I'm wondering why the operators for the Kochen-Specker theorem are 3-dimensional while they only produce two eigenvalues $\{0,1\}$. Is this degeneracy always needed regardless of the dimensionality of the Hilbert space or is it an artifact in the particular case of $d = 3$ (since there is no Kochen-Specker set when $d = 2$)?

If the former holds, then why is degeneracy needed?

  • $\begingroup$ What exact formal statement do you call the "Kochen-Specker" theorem? Sideways, do you know that if $A$, $B$, $C$ are hermitian matrices having only simple eigenvalues, and if $A$ commutes with $B$ and $B$ commutes with $C$, then $A$ commutes with $C$? $\endgroup$
    – Plop
    Jan 6, 2023 at 8:30
  • $\begingroup$ @Plop I'm referring to the Kochen-Specker theorem as stated in the paper I linked to, Eq. (6). $\endgroup$
    – Tfovid
    Jan 6, 2023 at 9:25
  • $\begingroup$ Then I don't understand your question: this statement does not talk about operators at all. $\endgroup$
    – Plop
    Jan 6, 2023 at 9:43
  • $\begingroup$ @Plop $\vec{u}$, $\vec{v}$, and $\vec{w}$ are mutually orthogonal vectors corresponding to rank-1 projectors. These vectors are 3-dimensional since they live in a 3-dimensional Hilbert space but the eigenvalues that result from the measurements are 0 and 1. $\endgroup$
    – Tfovid
    Jan 6, 2023 at 14:23
  • 1
    $\begingroup$ @Plop Vectors are equivalent to rank-1 projectors in this context. Also, tone down your attitude with "it's your choice". That kind of tone is a cancer in SE. $\endgroup$
    – Tfovid
    Jan 7, 2023 at 8:55

1 Answer 1


I'm not sure if this answers the question.

Let $SA$ denote the set of self-adjoint operators on a finite-dimensional Hilbert space. A subset $O$ of $SA$ is said to be (multiplicatively) $KS$ if there are no nonzero $v : O \rightarrow \mathbb{R}$ such that

  • for all $A,B \in O$, if $A$ and $B$ commute and if $AB \in O$, then $v(AB) = v(A)v(B)$;
  • for all $A,B \in O$, if $A$ and $B$ commute and if $A+B \in O$, then $v(A+B) = v(A)+v(B)$;

The statement you quote is equivalent to saying that the set of rank-1 orthogonal projectors is $KS$. Indeed, lany $v$ must map any orthogonal projector to $0$ or $1$; at least a projector is mapped to $1$ so the identity has nonzero idempotent image, so the identity is mapped to one, etc. The other direction is also easy: any map from vectors gives a map on rank-$1$ projectors.

Lemma 1: Let $O$ be a subset of pairwise commuting operators that contains a nonzero operator. Then $O$ is not $KS$.

Proof: Let $\phi$ be any common eigenvector of all elements of $O$ which correspond to a nonzero eigenvalue of at least one operator in $O$. Then the map $v$ sending each $A \in O$ to the eigenvalue of $\phi$ in $A$ satisfies the needed requirements.

Lemma 2: Let $O_1$ and $O_2$ be two non-$KS$ subsets, such that no element of $O_1$ commutes with no element of $O_2$. Then $O_1 \cup O_2$ is not $KS$.

Proof: Let $v_1$ and $v_2$ with the needed requirements, on $O_1$ and $O_2$. Then $v_1 \cup v_2$ (the common extension of $v_1$ and $v_2$ to $O_1 \cup O_2$) vacuously satisfies the needed requirements.

Lemma 3: Let $O$ be a subset of $SA$ such that the commutativity relation is an equivalence relation. Then $O$ is not $KS$.

Proof: Partition $O$ into classes: each class is non-$KS$ by Lemma 1, and so their union is non-$KS$ by (using inductively) Lemma 2.

Lemma 4: If $A$, $B$, $C$ are self-adjoint operators with simple spectrum, if $A$ commutes with $B$, if $B$ commutes with $C$, then $A$ commutes with $C$.

Proof: By the assumption of simplicity of spectrum, each diagonalizing basis for either one of $A$, $B$ and $C$ is essentially unique, that is, two diagonalizing bases are equal up to reordering their vectors and multiplying each vector by a complex number. Since $A$ and $B$ commute, they share a common diagonalization basis, and the same is true for $B$ and $C$. By what I've just said, we can assume that the two bases are the same. So $A$ and $C$ share a common diagonalization basis, so they commute.

Theorem: Any $KS$ subset must contain an operator with degenerate spectrum.

Proof: By Lemma 4, any subset $O$ containing only simple-spectrum operators is such that the commutativity relation is transitive, and therefore, an equivalence relation. Then, by Lemma 3, $O$ is not $KS$.


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