The spectral theorem states if $A$ is a Hermitian operator acting on an $n-$dimensional Hilbert space space $H$, and $\lambda_1, ... \lambda_m$ are $m \leq n$ distinct eigenvalues of $A$, then

$$ H=\oplus_{i=1}^{m} H_i,$$

where $H_i$ are the corresponding eigenspaces, and in turn

$$ A = \sum_{i=1}^{m}\lambda_i P_i,$$

where $P_i$-s are the corresponding projections. It is known that the spectrum of a Hermitian operator acting on a finite-dimensional Hilbert-space can be computed provided that its cardinality is known beforehand, or, alternatively, that for each pair of eigenvalues $\lambda_i, \lambda_j$, it is known whether $\lambda_i = \lambda_j$ or $\lambda_i \neq \lambda_j$. It is a crucial condition when we try to build a computable foundation of quantum mechanics since deciding which eigenvalue is actually measured essentially defines what eigenspace the system falls into after the measurement. The problem of degeneracy is described in Bridges and Svozil, 2000 in the following example:

enter image description here

Under this link, I came across this book of Ye where he explains the degeneracy problem as follows:

The representation in [when $\forall i,j. \lambda_i \neq \lambda_j \lor \lambda_i = \lambda_j$] above is exactly the same as the classical case. It seems that indistinguishable $\lambda_i$-s arise only in artificial constructions. In other words, we expect that the operators on finite-dimensional spaces appearing in natural physics contexts all satisfy the condition $\forall i,j. \lambda_i \neq \lambda_j \lor \lambda_i = \lambda_j$. Because, in quantum mechanics, the spectrum of an operator A is supposed to consist of the possible values of the observable corresponding to A. Any realistic observation can be performed only up to a finite precision. So, in a finite dimensional case, we can expect that the eigenvalues are all mutually distinguishable.

What is actually behind such a justification and what does it have to do with the fact that each measurement can be done up to a finite precision?

I also have a side question regarding the following theorem of Ziegler and Brattka (2001):

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where a matrix is called computable iff the corresponding linear map is computable. How does this fact enable us to reconstruct exactly the classical theorem? I can't see how we'd obtain the cardinality of the spectrum just from the condition that $A$ be computable (normal) matrix.

WARNING: there is no such a corollary in the preprint!


I think the point is this. It is impossible in general to decide computationally whether two computable real numbers $\alpha$ and $\beta$ are equal. If in fact they are not equal, by computing sufficiently good approximations you can determine that they are not equal, but no finite computation can determine that they are equal.

For a polynomial $P(\lambda)$ (in particular the characteristic polynomial of a matrix), the roots of $P$ are distinct iff the discriminant of $P$ is nonzero. This discriminant is a polynomial in the coefficients, so it is computable. But again, if the discriminant happens to be $0$ we can't determine that by a finite computation.

However, in "real-life" or "natural" situations, we should expect eigenvalues to be distinct unless they have a good reason (typically involving a symmetry) to be equal.

The situation in Corollary 15 is different: there is no requirement here for the $\lambda_j$ to be distinct.

EDIT: In fact, Corollary 15 is false. Consider the family of $2 \times 2$ matrices

$$ A(p) = \pmatrix{\min(p,0) & \max(p,0)\cr \max(p,0) & \max(-p,0)\cr} $$

where $p$ is a computable real number. Then $A(p)$ is a computable real symmetric matrix. If $p < 0$ it has orthonormal eigenvectors $$ \pmatrix{1/\sqrt{2}\cr 1/\sqrt{2}\cr}, \ \pmatrix{1/\sqrt{2}\cr -1/\sqrt{2}\cr}$$ while if $p > 0$ it has orthonormal eigenvectors $$ \pmatrix{1\cr 0\cr},\ \pmatrix{0\cr 1\cr}$$ The orthonormal eigenvectors $x_1,\; x_2$ mentioned in Corollary 15 can't be computable functions, because they can't be continuous at $p=0$.

  • $\begingroup$ Regarding the last paragraph: the answer is different because the question is different. We "avoid trichotomy" by not asking whether or not the $\lambda_i$ are distinct. $\endgroup$ Apr 10 '16 at 18:05
  • $\begingroup$ Well, how can we prove Corollary 15 without knowing which eigenvalues are distinct? $\endgroup$
    – Rubi Shnol
    Apr 10 '16 at 19:23
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    $\begingroup$ I don't know. The pdf file of Ziegler and Brattka you linked to has no Corollary 15. In fact they seem to indicate that Corollary 15 is false. $\endgroup$ Apr 10 '16 at 23:59
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    $\begingroup$ Corollary 15 does not state that the basis can be computed from the matrix. It is a non-uniform theorem: for every computable matrix there exist such-and-such basis, but there is no algorithm that passes from one to the other. So, Corollary 15 is in fact correct and you misread it - nowhere does it claim that the $x_i$ are computable functions. $\endgroup$ Apr 11 '16 at 8:45
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    $\begingroup$ I think we have to separate the theoretical question of computability from the practical question of whether our measuring apparatus is accurate enough to detect something. $\endgroup$ Apr 15 '16 at 16:49

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