# Do the interaction picture fields transform as free fields under boosts?

+ 8 like - 0 dislike
995 views

This post was originally written to ask about transformation properties of fields in the interaction picture of QFT under the Poincare transformations. Arnold Neumaier has pointed out that the question is vacuous, because there is no such thing as interaction picture in relativistic QFT. As far as my (not very educated, I must admit - so please correct me if I'm wrong) understanding goes, the problem is that interacting and non-interacting theories are in a way "infinitely different" and one cannot treat both representations of Poincare group and both field operators as well-defined operators acting in the same Hilbert space. This problem is perhaps one of the reasons why QFT is plagued with mathemathical problems. The approach applied by physicists who actually calculate outcomes of experiments is to neglect this problem, treating it as a technical difficulty and then cure resulting inconsistencies and infinities by renormalizaton procedure, leading to consistent results on the level of perturbation theory. In the remainder of this post I will play the game of assuming that interaction picture actually exists, and all calculations presented are purely formal manipulations.

First let's explain notations. Following development in Weinber's book vol. 1, chapters 2 and 3 I expect two representations of Poincare group to exist. $U_0(\Lambda,a)$ transforms asymptotic (in and out) states. $U(\Lambda, a)$ is representation of Poincare group of interacting theory. The relation between the generators of two representations is $H=H_0+V$, $\vec P = \vec P_0$, $\vec J = \vec J_0$, $\vec K = \vec K_0 + \vec W$, where $\vec W$ is correction term which needs to satisfy certain conditions to yield a Lorentz covariant theory (in the sense that $U_0$ commutes with the $S$ operator) - details are described in Weinberg. $\phi(x)$ is some quantum field in the Heisenberg picture. It can be a tensor, spinor or whatever and I neglect relevant indices. It transforms according to $$U(\Lambda, a) \phi(x) U(\Lambda,a)^{\dagger}= D(\Lambda ^{-1}) \phi (\Lambda x +a),$$ where $D$ is some finite dimensional representation of Lorentz group. This is the matrix which acts on Lorentz or spinor indices of $\phi$. Therefore saying what is $D$ is equivalent to specifying the algebraic type of the field $\phi$ - for example $D(\Lambda)=1$ corresponds to a scalar field, $D(\Lambda)=\Lambda$ is a vector field etc.

Now I choose an inertial frame and attempt to define fields in the interaction picture $\phi_I(x)$ as $$\phi_I(x) = U_0(x) \phi (0) U_0(x)^{\dagger}.$$ From the definition, $\phi_I(x)=\phi(x)$ if $x^0=0$ but not otherwise. This definition can be rewritten as $$\phi_I(x) = U_I(x) \phi(x) U_I(x)^{\dagger},$$ where $U_I(a)=U_0(a)U(a)^{\dagger}$. It follows easily from the definitions so far that if $R$ is a rotation and $a$ and arbitrary vector then $$U_0(R,a) \phi_I(x) U_0(R,a) = D(R^{-1}) \phi_I(R x +a),$$ and in this sense interaction picture fields $\phi_I(x)$ transform as free fields under the Poincare group. However, I fail to reproduce the same result if rotation $R$ is replaced by boost or some general Lorentz transformation. The best I can get is $$U_0(\Lambda, a) \phi_I(x) U_0(\Lambda,a)^{\dagger}=D_0(\Lambda^{-1}) \phi_I(\Lambda x +a),$$ where $D_0$ is some, but not necessarily the same representation of Lorentz group. This seems odd, because introduction of interactions shouldn't change the algebraic character of the fields.

My question can be put as follows: Is there a way to argue that (at least under some assumptions about interactions) the representations $D$ and $D_0$ are equal?

This post imported from StackExchange Physics at 2016-08-31 10:21 (UTC), posted by SE-user Blazej

asked Aug 23, 2016
edited Aug 31, 2016
Cannon and Jaffe discuss Lorentz boosts in a 1+1 dimensional field theory.

This post imported from StackExchange Physics at 2016-08-31 10:22 (UTC), posted by SE-user Keith McClary

+ 5 like - 0 dislike

Haag's theorem (the Wikipedia article linked to is poorly written but has useful references) says that the interaction picture for relativistic quantum fields does not exist - i.e., assuming one leads to a contradiction. The reason is that the spatial field $\phi(0,\mathbb{x})$ is ill-defined since $\phi(x)$ is only a distribution, so that one cannot match fields at a fixed finite time. The same holds for your $\phi(0)$.

