For a spin one-half particle

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Symmetry, angular momentum and spin · Symmetry, angular momentum and spin

The Structure of Quantum Mechanics, Ch Symmetries, generators and conservation laws

Problem

For a spin one-half particle, Θ^\hat\Theta acts on basis components as Θ^(c1,c2)=(−c2∗,c1∗)\hat\Theta(c_1,c_2) = (-c_2^*,c_1^*) in the {∣ ⁣↑⟩,∣ ⁣↓⟩}\{|\!\uparrow\rangle,|\!\downarrow\rangle\} basis, so Θ^∣ ⁣↑⟩=∣ ⁣↓⟩\hat\Theta|\!\uparrow\rangle=|\!\downarrow\rangle and Θ^∣ ⁣↓⟩=−∣ ⁣↑⟩\hat\Theta|\!\downarrow\rangle=-|\!\uparrow\rangle, and Θ^2=−1\hat\Theta^2=-1. Kramers' theorem states that whenever Θ^2=−1\hat\Theta^2=-1, every state satisfies ⟨ψ∣Θ^ψ⟩=0\langle\psi|\hat\Theta\psi\rangle=0.Someone claims to have found a counterexample. Their working, for ∣ψ⟩=12(∣ ⁣↑⟩+i∣ ⁣↓⟩)|\psi\rangle = \tfrac{1}{\sqrt2}\left(|\!\uparrow\rangle + i|\!\downarrow\rangle\right), is:Θ^∣ψ⟩=12(Θ^∣ ⁣↑⟩+i Θ^∣ ⁣↓⟩)=12(∣ ⁣↓⟩+i(−∣ ⁣↑⟩))=12(−i∣ ⁣↑⟩+∣ ⁣↓⟩),\hat\Theta|\psi\rangle = \tfrac{1}{\sqrt2}\left(\hat\Theta|\!\uparrow\rangle + i\,\hat\Theta|\!\downarrow\rangle\right) = \tfrac{1}{\sqrt2}\left(|\!\downarrow\rangle + i(-|\!\uparrow\rangle)\right) = \tfrac{1}{\sqrt2}\left(-i|\!\uparrow\rangle + |\!\downarrow\rangle\right),from which they compute ⟨ψ∣Θ^ψ⟩=12[1⋅(−i)+(−i)⋅1]=−i≠0\langle\psi|\hat\Theta\psi\rangle = \tfrac12\left[1\cdot(-i) + (-i)\cdot 1\right] = -i \ne 0, and conclude that Kramers' theorem is violated. ((a)) Identify the single step in the working that is not valid, and say what property of Θ^\hat\Theta it ignores. ((b)) Redo the calculation of Θ^∣ψ⟩\hat\Theta|\psi\rangle correctly. ((c)) Recompute ⟨ψ∣Θ^ψ⟩\langle\psi|\hat\Theta\psi\rangle and check it against Kramers' theorem.

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