# Haag's theorem in a box

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 $$\otimes_{k\geq 0} \frac{1}{u_k}\sum_n \left(\frac{v_k}{u_k}\right)^n |{n}\rangle_k|{n}\rangle_{-k}$$

• 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 ($\sum |a_i|^2 < \infty$). – Prof. Legolasov Dec 19 '16 at 14:26
• I added the the expressions in an 'appendix' – octonion Dec 19 '16 at 19:36
• @octonion BTW you might want to read about the van Hove phenomenon. – AccidentalFourierTransform Dec 19 '16 at 19:45
• 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). – AccidentalFourierTransform Dec 19 '16 at 19:52
• 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. – Arnold Neumaier Dec 20 '16 at 17:35