Why is the bandwidth for this QPSK system larger than expected?
In gnuradio, the throttle block only determines the speed at which your computer is processing samples - it does not determine what the actual sample or baud rates are. As an example - consider that you generate random PSK symbols at 1 sample per symbol, and then upsample by 2 with a root raised cosine w/ 35% exccess BW so we're now at 2 samples per symbol (this is what the constellation modulator block does in your flowgraph). What is the absolute baudrate that you're running at? Answer: Whatever you want it to be. If I hook that I/Q stream up to an RF front end, I could theoretically transmit at any baud rate using those exact same samples. E.g. if the DAC is running at 1MSPS (assuming this is an I/Q DAC for simplicity like you might consider the ad9361) then my baud rate would be 1MSPS/(2 samples/symbol) = 500k symbols/sec. Assuming my PC could keep up (and I increased my throttle block so that my PC isn't slowing me down from producing samples), I could instead run the DAC at 10MSPS to transmit at 5M symbols/sec.
So - your question will probably be - how do I simulate a certain baud rate without a DAC/front-end? Well - it's whatever you choose it to be. In your flowgraph, you set the FFT sink to think that the sample rate is 500 kHz. The data you have hooked up to that sink is coming in at 2 samples/symbol, but factoring in the excess bw, the relative bandwidth will be about (1.35/2.0)*500e3 = 337.5 kHz which is exactly what the spectrum is showing you. If you did further upsampling and then looked at the FFT, the signal would appear narrower and vice versa.