Physics

Physics · modern-physics

Problem

Calculate the following expressions, (i) (c⋅σ)n(\mathbf{c} \cdot \boldsymbol{\sigma})^{n}, and then (ii) (bI+c⋅σ)n(b \mathrm{I}+\mathbf{c} \cdot \boldsymbol{\sigma})^{n}, for the scalar product c⋅σ\mathbf{c} \cdot \boldsymbol{\sigma} of the Pauli matrix vector σ≡nxσx+nyσy+nzσz\boldsymbol{\sigma} \equiv \mathbf{n}_{x} \sigma_{x}+\mathbf{n}_{y} \sigma_{y}+\mathbf{n}_{z} \sigma_{z} by an arbitrary cc-number vector c\mathbf{c}, where n⩾0n \geqslant 0 is an integer, and bb is an arbitrary scalar cc number. Hint: For task (ii), you may like to use the binomial theorem 2{ }^{2}, and then transform the result in a way enabling you to use the same theorem backwards.
Intuitive explanation

The key insight is that Pauli matrices satisfy σi2=I\sigma_i^2 = I and anticommute when different, which means (c⋅σ)2=∣c∣2I(\mathbf{c} \cdot \boldsymbol{\sigma})^2 = |\mathbf{c}|^2 I—the product collapses to a scalar times the identity. This alternating pattern (odd powers give c⋅σ\mathbf{c} \cdot \boldsymbol{\sigma}, even powers give ∣c∣2I|\mathbf{c}|^2 I) is your ticket to evaluating any power. For part (ii), the hint suggests expanding with the binomial theorem, then recognizing that the mixed terms reorganize into the same alternating structure, letting you collapse the sum back into a closed form using the binomial theorem in reverse.

Intuitive explanation

The key insight is that Pauli matrices satisfy σi2=I\sigma_i^2 = I and anticommute when different, which means (c⋅σ)2=∣c∣2I(\mathbf{c} \cdot \boldsymbol{\sigma})^2 = |\mathbf{c}|^2 I—the product collapses to a scalar times the identity. This alternating pattern (odd powers give c⋅σ\mathbf{c} \cdot \boldsymbol{\sigma}, even powers give ∣c∣2I|\mathbf{c}|^2 I) is your ticket to evaluating any power. For part (ii), the hint suggests expanding with the binomial theorem, then recognizing that the mixed terms reorganize into the same alternating structure, letting you collapse the sum back into a closed form using the binomial theorem in reverse.

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