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Let $\mathscr{H}$ be a complex Hilbert space and $A$ be an Hermitian operator $A: \mathscr{H}\to \mathscr{H}$. Suppose, for a moment, that $A$ has a set of discrete eigenvalues $\{\lambda_{n}\}_{n\in \mathbb{N}}$ with corresponding normalized eigenvectors $\{e_{n}\}_{n\in \mathbb{N}}$, which are assumed to form a complete orthonormal set. Then, every $\psi\in \mathscr{H}$ can be expressed as: $$\psi = \sum_{n\in \mathbb{N}}\langle e_{n},\psi\rangle e_{n} $$

Using Dirac's notation, $e_{n}$ becomes $|n\rangle$ and $\psi$ becomes $|\psi\rangle$. Now, I was reading Nakahara's book and, at some point, he states:

If $A$ has a continuous spectrum $a$, the state is expanded as: \begin{eqnarray} |\psi\rangle = \int da \psi(a) |a\rangle \tag{1}\label{1} \end{eqnarray} and the completeness relation now takes the form: $$\int da |a\rangle \langle a| = I$$

In (\ref{1}), there seems to be an underlying assumption which is that the set of eigenvectors of an Hermitian operator on a Hilbert space spans the whole space. Although this is usually used by physicists, this is not as simple as it seems, because sometimes we have to deal with Rigged Hilbert spaces instead. This is briefly discussed here but the answers to the linked post do not provide much details.

I find it very difficult to understand the physicist point of view using what I know of functional analysis. Moreover, there seems to be quite a few books discussing these topics.

So, my question is: what is the true mathematical meaning of (\ref{1})? Do we need Rigged Hilbert spaces? Do we need the spectral theorem?

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    $\begingroup$ If you want to make sense of the kinds of expressions physicists like to use in connection with observables with continuous spectrum, you will probably want to think in terms of the spectral theorem. I think you should interpret your first equation in terms of the direct integral formulation of the spectral theorem for the position operator on $L^2(\mathbb{R})$. $\endgroup$ – Christopher Beem Jan 25 at 20:49
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Perhaps an example will help? Consider the momentum operator $A=-id/dx$, the eigenstates $|a\rangle$ are plane waves $\langle x|a\rangle=(2\pi)^{-1/2} e^{iax}$, the expansion (1) of an arbitrary state $|\psi\rangle$ in the $|a\rangle$ basis is the Fourier integral $$\langle x|\psi\rangle=\int_{-\infty}^\infty da\,\psi(a)\langle x|a\rangle=(2\pi)^{-1/2}\int_{-\infty}^\infty da\,e^{iax}\psi(a),$$ and the completeness relation reads $$\int_{-\infty}^\infty da\,\langle x|a\rangle\langle a|x'\rangle=(2\pi)^{-1}\int_{-\infty}^\infty da\,e^{ia(x-x')}=\delta(x-x')=\langle x|I|x'\rangle,$$ with $\delta(x-x')$ the Dirac delta function distribution.

So yes, you need to equip the Hilbert space with a distribution theory --- as for any unbounded observable with a continuous spectrum a rigged Hilbert space is needed.

I found this discussion of the Dirac bra-ket formalism in a continuous spectrum instructive.

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