Non-equivalent definitions of diagonalizable linear operator on Hilbert space?

Question

While reading Nielsen & Chuang's Quantum Computation and Quantum Information, I was troubled by their definition of a diagonalizable linear operator, which requires the eigenvectors to be orthogonal, thus implying that diagonalizable is equivalent to normal. Are there good reasons to add the restriction of orthogonality for the eigenvectors?

Answer 1

In Quantum Mechanics, measurements of a state variable $A$ for a state $x$ of the system is made through a quadratic form $\langle Ax,x\rangle$, and this value must be real. So you are you working with a symmetric linear operator $A$. The requirement that $A$ be self-adjoint, and not just symmetric, is made for various reasons. Then you're dealing with eigenfunction expansions of self-adjoint linear operators on a Hilbert space. Eigenvectors for such an $A$ are automatically orthogonal if they associated with different eigenvalues. It's a little more difficult to keep thinking in terms of orthogonal eigenfunction expansions in the general case, but the notation is compelling enough that they keep it, and they adapt it to the more general case. Mathematicians abstract to a spectral measure $E$ in order to representat $A$: $$ Ax=\int_{\sigma(A)}\lambda dE(\lambda)x $$ Here $E(S)x \perp E(T)x$ if $S\cap T=\emptyset$. So the orthogonality is built in, but it has been generalized beyond trying to keep to one-dimensional eigenfunction expansions of the form $$ x = \int_{\sigma(A)}\langle \phi_{\lambda}|x\rangle\phi_{\lambda} d\lambda $$ (Physicsts use inner products that are linear in the second coordinate and conjugate linear in the first, which accounts for $x$ being on the right in their inner product.) The Mathematicians version of orthogonality is that $E(S)x\perp E(T)x$ if $S\cap T=\emptyset$. The Physicists push this notation to an infinitesimal limit, and try to think of $\phi_{\lambda}d\lambda$ in these terms. I personally do not like that notation much, but it has a long History, and it seems to be here to stay. For simple cases such as ordinary Fourier transform, this is just fine: $$ f = \int_{-\infty}^{\infty}\langle e_{\lambda},f\rangle e_{\lambda} d\lambda. $$ In this case, the above is a simple recasting of the Fourier transform and it's inverse: $$ f = \int_{-\infty}^{\infty}\langle f,\frac{1}{\sqrt{2\pi}}e^{i\lambda x'}\rangle \frac{1}{\sqrt{2\pi}} e^{i\lambda x} d\lambda, \\ \frac{1}{i}\frac{d}{dx}f = \int_{-\infty}^{\infty}\lambda\langle f,\frac{1}{\sqrt{2\pi}}e^{i\lambda x'}\rangle \frac{1}{\sqrt{2\pi}}e^{i\lambda x}d\lambda. $$ Once you veer away from the simple cases, it is hard to keep the Physicist's notation straight, and to keep it accurate because the eigenspaces are not spanned by one-dimensional functions $\lambda\mapsto\phi_{\lambda}$. That's why von Neumann came up with his Spectral Integral representation of a self-adjoint linear operator on a Hilbert Space: $A = \int_{\sigma(A)}\lambda dE(\lambda)$. This works in the general case, and it retains much of the original character without having to try to fit it back into Dirac notation.