Much (All?) of quantum theory can be done in separable Hilbert spaces with a countable basis.
How about quantum field theory? Is it “quite happy” (mathematically consistent) if everything is countable, or does it “need” to use an uncountable, continuous space (e.g. rigged Hilbert space) for mathematical consistency, or some other reason?


1 Answer 1


Rigged Hilbert spaces have no special relationship to quantum field theory. We can talk about more general elements of "the" Hilbert space such as wave functions looking like distributions. They don't have a finite norm but they're still useful to talk about.

Truly physical states that may be realized in practice are normalizable - their norm may be chosen to be 1, in fact, so rigged Hilbert spaces represent a mathematically convenient but physically unacceptable extension of the Hilbert space.

Quite generally, we want the Hilbert space to be separable in QFT, too. After all, the Hilbert space of a QFT isn't much different from the direct sum of the $N$-particle Hilbert spaces in (non-relativistic) QM (summed over non-negative integers $N$) and those are separable.

In QFT, we may encounter superselection sectors, i.e. separable Fock-style (but interacting) Hilbert spaces built around a vacuum that isn't unique but depends on a parameter. In this case, the total Hilbert space is the direct sum over the (uncountably many) values of the parameters labeling the vacua ("moduli"). But it's still true that only one superselection sector is ultimately relevant for the description of real physical phenomena, so even in this case, one may say that the ultimate fully allowed Hilbert space is separable.

QFT gives us new ways to organize the states and define the operators on the Hilbert space but at the end, it's right to think about a QFT as another quantum mechanical theory with countable bases. The same remark applies to string theory.

  • $\begingroup$ One minor correction: In the rigged Hilbert space formalism, the space of states is a subspace of the usual Hilbert space. It is of course physically unacceptable to use the eigenvectors of operators with continuous spectrum as states, but no one who thinks in terms of this formalism actually does that. $\endgroup$
    – user1504
    Commented May 22, 2013 at 11:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.