Let $|\psi\rangle \to |\psi'\rangle = \hat{T}(\delta x)|\psi\rangle$ for infinitesimal $\delta x.$ Show that $\langle x \rangle' = \langle x \rangle + \delta x$ and $\langle p_x \rangle' = \langle p_x\rangle.$
I am confused. Why would $\langle x \rangle = \langle x \rangle + \delta x?$ Shouldn't it equal $\langle x \rangle?$ Since, $\langle x\rangle' = \langle \psi'|\hat{x}|\psi'\rangle = \langle \psi'|x\hat{T}(\delta x)|\psi\rangle$ and using $\hat{T}(\delta x) = e^{-i\hat{p}_x\delta x/\hbar}$ then $\langle \psi |\hat{T}^{\dagger}(\delta x)\hat{T}(\delta x)x|\psi\rangle = \langle x\rangle.$