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Every density operator can be written asTherefore $⟨\varphi |\rho |\varphi ⟩={\sum }_j{p}_j|⟨{\psi }_j|\varphi ⟩{|}^2\ge 0$, so $\rho$ is a positive semidefinite operator, and $\mathrm{Tr}\rho ={\sum }_j{p}_j=1$. A pure quantum state has a rank-one density operator $|\psi ⟩⟨\psi |$, equivalently ${\rho }^2=\rho$ and $\mathrm{Tr}({\rho }^2)=1$. A mixed quantum state is not pure and has $\mathrm{Tr}({\rho }^2)<1$ in finite dimension.

The Von Neumann equation and the Pauli matrix commutator identity giveso the Bloch vector obeysWriting $\omega =|\mathbf{\omega }|$ and $\hat{\mathbf{\omega }}=\mathbf{\omega }/\omega$, the assumption $\mathbf{a}\cdot \mathbf{\omega }=0$ givesand henceUnitary evolution rotates the Bloch vector and preserves its length. Since $|\mathbf{a}|<1$, the state remains mixed, whereas $|{\uparrow }_x⟩$ has the unit Bloch vector $(1,0,0)$. The system is therefore never definitely in $|{\uparrow }_x⟩$.

A subset $A$ of a metric space is nowhere dense when the interior of its closure is empty. One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of nowhere dense closed sets; equivalently, a countable intersection of dense open sets is dense.

Fix $\epsilon >0$ and work in the complete interval $[1,2]$. For $N\ge 1$ define the closed setThe sets are closed by the continuity of $f$, and the assumed pointwise convergence gives $[1,2]={\bigcup }_N{E}_N$. The Baire category theorem therefore gives an $N$ and a nonempty open interval $(a,b)\subseteq [1,2]$ contained in $E_N$.

The intervals $[na,nb]$ overlap once $(n+1)a\le nb$, which holds for every sufficiently large natural number $n$. Their union consequently contains a half-line. Every sufficiently large $t$ can therefore be written as $t=nx$ with $n\ge N$ and $x\in (a,b)$, and then $|f(t)|=|f(nx)|\le \epsilon$. Since $\epsilon$ was arbitrary, $f(t)\to 0$ as $t\to \infty$.

Write $\ell (x,\theta )=\log f(x,\theta )$ and let ${\dot{\ell }}_{{\theta }_0}(x)$ be its score function. A second-order Taylor expansion around ${\theta }_0$ gives, for fixed $h$,where every ${\theta }_{i,n}$ lies between ${\theta }_0$ and ${\theta }_0+h/\sqrt{n}$. The mean-zero score identity, the Fisher information matrix, and the multivariate central limit theorem giveThe assumed regularity and a uniform law of large numbers giveConsequently, by the Slutsky theorem,

For the stated Gaussian shift model, direct expansion of the two normal distribution densities givesSince $I({\theta }_0)X\sim N(0,I({\theta }_0))$, this has exactly the limiting distribution above. Equivalently, both limits are normal with mean $-\frac{1}{2}{h}^{T}I({\theta }_0)h$ and variance $h^{T}I({\theta }_0)h$. This is the local asymptotic normality of the model at ${\theta }_0$.