Momentum and Translation operator.

  • #1

Main Question or Discussion Point

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

pUIINX.gif


where
gif.latex?T(0)=1.gif
and
pQzIHQ.gif


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

  • #2
dextercioby
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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.
 
  • #3
vanhees71
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I don't understand your notation. Anyway, in position representation, the translation operator is given by

[tex]\hat{U}(\vec{\xi})=\exp(\mathrm{i} \hat{\vec{p}} \cdot \vec{\xi}),[/tex]

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

[tex]\hat{U}(\vec{\xi}) \psi(\vec{x}):=\psi(\vec{x}+\vec{\xi}).[/tex]

Comparing this to the Taylor-expansion formula,

[tex]\psi(\vec{x}+\vec{\xi})=\exp(\vec{\xi} \cdot \vec{\nabla}) \psi(\vec{x}),[/tex]

immediately gives the well-known expression

[tex]\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}[/tex]

for the momentum operator in the position representation.
 
  • #4
@dextercioby: That's exactly what I want as an answer. Thanks.
@vanhees71: Thanks for that reply :)
 

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