= Solution
Let $U_1=e^{-iH(t-t_1)/\hbar}$ and $U_2=e^{-iH(t_2-t)/\hbar}$ describe <unitary time evolution>. For an ideal <projective measurement> with the <Lüders rule>, its unconditioned outcome probability is $p_j=\|P_jU_1|\psi\rangle\|^2$. After that outcome, the normalized state is $P_jU_1|\psi\rangle/\sqrt{p_j}$. The <Born rule> probability of successful <postselection> is then $|\langle\psi'|U_2P_jU_1|\psi\rangle|^2/p_j$. Multiplying gives the joint probability
$$
\Pr(j,\psi'\mid\psi)=|\langle\psi'|U_2P_jU_1|\psi\rangle|^2.
$$
<Conditional probability> therefore gives the <Aharonov-Bergmann-Lebowitz rule>:
$$
\boxed{\Pr(j\mid\psi,\psi',\{P_k\})=
\frac{|\langle\psi'|U_2P_jU_1|\psi\rangle|^2}
{\sum_k|\langle\psi'|U_2P_kU_1|\psi\rangle|^2}.}
$$
The denominator must be positive; otherwise the selected subensemble does not occur. Define the forward-evolved ket $|a\rangle=U_1|\psi\rangle$ and backward-evolved ket $|b\rangle=U_2^\dagger|\psi'\rangle$. The numerator becomes $|\langle b|P_j|a\rangle|^2$, which is unchanged by interchanging $a,b$. Equivalently, with $\rho_a=|a\rangle\langle a|$ and $\rho_b=|b\rangle\langle b|$,
$$
\Pr(j\mid a,b)=\frac{\operatorname{Tr}(\rho_bP_j\rho_aP_j)}{\sum_k\operatorname{Tr}(\rho_bP_k\rho_aP_k)}.
$$
This expresses the boundary-state symmetry explicitly. It follows from the ordinary time-asymmetric preparation, <Born rule>, and state update; it does not posit an additional backward dynamical collapse.
Restore the post-selected vector omitted entirely from the TeX aid by reading the original PDF. Write $D=N^2-N+1$ and
$$
|\psi'\rangle=\frac{\sum_{i=1}^N|i\rangle-(N-1)|N+1\rangle}{\sqrt D}.
$$
Its norm is one because $N+(N-1)^2=D$. For the uniform prestate and $H=0$, put $c=1/\sqrt{(N+1)D}$. The individual transition amplitudes are
$$
\langle\psi'|P_i|\psi\rangle=c\quad(1\leq i\leq N),\qquad
\langle\psi'|P_{N+1}|\psi\rangle=-(N-1)c,
\qquad \langle\psi'|\psi\rangle=c.
$$
In experiment $E_i$, the complement amplitude is $c-c=0$. Thus for every $i\leq N$,
$$
\boxed{\Pr(P_i\mid E_i,\psi,\psi')=1,\qquad
\Pr(I-P_i\mid E_i,\psi,\psi')=0.}
$$
The successful <postselection> rate in this experiment is $c^2=1/((N+1)D)$.
In the fully resolved experiment $E_0$, the <ABL rule> squares the individual amplitudes before summing. Their squared sum is $Dc^2=1/(N+1)$, giving
$$
\boxed{\Pr(P_i\mid E_0,\psi,\psi')=\frac1D\quad(1\leq i\leq N),\qquad
\Pr(P_{N+1}\mid E_0,\psi,\psi')=\frac{(N-1)^2}{D}.}
$$
These probabilities sum to one. For $N=1$ they reduce to a single certain outcome in either <quantum measurement>. For $N\geq2$, the <N-box pre- and post-selection paradox> is that each separate binary question can be answered affirmatively with certainty, although the fully resolved <quantum measurement> cannot give all those outcomes at once.
There is no inconsistency. In $E_i$, the unresolved complement preserves coherent cancellation between its basis contributions under the <Lüders rule>. In $E_0$, those alternatives are resolved, so their squared amplitudes add instead. The different <projective measurements> disturb the state differently and have different <postselection> success rates. Merely merging the recorded $E_0$ outcomes afterwards does not reproduce $E_i$. This is the <context dependence of pre- and post-selected measurements>; certainties in mutually alternative experiments do not describe simultaneous measurement-independent properties.
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