the relationship between bandwidth and the datarate
Ideal symbol pulses are sequences of rectangular pulses.  Rectangular pulses, unfortunately, have horrible frequency profiles, as shown below.


The main lobe is where most of the pulse power is, and the other lobes are called side lobes.  Though they do diminish in power the farther out they go, they don't diminish very quickly.  The main purpose of pulse shaping is to dramatically reduce the power of those side lobes.

The width of the primary lobe is $\frac{2}{T}$, where $T$ is the rectangular pulse width, which is equivalent to the symbol period.  Thus, once pulse shaping is employed, the signal bandwidth is $\frac{2}{T}$.

Since BPSK transmits one bit per symbol, and the symbol rate is $\frac{1}{T}$, BPSK can transmit $\frac{1}{T}$ bits per second.  Therefore, the bit rate divided by the bandwidth is $\frac{\frac{1}{T}}{\frac{2}{T}}$, which reduces to $\frac{1}{2}$.  Thus, the bit rate is $\frac{1}{2}$ the bandwidth.

When the signal is at baseband half of the mainlobe is in the positive frequencies and half is in the negative frequencies.  Given that it is a real signal, the two halves mirror each other so some people consider it the bandwidth to be halved, or $\frac{1}{T}$.  Thus, at baseband the bit rate would be equal to the bandwidth.

QPSK transmits two bits per symbol, so the bit rate for QPSK is $\frac{2}{T}$.  It follows that QPSK can transmit $2$ bits per Hz of bandwidth at baseband, and $1$ bit per Hz at passband.