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In Nielsen and Chuang, they write: Let $A = \sum_a a|a\rangle \langle a|$ be the spectral decomposition of $A$. Define $f(A) = \sum_a f(a) |a \rangle \langle a|$. Apparently this is uniquely defined. I'm having trouble seeing why this is.

If we used some other orthonormal basis of eigenvectors of $A$, say $A = \sum_b b|b\rangle \langle b|$, then why is $\sum_a f(a)|a\rangle \langle a| = \sum_b f(b)|b\rangle \langle b|$? I think there must be some property about eigenvectors sharing the same eigenvalues, but I'm unsure about what I'm missing.

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2 Answers 2

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It's a standard result of linear algebra that the spectral decomposition of an operator is unique (up to a trivial reordering of the eigenvalues/vectors). If you could write $A$ as both $\sum_a a | a \rangle \langle a |$ and $\sum_b b | b \rangle \langle b |$, with the $a$s and $b$s nontrivially related, then this would violate the uniqueness of the eigendecomposition.

If you were to work in an orthonormal basis other than the eigenbasis, then the operator wouldn't be diagonal, but would look like $A = \sum_{n,m} c_{nm} | n \rangle \langle m |$. You are correct that letting $c_{nm} \to f(c_{nm})$ would be basis-dependent and generally not particularly interesting or meaningful.

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  • $\begingroup$ This makes sense, but I would like to convince myself of your unique spectral decomposition claim. In this text, the theorem is stated as: "Any normal operator $M$ on a vector space $V$ is diagonal with respect to some orthonormal basis for $V$. Conversely, any diagonalizable operator is normal." I don't see where the uniqueness is coming from exactly... $\endgroup$
    – theQman
    Oct 18, 2016 at 19:32
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    $\begingroup$ This answer seems to completely ignore the issue degenerate eigenvalues. There you can choose your basis in the eigenspace freely, its not just a reordering of the eigenvalues. $\endgroup$ Oct 18, 2016 at 19:36
  • $\begingroup$ Yes, I think that is what I'm trying to clarify. For example the 2x2 identity matrix has both $\{[1,0], [0,1]\}$ and $\{[1/\sqrt{2},1/\sqrt{2}], [-1/\sqrt{2},1/\sqrt{2}]\}$ as orthonormal basis using eigenvectors. It isn't clear to me how to rigorously clean up these sort of cases. $\endgroup$
    – theQman
    Oct 18, 2016 at 19:43
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    $\begingroup$ Indeed, you need to use the orthogonal projectors $P_a$ on the eigenspaces (supposing $A$ is Hermitian), and for $A = \sum_a a P_a$, you get $f(A) = \sum_a f(a) P_a$. If the eigenvalue is simple, then $P_a = |a\rangle \langle a|$. $\endgroup$
    – user130529
    Oct 18, 2016 at 20:43
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    $\begingroup$ @theQman I was hoping that no one would bring up degeneracy. You're right that within a degenerate eigenspace, you're free to choose any basis, so in that case the eigendecomposition is not unique. But as claude chuber points out, for any basis $\{ | n \rangle \}$ of the degenerate eigenspace, the sum $\sum_n | n \rangle \langle n |$ is actually basis independent - it corresponds to the orthogonal projection operator $\hat{P}$ onto the degenerate eigenspace. So $\sum_n f(n) | n \rangle \langle n | = f(\lambda) \hat{P}$ (where $\lambda$ is the degenerate eigenvalue) is basis-independent ... $\endgroup$
    – tparker
    Oct 19, 2016 at 0:55
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This stems from a misformulation of the spectral theorem. In a proper mathematical text one never sees it stated using a basis precisely because that is not unique. Physicists often assume "for the sake of simplicity" a nondegenerate spectrum, where every eigenvalue has an algebraic multiplicity of 1, and consider nontrivial multiplicities a degenerate case that can be covered if needed.

The spectral theorem gives a unique decomposition using projectors instead of $|n〉〈n|$, which for finite-dimensional cases looks like $$A = \sum_{α ∈ σ(A)} α P_α,$$ where $\{P_α\}$ are mutually orthogonal projectors. The application of an analytic function $f$ is then defined as $$f(A) = \sum_{α ∈ σ(A)} f(α) P_α$$ and the uniqueness naturally follows: indeed, there's no point where this could become non-unique.

If you went one step further and decomposed each of the projectors $$P_α = \sum_{k=1}^{ν_α} |v_α^{(k)}〉〈v_α^{(k)}|$$ in some orthonormal basis of its range, you'd obtain the decomposition used by Nielsen and Chuang, but as you noted, it would no longer be unique. (The eigenvalues still are, and their ordering does not matter.) The trick why this also works is then to note that if you group the summands putting ones using the same eigenvalue in an inner sum, $$A = \sum_a a|a〉〈a| = \sum_{α ∈ σ(A)} α \sum_{k=1}^{ν_α} |a_α^{(k)}〉〈a_α^{(k)}|,$$ where $\{|a_α^{(k)}〉\}$ are $ν_α$ of the original eigenvectors $|a〉$ all corresponding to the eigenvalue $α$, then also $$f(A) = \sum_a f(a)|a〉〈a| = \sum_{α ∈ σ(A)} f(α) \sum_{k=1}^{ν_α} |a_α^{(k)}〉〈a_α^{(k)}|.$$ By obtaining the same $f(a)$ for all $a$ which corresponded to the same eigenvalue $α$, the grouping is not violated, and once we agree that $$\sum_{k=1}^{ν_α} |a_α^{(k)}〉〈a_α^{(k)}|$$ is the projector $P_α$, then $$f(A) = \sum_{α ∈ σ(A)} f(α) P(α),$$ which as we know from above is the unambiguous definition.

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