Proof of Kochen--Specker Theorem: Conversion of Product Rule to ...

CHIN. PHYS. LETT. Vol. 26, No. 7 (2009) 070305

Proof of Kochen–Specker Theorem: Conversion of Product Rule to Sum Rule

*

S.P.Toh1,2** , Hishamuddin Zainuddin2 1

Faculty of Applied Science, Inti International University College, Persiaran Perdana BBN, Putra Nilai, 71800 Nilai, Negeri Sembilan, Malaysia 2 Laboratory of Computational Science and Informatics, Institute for Mathematical Research, Universiti Putra Malaysia, 43400 UPM Serdang, Selangor, Malaysia

(Received 31 March 2009) Valuation functions of observables in quantum mechanics are often expected to obey two constraints called the sum rule and product rule. However, the Kochen–Specker (KS) theorem shows that for a Hilbert space of quantum mechanics of dimension 𝑑 ≥ 3, these constraints contradict individually with the assumption of value definiteness. The two rules are not irrelated and Peres [Found. Phys. 26 (1996) 807] has conceived a method of converting the product rule into a sum rule for the case of two qubits. Here we apply this method to a proof provided by Mermin based on the product rule for a three-qubit system involving nine operators. We provide the conversion of this proof to one based on sum rule involving ten operators.

PACS: 03. 65. Ta Quantum mechanical states impose statistical restrictions on the results of measurements. Some physicists think that the restriction is due to the incomplete description of the quantum system by the states and hence they propose hidden variable theories in the hope that these would give a complete description. The Kochen–Specker (KS) theorem provides a strong argument showing that even if hidden variables do exist and can be used to interpret quantum mechanics, the value assignments made must be contextual. A value of an observable is said to be contextual if the value measured depends on measurement context. Apart from noncontextuality, value definiteness is another old belief held by physicists which states that all (compatible) observables of a physical system have definite values at all times. Unfortunately, applying noncontextuality and value definiteness to a quantum mechanical system would give rise to a contradiction.[1−3] In quantum mechanics, observables are represented by Hermitian operators that have real eigenvalues. Quantum mechanics requires that the results of measuring an observable be eigenvalues of the corresponding Hermitian operator. Quantum mechanics further requires that if observables 𝐴, 𝐵, 𝐶, . . . belong to mutually commuting subsets of the observables and satisfy 𝑓 (𝐴, 𝐵, 𝐶, . . .) = 0, then the only allowed results of a simultaneous measurement of 𝐴, 𝐵, 𝐶, . . . are the set of simultaneous eigenvalues 𝑣(𝐴), 𝑣(𝐵), 𝑣(𝐶), . . . constrained by 𝑓 (𝑣(𝐴), 𝑣(𝐵), 𝑣(𝐶), . . .) = 0. In particular, this can be expressed as a sum rule: (a) If 𝐴, 𝐵, 𝐶 are the compatible observables and 𝐶 = 𝐴 + 𝐵, then 𝑣(𝐶) = 𝑣(𝐴) + 𝑣(𝐵). Alternatively, one can have the product rule, i.e. (b) if 𝐴, 𝐵, 𝐶 are the compatible observables and 𝐶 = 𝐴 · 𝐵, then

𝑣(𝐶) = 𝑣(𝐴) · 𝑣(𝐵).[1] The KS theorem states that for Hilbert space of quantum mechanical state vectors of dimension > 2, assumptions of value definiteness contradict either the sum rule or product rule. To avoid contradiction and to maintain the value definiteness, the measured values of observables must be contextual. Consider a pair of spin-1/2 particles and their spin observables. The following magic square in Table 1 consists of nine tensor-product spin operators given by Mermin.[3] As usual, 𝜎𝑥 , 𝜎𝑦 , 𝜎𝑧 are the Pauli matrices related to the individual spin observables. Table 1. Mermin’s magic square. 𝐼 ⊗ 𝜎𝑧 𝜎𝑥 ⊗ 𝐼 𝜎𝑥 ⊗ 𝜎𝑧

