Momentum operator in QM I am reading a passage from the book "Decoherence and the Quantum to Classical Transition" which describes a scattering process in which a light environmental particle with initial wavefunction $|\chi_i\rangle$ bounces off a heavy particle in position eigenfunction $|x\rangle$ and transition to a quantum state $|\chi_i(x)\rangle$, so that the overall wavefunction undergoes the evolution $|\chi_i\rangle|x\rangle\rightarrow|\chi_i(x)\rangle|x\rangle$. However, the book then threw the following line in:

The state $|x\rangle$ can be thought of as the state $|x=0\rangle$ (corresponding to the scattering center being located at the origin) translated by the action of the momentum operator $\hat p$:$$|x\rangle=e^{-i\hat p\cdot x/\hbar}|x=0\rangle.$$

However, I don't have any intuitive understanding for why the operator $e^{-i\hat p\cdot x/\hbar}$ should map a position eigenstate $|0\rangle$ to a position eigenstate $|x\rangle$. Could anyone point me to what this theorem is called and where I could find a derivation of it?
 A: This is representation of a translation operator:
$$
e^{ip_x a/\hbar}=e^{a\partial_x}
$$
Acting on a function (which has derivatives of infinite order) we have
$$
e^{ip_x a/\hbar}f(x)=e^{a\partial_x}f(x)=
\sum_{n=1}^{+\infty}\frac{a^n\partial_x^n}{n!}f(x)=
\sum_{n=1}^{+\infty}\frac{a^n}{n!}f^{(n)}(x)=f(x+a),
$$
since the last sum is just Taylor expansion of $f(x+a)$ near point $x$.
This is obviously not a coincidence, since momentum operator is the generator of infinitesimally small translations, and its conservation follows from translational invariance (see Noether theorem.)
A: Quantum mechanics replaces canonical variables with operators and then preserves the classical Lie algebra of Hamiltonian mechanics by replacing the Poisson brackets with a commutation relation. Typically,
$$[\hat{x},\hat{p}]=i\hbar.$$
This defines an algebra and a phase space associated to your Hilbert space in the sense that this relation fixes the structure of the spectrum of those operators
$$\hat{x}\vert x \rangle = x\vert x \rangle,\\
 \hat{p}\vert p \rangle = p\vert p \rangle,$$ where x and p are continuous variables. How do you prove that?
You build a displacement operator $$\hat{D}(x)=e^{-i\hat{p}x},$$ such that
$$\hat{D}^{\dagger}(x)\hat{x}\hat{D}(x)=e^{i\hat{p}x_0}\hat{x}e^{-i\hat{p}x_0}=\hat{x}+x_0,$$ which derives from the Baker-Campbell-Haussdorf identity and your algebra (commutation relation).
This expression holds for any real $x_0$ leading to the conclusion that the spectrum of $\hat{x}$ is the whole real line. And of course you can do the same with $\hat{p}$.
Then you can define the eigenstates as $$\vert x \rangle =\hat{D}(x)\vert 0 \rangle,$$
where $\vert 0 \rangle$ is your original reference state.
