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The operator $P$ in quantum mechanics is the generator for the translation transformation. We have:

$$\exp(iPa)|x\rangle=|x+a\rangle$$

Similarly, I think the operator $X$ is the generator for the Galileo transformation:

$$\exp(-iXq)|p\rangle=|p+q\rangle$$

Is this ultimately not consistent with the Lorentz transformations? Did we base on Galileo transformations somewhere when we construct quantum mechanics?

EDIT: Sorry I just found out that the summation of momentum has nothing to do with Galileo transformations... Anyway if you can refer me to some more informations or give me some insights about this it would be appreciated. Like:

What is the transformation that the position operator generates?

The Lorentz boost is an unitary transformation. What is the associated oversable? (What happens in case we consider Galileo transformation?)

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    $\begingroup$ I'm not exactly sure what you're asking. If you're asking whether quantum mechanics is consistent with special relativity, then the answer is no. The standard version of quantum mechanics is non-relativistic. You have to go to relativistic quantum mechanics, or, rather, quantum field theory, to do quantum mechanics and relativity together. $\endgroup$
    – ACuriousMind
    Jan 16, 2016 at 18:16
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    $\begingroup$ That. And a Lorentz boost is not a unitary transformation. $\endgroup$
    – Daniel
    Jan 16, 2016 at 18:17
  • $\begingroup$ The Dirac equation is an intermediate step between Schrodinger's equation and QFT. $\endgroup$ Jan 16, 2016 at 18:57

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In non relativistic quantum mechanics, referring to an irreducible projective unitary representation of Galileo group, up to a multiplicative factor (the mass) the position operator naturally arises as the generator of boost transformation. This is equivalent to the standard translation in the space of momenta.

In relativistic QM the standard translation in the space of momenta ceases to be a symmetry and the generator of the boost has another form.

The relativistic definition of the position operator is more involved. It is possible but uses a different approach, technically based on the so called imprimitivity structures. It is possible to prove that, for elementary systems (unitary irreducible representations of Poincaré group), the position observable is uniquely defined for massive systems, otherwise it is not always well-defined, depending on the value of the spin. The position operator is also known as Newton-Wigner position operator.

A good reference is the book by Varadarajan Geometry of Quantum Theory, in one of the last chapters it studies the problem in details. Also in Barut Raczka's textbook Theory of group representations and applications there is a detailed, but less rigorous, discussion in one of the last chapters.

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  • $\begingroup$ Thank you for your answer! And also in relativistic QM, is there an unitary representation of the Lorentz group? If yes, is the generator of the boost an observable? $\endgroup$ Jan 16, 2016 at 18:23
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    $\begingroup$ Yes it is. It is an operator depending explicitly on time. Its conservation rule gives rise to the so called theorem of the centre of mass which, in relativistic theory it is not a trivial consequence of momentum conservation. Essentially because the notion of mass includes part of the energy... $\endgroup$ Jan 16, 2016 at 18:25
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    $\begingroup$ For massive spinless elementary systems, if $K^j(t)$ is the boost generator along the direction $x^j$, we have $K^j(0)= \frac{1}{2}(X^jP^0+P^0X^j)$ where $X^j$ is the position operator... You see that in the non relativistic limit $P^0 \sim mI$ so that $K^j(0)= mX^j$. $\endgroup$ Jan 16, 2016 at 18:30
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    $\begingroup$ Momenta translations are symmetries in the sense that they are described by unitary operators, but they are not dynamical symmetries in the sense that they do not give rise to conserved quantities by means of a quantum version of Noether theorem, just because they are not elements of the self-adjoint representation of the Lie algebra of Poincaré group. $\endgroup$ Jan 16, 2016 at 18:34
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    $\begingroup$ They are, by definition the position operators! $\endgroup$ Jan 16, 2016 at 18:39

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