𝜎𝑧 ⊗ 𝐼 𝐼 ⊗ 𝜎𝑥 𝜎𝑧 ⊗ 𝜎𝑥

𝜎𝑧 ⊗ 𝜎𝑧 𝜎𝑥 ⊗ 𝜎𝑥 𝜎𝑦 ⊗ 𝜎𝑦

Each row and each column is a triad of commuting operators. Each entry of the operator has eigenvalue 1 or −1. We can easily check the validity of following six functions, (𝐼 ⊗ 𝜎𝑧 )(𝜎𝑧 ⊗ 𝐼)(𝜎𝑧 ⊗ 𝜎𝑧 ) = 𝐼 ⊗ 𝐼, (𝜎𝑥 ⊗ 𝐼)(𝐼 ⊗ 𝜎𝑥 )(𝜎𝑥 ⊗ 𝜎𝑥 ) = 𝐼 ⊗ 𝐼, (𝜎𝑥 ⊗ 𝜎𝑧 )(𝜎𝑧 ⊗ 𝜎𝑥 )(𝜎𝑦 ⊗ 𝜎𝑦 ) = 𝐼 ⊗ 𝐼, (𝐼 ⊗ 𝜎𝑧 )(𝜎𝑥 ⊗ 𝐼)(𝜎𝑥 ⊗ 𝜎𝑧 ) = 𝐼 ⊗ 𝐼, (𝜎𝑧 ⊗ 𝐼)(𝐼 ⊗ 𝜎𝑥 )(𝜎𝑧 ⊗ 𝜎𝑥 ) = 𝐼 ⊗ 𝐼, (𝜎𝑧 ⊗ 𝜎𝑧 )(𝜎𝑥 ⊗ 𝜎𝑥 )(𝜎𝑦 ⊗ 𝜎𝑦 ) = −𝐼 ⊗ 𝐼. According to the product rule, we have 𝑣(𝐼 ⊗ 𝜎𝑧 )𝑣(𝜎𝑧 ⊗ 𝐼)𝑣(𝜎𝑧 ⊗ 𝜎𝑧 ) = 𝑣(𝐼 ⊗ 𝐼), 𝑣(𝜎𝑥 ⊗ 𝐼)𝑣(𝐼 ⊗ 𝜎𝑥 )𝑣(𝜎𝑥 ⊗ 𝜎𝑥 ) = 𝑣(𝐼 ⊗ 𝐼), 𝑣(𝜎𝑥 ⊗ 𝜎𝑧 )𝑣(𝜎𝑧 ⊗ 𝜎𝑥 )𝑣(𝜎𝑦 ⊗ 𝜎𝑦 ) = 𝑣(𝐼 ⊗ 𝐼),

* Supported

𝑣(𝐼 ⊗ 𝜎𝑧 )𝑣(𝜎𝑥 ⊗ 𝐼)𝑣(𝜎𝑥 ⊗ 𝜎𝑧 ) = 𝑣(𝐼 ⊗ 𝐼),

by SAGA under Grant No P55c, Ministry of Science, Technology & Innovations, Malaysia. [email protected] c 2009 Chinese Physical Society and IOP Publishing Ltd ○ ** Email:

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(1)


𝑣(𝜎𝑧 ⊗ 𝐼)𝑣(𝐼 ⊗ 𝜎𝑥 )𝑣(𝜎𝑧 ⊗ 𝜎𝑥 ) = 𝑣(𝐼 ⊗ 𝐼), 𝑣(𝜎𝑧 ⊗ 𝜎𝑧 )𝑣(𝜎𝑥 ⊗ 𝜎𝑥 )𝑣(𝜎𝑦 ⊗ 𝜎𝑦 ) = −𝑣(𝐼 ⊗ 𝐼).

(2)

Each operator appears in two contexts in Eq. (1), for example, 𝐼 ⊗ 𝜎𝑧 appears in measurement contexts given in the first and fourth equations in Eq. (1). According to value definiteness, if 𝑣(𝐼 ⊗ 𝜎𝑧 ) = 1 in the first one of Eq. (2), then it must carry the same value in the fourth one of Eq. (2). However, based on the value definiteness and the product rule in Eq. (2), the product of values on both the sides of Eq. (2) will not tally. This proves the KS theorem. If the definite values of the operators are allowed to be contextual, then the contradiction will be removed. The above simple KS theorem proof shows the contradiction arising from the value definiteness in different contexts and the product rule. Peres[2] has argued that the sum rule interpretation-wise could be better. He conceived a trick to convert the product rule of Mermin’s square into a sum rule and again proved the contradiction. In this Letter, we adopt the idea of Peres and apply it to prove the KS theorem for Mermin’s pentagram for the system of three qubits. Mermin[3] constructed an eight-dimensional set of observables out of three independent spin-1/2 particles, and considered the set of ten observables as shown in Fig. 1.

