Haag's theorem

Prerequisites: Axiomatic Quantum Field Theory. Full path (26 pages)

• Intuitive
• Concrete
• Abstract
• Why is it interesting?
• FAQ

Intuitive

In ordinary quantum mechanics we often split the Hamiltonian into a free part that we can solve and an interaction that we treat as a perturbation. The states of the interacting system are then described as combinations of the states of the free system, so both live in the same space of states. Textbook quantum field theory does the same thing in the interaction picture: we start from free particles and switch on the interaction.

Haag's theorem says that in relativistic quantum field theory, in infinite space, this cannot be done literally. The vacuum of an interacting theory differs from the free vacuum at every point of space. Each small region contributes a small factor to the overlap between the two vacua, and in infinite space infinitely many such factors multiply to zero. The interacting vacuum is then not a state in the space of states of the free theory, and the two theories need different, incompatible spaces of states. Haag showed that if one insists on using the same space with the usual symmetry properties, the "interacting" theory is forced to be free.

This does not mean that perturbative calculations in quantum field theory are wrong. They are done with a finite volume or other regularisation, where the problem does not arise, and the physical predictions (scattering amplitudes, after renormalisation) have a well-defined limit. The theorem shows that the interaction picture is a calculational tool, and that a mathematically rigorous interacting theory has to be constructed by other means.

Concrete

Statement. Let $\phi_0$ be a free scalar field and $\phi$ another Hermitian scalar field, both Poincaré covariant Wightman fields whose vacua are the unique states invariant under the Euclidean group (spatial translations and rotations) at a fixed time $t$. Suppose there is a unitary operator $V$ with $$\phi(t,\vec x) = V \phi_0(t, \vec x) V^{-1}, \qquad \pi(t,\vec x) = V \pi_0(t, \vec x) V^{-1} ,$$ which is what the interaction picture assumes. Then $V$ maps the free vacuum to the vacuum of $\phi$ (up to a phase), and the Wightman functions of the two fields agree up to the four-point function. Since a field whose two-point function equals that of a free field is itself a free field, $\phi$ is a free field. An interaction picture with a unitary $V$ in infinite volume therefore only exists for free theories.

The simplest example: two masses. Already two free fields with different masses $m$ and $m'$ cannot be represented on the same Fock space in infinite volume. At time $t = 0$ write both in terms of the same canonical fields $\phi(\vec x)$, $\pi(\vec x)$. With $\omega_k = \sqrt{\vec k^2 + m^2}$ and $\omega'_k = \sqrt{\vec k^2 + m'^2}$, the annihilation operators of the two theories are related by a Bogoliubov transformation, $$a'_{\vec k} = \cosh\theta_k\, a_{\vec k} + \sinh\theta_k\, a^\dagger_{-\vec k}, \qquad \cosh\theta_k = \frac{1}{2}\left( \sqrt{\frac{\omega'_k}{\omega_k}} + \sqrt{\frac{\omega_k}{\omega'_k}} \right), \qquad \sinh\theta_k = \frac{1}{2}\left( \sqrt{\frac{\omega'_k}{\omega_k}} - \sqrt{\frac{\omega_k}{\omega'_k}} \right) .$$ Each pair of modes $(\vec k, -\vec k)$ is squeezed, and in a box of volume $V$ the overlap of the two vacua is $$\langle 0 | 0' \rangle = \prod_{\vec k} (\cosh\theta_k)^{-1/2} = \exp\left( - \frac{V}{2} \int \frac{d^3k}{(2\pi)^3} \ln\cosh\theta_k \right) .$$ For large $|\vec k|$, $\theta_k \approx (m'^2 - m^2)/(4\vec k^2)$, so the integral converges. The exponent is proportional to the volume, and $\langle 0|0'\rangle \to 0$ as $V \to \infty$. The same happens for the overlap of $|0'\rangle$ with any state with finitely many particles, so in infinite volume $|0'\rangle$ is not a vector in the Fock space of the field with mass $m$: the two representations of the canonical commutation relations are unitarily inequivalent. For an interacting theory the same mechanism makes the interacting vacuum orthogonal to the whole free Fock space.

How practice deals with it. Perturbation theory is set up with an infrared regulator (finite volume, or an adiabatic switching of the coupling) and an ultraviolet cutoff. In the regulated theory the interaction picture is well defined. Physical quantities such as renormalised scattering amplitudes, computed through the LSZ formula, have finite limits when the regulators are removed, even though the operator $V$ does not. In rigorous treatments, scattering is described by Haag-Ruelle theory, which constructs asymptotic particle states directly from the interacting fields without an interaction picture. In two and three spacetime dimensions constructive field theory has built interacting models, such as $\phi^4$ theory, whose Hilbert space representation is inequivalent to the free one, consistent with Haag's theorem.

For the best explanation, see the section "How to stop worrying about Haag's theorem" in the book "The Conceptual Framework of Quantum Field Theory" by A. Duncan

Abstract

In particular Haag's theorem, that one cannot consistently define the vacuum in interacting theories, is now considered to have been the wrong way to look at the problem, and largely disappears with considerations of RG flow.A Practical Introduction to Quantum Field Theory by Felix Flicker

Why is it interesting?

According to something called Haagís theorem there appears to be no known consistent formalism within which interacting quantum field theory can be expressed.

p. 115 in "An interpretative introduction to quantum Field theory" by Teller (1995)

FAQ

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