What is the theoretical bit error rate for QPSK Matlab simulation when constellation points are transmitted (no symbol duration)?
TL;DR The theoretical BER of an ideal QPSK system is $\displaystyle Q\left(\sqrt{\frac{2E_b}{N_0}}\right)$ where $E_b$ is the energy per bit and $\frac{N_0}{2}$ is the two-sided power spectral density of the additive white Gaussian noise channel.



The receiver in a standard QPSK system is essentially two BPSK receivers using phase-orthogonal carriers. With a standard matched filter, the signal output in each BPSK receiver has value $E_b$ while the noise variance is $\sigma^2 = E_b N_0/2$.  Putting in the complex-value stuff, the signal output in the QPSK system is a complex number taking on values $(\pm E_b \pm jE_b)$ perturbed by complex Gaussian noise $N_i+jN_Q$ whose real and imaginary parts are independent zero-mean Gaussian random variables with variance $E_b N_0/2$. The decisions that each BPSK receiver makes are independent (because the noise in the receivers are independent), and since each receiver needs to decide between $E_b+N$ and $-E_b+N$, each receiver makes errors with probability 
$\displaystyle Q\left(\frac{E_b}{\sqrt{E_bN_0/2}}\right) = Q\left(\sqrt{\frac{2E_b}{N_0}}\right)$. Notice the complete lack of information about the symbol duration $T$ or the channel bandwidth $B$.

"But, but, but..." you splutter, "All this is nonsense.  My receiver
has signal output $(\pm A \pm jA)$ and the noise variance is measured to be $(12+j12)$, not this $E_b$ stuff that you keep on yammering about, and what I want to know is how do I get $E_b$ and $N_0$ when I simulate the system!"

Not to worry. The BPSK receiver is a linear receiver (except for the sampler and decision device), and so the output that you are measuring is really some constant $K$ times $E_b$, and the noise variance $12$ is just
$K^2E_bN_0/2$. So, when you run your simulations, generating gazillions of complex numbers $(\pm A \pm jA)$, perturbing them with complex Gaussian noise of variance $(12+j12)$, your simulated BER should be quite close to $$Q\left(\frac{A}{\sqrt{12}}\right) = Q\left(\frac{KE_b}{\sqrt{K^2E_bN_0/2}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right).$$
In short, you don't need to know the value of $K$ or to calculate the value of $E_b$ or $N_0$ explicitly; just use your measured values and go from there; it all comes out in the wash.