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It is very common (see e.g. page 18 of Ballentine's Quantum Mechanics: A Modern Development) for the following development to take place. We couch the discussion in Dirac's bra-ket notation noting that (as I am not really capable of) it is possible to make this precise in the rigged Hilbert space formalism (for operators with continuous spectra etc.).

It is a fact that self-adjoint operator $A$ admits a complete eigenbasis, so that it is a subsequent theorem that we can write $A$ as $$A = \sum a_i |a_i \rangle \langle a_i |.$$ It is then very common to say that this motivates the definition of a function of such an operator, $f(A)$, by $$f(A) = \sum f(a_i)|a_i \rangle \langle a_i |.$$

My question surrounds the definition of the function of an oeprator. My suspicion is that there is a different definition of $f(A)$ from operator theory, and that the definition I have quoted above is equivalent to said definition via some theorem. Is that indeed the case?

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    $\begingroup$ Most take this Sylvester's formula as a definition. The projection operators you resolved the identity to are the analog of Frobenius covariants for matrices. You have qualms? $\endgroup$ Commented Jan 16, 2023 at 20:19
  • $\begingroup$ @CosmasZachos But doesn't even writing the formula presuppose that the expansion converges? Also, surely it is possible to define functions of (some) non-self adjoint operators -- I would then expect that in the particular case of a self-adjoint operator that definition would reduce to this one, no? $\endgroup$
    – EE18
    Commented Jan 16, 2023 at 20:22
  • $\begingroup$ @TobiasFünke I suppose (1) because this definition only works for self-adjoint operators and I have to imagine that functions of non-self-adjoint operators are sometimes well-defined and (2) there is no establishment of existence (convergence) in the definition given. The Hilbert space requirement is that the sum of moduli squared of coefficients be finite but is that clear? Again, this is just a suspicion from someone who only knows real analysis at the level of Baby Rudin, and certainly no functional analysis. $\endgroup$
    – EE18
    Commented Jan 16, 2023 at 20:34
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    $\begingroup$ Sylvester's theorem works for diagonalizable operators. Convergence issues belong to MSE, I fear. $\endgroup$ Commented Jan 16, 2023 at 20:44
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    $\begingroup$ No, I don't think it is defined like this; note that $H$ is usually unbounded. Usually it is defined in terms of spectral theory or discussed in the context of Stone's theorem. $\endgroup$ Commented Jan 16, 2023 at 20:53

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There are a number of ways to define functions of operators. A common example is via a power series, e.g. $$\exp(A) := \sum_{k=0}^\infty \frac{1}{k!} A^k$$

One can show that if the radius of convergence of the series of the original function $f$ is $R$, then the corresponding operator $f(A)$ is well-defined iff the operator norm $\Vert A\Vert_{op} < R$.

The requirement that $A$ be bounded (and that $\Vert A\Vert_{op}$ be within the radius of convergence of the Taylor series) is very limiting. There are several important alternatives (e.g. the holomorphic functional calculus, the continuous functional calculus, the polynomial functional calculus), but the most important is the Borel functional calculus which works via the spectral theorem. If $A$ is a normal operator, then it can be expressed as $$A = \int_{\sigma(A)}\lambda \ \ \mathrm dP^A(\lambda) $$ where $\sigma(A)\subseteq \mathbb C$ is the spectrum of $A$ and $P^A$ is its projection-valued measure. Note that this reduces to the familiar expression $$A = \sum_{\lambda \in \sigma(A)} \lambda |\lambda \rangle\langle \lambda|$$ if $\sigma(A)$ is purely discrete and its eigenspaces are one-dimensional. In this language, given a Borel-measurable function $f:\mathbb C\rightarrow \mathbb C$, we have that

$$f(A) := \int_{\sigma(A)} f(\lambda) \ \mathrm dP^A(\lambda) \rightsquigarrow \sum_{\lambda\in \sigma(A)} f(\lambda) |\lambda\rangle\langle \lambda|$$

This is the definition of a function of a possibly-unbounded operator which is most commonly used in quantum mechanics (e.g. in Stone's theorem).

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    $\begingroup$ In your last equation, is $:=$ meant to say "is defined by"? I suppose given your last written line but wanted to make sure. In that case I guess the "definition" I noted is just that, a definition. $\endgroup$
    – EE18
    Commented Jan 16, 2023 at 21:22
  • $\begingroup$ @EE18 Yes, that is meant to be a definition (as is the first expression I wrote at the top). $\endgroup$
    – J. Murray
    Commented Jan 16, 2023 at 21:44

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