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[X,Y]=imℏ b⋅v 1[X,Y]=\frac{im}{\hbar}\,\mathbf b\cdot\mathbf v\,\mathbb 1

Why this formula appears here

That triviality is what makes the loop phase exact rather than approximate. Write X=-ib\mathbf b⋅\cdotP\mathbf P/ℏ\hbar and Y=-iv\mathbf v⋅\cdotK\mathbf K/ℏ\hbar . Their commutator is [X,Y]=imℏ b⋅v 1[X,Y]=\frac{im}{\hbar}\,\mathbf b\cdot\mathbf v\,\mathbb 1. Because [X,Y] is itself proportional to the identity, it commutes with both X and Y , so every higher term in the Baker–Campbell–Hausdorff expansion of eXe^XeYe^Ye−Xe^{-X}e−Ye^{-Y} vanishes identically, at all orders, with no small-loop approximation required:

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bb

Symbol b

b occurs below the fraction bar. The quantity above the bar is divided by this expression; zero is excluded as a denominator.

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With a fixed numerator, increasing a nonzero denominator reduces the fraction. Read it with the definitions, units, and assumptions supplied by the article.

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[X,Y]=imℏ b⋅v 1.[X,Y]=\frac{im}{\hbar}\,\mathbf b\cdot\mathbf v\,\mathbb 1.

Equation 33 · Evolutionary Physics

The Clock That Comes Back Wrong by Exactly Its Mass

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

That triviality is what makes the loop phase exact rather than approximate. Write X=-ib\mathbf b⋅\cdotP\mathbf P/ℏ\hbar and Y=-iv\mathbf v⋅\cdotK\mathbf K/ℏ\hbar . Their commutator is [X,Y]=imℏ b⋅v 1[X,Y]=\frac{im}{\hbar}\,\mathbf b\cdot\mathbf v\,\mathbb 1. Because [X,Y] is itself proportional to the identity, it commutes with both X and Y , so every higher term in the Baker–Campbell–Hausdorff expansion of eXe^XeYe^Ye−Xe^{-X}e−Ye^{-Y} vanishes identically, at all orders, with no small-loop approximation required:

Meanings in this article

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