# Is degeneracy of eigenvalues required for the Kochen-Specker theorem?

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?

• 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 $C$?
– Plop
Jan 6, 2023 at 8:30
• @Plop I'm referring to the Kochen-Specker theorem as stated in the paper I linked to, Eq. (6). Jan 6, 2023 at 9:25
• Then I don't understand your question: this statement does not talk about operators at all.
– Plop
Jan 6, 2023 at 9:43
• @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. Jan 6, 2023 at 14:23
• @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. Jan 7, 2023 at 8:55

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 $$KS$$.