There are many posts related to this issue on this site, but I have found none that answer my specific questions about this matter.

I review my understanding of Weinbergs approach. There are probably some misconceptions throughout and I am very thankful if someone can clear up any of them. Also I will ask some more concrete questions.

So we have the full Hamiltonian operators $H$ and the free Hamiltonian $H_0$ and the difference is some potential $V$ $$ H = H_0 + V.\tag{3.1.8} $$ Also we have corresponding Hilbert spaces $\mathcal H$ and $\mathcal H_0$, where $\mathcal H_0$ is a familiar Fock-space which is somehow embedded in $\mathcal H$, in other words $\mathcal H_0$ is sub-Hilbert-space of $\mathcal H$. A fundamental assumption is that the eigenvalue-spectra $E_{\alpha}$ of $H$ and $H_0$ coincide. Although this is mysterious to me, I can accept this.
Now $\mathcal H_0$ is well-defined. It is spanned by the orthonormal energy eigenstates $\Phi_{\alpha}$, $$H_0 \Phi_{\alpha} = E_{\alpha} \Phi_{\alpha},\tag{3.1.9}$$ that is any $\Phi \in \mathcal H_0$ can be written as $$ \Phi = \int d\alpha~ g(\alpha) \Phi_{\alpha}, $$ where $g$ is a (momentum-space) wave-function (for the explanantion of the $\alpha$ notation I refer to the book). I presume that we can not write any $\Psi \in \mathcal H$ in this way, otherwise we would simply have $\mathcal H \simeq \mathcal H_0$, which is of course trivial, right?

Now consider the energy eigenstates of $H$. Since its spectrum is the same as $H_0$ we can also label them by $\alpha$. However there seems to be another fundamental assumption. Namely that there exist two sets of eigenstates $\Psi_{\alpha}^{\pm}$, with $$H\Psi_{\alpha}^{\pm} = E_{\alpha} \Psi_{\alpha}^{\pm}\tag{3.1.11},$$ and such that they obey the following identity $$ \lim_{t \rightarrow \mp \infty} \int d\alpha~ e^{-iE_{\alpha} t} g(\alpha) \Psi_{\alpha}^{\pm} = \lim_{t \rightarrow \mp \infty} \int d\alpha~ e^{-iE_{\alpha} t} g(\alpha) \Phi_{\alpha} \tag{3.1.12} $$ for all (smooth?) wave-functions $g(\alpha)$. Weinberg shows that the $\Psi_{\alpha}^{\pm}$ are also orthonormal. This seems to imply that the $\Psi_{\alpha}^{\pm}$ each also span $\mathcal H_0$ or at least two copies $\mathcal H_0^+ \simeq \mathcal H_0^- \simeq \mathcal H_0$ which are in some way embedded in $\mathcal H$.

Q1: Do $\Psi_{\alpha}^+, \Psi_{\alpha}^-, \Phi_{\alpha}$ actually span the same subspace $\mathcal H_0 \subset \mathcal H$? This seems to be necessary in order to expand them in terms of one another which is done repeatedly throughout the chapter. Can this be shown? Is it obvious? Or does it need to be assumed?

I am also confused about the physical interpretation of these definitions. I take the Heisenberg picture viewpoint. Say the quantum system is described be the state $\Psi \in \mathcal H$ as seen from some reference frame. Say the hermitian operator $\mathcal O(t)$ measures some properties related the particle content of a given state. An observer in the far past will measure the following expectation value $$ \langle \mathcal O(-\infty) \rangle = \lim_{t \rightarrow - \infty} (e^{-iHt}\Psi, \mathcal O(0) e^{-iHt}\Psi) $$ whereas an observer in the far future will measure $$ \langle \mathcal O(+\infty) \rangle = \lim_{t \rightarrow + \infty} (e^{-iHt}\Psi, \mathcal O(0) e^{-iHt}\Psi). $$ Then it makes sense to me to define the corresponding in-state $\Psi^-$ and out-state $\Psi^+$ as $$ \Psi^{\pm} \equiv \lim_{t \rightarrow \mp \infty} e^{-i H t} \Psi. $$ Q2: It also makes sense to claim that $\Psi^{\pm} \in \mathcal H_0$. Can this be shown from the previous assumptions? Is it obvious? Or does it need to be assumed?

Then, if my assertion in Q1 is valid, we may expand $\Psi^{\pm}$ in either $\Psi_{\alpha}^+, \Psi_{\alpha}^-$ or $\Phi_{\alpha}$.