σy1σy2σx3 σy1σx2σy3 σ3x

ˆ = −𝑃ˆ1 − 𝑃ˆ2 − 𝑃ˆ3 − 𝑃ˆ4 + 𝑃ˆ5 + 𝑃ˆ6 + 𝑃ˆ7 + 𝑃ˆ8 , (4) Ω where the eigenvalues 𝜆 of the ten operators in the pentagram is either 1 or −1, and 𝑃𝑖 is the mutually orthogonal projection operators on the Hilbert space. ˆ = ∑︀𝑑 𝐹 (𝜆)𝑃ˆ𝜆 where 𝑑 = 8 in our Applying 𝐹 (Ω) 𝑖=1 case, with ln(−1) = 𝑖𝜋 and ln(1) = 0, we obtain ˆ = 𝑃ˆ1 + 𝑃ˆ2 + 𝑃ˆ3 + 𝑃ˆ4 . (1/𝑖𝜋) ln(Ω)

(5)

As each term on the right of Eq. (5) is a projection ˆ is indeed a projection operator operator, (1/𝑖𝜋) ln(Ω) with eigenvalues 1 or 0. On the other hand, subtracting Eq. (4) from the ∑︀𝑑 resolution of identity matrix, 𝐼 = 𝑖=1 𝑃ˆ𝑖 we obtain ˆ = 𝑃ˆ1 + 𝑃ˆ2 + 𝑃ˆ3 + 𝑃ˆ4 . (1/2)(𝐼 − Ω)

(6)

Finally, we have ˆ = 𝑃ˆ1 + 𝑃ˆ2 + 𝑃ˆ3 + 𝑃ˆ4 = (1/2)(𝐼 − Ω). ˆ (7) (1/𝑖𝜋) ln(Ω)

σy1 σx1 σx2 σx3

basis in which the operator is diagonal, take the logarithms of its eigenvalues, and then transform it back in the original basis. If the eigenvalues are 1 and −1, they become 0 and 𝑖𝜋, respectively. Dividing the result by 𝑖𝜋 then we finally obtain a projection operator (see the proof below). We take the logarithm of an ˆ divided by 𝑖𝜋 tantamount to changing Ω ˆ operator Ω ˆ into (1/2)(𝐼 − Ω), which is a projection operators, as shown in the following.[2] By spectral decomposition, we have

Now we take 𝜎 1𝑦 as an example, where (︂ )︂ 04 −𝑖𝐼4 𝜎 1𝑦 = , 𝑖𝐼4 04

σx1σy2σy3

σ3y σ

1 x

σ2y

σ2x

Fig. 1. Mermin’s pentagram.

The four operators on each line of the pentagram are mutually commuting. It is easily shown that (𝜎 1𝑦 )(𝜎 2𝑦 )(𝜎 3𝑥 )(𝜎 1𝑦 𝜎 2𝑦 𝜎 3𝑥 )

(𝜎 1𝑦 )(𝜎 2𝑥 )(𝜎 3𝑦 )(𝜎 1𝑦 𝜎 2𝑥 𝜎 3𝑦 ) = 𝐼, (𝜎 1𝑥 )(𝜎 2𝑦 )(𝜎 3𝑦 )(𝜎 1𝑥 𝜎 2𝑦 𝜎 3𝑦 ) = 𝐼, (𝜎 1𝑥 )(𝜎 2𝑥 )(𝜎 3𝑥 )(𝜎 1𝑥 𝜎 2𝑥 𝜎 3𝑥 ) = 𝐼, (𝜎 1𝑥 𝜎 2𝑦 𝜎 3𝑦 )(𝜎 1𝑦 𝜎 2𝑥 𝜎 3𝑦 )(𝜎 1𝑦 𝜎 2𝑦 𝜎 3𝑥 )(𝜎 1𝑥 𝜎 2𝑥 𝜎 3𝑥 ) = −𝐼.

with the subscript denoting the size of block square matrices. Its eigenvalues are −1 or 1. Taking their corresponding eigenvectors as columns, we obtain the unitary matrix ⎛ −1 0 ⎞ 0 0 0 0 0 −1 ⎜ ⎜ ⎜ 1 ⎜ 𝑈=√ ⎜ 2⎜ ⎜ ⎜ ⎝

= 𝐼,

(3)

Mermin[3] again proved the KS theorem by similarly showing the contradiction arising from adopting value definiteness in different contexts and the product rule. Adopting the idea of Peres,[2] we will show in the following the contradiction from value definiteness in different contexts with the corresponding sum rule. The first step towards converting product rule into sum rule is to take the logarithms of each operator in the pentagram. To this end, we first transform it to a

(8)

