13
$\begingroup$

Haag's theorem states that the interaction picture does not exist in a rigorous way in relativistic quantum field theory. Some ways in which the interaction picture can fail are that the eigenstates of $H=H_0+V$ can not be expanded in terms of the eigenstates of $H_0$ in a normalizable way, or that the eigenstates of $H$ are not in the domain of definition of $H_0$ and vice versa (they have 'infinite energy').

Maybe the simplest example is if we take $H_0$ to be the Hamiltonian of a free scalar particle of mass $m$ in 1 spatial dimension. If $V=\lambda \int:\phi^2 :dx$, then the full Hamiltonian should just describe a free scalar particle with some shifted mass $m'$. Haag's theorem then states that the Fock states of particles of mass $m'$ can't be expanded in terms of the Fock states of mass $m$.

However, if we simply put the scalar particle on a finite volume L, then we can find the Fock states of one mass in terms of the other. To show this you'll end up doing a Bogoliubov transformation to diagonalize $H$ and you'll find the $m'$ particle vacuum can be expanded in terms of the $m$ particle Fock states.

And what's more, the expectation value of $H_0$ and $N_0$ in the $m'$ particle vacuum appears to be finite. I ended up getting expressions like $\langle H_0\rangle=L\int dk O(k^{-2})$.

So Haag's theorem in this case just seems to be the statement that the $m'$ particle vacuum has non-zero energy and number density in terms of the $m$ particles. So the total energy of the vacuum goes to infinity as $L\rightarrow\infty$, and $H_0$ is not defined on the $m'$ particle Fock states.

My question is whether this is typical? Is Haag's theorem just the statement that the interacting vacuum has non-zero but finite energy density with respect to $H_0$, and so the interaction picture does indeed exist if you have an infrared cutoff? Or are there well-known examples where the interaction picture doesn't exist unless you have a ultraviolet cutoff too? (Note that I'm using normal ordering in $V$ to get rid of harmless divergences)

Appendix

I don't want to type up a full derivation, but my calculation for the $m$ particle annihilation operator $a$ in terms of the $m'$ particle operators $b$ is $$a_k=u_k b_k + v_k b^\dagger_{-k}$$ $$u_k=\cosh \phi_k\qquad v_k=\sinh \phi_k$$ $$\exp(-2\phi_k)=\omega_k'/\omega_k$$ where $\omega'_k$ is the energy of the $k$ mode assuming it has mass $m'$.

The $m'$ particle vacuum in terms of the $m$ particle Fock states is $$\otimes_{k\geq 0} \frac{1}{u_k}\sum_n \left(\frac{v_k}{u_k}\right)^n |{n}\rangle_k|{n}\rangle_{-k}$$

$\endgroup$
5
  • 3
    $\begingroup$ Could you provide your calculations? I'd like to see the expressions for the Bogolubov coefficients. Note that not any expansion of a state in the Fock basis of $m$ particles belongs to this Fock space, for a state to belong to the space the coefficients have to be square-summable ($\sum |a_i|^2 < \infty $). $\endgroup$ Dec 19, 2016 at 14:26
  • $\begingroup$ I added the the expressions in an 'appendix' $\endgroup$
    – octonion
    Dec 19, 2016 at 19:36
  • 1
    $\begingroup$ @octonion BTW you might want to read about the van Hove phenomenon. $\endgroup$ Dec 19, 2016 at 19:45
  • 2
    $\begingroup$ Also, see Haag's Theorem and the Interpretation of Quantum Field Theories with Interactions by Fraser (e.g., page 5). The author defends that a finite volume circumvents Haag's theorem. (I have read some of the thesis some time ago, I don't remember much of it, but I'm sure that it covers some of your concerns). $\endgroup$ Dec 19, 2016 at 19:52
  • 4
    $\begingroup$ Yes, Haag's theorem is due to the infinite volume limit and the associated infrared phenomena. Note that the infinite volume limit is also essential to have a well-defined S-matrix. Thus fields in a box give a severe mutilation of relativistic QFT. $\endgroup$ Dec 20, 2016 at 17:35

2 Answers 2

2
$\begingroup$

This is an answer to a very old question, but I found this while browsing the site and have been studying the topic recently. The answer to your question is that what you're describing is non-generic. It is a feature of perturbations to a Hamiltonian that are very UV-regular. As you noted, it occurs for quadratic perturbations generically. An effect like this also occurs in the construction of the 2d $P(\phi)_2$ theories from the 70s. Similarly to the quadratic perturbation case, a general interaction term in 2D can be made to be UV finite after normal-ordering. The resulting interaction (with IR cutoff) is a well-defined self-adjoint operator on Fock space, and the change-of-representation occurs only as the IR cutoff is removed. This reflects the fact that the vacuum representations for such theories are locally Fock, meaning a unitary equivalence exists so long as you only consider bounded regions.

However, this immediately fails in 2+1D. With the $\phi_3^4$ theory, as you remove the UV cutoff while maintaining the IR one, you are already forced to leave the Fock representation by what I think is essentially a nonperturbative, infinite wavefunction renormalization. So already in 2+1D, the situation you described is not at all typical. UV singularities get worse in higher dimensions, so any hypothetical 4D QFT will be even wilder in this regard.

A reference for the 2D construction using a Hamiltonian approach is Quantum Field Theory and Statistical Mechanics by Glimm and Jaffe. I have not read the 3d construction in detail, but it's significantly more difficult and I am not aware of a pedagogical reference that goes through it. I think you'd have to go through the original papers. The specific discussion of the finite-volume change of representation can be found in this paper of Glimm's: https://link.springer.com/article/10.1007/BF01654131

I'd love to hear if you find something more modern, as I'd like to work through it myself.

$\endgroup$
0
$\begingroup$

One may read Haags theorem in two ways: In a field theory with uncountable degrees of freedom, the representation spaces of observables in normed spaces over the spacelike Cauchy surface foliation of space-time, cannot be guaranteed to be unitarly equivalent to the Fock space representations, ie. free tensor product of bounded linear operators on Hilbert spaces. There will be uncountable many inequivalent representations.

On the the other hand, Haag opens the way to approximations following different paths of deformation the free algebra, that only preserves some measureable expectations of central physical importance, leading to a ladder of better and wider applicable approximations.

That's the picture we see today: Stick to your school of thoughts and methods, perhaps it works and yields some results, some steps further towards or away from a TOE.

As a receipt for an academic career, it has a statistically very low expectation.

$\endgroup$

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.