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Equation 80 · Part 4 · A Reference Frame Becomes Classical by Publishing Its Orientation

Symbol σ^(n)

Hint(n)=ℏΩ J⋅σ(n)2j,Un=exp⁡ ⁣(−iκJ⋅σ(n)2j),H_{\mathrm{int}}^{(n)} =\hbar\Omega\, \frac{\mathbf J\cdot\boldsymbol\sigma^{(n)}}{2j}, \qquad U_n=\exp\!\left(-i\kappa \frac{\mathbf J\cdot\boldsymbol\sigma^{(n)}}{2j}\right),
σ(n)\sigma^{(n)}

What this part means

σ^(n) occurs above the fraction bar. The numerator is divided by the entire denominator below it.

Its job in the formula

σ^(n) occurs above the fraction bar. The numerator is divided by the entire denominator below it.

The passage around this formula

Let J\mathbf J be the spin- j angular-momentum operator and σ(n)\boldsymbol\sigma^{(n)} the Pauli vector for probe n . A candidate rotationally invariant interaction is Hint(n)=ℏΩ J⋅σ(n)2j,Un=exp⁡ ⁣(−iκJ⋅σ(n)2j)H_{\mathrm{int}}^{(n)} =\hbar\Omega\, \frac{\mathbf J\cdot\boldsymbol\sigma^{(n)}}{2j}, \qquad U_n=\exp\!\left(-i\kappa \frac{\mathbf J\cdot\boldsymbol\sigma^{(n)}}{2j}\right). where Ω\Omega has units of inverse seconds, interaction time τ\tau has seconds, and κ\kappa=Ω\Omegaτ\tau is dimensionless. The factor 2j is a declared normalization choice, not an inherited law. Alternative normalizations must be compared because they change the large- j limit if coupling strength is held fixed under different conventions.

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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