Momentum and Translation operator.

[Black Integra]

In the concept of translation operator, there's a line said that

where

and

How could they interpret that this parameter ρ is really a momentum operator?

 

[dextercioby]

In the position representation, the momentum along axis x is proportional to the derivative operator wrt x. This is a consequence of the Stone-von Neumann uniqueness theorem.

Since the infinitesimal generator of translations along x-axis is proportional to the first derivative, again wrt x, it follows simply that the momentum along x is the generator of translations along the same axis, i.e. x.

 

[vanhees71]

I don't understand your notation. Anyway, in position representation, the translation operator is given by

$$\hat{U}(\vec{\xi})=\exp(\mathrm{i} \hat{\vec{p}} \cdot \vec{\xi}),$$

where I've set $\hbar=1$ for simplification of notation. By definition it acts on a wave function in the domain of the momentum operator by

$$\hat{U}(\vec{\xi}) \psi(\vec{x}):=\psi(\vec{x}+\vec{\xi}).$$

Comparing this to the Taylor-expansion formula,

$$\psi(\vec{x}+\vec{\xi})=\exp(\vec{\xi} \cdot \vec{\nabla}) \psi(\vec{x}),$$

immediately gives the well-known expression

$$\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}$$

for the momentum operator in the position representation.

 

[Black Integra]

@dextercioby: That's exactly what I want as an answer. Thanks.
@vanhees71: Thanks for that reply :)