2
$\begingroup$

$\mathbf{Background:}$ Consider a free scalar field $\phi$ ($\mathcal{L}_0 = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi + \frac{1}{2}m^2 \phi^2$).

In the Hamiltonian viewpoint, this system has a Hilbert space $\mathcal{H}_0$ (the Fock space). We can write down a resolution of the identity in $\mathcal{H}_0$ (in the Schrodinger picture):

$\begin{align} I = |0 \rangle \langle 0 | + \int dp\ |p \rangle \langle p | + \frac{1}{2!} \int dp_1 dp_2\ |p_1 p_2 \rangle \langle p_1 p_2 | +\ \cdots \end{align}$

$\mathbf{Question:}$ If we add an interaction term to the Lagrangian (such as $\lambda \phi^4$), is $I$ still the identity operator in the new Hilbert space $\mathcal{H}$?

(To be clear, I mean exactly the same $I$, with the free vacuum and free particle states.)

$\mathbf{Motivation:}$ The analogous thing holds in QM. A non-relativistic particle moving in a potential $V$ has a Hilbert space $L^2({\mathbb{R}})$, so by Fourier transform, $\int dp\ |p \rangle \langle p |$ is a resolution of the identity. (Where $\langle x | p \rangle = \frac{1}{\sqrt{2\pi}} e^{i p x}$ are free particle states.) The Hilbert space is the same regardless of $V$.

$\endgroup$
1
$\begingroup$

Yes, that's still a resolution of the identity.

But knowing this isn't likely to make your life easier. The observable algebra for the free theory and the observable algebra for the interacting theory are generated by unbounded operators. These algebras do not have the same domains of definition. The 'free-particle' resolution of the identity you wrote down is convenient when you're dealing with the domain where the free field operators have a well-defined action, but it's generally singular when you're trying to study vectors which live in the domain of the interacting field operators.

$\endgroup$
  • $\begingroup$ Thanks. Can you give a simple example? If it is indeed a resolution of the identity, doesn't it follow that the free-field operators have a well-defined action on any given state in the full interacting Hilbert space? $\endgroup$ – marlow Oct 30 '13 at 21:16
  • $\begingroup$ Do you know of any references for your various claims? $\endgroup$ – joshphysics Oct 30 '13 at 21:23
  • $\begingroup$ @marlow I'd actually have to describe an interacting QFT in sufficient detail. Do you have a simple example you like? $\endgroup$ – user1504 Oct 31 '13 at 0:29
  • $\begingroup$ @marlow What do you mean by resolution of the identity? I claim that the expression you wrote down is only meaningful in the context of rigged Hilbert spaces. (For some explanation of the concept, see physics.stackexchange.com/questions/43515/…) $\endgroup$ – user1504 Oct 31 '13 at 0:40
  • $\begingroup$ @joshphysics Not really. This is the sort of thing one learns by thinking about constructive field theory -- in particular, in claiming that the sum of integrals above is a resolution of the identity, I'm just restating the fact that the action of free fields is defined only a certain nuclear subspace of the usual Hilbert space -- but I don't know of anywhere it's written down particularly well. $\endgroup$ – user1504 Oct 31 '13 at 0:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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