Are lattice translations generated by crystal momentum or real momentum?

Translation operator have the form $$$$T_R = e^{i\frac{\hat{p}}{\hbar}\cdot R}$$$$ and this can be easily proved, with $$\hat{p}$$ being the total momentum operator; but in mamy textbooks is stated that the generator for lattice translation is the Crystal momentum operator and that is not related in any way to the total momentum $$\hat{p}$$ (Ashcroft - Appendix M). How can the lattice translation operator have two form different form?

• The translation symmetry of a crystal is a discrete symmetry. Momentum is the generator of continuous translation symmetry. Where did you get the above form for $T_R$? – SRS Jul 13 '20 at 15:44
• If you apply $T_R=e^{i\frac{\textbf{p}}{\hbar}R}$ to a generic wave function you would obtain $\psi(r+R)$ but if you define $T_R=e^{i\frac{\textbf{K}}{\hbar}R}$, with $\textbf{K} \psi = \hbar k \psi$, and apply it to a generic wave function you also obtain $\psi(r+R)$ due to the Bloch Theorem; so i am a little confuse, which one Is the translation operator? My source physics.stackexchange.com/questions/536039/… and physics.stackexchange.com/questions/538516/… – user2519740 Jul 13 '20 at 17:49

To incorporate periodicity of lattice, we use Bloch wavefunction which has extra factor $$u(x)$$ multiplied to the translation operator and is periodic. But this wavefunction is not momentum eigenstate and therefore its eigenvalue is not the momentum, however it tells how the state transforms under translations.
• My question was about how it is possibile that $T_R=e^{i\frac{p}{\hbar}R}$ and at the same time $T_R=e^{i\frac{K}{\hbar}R}$ where K Is the Crystal momentum operator – user2519740 Jul 13 '20 at 7:15