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?

  • $\begingroup$ I am not mathematically polished enough to answer your original question but I think I can point out some confusion expressed in your last paragraph. 1. Integration is done only over continuous eigenstates--so for example, you never integrate over spin-eigenstates with any spin index. You always sum over spin indices. $\endgroup$ – Dvij Mankad May 22 at 16:00
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    $\begingroup$ 2. When one says that eigenstates of an observable form a complete set, it is a subtle statement. It, for example, doesn't mean that a spin eigenstate can be expressed as a linear combination of position eigenstates. But, rather, each position eigenvalue has a degenerate eigensubspace and the set of spin eigenstates form a basis for that eigensubspace. $\endgroup$ – Dvij Mankad May 22 at 16:00
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    $\begingroup$ 3. Finally, there is nothing unusual from a physicist's perspective that a vector space spanned by an uncountably infinite set of eigenstates can also be spanned by a countably infinite set of eigenstates. For example, the Hilbert space of a particle on a ring can be spanned by countably infinite momentum eigenstates as well as by uncountably infinite position eigenstates. But that is not what is happening with the spin eigenstates and the position eigenstates. They share common eigenstates as the spin and the position operators fully commute. $\endgroup$ – Dvij Mankad May 22 at 16:07
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    $\begingroup$ To do this stuff rigorously requires a lot of functional analysis and operator theory. (Your text probably says something in the preface or introduction like "appropriate level of rigour".) Summing over both discrete and continuum states is done using a "projection valued measure". $\endgroup$ – Keith McClary May 22 at 17:19
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    $\begingroup$ you also you want look up "rigged hilbert space", a cursory understanding of this might be a middle ground. $\endgroup$ – lalala May 22 at 17:36

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.

  • $\begingroup$ Good answer, although the right hand side should flip all bra and kets such that the x are on the left (by complex conjugation). What I mean Eigenstates of x are bras and cannot be kets. $\endgroup$ – lalala May 22 at 18:47
  • $\begingroup$ Eigenstates of $X$ do no exist at all. At most they can be defined as Schwartz distributions in Gelfand formalism...interpreting in distributional sense the eigenvalue problem. $\endgroup$ – Valter Moretti May 22 at 19:16
  • $\begingroup$ I know. I was sloppy in the terminology. What I mean: <x|f> can be interpreted/is as f(x). So instead of writing <f|x> it is better to write <x|f>^* (which the is equal to f(x)^*) $\endgroup$ – lalala May 22 at 19:35
  • $\begingroup$ @Valter Moretti Thank you, your answer helped me a lot. More generally, given any complex function f : R --> C, could we define integral( |x> f(x) dx ) to be the unique |v> satisfying <phi | v > = integral( <phi | x> f(x) dx) for all <phi|? Could we do something similar with bras, and say that integral(<x| f(x) dx) is the unique <v| satisfying <v | phi> = integral(<x | phi> f(x) dx) for all |phi>? Sorry for the messy notation, these comments don't seem to process LaTeX. $\endgroup$ – FlyingPiper May 22 at 19:51
  • $\begingroup$ The answer is yes for both questions! $\endgroup$ – Valter Moretti May 22 at 20:15

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