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I found this in the book Geometric Phase in Quantum Systems by A. Bohm et al.

Where the position space representation of the momentum operator carries a (Where exactly my doubt is) coefficient of 1-form with the condition

$$\partial_i \omega_j - \partial_j \omega_i =0 \implies d\omega=0$$

The author(s) argued about $Poincare \ lemma$ and how, for $\mathbb{R}^m$ configuration space the term can be $gauged \ away$.

I understand usual momentum operator representation without this 1-form, and this is very non-trivial for me.

Can someone please explain me how this comes and what it means in details?

Qmechanic
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prikarsartam
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1 Answers1

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  1. For $M=\mathbb{R}^m$, starting from the canonical commutation relations (CCRs), we have the Stone-von Neumann theorem, which proves the existence of the standard Schrödinger position representation (i.e. without the 1-form $\omega$) up to unitary equivalence, cf. e.g. this Phys.SE post.

  2. However conjugating with unitary multiplication operators naturally induces an exact $\omega$ one-form, cf. e.g. my Phys.SE answer here.

  3. Therefore it is quite natural to consider a 1-form $\omega$ from the onset, even for a topologically non-trivial $m$-dimensional configuration space $M$.

  4. The CCRs then imply that the 1-from $\omega$ must be a closed one-form. The cohomology class $[\omega]$ is conserved under conjugating with unitary multiplication operators.

  5. Since $M=\mathbb{R}^m$ is a contractible space, we may use Poincare Lemma to show that the cohomology class $[\omega]=[0]$ is trivial.

Qmechanic
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