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

What does this equation mean?

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

Read the formula alongside the article passage below. Each part has a deeper page with its role in the equation, the supporting passage and nearby citations.

Start withim
Divide byhbar
This relates to[X,Y]
How to read the two sides of this formula. Follow the article passage for the meaning of each quantity.

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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XX

Symbol X

X is part of the quantity the equation computes from the expression on the right.

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YY

Symbol Y

Y is part of the quantity the equation computes from the expression on the right.

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ii

Symbol i

i occurs above the fraction bar. The numerator is divided by the entire denominator below it.

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mm

Symbol m

the mass.

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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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vv

Symbol v

v is one factor in the product that computes the quantity on the left.

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=

=

The expressions on both sides represent the same quantity under the stated assumptions.

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fraction

fraction

Divide the expression above the line by the one below it.

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multiplication

multiplication

Multiply the quantities on either side.

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imim

Numerator: im

The complete quantity above the fraction bar.

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ℏ\hbar

Denominator: hbar

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

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How to interpret it

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

What the article says around this equation

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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