Q3: The "correct" basis to choose for the expansion seems to be $\Psi_{\alpha}^+$ for $\Psi^+$ and $\Psi_{\alpha}^-$ for $\Psi^-$. $$ \Psi^{\pm} = \int d\alpha~(\Psi_{\alpha}^{\pm}, \Psi^{\pm}) \Psi_{\alpha}^{\pm}. $$ What I mean by "correct" is that $|(\Psi_{\alpha}^{\pm}, \Psi^{\pm})|^2$ can then be viewed as the probability distribution of measuring for $\mathcal O$ the particle content $\alpha$ at $t \rightarrow \mp \infty$. The reason should have something to do with the interpretation that $\Psi_{\alpha}^{\pm} $ describes a state with particle content $\alpha$ only when $\mathcal O$ is measured at $t \rightarrow \mp \infty$, which should somehow be equivalent to (3.1.12). Is this correct? If yes, how can this be seen from (3.1.12)? If no, how are $\Psi^{\pm}$ to be related to the $\Psi_{\alpha}^{\pm}$? The problem I have in trying to put this on a more rigorous footing is the lack of characterization of the state $\Psi$.

Q4: Weinberg explicitly states (p.109) that the in- and out-states can not be written as the limits of some state $\Psi(t)$ for $t \rightarrow \mp \infty$. However defining $\Psi(t) \equiv e^{-iHt} \Psi$ does exactly that. This is of course just the familiar Schrödinger picture formulation. Is this viewpoint not valid? Or does Weinberg mean something different here?


1 Answer 1



To set the stage, I'll address a couple of preliminary issues first, and then I'll answer the numbered questions Q1-Q4.

A fundamental assumption is that the eigenvalue-spectra ... of $H$ and $H_0$ coincide. Although this is mysterious to me, I can accept this.

From the context, I suspect that Weinberg (reference 1) might be using the word spectrum with a nuance that exceeds its standard meaning in the mathematical literature, implying something about the states as well as about the spectrum itself, but I'll explain why the statement is true if we define spectrum in the standard way. Namely: the spectrum of an operator $H$ is the set of complex numbers $z$ for which $H-z$ is not invertible (reference 4). Suppose that

  • The spectrum of $H$ has a gap, meaning that all states are separated in energy from the vacuum by a finite gap $\geq M$.

  • The model whose Hamiltonian is $H$ has at least one family of single-particle states among which the energy can be arbitrarily close to $M$.

Then $H$ and $H_0$ definitely do have the same spectrum. This is clear because $H_0$ is defined to have the same single-particle states as $H$, and because single-particle states can have arbitrary energy $>M$ if the lower bound is $M$. Therefore both $H$ and $H_0$ have the spectrum $\{0\}\cup (M,\infty)\subset\mathbb{R}$, no matter what the multi-particle states look like.

But here's the important point, which might relate to what Weinberg really meant by spectrum: equality of the spectra (with the standard definition I used above) does not imply the existence of multiparticle states satisfying Weinberg's equation (3.1.12). The real justification for (3.1.12) is that in QFT, if single-particle states exist, then we can also construct multi-particle states in which the particles are so widely separated that they might as well be non-interacting. As they get farther away from each other (in the infinite future or infinite past), we can take them to approach energy eigenstates. For a more careful treatment of this, see theorem 4.2.1 in reference 2.

I presume that we can not write any $\newcommand{\cH}{\mathcal{H}}\Psi\in \cH$ in this way, otherwise we would simply have $\cH\simeq \cH_0$, which is of course trivial, right?

The Hilbert spaces $\cH$ and $\cH_0$ are isomorphic ($\simeq$) to each other, of course, regardless of the Hamiltonians $H$ and $H_0$. You're really asking whether they're equal to each other (whether the subset $\cH_0\subseteq\cH$ is all of $\cH$). Well, the operators $H=H_0+V$ and $H_0$ both act on $\cH$, and since $H_0$ is a free-field theory, we know that its "in/out states" do span all of $\cH$, so $\cH_0=\cH$.

I'm not sure what you were getting at with "...which is of course trivial, right?" but I'll add this comment just in case. A Hilbert space has no physical significance by itself. It's just a vector space over $\mathbb{C}$ with an inner product, satisfying some conditions. In quantum physics, a model (or theory) is a Hilbert space together with a map that says which operators represent which measurable things. Example: QCD and nonrelativistic single-particle QM are wildly different models, but their Hilbert spaces are isomorphic to each other. I wrote another answer to help clear this up, because sometimes people say Hilbert space when they really mean model (or theory).

Question Q1

According to page 3 in reference 3, this hasn't been strictly proven one way or the other. One difficulty is that the interaction $V$ can't be neglected when the $H$-particles are close together, and $\cH_0$ includes states in which the $H_0$-particles are close together. In leiu of a proof, I'll give a heuristic argument.

