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In nonrelativistic quantum mechanics, are different operators possible as a candidate for the momentum operator, given that one has fixed one position operator and a hilbert space that this position operator acts uppon?

If I choose the Hilbert space to be the space of square integrable functions, and the position operator to be $x$, then the usual choice for the momentum operator is $- i \frac{\partial}{\partial x}$. However, there are other choices, that would also satisfy the commutation relations:

\begin{align} -i \frac{\partial}{\partial x} + f(x) \end{align}

With an abitrary function $f$ would satisfy the communtation relations just as well. Is there some kind of "gauge freedom" that is fixed by choosing $\hat{p} = - \frac{\partial}{\partial x}$, or are there additional assumptions that $\hat{p}$ has to satisfy?

EDIT Iread in other answers that the momentum operator has to generate translations, and this answer mentions that the property of being the generator of translations is linked to the property of the commutation relation. It however also states that the choice of the momentum operator is a unique choice. Am I missing something? If the ccr's are satisfied, then the given formula of a unitary shift of the position operator can be used, resulting in the said translation.

Qmechanic
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Quantumwhisp
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  • Possible duplicates: https://physics.stackexchange.com/q/70203/2451 , https://physics.stackexchange.com/q/41880/2451 , https://physics.stackexchange.com/q/45248/2451 , https://physics.stackexchange.com/q/80357/2451 and links therein. – Qmechanic Aug 18 '18 at 06:07
  • Yes, there is a well-known and utilized "gauge freedom". For $g(x)$ the antiderivative of f, so $f=\partial_x g$, note $(x,-i\partial_x) \mapsto e^{-ig(x)}(x,-i\partial_x) e^{ig(x)}= (x,-i\partial_x +f)$. – Cosmas Zachos Mar 05 '21 at 22:07

1 Answers1

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The canonical commutation relations in the Lie algebra form $[x,p]=i$ do not generate necessarily the Heisenberg group, and therefore in that form there is no uniqueness result.

There are explicit counterexamples of densely defined self-adjoint operators, with a common core on which they satisfy the CCR in the Lie algebra form, that do not generate the Heisenberg group. It is possible to find one such example in the book by Reed and Simon, for example.

The counterexamples are trickier than the one suggested by the OP, because both operators should be self-adjoint (to be physically relevant observables), with a common core on which the CCR are satisfied.

There is uniqueness instead, up to (unitary) group homomorphisms, of the irreducible representations of the Heisenberg group, i.e. of the exponential version of the CCR. In other words, if one requires the CCR in exponential form (the so-called Weyl relations), and fixes one of the two generators (e.g. the position operator), then the other is uniquely determined. This uniqueness result is called Stone-von Neumann uniqueness theorem (and it is a particular case of Mackey's systems of imprimitivity).

yuggib
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