I'm having trouble understanding the completeness condition for bra and ket vectors in Hilbert space, especially in the continuous case. The discrete case makes a fair amount of sense; given any observable corresponding to a linear operator $Q$ with countably many eigenvalues, any quantum state $| \psi \rangle$ can be written in the $Q$ eigenbasis as
$$|\psi \rangle = \sum_{n = 1}^{\infty} |e_n \rangle \langle e_n | \psi \rangle$$
…where $\langle e_n |$ and $| e_n \rangle$ are the bra and ket corresponding to the $n^{\text{th}}$ eigenvalue of $Q$. But say that $Q$ has uncountably many eigenvectors; say, $\{ | e_\alpha \rangle \}$ indexed over a subset $\mathcal{A} \subseteq \mathbb{R}$. Then, the above equation becomes
$$| \psi \rangle = \int_\mathcal{A} | e_\alpha \rangle \langle e_\alpha | \psi \rangle \, d\alpha$$
How do we define an integral over ket vectors? Both the Lebesgue and Riemann definitions of the integral require us to create "test functions" (i.e. Riemann sums or integrable simple functions) that are bounded above by the integrand, which requires an order relation. I see no reason why such an order relation should exist on the space of ket vectors!
The best solution I have so far is to say that each ket vector corresponds to a complex valued wavefunction; that is, we identify $| e_\alpha \rangle$ with its representation in the position basis, which is just a function $\mathbb{R} \to \mathbb{C}$. Then, the integral above is simply an integral over complex valued functions, which I'm quite comfortable with. But this only works if every quantum state is spanned by the position basis, so that such a representation exists. For example, I have a hard time believing that the spin states $|\uparrow \rangle$ and $| \downarrow \rangle$ are expressible in position basis; why would position also encode information about spin?
"Introduction to Quantum Mechanics" by David Griffiths seems to resolve this by declaring that "the eigenfunctions of an observable are complete: any function (in Hilbert space) can be expressed as a linear combination of them." This suggests the even more uncomfortable scenario where the spin eigenstates (which are countable) span all of the position eigenstates (which are uncountable), so I have a feeling that something more is at work here.
Do you have any thoughts on this?
Best Answer
All that question can be managed in terms of rigged Hilbert spaces defined by Gelfand. However, that approach is so complicated than, for instance, von Neumann's one relying on the notion of projection-valued measure, that ii is more convenient using those manipulations just to grasp some plausible result. Finally that result can be proved using less cumbersome technologies.
A convenient theoretical idea is however to define $$|\Psi\rangle = \int |x \rangle \langle x| \psi \rangle dx $$ as the unique vector (via Riesz' lemma) such that $$\langle \Phi|\Psi\rangle = \int \langle \Phi |x \rangle \langle x| \psi \rangle dx$$ for every $|\Phi\rangle$. Notice that the integral in the right-hand side is now understood in the standard way.