The orthogonal/unitary matrix/operator can be seen as an isometric operation or transformation, which preserves the inner product ("length" and "angle"). But I have difficulties to intuitively understand the symmetric/Hermitian matrix/operator. I mean what kind of operation or transformation can be interpreted as the symmetric/Hermitian matrix/operator?



In some sense, getting "intuition" for this kind of thing is a central and generally unresolvable issue in operator theory (ie, it depends on whatever applications of operator theory you have in mind, and to some extent on the actual operator you are studying).

In another sense, a reasonably complete answer is provided by some version of the spectral theorem.

Self-adjoint operators are multiplication operators by real-valued functions.

Similarly, normal operators (ones commuting with their Hermitian adjoints) are multiplication operators by complex-valued functions. In a very reductive sense, "that's all there is to it."

Many textbook treatments of the spectral theorem for (finite) Hermitian matrices are rather matrix-theoretic in nature, which tends to obscure some of the underlying conceptual stuff. Basis-free treatments of the finite dimensional spectral theorem given in books like Sheldon Axler's Linear algebra done right point the way to a more abstract understanding.

A next step would be to understand the spectral theorem for operators on separable, but not necessarily finite dimensional, Hilbert spaces. This requires a much more substantial investment in analysis, but a good introduction is Paul Halmos' article What does the spectral theorem say? (I have linked to the journal version of this article; scanned PDFs are also available via Google).

After that, you might want to understand the notion of self-adjointness for unbounded operators on Hilbert space (this is the case of central interest in quantum mechanics). A good resource there is Volume 1 of Reed and Simon's Methods of modern mathematical physics.

  • $\begingroup$ It's very helpful and inspiring. Thanks a lot! $\endgroup$ – chaohuang May 25 '12 at 22:29
  • $\begingroup$ That's a great article you posted. Thank you! $\endgroup$ – Alex Lapanowski Apr 14 '13 at 19:41
  • $\begingroup$ Please note that the validity of the statement in yellow and the one following it are contested in another answer on this page, left by a user without enough reputation to directly comment on your answer. $\endgroup$ – Alex M. Jun 11 '15 at 13:37

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