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Consider the position and momentum operator on $L^2(\mathbb{R})$ defined on a dense domain, say $D(X) = D(P)=\mathscr{S}(\mathbb{R})$ with the position operator being defined as $$ X\phi=x\phi(x) \qquad \forall \phi \in \mathscr{S}(\mathbb{R}) $$ and the momentum operator is defined by $$ P\phi=-i \frac{d}{dx}\phi(x) \qquad \forall \phi \in \mathscr{S}(\mathbb{R}) $$ Then we have the canonical commutation relations $$ [X,P]\phi=i\phi \qquad \forall \phi \in \mathscr{S}(\mathbb{R}) \tag{1}$$ For any unitary operator $U$ (for example the Fourier transform), it is easy to see that $U^{\dagger}PU$ and $U^{\dagger}XU$ satisfy the same commutation relations.

Question: If $2$ self-adjoint, densely defined linear operators satisfy (1), are they unitarily equivalent to the position and momentum operator?

In other words: Do the canonical commutation relations already determine position and momentum operator up to unitary equvialence?

I am asking this question out of curiousity, since it stuck in my head for quite some time. I have thought about (formally) taking operator exponentials, since $\exp(-iX)\phi=e^{ix}\phi(x)$ and $\exp(iaP)\phi=\phi(x+a)$ and then use the Stone-von Neumann theorem for the Heisenberg group, however, the issue with the domain remains and also taking operators exponential is non-trivial for this reason. Feel free to modify my domain to e.g. $D(\cdot)=W^{1,2}(\mathbb{R}),C_c^{\infty}(\mathbb{R}),..$ or whatever you see fit.

I am somewhat familiar with the basic spectral theory for unbounded operators and I have already encountered the concept of a rigged Hilbert space, so feel free to use them without elaborating every detail.

FD_bfa
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F. Conrad
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    I think you're probably looking for the notion of an irreducible representation. You could have many copies of the same operators and end up with the same commutation relations. You want to look more into that. – Disintegrating By Parts Sep 12 '21 at 21:39
  • I've already tried to look into that notion by trying to apply the Stone von-Neumann theorem - but I didn't get anywhere without putting in countless hours, as it is just some random question I cam up with recently in my head. – F. Conrad Sep 12 '21 at 21:49

1 Answers1

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Here's a starting point that may help: $$ \left[X,P\right]=iI \\ (XP-PX)=iI \\ X(P-\lambda I)-(P-\lambda I)X = iI \\ (P-\lambda I)^{-1}X-X(P-\lambda I)^{-1}=i(P-\lambda I)^{-2}=\frac{d}{d\lambda}i(P-\lambda I)^{-1} \\ [(P-\lambda I)^{-1},X]=i\frac{d}{d\lambda}(P-\lambda I)^{-1}. $$ Then you can use Stone's Formula for the Spectral Measure $E$ associated with $P$: $$ s-\lim_{v\downarrow 0}\frac{1}{2\pi i}\int_{a}^{b}(P-(u+iv)I)^{-1}-(P-(u-iv)I)^{-1}du = \frac{1}{2}(E[a,b]+E(a,b)) $$ (The $s-\lim$ refers to convergence in the strong operator topology.)

Disintegrating By Parts
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