Symbol rate and optical bandwidth relation
In the paper you have supplied, the optical carrier is modulated with an electrical signal generated by an AWG. The bandwidth of the resulting optical signal will be identical to that of the electrical signal of the AWG and do indeed depend on the waveform. 

The paper further states that the AWG employs a sample rate (not symbol rate) of 25 GSa/s. The generated electrical signal is an OFDM signal, which in this case uses an FFT length of 4096, yielding as many subcarriers. However, only 2184 subcarriers are used.

The spacing \$\Delta f\$ between two adjacent subcarriers is 
$$
\Delta f = \frac{f_s}{N}
$$
with \$f_s\$ the sample rate and \$N\$ the FFT length used.
(see this dsp.stackexchange answer)

In this case, the subcarrier spacing is thus approximately 6.1 MHz. When 2184 adjacent subcarriers are used, the spacing between the first and last subcarrier is \$2183 \cdot 6.1\text{ MHz} \approx 13.3\text{ GHz}\$. With the bandwidth of a subcarrier negligible compared to this, the bandwidth of the resulting optical signal is also approximately 13.3 GHz.