0 0 0 𝑖 0 0 0

0 −1 0 0 0 𝑖 0

−𝑖 0 0 0 −1 0 0

0 0 −𝑖 0 0 0 −1

−𝑖 0 0 0 1 0 0

0 −1 0 0 0 −𝑖 0

0 0 −𝑖 0 0 0 1

0 ⎟ 0 ⎟ ⎟ 0 ⎟ ⎟. −𝑖 ⎟ ⎟ 0 ⎟ ⎠ 0 0 (9)

We perform the transformation 𝑈 −1 𝜎 1𝑦 𝑈 on 𝜎 1𝑦 and then turn it into a diagonal matrix. Taking its logarithm and dividing by 𝑖𝜋 will produce another diagonal matrix 𝐵. Finally, transforming it back in the original basis by 𝑈 𝐵𝑈 −1 one obtains (︂ )︂ 𝐼4 𝑖𝐼4 1 (1/𝑖𝜋) ln(𝜎 𝑦 ) = (1/2) , (10) −𝑖𝐼4 𝐼4 which is a projection operator. By virtue of Eq. (7), this matrix can also be obtained by taking summation of four projection operators constructed from the

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𝑣(𝑃ˆ𝜎1𝑦 ) + 𝑣(𝑃ˆ𝜎2𝑥 ) + 𝑣(𝑃ˆ𝜎3𝑦 ) + 𝑣(𝑃ˆ𝜎1𝑦 𝜎2𝑥 𝜎3𝑦 ),

outer product of eigenvectors whose eigenvalue is −1 or through (1/2)(𝐼 ⊗ 𝐼 ⊗ 𝐼 − 𝜎 1𝑦 ⊗ 𝐼 ⊗ 𝐼). Using the steps sketched above, we can now construct ten projection operators by three different means, and from these we can form five summations according to the five lines of pentagram in Fig. 1, namely, (1/𝑖𝜋)[ln(𝜎 1𝑦 + ln(𝜎 2𝑦 ) + ln(𝜎 3𝑥 ) + ln(𝜎 1𝑦 𝜎 2𝑦 𝜎 3𝑥 ))], (1/𝑖𝜋)[ln(𝜎 1𝑦 ) + ln(𝜎 2𝑥 + ln(𝜎 3𝑦 ) + ln(𝜎 1𝑦 𝜎 2𝑥 𝜎 3𝑦 ))], (1/𝑖𝜋)[ln(𝜎 1𝑥 ) + ln(𝜎 2𝑥 ) + ln(𝜎 3𝑥 ) + ln(𝜎 1𝑥 𝜎 2𝑥 𝜎 3𝑥 )], (1/𝑖𝜋)[ln(𝜎 1𝑥 ) + ln(𝜎 2𝑦 + ln(𝜎 3𝑦 ) + ln(𝜎 1𝑥 𝜎 2𝑦 𝜎 3𝑦 ))], (1/𝑖𝜋)[ln(𝜎 1𝑥 𝜎 2𝑦 𝜎 3𝑦 ) + ln(𝜎 1𝑦 𝜎 2𝑥 𝜎 3𝑦 ) + ln(𝜎 1𝑦 𝜎 2𝑦 𝜎 3𝑥 ) + ln(𝜎 1𝑥 𝜎 2𝑥 𝜎 3𝑥 )],

(11)

𝑃ˆ𝜎1𝑦 + 𝑃ˆ𝜎2𝑦 + 𝑃ˆ𝜎3𝑥 + 𝑃ˆ𝜎1𝑦 𝜎2𝑦 𝜎3𝑥 , 𝑃ˆ𝜎1𝑦 + 𝑃ˆ𝜎2𝑥 + 𝑃ˆ𝜎3𝑦 + 𝑃ˆ𝜎1𝑦 𝜎2𝑥 𝜎3𝑦 , 𝑃ˆ𝜎1𝑥 + 𝑃ˆ𝜎2𝑥 + 𝑃ˆ𝜎3𝑥 + 𝑃ˆ𝜎1𝑥 𝜎2𝑥 𝜎3𝑥 , 𝑃ˆ𝜎1𝑥 + 𝑃ˆ𝜎2𝑦 + 𝑃ˆ𝜎3𝑦 + 𝑃ˆ𝜎1𝑥 𝜎2𝑥 𝜎3𝑦 , 𝑃ˆ𝜎1𝑥 𝜎2𝑦 𝜎3𝑦 + 𝑃ˆ𝜎1𝑦 𝜎2𝑥 𝜎3𝑦 + 𝑃ˆ𝜎1𝑦 𝜎2𝑦 𝜎3𝑥 + 𝑃ˆ𝜎1𝑥 𝜎2𝑥 𝜎3𝑥 ,

