Originally Posted by

**Ackbeet** Well, the Hermitian adjoint is defined, at least in finite-dimensional Hilbert spaces, as the complex conjugate transpose. Scalars you can think of as 1-dimensional vectors. Transposing a scalar doesn't change it, but the complex conjugate part does (if it's complex, of course). *Hence, the Hermitian adjoint of a scalar is just the complex conjugate.* The fact that I'm right-multiplying by the scalar is unimportant, as scalars can pass through vectors no problem (that is, scalar multiplication of a vector is commutative).

I think you're confusing two different proofs.

1. If you're trying to show that the eigenvalues of an Hermitian operator are real, then you play around with

$\displaystyle \langle a'|A|a'\rangle,$ not $\displaystyle \langle a'|A|a''\rangle.$

The reason is that you have to know that $\displaystyle \langle a'|a'\rangle\not=0, $ which is true because eigenvectors, by definition, are nonzero.

2. If you're trying to prove that the eigenvectors of differing eigenvalues are orthogonal, then you play around with

$\displaystyle \langle a'|A|a''\rangle,$ where the eigenvalues of the two eigenvectors there are different.

I would definitely prove that the eigenvalues are real before proving that the eigenvectors of differing eigenvalues are orthogonal.