Objectivity in Quantum Measurement

Abstract

The objectivity is a basic requirement for the measurements in the classical world, namely, different observers must reach a consensus on their measurement results, so that they believe that the object exists “objectively” since whoever measures it obtains the same result. We find that this simple requirement of objectivity indeed imposes an important constraint upon quantum measurements, i.e., if two or more observers could reach a consensus on their quantum measurement results, their measurement basis must be orthogonal vector sets. This naturally explains why quantum measurements are based on orthogonal vector basis, which is proposed as one of the axioms in textbooks of quantum mechanics. The role of the macroscopicality of the observers in an objective measurement is discussed, which supports the belief that macroscopicality is a characteristic of classicality.

A Proof for the Proposition 1

Proposition 1

For a tripartite density matrix $\rho _{ABC}$, if its reduced matrices $\rho _{AB}=\mathrm {tr}_{C}[\rho _{ABC}]$ and $\rho _{AC}=\mathrm {tr}_{B}[\rho _{ABC}]$ have the forms of

$$\begin{aligned} \rho _{AB}&=\sum _{n}p_{n}|\mathsf {a}_{n},\,\mathsf {b}_{n}\rangle \langle \mathsf {a}_{n},\,\mathsf {b}_{n}|, \end{aligned}$$

(20)

$$\begin{aligned} \rho _{AC}&=\sum _{n}p_{n}|\mathsf {a}_{n},\,\mathsf {c}_{n}\rangle \langle \mathsf {a}_{n},\,\mathsf {c}_{n}|, \end{aligned}$$

(21)

then there exists an orthonormal vector set $\{|\varPhi _{i}\rangle \}$, such that the tripartite $\rho _{ABC}$ can be written as

$$\begin{aligned} \rho _{ABC}&=\sum _{i}\lambda _{i}|\varPhi _{i}\rangle \langle \varPhi _{i}|,\qquad \lambda _{i}\ge 0\nonumber \\ |\varPhi _{i}\rangle&=\sum _{n}\mathsf {C}_{n}^{(i)}|\mathsf {a}_{n},\,\mathsf {b}_{n},\,\mathsf {c}_{n}\rangle . \end{aligned}$$

(22)

Here $\{|\mathsf {a}_{n}\rangle \}$, $\{|\mathsf {b}_{n}\rangle \}$ and $\{|\mathsf {c}_{n}\rangle \}$ are complete basis sets for the Hilbert space ${{\mathcal {H}}}_{A}$, ${{\mathcal {H}}}_{B}$ and $\mathcal{H}_{C}$ respectively, but not necessarily orthogonal ones.

For clarity, we use $A,\,B,\,C$ here to replace the $S,\,D,\,D'$ in the main text. To prove this proposition, we need the following lemma:

Lemma

Let $\mathbf {P}$ be a positive definite matrix and $\mathbf {C}$ a semi-positive one. If $\mathrm{tr}[\mathbf {C}\cdot \mathbf {P}]=0$, then $\mathbf {C}$ is a zero matrix.

Proof

We decompose the positive matrix $\mathbf {P}$ in its eigen basis as $\mathbf {P}=\sum _{n}\lambda _{n}|n\rangle \langle n|$, where all $\lambda _{n}>0$. Then we have $\mathrm {tr}[\mathbf {C}\cdot \mathbf {P}]=\sum _{n}\lambda _{n}\langle n|\mathbf {C}|n\rangle =0$. To make sure $\langle n|\mathbf {C}|n\rangle =0$ for all the basis $\{|n\rangle \}$, $\mathbf {C}$ must be a zero matrix. $\square$

With the help of the above lemma, the proof of Proposition 1 lies as follows.

Proof

For the tripartite density matrix $\rho _{ABC}$, we can always write it as the eigen spectrum decomposition $\rho _{ABC}=\sum _{i}\lambda _{i}|\varPhi _{i}\rangle \langle \varPhi _{i}|$, where $|\varPhi _{i}\rangle$ are orthonormal basis, and $\lambda _{i}>0$ are the non-zero eigenvalues respectively. But now we could only write down $|\varPhi _{i}\rangle$ in a general form

$$\begin{aligned} |\varPhi _{i}\rangle =\sum _{m=1}^{M}\sum _{n=1}^{N}\sum _{l=1}^{L}\mathsf {C}_{mnl}^{(i)}|\mathsf {a}_{m},\,\mathsf {b}_{n},\,\mathsf {c}_{l}\rangle , \end{aligned}$$