(12)

or (1/2)[4𝐼 − (𝜎 1𝑦 𝜎 2𝑦 + 𝜎 3𝑥 + 𝜎 1𝑦 𝜎 2𝑦 𝜎 3𝑥 )], (1/2)[4𝐼 − (𝜎 1𝑦 + 𝜎 2𝑥 + 𝜎 3𝑦 + 𝜎 1𝑦 𝜎 2𝑥 𝜎 3𝑦 )], (1/2)[4𝐼 − (𝜎 1𝑥 + 𝜎 2𝑥 + 𝜎 3𝑥 + 𝜎 1𝑥 𝜎 2𝑥 𝜎 3𝑥 )], (1/2)[4𝐼 − (𝜎 1𝑥 + 𝜎 2𝑦 + 𝜎 3𝑦 + 𝜎 1𝑥 𝜎 2𝑦 𝜎 3𝑦 )], (1/2)[4𝐼 − (𝜎 1𝑥 𝜎 2𝑦 𝜎 3𝑦 + 𝜎 1𝑦 𝜎 2𝑥 𝜎 3𝑦 + 𝜎 1𝑦 𝜎 2𝑦 𝜎 3𝑥 + 𝜎 1𝑥 𝜎 2𝑥 𝜎 3𝑥 )].

(13)

In Eqs. (11), (12) and (13), each of the summations gives rise to a matrix whose eigenvalues are 0, 2, 4, apart from the fourth matrix which carries eigenvalues 1, 3. Hence, the sum of eigenvalues of each set of matrices must be an odd number. Based on the sum rule and Eq. (12), we have the valuations 𝑣(𝑃ˆ𝜎1𝑦 ) + 𝑣(𝑃ˆ𝜎2𝑦 ) + 𝑣(𝑃ˆ𝜎3𝑥 ) + 𝑣(𝑃ˆ𝜎1𝑦 𝜎2𝑦 𝜎3𝑥 ),

𝑣(𝑃ˆ𝜎1𝑥 ) + 𝑣(𝑃ˆ𝜎2𝑥 ) + 𝑣(𝑃ˆ𝜎3𝑥 ) + 𝑣(𝑃ˆ𝜎1𝑥 𝜎2𝑥 𝜎3𝑥 ), 𝑣(𝑃ˆ𝜎1 ) + 𝑣(𝑃ˆ𝜎2 ) + 𝑣(𝑃ˆ𝜎3 ) + 𝑣(𝑃ˆ𝜎1 𝜎2 𝜎3 ), 𝑥

𝑦

𝑦

𝑥

𝑥

𝑦

𝑣(𝑃ˆ𝜎1𝑥 𝜎2𝑦 𝜎3𝑦 ) + 𝑣(𝑃ˆ𝜎1𝑦 𝜎2𝑥 𝜎3𝑦 ) + 𝑣(𝑃ˆ𝜎1𝑦 𝜎2𝑦 𝜎3𝑥 ) + 𝑣(𝑃ˆ𝜎1𝑥 𝜎2𝑥 𝜎3𝑥 ).

(14)

The values that can be assigned for each projection operator in (14) are either 0 or 1. For each summation in Eq. (14), one and only one projection operator can be assigned the value 1. Lastly, due to assumption of value definiteness, those projection operators that have been assigned value 1(0) in a context (summation it appears), they must carry the same value 1(0) in other contexts. Restricted by these constraints, it is obvious that the sum of values assigned in Eq. (14) must be an even number, as each value function 𝑣(·) of the operator appears twice. This brings us to the contradiction and proves the KS theorem. If the definite values of the assignment are allowed to be contextual, the contradiction can be removed. In summary, we have explained the proof of the KS theorem for Mermin’s pentagram using the sum rule instead of the product rule as carried out by Mermin.[3] The conversion of the product rule into the sum rule by Peres[2] is easily generalized for threequbit system albeit in a more complex case. The example here amply demonstrates that Peres’ conversion trick is not tied to the nine operators in fourdimensional space given in Ref. [2]. We thank L.C. Kwek for his encouragement and support in this work.

References [1] Held C 2008 Stanford Encyclopedia of Philosophy URL http://plato.stanford.edu/archives/win2008/entries/kochenspecker/ [2] Peres A 1996 Found. Phys. 26 807 [3] Mermin N D 1993 Rev. Mod. Phys. 65 803

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