Equation 33 · Part 11 · The Clock That Comes Back Wrong by Exactly Its Mass

Denominator: hbar

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

\hbar

What this part means

The complete quantity below the fraction bar; it must be nonzero for this division.

Its job in the formula

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

Full expressionDenominator: hbarArticle meaning

The passage around this formula

That triviality is what makes the loop phase exact rather than approximate. Write X=-i $\mathbf b$ $\cdot$ $\mathbf P$ / $\hbar$ and Y=-i $\mathbf v$ $\cdot$ $\mathbf K$ / $\hbar$ . Their commutator is $[X,Y]=\frac{im}{\hbar}\,\mathbf b\cdot\mathbf v\,\mathbb 1$ . 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 $e^X$ $e^Y$ $e^{-X}$ $e^{-Y}$ vanishes identically, at all orders, with no small-loop approximation required:

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Learn the underlying idea

A fraction a/b means a divided by b. The top number is the numerator; the bottom number is the denominator, and it cannot be zero.

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