Published equation contexts

$G(\mathbf v)=\exp(-i\mathbf v\cdot\mathbf K/\hbar)$

Why this formula appears here

Represent the loop $\mathcal L_{\rm G}$ on a Hilbert space. A spatial translation by $\mathbf b$ is the unitary T( $\mathbf b$ )= $\exp(-i\mathbf b\cdot\mathbf P/\hbar)$ , built from the momentum operator $\mathbf P$ ; a Galilei boost by $\mathbf v$ is G( $\mathbf v$ )= $\exp(-i\mathbf v\cdot\mathbf K/\hbar)$ , built from the boost generator $\mathbf K$ . The classical loop is T( $\mathbf b$ )G( $\mathbf v$ )T(- $\mathbf b$ )G(- $\mathbf v$ )= $\mathbb 1$ . The operator loop is not automatically the identity, because $\mathbf K$ and $\mathbf P$ need not commute as operators even when translations and boosts commute as group elements.

Read the full article-specific guide →

Read the representative guide

G

Symbol G

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

Read this term in its guide →

v

Symbol v

v is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

Read this term in its guide →

i

Symbol i

i is one of the signed contributions combined to compute the quantity on the left.

Read this term in its guide →

K

Symbol K

K is one of the signed contributions combined to compute the quantity on the left.

Read this term in its guide →

How to interpret it

Read it with the definitions, units, and assumptions supplied by the article.

Published contexts (1)

A symbol can carry a different meaning in another article. Each occurrence keeps its own guide and term definitions.

G(\mathbf v)=\exp(-i\mathbf v\cdot\mathbf K/\hbar)

Equation 22 · 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.

Represent the loop $\mathcal L_{\rm G}$ on a Hilbert space. A spatial translation by $\mathbf b$ is the unitary T( $\mathbf b$ )= $\exp(-i\mathbf b\cdot\mathbf P/\hbar)$ , built from the momentum operator $\mathbf P$ ; a Galilei boost by $\mathbf v$ is G( $\mathbf v$ )= $\exp(-i\mathbf v\cdot\mathbf K/\hbar)$ , built from the boost generator $\mathbf K$ . The classical loop is T( $\mathbf b$ )G( $\mathbf v$ )T(- $\mathbf b$ )G(- $\mathbf v$ )= $\mathbb 1$ . The operator loop is not automatically the identity, because $\mathbf K$ and $\mathbf P$ need not commute as operators even when translations and boosts commute as group elements.

Equation guide → · Article →

G(v)=exp⁡(−iv⋅K/ℏ)G(\mathbf v)=\exp(-i\mathbf v\cdot\mathbf K/\hbar)G(v)=exp(−iv⋅K/ℏ)

Why this formula appears here

Read the representative guide

Symbol G

Symbol v

Symbol i

Symbol K

How to interpret it

Learn the mathematics behind it

Published contexts (1)

The Clock That Comes Back Wrong by Exactly Its Mass

$G(\mathbf v)=\exp(-i\mathbf v\cdot\mathbf K/\hbar)$