What is the minimum sampling rate for QPSK or what is the bandwidth of QPSK signal? Consider  we have created a QPSK signal with a carrier frequency 100kHz. Before applying matched filter or demodulation we need to take samples from the received signal. So what is the minimum sampling rate we must have? Consider that somehow we have perfect synchronization and there is no drift of the carrier.

Any help would be appreciated.

I have asked this question (for more attention) at stackexchange .
As best as I can tell from the terminology, QPSK a.k.a. 4-PSK or 4-QAM can convey twice the number of bits of information as BPSK in the same bandwidth.

Given a one-sided bandwidth of $B$ and assuming sufficient S/N ratio so that no bits are corrupted, BPSK can transfer bits at a rate of $2B$.  That means, given the same one-sided bandwidth, QPSK can transfer bits at a rate of $4B$.

But the sample rate must be at least $2B$.  If the receiver is perfectly synchronized to the transmitter, there would be one sample per bit with no bit error, if the transmitter sends a bipolar $\operatorname{sinc}(\cdot)$ for each bit.

But for QPSK, each sample is a complex value with a real part and imaginary part, each part containing one bit sample if this is 4-PSK.

So if you use quadrature demodulation to bump the IF or RF down to baseband $i(t)$ and $q(t)$, and sample those signals, you will need one complex-valued sample per pair of bits.