In particular, Haag's theorem states that the representations of the commutation relations of the interacting theory and the free theory are inequivalent. This more or less answers what your final question seems to be asking. (Either one keeps the Poincare representation intact and does not get a free field representation from the interacting one - what you apparently try to do -, or one keeps the free commutation relations intact - what Weinberg is doing - and does not get a valid interaction picture.)

This is not odd, as the interactions are too singular to define a self-adjoint operator in the noninteracting Hilbert space. Renormalization is needed just because of this fact. See my tutorial paper Renormalization without infinities - a tutorial for explanations why singular interactions change the Hilbert space, already in much simpler situations.

The fact that the interacting fields carry a representation of the Poincare group implies that their transformation behavior is the same as for the corresponding free fields - for translations, rotations, and boosts. Indeed, what you quoted from Weinberg holds for any relativistic quantum field. What is different from the free case is that they neither satisfy the free field equations nor the canonical commutation rules. They only satisfy the causality requirement of spacelike (anti)commutativity.

There is no reason why your $\phi_I$ should have nice Lorentz properties as it is a noncovariant object, due to having fixed a time direction (in the construction of the two representations $D$ and $D_0$). The correspondence between the asymptotic fields $\phi_\pm(x)$ and the interacting field $\phi(x)$ is not given by your relation (that matches times at $t=0$) but by a relation that matches times at $t=\pm\infty$. The correct correspondence is given by Haag-Ruelle theory.

In the less rigorous treatment by Weinberg (which pretends that one can get the interacting fields from the free field by switching on interactions), you can find the relation discussed in the context where he proves the Lorentz invariance of the S-matrix.

This post imported from StackExchange Physics at 2016-08-31 10:22 (UTC), posted by SE-user Arnold Neumaier
answered Aug 28, 2016 by (15,747 points)
I must admit that I didn't know about Haag's theorem. From reading QFT textbooks one gets the impression that everything is all right with interaction picture. I don't know what to think about it, so I will leave this question open for now.

This post imported from StackExchange Physics at 2016-08-31 10:22 (UTC), posted by SE-user Blazej
@Blazej: Well, as long as a cutoff is used, or only formal manipulations are done, one can pretend that the interaction picture exists. Its nonexistence leads to the UV divergences that must be cured by renormalization, which destroys the interaction picture but saves the perturbative computability. I'll add a reference to Haag's theorem. -- Independent of that, your first formula is meaningless as it stands since the right hand side depends on $t$ but the left side shows no sign of this.

This post imported from StackExchange Physics at 2016-08-31 10:22 (UTC), posted by SE-user Arnold Neumaier
Note that the Wikipedia article I linked to is poorly written but has useful references.

This post imported from StackExchange Physics at 2016-08-31 10:22 (UTC), posted by SE-user Arnold Neumaier
Both sides depend on $t$. What I meant (an forgot to write) is that $t=x^0$.

This post imported from StackExchange Physics at 2016-08-31 10:22 (UTC), posted by SE-user Blazej
@Blazej: Then it is still not ok, as the right hand $x$ must be taken at $x^0=0$. Please correct your formula, as I want to further comment on it.

This post imported from StackExchange Physics at 2016-08-31 10:22 (UTC), posted by SE-user Arnold Neumaier
I rewrote my post from the scratch. I hope that now it is clearer what I meant. As far as I understand this approach I am trying to use (and seems to be prohibitted by Haag's theorem) is what is usually done in perturbative calculations and at this level it works after renormalization. Please excuse me if I am completely wrong, these are merely my baby steps in studying QFT.

This post imported from StackExchange Physics at 2016-08-31 10:22 (UTC), posted by SE-user Blazej
@Blazej: No need to excuse yourself. I updated my answer.

This post imported from StackExchange Physics at 2016-08-31 10:22 (UTC), posted by SE-user Arnold Neumaier
+ 0 like - 0 dislike

Yes, they do.

As to "UV problem" in QFT, it depends entirely on how we construct the interaction. As far as our field products are not proven to be a unique way of interaction, we must admit other ways, which may be free from undesirable perturbative corrections.

answered Aug 31, 2016 by (92 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOve$\varnothing$flowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.