In a theory with no interactions, even if we started with a state with $N$ particles sitting right on top of each other, dispersion (because of the usual "uncertainty principle") would eventually make their wavefunctions spread out so thinly that the state is well-approximated by a superposition of states in which the particles are all far away from each other.

Now consider the interacting theory, with Hamiltonian $H$. Start with an arbitrary state $\Psi$ and consider what happens in the infinite past/future, like in your Q3. Heuristically, we expect the same thing to happen. The possibility of bound states is not a problem, because if some of the particles remain permanently bound together, then we have already included that bound state as one of the things we're calling a particle in $\cH_0$.

Altogether, this suggests that even though we started with an arbitrary state $\Psi$, it asymptotically (in the past/future) approaches a superposition of states in which all of the particles are widely separated. If this heuristic argument is correct, then the answer to question Q1 is yes.

Question Q2

Same answer as Q1.

Question Q3

If the answer to Q1 really is yes, then we can expand any state $\Psi$ in terms of either in-states or out-states. Expanding it in terms of in-states (respectively out-states) is more useful if we want to consider observables that are well-localized in the distant past (respectively future), like you suggested. I don't know how this can be "seen from (3.1.12)" by itself, but it can be seen from my heuristic answer to Q1.

Question Q4

Weinberg says we don't, not that we can't. His statement is merely meant to remind us that he's using the Heisenberg picture.

  1. Weinberg (1995), The Quantum Theory of Fields (Volume I: Foundations) (Cambridge University Press)

  2. Haag (1996), Local Quantum Physics (Springer)

  3. Buchholz and Summers (2005), Scattering in Relativistic Quantum Field Theory: Fundamental Concepts and Tools (https://arxiv.org/abs/math-ph/0509047), which was brought to my attention by another answer

  4. Page 6 in Murphy (1990), $C^*$-Algebras and Operator Theory (Academic Press), and also page 180 in Debnath and Mikusiński (2005), Introduction to Hilbert Spaces with Applications (Academic Press)

  • $\begingroup$ Thanks for your nice answer! I have a few follow-up questions. I presume that if $\mathcal H = \mathcal H_0$ isnt then $\mathcal H$ also a Fock space? How is this then reconcilled with the often mentioned statement that the vacuum states of $\mathcal H_0$ and $\mathcal H$ are not the same? Is everything about the quantum theory characterized by how the same Hilbert-space is mapped into itself by Poincare transformations, and the bound states then just expressions of the fact that due to non-trivial Hamiltonian $H$ certain configurations of particles do not dissipate? $\endgroup$
    – jkb1603
    Commented Nov 15, 2021 at 10:45
  • $\begingroup$ @jkb1603 "Fock space" is just a particular way of constructing a Hilbert space. All infinite-dimensional separable Hilbert spaces (over $\mathbb{C}$) are the same as far as their Hilbert-space structure is concerned, no matter how we constructed it. A Hilbert space doesn't know or care which of its vectors represents the vacuum state. $\endgroup$ Commented Nov 15, 2021 at 13:51
  • 1
    $\begingroup$ @jkb1603 In this particular case, the two models do have the same vacuum state by construction. When people say that the interacting and non-interacting models don't have the same vacuum state, they're typically talking about a different non-interacting model, namely the one we get by discarding all non-quadratic terms in the bare Hamiltonian. That's very different than what Weinberg calls the non-interacting model, which includes all of the effects of the interactions that go into determining which states are single-particle states (and the vacuum state). $\endgroup$ Commented Nov 15, 2021 at 14:28
  • 1
    $\begingroup$ @Mr.Feynman I don't remember where I've seen this explained directly, but we can construct examples using lattice QFT, where the math is unambiguous without being dependent on perturbation theory. Consider the $\phi^4$ model, with potential $V(\phi)=\mu\phi^2+g\phi^4$, where $\mu$ and $g$ are the bare coefficients (not renormalized). The vacuum state with $\mu=g=0$ has infinite correlation length, but the vacuum state with $\mu=0$ and $g>0$ has finite correlation length, which can be made arbitrarily short by making $g$ arbitrarily large. $\endgroup$ Commented Apr 25, 2023 at 2:26
  • 1
    $\begingroup$ @Mr.Feynman (The assertions about correlation length in my previous comment can be deduced using the intuition that I described in another answer.) The point is that although the Hamiltonian determines which state qualifies as the vacuum (lowest-energy) state, the correlation length of a given state is independent of the Hamiltonian. So if the correlation length of the vacuum state changes, that can only be because we changed which state is the vacuum state. $\endgroup$ Commented Apr 25, 2023 at 2:27

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.