(23)

where $\mathsf {C}_{nml}^{(i)}$ are complex numbers. It then follows that

$$\begin{aligned}&\rho _{ABC}=\sum _{\begin{array}{c} mnl \\ m'n'l' \end{array}}\varrho _{mnl,\,m'n'l'}|\mathsf {a}_{m},\mathsf {b}_{n},\mathsf {c}_{l}\rangle \langle \mathsf {a}_{m'},\mathsf {b}_{n'},\mathsf {c}_{l'}|,\nonumber \\&\quad \varrho _{mnl,\,m'n'l'}:=\sum _{i}\lambda _{i}\cdot \mathsf {C}_{mnl}^{(i)}\overline{\mathsf {C}}_{m'n'l'}^{(i)}, \end{aligned}$$

(24)

and the reduced density matrix $\rho _{AB}$ becomes

$$\begin{aligned} \rho _{AB}=\mathrm{Tr}_{C}[\rho _{ABC}]=\sum _{\begin{array}{c} m,n \\ m',n' \end{array}}\Big (\sum _{l,l'}\varrho _{mnl,\,m'n'l'}\langle \mathsf {c}_{l'}|\mathsf {c}_{l}\rangle \Big )|\mathsf {a}_{m},\mathsf {b}_{n}\rangle \langle \mathsf {a}_{m'},\mathsf {b}_{n'}|. \end{aligned}$$

(25)

Comparing this with the required form of $\rho _{AB}$ [Eq. (20)], we come to the following equation

$$\begin{aligned} \sum _{l,l'}\varrho _{mnl,\,m'n'l'}\langle \mathsf {c}_{l'}|\mathsf {c}_{l}\rangle =\delta _{mm'}\delta _{nn'}\cdot \delta _{mn}p_{n}. \end{aligned}$$

(26)

Now we introduce two $L\times L$ matrices $\mathbf {C}^{(mn;m'n')}$ and $\mathbf {P}$, which are defined by

$$\begin{aligned}{}[\mathbf {C}^{(mn;m'n')}]_{l,l'}=\varrho _{mnl,\,m'n'l'},\quad \mathbf {P}_{l',l}=\langle \mathsf {c}_{l'}|\mathsf {c}_{l}\rangle . \end{aligned}$$

(27)

With their help, Eq. (26) can be written in a compact form

$$\begin{aligned} \mathrm {tr}[\mathbf {C}^{(mn;m'n')}\cdot \mathbf {P}]=\delta _{mm'}\delta _{nn'}\cdot \delta _{mn}p_{n}. \end{aligned}$$

(28)

One notices that when $m=m'$, $n=n'$, $m\ne n$, we have

$$\begin{aligned} \mathrm {tr}[\mathbf {C}^{(mn;mn)}\cdot \mathbf {P}]=0. \end{aligned}$$

(29)

It is easy to verify that $\mathbf {C}^{(mn;mn)}$ is a semi-positive matrix,¹ and $\mathbf {P}$ is positive definite.² Therefore, according to above lemma, we know that $\mathbf {C}^{(mn;mn)}$ is a zero matrix when $m\ne n$. Thus we obtain

$$\begin{aligned}{}[\mathbf {C}^{(mn;mn)}]_{l,l}=\varrho _{mnl,\,mnl}=\sum _{i}\lambda _{i}\cdot |\mathsf {C}_{mnl}^{(i)}|^{2}=0. \end{aligned}$$

(30)

Since all the $\lambda _{i}>0$ in the above summation, that leads to

$$\begin{aligned} \mathsf {C}_{mnl}^{(i)}=0,\quad \forall \,i,\,l,\,m\ne n. \end{aligned}$$

(31)

In the same way, by comparing with $\rho _{AC}$ [Eq. (21)], we can prove

$$\begin{aligned} \mathsf {C}_{mnl}^{(i)}=0,\quad \forall \,i,\,n,\,m\ne l. \end{aligned}$$

(32)

Therefore, the only possible non-zero coefficients $\mathsf {C}_{mnl}^{(i)}$ are those satisfying $m=n=l$, thus, we write the coefficients as $\mathsf {C}_{mnl}^{(i)}=\delta _{mn}\delta _{ml}\cdot \mathsf {C}_{n}^{(i)}$, then we obtain the expression

$$\begin{aligned} |\varPhi _{i}\rangle =\sum _{n}\mathsf {C}_{n}^{(i)}|\mathsf {a}_{n},\,\mathsf {b}_{n},\,\mathsf {c}_{n}\rangle \end{aligned}$$

(33)

and complete the proof. $\square$