Two-level measurement and formal-coefficient boundary

The displayed pairing is linear in the first variable by coordinatewise addition and scalar multiplication, is conjugate-symmetric by complex conjugation, and satisfies $⟨x,x⟩=\mathrm{\mid }{x}_0{\mathrm{\mid }}^2+\mathrm{\mid }{x}_1{\mathrm{\mid }}^2\ge 0$, with equality only for $x=0$. Thus it is a positive-definite complex inner product with norm $\mathrm{\parallel }x\mathrm{\parallel }=(\mathrm{\mid }{x}_0{\mathrm{\mid }}^2+\mathrm{\mid }{x}_1{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$.

Take the coefficient ring to be the repository real field, which is a commutative ring by [F6], and define $a_n:=n!$ for each $n\in {\mathbb{N}}_0$. Each coefficient is a finite product of real integers, so $(a_n{)}_{n\ge 0}$ is a coefficient function and hence $F(g)\in \mathbb{R}\mathrm{⟦}g\mathrm{⟧}$ by [F4]. This formal membership does not assert convergence.

By [F4], a formal coefficient is an entry in a coefficient function and does not itself select a numerical argument or a model. By [A1], the postulate assigns outcome probabilities only through its model Born rule; by [A5] and [A7], this formal example supplies no state, instrument, or map from the coefficient sequence to such data. Therefore the postulate assigns no exact probability to an isolated coefficient; this does not exclude a separate model that supplies such a map.

Let $(x_n)$ be Cauchy in the norm of step 1.1, with coordinates $x_n=(z_{n,0},z_{n,1})$. For each $j$, the estimate $\mathrm{\mid }{z}_{n,j}-z_{m,j}\mathrm{\mid }\le \mathrm{\parallel }{x}_n-x_m\mathrm{\parallel }$ makes $(z_{n,j})$ Cauchy in $\mathbb{C}$. By [F3] it converges to some $z_j$. Then $\mathrm{\parallel }{x}_n-(z_0,z_1){\mathrm{\parallel }}^2=\mathrm{\mid }{z}_{n,0}-z_0{\mathrm{\mid }}^2+\mathrm{\mid }{z}_{n,1}-z_1{\mathrm{\mid }}^2\to 0$. So ${\mathbb{C}}^2$ is complete, and is a complex Hilbert space by [F2]; it is nonzero since $e_0\mathrm{\ne }0$.

At $g=0$, the numerical sum equals $1$ by [A3]. Fix any nonzero complex $g$ and let $u_n:=n!g^n$. By [F5] choose a natural $N>2\mathrm{/}\mathrm{\mid }g\mathrm{\mid }$. For every $n\ge N$, $\mathrm{\mid }{u}_{n+1}\mathrm{\mid }\mathrm{/}\mathrm{\mid }{u}_n\mathrm{\mid }=(n+1)\mathrm{\mid }g\mathrm{\mid }>2$, so $\mathrm{\mid }{u}_n\mathrm{\mid }\ge \mathrm{\mid }{u}_N\mathrm{\mid }>0$. Thus the terms do not tend to zero. If the partial sums converged, their consecutive differences $u_n$ would tend to zero, a contradiction. Therefore the numerical series diverges for every nonzero $g$.

For $P_0$, both inner products in the symmetry identity equal $x_0\overline{y_0}$; for $P_1$, both equal $x_1\overline{y_1}$. Thus $P_i^{\ast }=P_i$. Multiplying the diagonal matrices gives $P_i^2=P_i$ and $P_0+P_1=I_2$. For $x=(x_0,x_1)$, $\mathrm{\parallel }{P}_{0}x{\mathrm{\parallel }}^2=\mathrm{\mid }{x}_0{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}^2$ and $\mathrm{\parallel }{P}_{1}x{\mathrm{\parallel }}^2=\mathrm{\mid }{x}_1{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}^2$, so both operators are bounded. With two outcomes and one Kraus operator each, ${\sum }_{i=0}^1{M}_{i,1}^{\ast }{M}_{i,1}=P_0+P_1=I_2$; [L1] therefore identifies the data as a normalized finite instrument on the Hilbert space of step 2.1.

The normalization of $v$ gives $⟨v,v⟩=\mathrm{\mid }\alpha {\mathrm{\mid }}^2+\mathrm{\mid }\beta {\mathrm{\mid }}^2=1$. By [L3], $T_{v,v}x=⟨x,v⟩v$; substituting $⟨x,v⟩=x_0\bar{\alpha }+x_1\bar{\beta }$ gives $\rho =T_{v,v}=\left(\begin{array}{cc}\mathrm{\mid }\alpha {\mathrm{\mid }}^2& \alpha \bar{\beta }\\ \beta \bar{\alpha }& \mathrm{\mid }\beta {\mathrm{\mid }}^2\end{array}\right)$. The formula in [L3] gives the displayed matrix, and [L2] identifies this unit-vector rank-one operator as a density operator under ${\mathrm{A}\mathrm{C}}_{\omega }$.

By the radius definition [A4], for every $r>0$ the nonzero real number $g=r\mathrm{/}2$ lies inside that disk. Step 2.2 proves divergence at this $g$, so no positive $r$ belongs to the defining set. The value $r=0$ belongs vacuously because its disk is empty; hence $R=0$.

The actions are $P_{0}v=\alpha e_0$ and $P_{1}v=\beta e_1$, so their squared norms are $\mathrm{\mid }\alpha {\mathrm{\mid }}^2$ and $\mathrm{\mid }\beta {\mathrm{\mid }}^2$. The pure-state formula in [L2] gives $p_0=\mathrm{\mid }\alpha {\mathrm{\mid }}^2$ and $p_1=\mathrm{\mid }\beta {\mathrm{\mid }}^2$, whose sum is one. By [A1] these are the adopted model probabilities; each lies in $[0,1]$.

From [L1], ${\mathcal{J}}_0(\rho )=P_0\rho P_0$ and ${\mathcal{J}}_1(\rho )=P_1\rho P_1$. Using the matrix in step 3.2, $P_0\rho P_0=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}\mathrm{\mid }\alpha {\mathrm{\mid }}^2& \alpha \bar{\beta }\\ \beta \bar{\alpha }& \mathrm{\mid }\beta {\mathrm{\mid }}^2\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)=\mathrm{\mid }\alpha {\mathrm{\mid }}^2{P}_0,$ while $P_1\rho P_1=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\mathrm{\mid }\alpha {\mathrm{\mid }}^2& \alpha \bar{\beta }\\ \beta \bar{\alpha }& \mathrm{\mid }\beta {\mathrm{\mid }}^2\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)=\mathrm{\mid }\beta {\mathrm{\mid }}^2{P}_1\mathrm{.}$ Also $P_i=T_{e_i,e_i}$ by [L3], so [L2] makes $P_i$ a density operator. If $p_0>0$, ${\mathcal{J}}_0(\rho )\mathrm{/}{p}_0=P_0$; if $p_0=0$, division is undefined. The same argument gives $P_1$ when $p_1>0$ and no normalized update when $p_1=0$, exactly as adopted in [A6].