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ID: past-exam-of-the-advanced-quantum-condensed-matter-physics-course-of-the-university-of-cambridge/2013/paper-1/2/c/solution

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Past exam of the advanced Quantum Condensed Matter Physics course of the University of Cambridge / 2013 / Paper 1 / 2 / c / Solution by

Codex  0 2026-10-20

With opposite magnetic moments, a fixed global spin is the low-resistivity species in one ferromagnetic layer and the high-resistivity species in the other. Thus both spin channels have the same four-layer averaged resistivity

\overset{ρ}{ˉ}_{A P} = \frac{t _{1} ρ _{FM}^{L} + t _{1} ρ _{FM}^{H} + 2 t _{2} ρ _{NM}}{2 D} = \frac{a + b}{2 D} .

(1)

Each channel's square resistance is

(a + b) / (4 D^{2})

, and the two channels run in parallel. Therefore

\frac{1}{R _{↑, ↓}} = \frac{8 D ^{2}}{a + b} .

(2)

Both channels encounter the strongly scattering magnetic layer; neither retains the uniformly favourable path available for parallel magnetizations.

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