Past exam of the quantum physics course of the University of Oxford / 2016 / 7

The spin Hilbert space is a tensor product of two two-dimensional spaces and therefore has dimension four. Angular momentum addition decomposes it asThese are three orthogonal triplet states and one singlet state. For identical electrons the complete wavefunction must also satisfy fermionic exchange symmetry: a symmetric triplet spin factor accompanies an antisymmetric spatial factor, whereas the antisymmetric singlet spin factor accompanies a symmetric spatial factor. If both electrons occupy the same spatial orbital, only the singlet spin factor is allowed. The four-dimensional spin space does not assert four allowed spin states for that particular spatial occupation.

Expanding the total angular momentum and replacing transverse components by ladder operators givesThe particle exchange operator obeys $P_{12}^2=I$ and $P_{12}{s}_{1\alpha }{P}_{12}=s_{2\alpha }$, with the reverse relation for index $2$. These statements also apply to the ladder operators. Exchanging the indices preserves $s_1^2+s_2^2$ and $s_{1z}+s_{2z}$; it interchanges the last two terms in $S^2$. Operators on different tensor factors commute. HenceThe full particle exchange includes the spin coordinates as well as the spatial coordinates.

Write $\Psi =(|\downarrow \uparrow ⟩+|\uparrow \downarrow ⟩)/\sqrt{2}$. Each term has cancelling $z$ components, so $S_z\Psi =0$. In the formula for $S^2$, the first three terms contribute $(3/2-1/2){\hslash }^2\Psi ={\hslash }^2\Psi$. The two ladder products exchange the two terms and contribute another ${\hslash }^2\Psi$. ThereforeThis spin state is symmetric; the accompanying spatial state must be antisymmetric for two electrons. A labelled product $|{\downarrow }_1{\uparrow }_2⟩$ alone is not an eigenstate of spin exchange: its exchanged vector is orthogonal to it. Thus it cannot supply a definite exchange symmetry when multiplied by a fixed symmetric spatial factor.

This does not prohibit opposite spins in different spatial orbitals. For example, with orthonormal orbitals ${\varphi }_a,{\varphi }_b$, the Slater determinantis a valid antisymmetric two-electron state. What fails is treating persistent particle labels as physically distinguishable identities without the required antisymmetrization.

For a conditional spin measurement of identical fermions addressing the $z$ component in one spatial mode, the Born rule givesConditioned on this result, the mode spin state is $|{\uparrow }_a{\downarrow }_b⟩$; the other outcome produces $|{\downarrow }_a{\uparrow }_b⟩$. The spin product is not a spin-exchange eigenstate. The measurement has selected an addressed mode and recorded an outcome, so this conditional spin description distinguishes the measured subsystem.

The full electron wavefunction still obeys fermionic exchange symmetry. For example, the conditional spin-up outcome in spatial mode $a$ is represented in particle coordinates byIt is antisymmetric under full particle exchange. Thus the loss of a definite symmetry of the conditional spin product does not mean that measurement turns identical electrons into distinguishable particles. If the outcome is discarded, the mode spin state is instead the equal mixture of the two opposite-spin products. The measurement must specify which spatial mode is addressed for the conditional statement to be operationally meaningful.