Quantum measurenment for incomplete basis?

Question

Standard view of quantum measurement requires a complete basis of possibilities, however some of them might make no sense - can we just normalize probability without these possibilities? To better express this question, let me describe the situation where I got it.

In page 9 of this Preskill lecture we can find simple form of Bell inequalities for three binary variables: $$ P(A=B) + P(A=C) + P(B=C) \geq 1 $$ which is kind of obvious: "This is satisfied by any probability distribution for the three coins because no matter what the values of the coins, there will always be two that are the same." This lecture also contains example of state for which QM gives P(A=B) + P(A=C) + P(B=C) =3/4 violation.

For Bell violation by MERW, I wanted to construct an example of violation of this inequality using only real non-negative amplitudes, what lead me to this candidate ($|ABC\rangle$): $$\psi=(|001\rangle + |010\rangle +|100\rangle +|011\rangle +|101\rangle +|110\rangle)/\sqrt{6}$$ The question is: what is the probability distribution when measuring two out of three such variables?

The interesting basis for measuring first two variables (AB, taking trace over C), is: $$n_{00}=(|000\rangle+|001\rangle) /\sqrt{2}\qquad n_{01}=(|010\rangle+|011\rangle) / \sqrt{2}$$ $$n_{10}=(|100\rangle+|101\rangle ) / \sqrt{2}\qquad n_{11}=(|110\rangle+|111\rangle) / \sqrt{2}$$ getting $$|\langle n_{00}|\psi\rangle|^2=1/12\quad |\langle n_{01}|\psi\rangle|^2=4/12 \quad |\langle n_{10}|\psi\rangle|^2=4/12 \quad |\langle n_{11}|\psi\rangle|^2=1/12 $$ However, as this basis is not complete, instead of 1, these four possibilities sum up to 10/12.

The remaining (1+1)/12 are e.g. in $n'_{00}=(|000\rangle-|001\rangle) /\sqrt{2}$, which seems to make no sense (?) while measuring just first two variables. If it does, wheat would such state mean?

Can we normalize dividing by this 10/12?

Getting probability of 00 as 1/10, and finally violation to $P(A=B) + P(A=C) + P(B=C) = 6/10$?

Answer 1

Generally you can't just drop out certain states without sufficient justification (like in case of superselection sectors). And in you case "makes no sense" signifies more your misunderstanding, I'm afraid.

The eigenspace of the observables may have more than one dimension. I.e. there can actually be two (and more) linearly independent eigenvectors $|\lambda,n\rangle$ corresponding to the same eigenvalue $\lambda$. The probability then is given by, \begin{equation} P(\Lambda=\lambda)=\sum_n |\langle\lambda,n|\psi\rangle|^2=\langle\psi|\hat{P}_{\lambda}|\psi\rangle,\quad \hat{P}_{\lambda}=\sum_n |\lambda,n\rangle\langle\lambda,n| \end{equation} where $\hat{P}_{\lambda}$ is a projection operator on the corresponding eigenspace.

In case of the multipartite systems the observables relevant only to the one subsystems (like your case of three subsystems $A$,$B$ and $C$) can be written as, \begin{equation} \hat{O}=\hat{O}_A\otimes \hat{\mathbb{I}}_{BC} \end{equation} where $\hat{\mathbb{I}}$ is simply a unit operator. If $\hat{O}_A$ has eigenvectors $|\lambda_n\rangle$ than eigenvectors for the full $\hat{O}$ are, \begin{equation} |\lambda_n,e_m\rangle=|\lambda_n\rangle\otimes |e_m\rangle_{BC} \end{equation} where $|e_m\rangle_{BC}$ are vectors from the arbitrary full basis in $BC$. So the probability is given instead by, \begin{equation} P_{\lambda_n}=\sum_m |\langle \lambda_n,e_m|\psi\rangle|^2 \end{equation} The formula is connected to the properly done partial trace in the density matrix formalism (which converts entangled pure states into mixed reduced density matrices for subsystems)

So in your example both $n_{ab}$ and $n'_{ab}$ will be eigenvectors for any observable constructed from $\hat{A}$ and $\hat{B}$. They can't be distinguished by measuring only those two observables however that doesn't mean that any one of them can be omitted. The probability thus given by, \begin{equation} P(a,b)=|\langle n_{ab}|\psi\rangle|^2+|\langle n'_{ab}|\psi\rangle|^2 \end{equation}

And you could guess it from the start if instead considered, \begin{equation} \frac{1}{\sqrt{2}}n_{ab}+\frac{1}{\sqrt{2}}n'_{ab}=|ab0\rangle,\quad \frac{1}{\sqrt{2}}n_{ab}-\frac{1}{\sqrt{2}}n'_{ab}=|ab1\rangle \end{equation}