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In his QFT book, Weinberg claims what follows (Vol I, pag. 144-145): given a (free) representation of the Poincaré group with generators $\bf P$ (spatial translations), $H_0$ (time translations), $\bf J$ (rotations), and ${\bf K}_0$ (boosts), one obtains another (interacting) reprentation by replacing $H_0$ and ${\bf K}_0$ with $$ H = H_0 + \int \mathcal{H}({\bf x}, 0) d^3x \; \; \text{ and } \; \; {\bf K} = {\bf K}_0 - \int {\bf x} \mathcal{H}({\bf x}, 0) d^3x, $$ respectively, where $\mathcal{H}(x)= \mathcal{H}({\bf x}, t)$ is a self-adjoint operator density which is covariant and causal, that is: $$ U_0(\Lambda, a) \mathcal{H}(x) U_0(\Lambda, a)^{-1}= \mathcal{H}(\Lambda x + a) \; \; \text{ and } \; \; [\mathcal{H}(x), \mathcal{H}(y )] = 0 \text{ for } (x - y)^2 \geq 0, $$ where $U_0(\Lambda, a)$ is the free representation.

Weinberg provides only a partial proof of this claim. Does somebody know if this is a rigorous result, and where one can possibly find a complete/rigorous proof?

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What level of rigorousness do you want ? I think arguments given by Weinberg in section 3.3. make a fairly complete proof of his statements (at least at heuristic level). –  user10001 Nov 18 '12 at 13:18
    
The same level of rigorousness at which non-interacting representation can be defined, which is the level of a mathematical theorem. –  BGal Nov 18 '12 at 18:00
    
then I guess (as @Arnold says) its still an unsolved problem :) –  user10001 Nov 18 '12 at 18:07
    
I think the problem can be split into two parts: 1) given a density H(x) with suitable properties (what are the exact properties?), can an interacting representation be constructed as shown above? 2) to find a density H(x) with the required properties. My original question was about point 1. –  BGal Nov 18 '12 at 18:28
    
Hmm.. the conditions on H are I think given by equations 3.3.19, and 3.3.21 in Weinberg's book (note that in these equation he denotes (integral of) H(x) as V). –  user10001 Nov 18 '12 at 18:45

2 Answers 2

up vote 3 down vote accepted

Using the notation:

$ H = H_0 + V$ and $ \mathbf{K} = \mathbf{K_0} + \mathbf{Z}$

The requirement that both free and interacting generators satisfy the Poincaré algebra commutaion relations, lead to the following requirements: (Please see the following book by Eugene Stefanovich (published in the arXiv) equations 6.22-6.26 (page 179):

$[\mathbf{J} , V] = [\mathbf{H_0} , V] = 0$

$[P_i , Z_j] = i \delta_{ij} V, [J_i, Z_j] = i \epsilon_{ijk} Z_k$

$[K_{0[i}, Z_{j]}] +[Z_i,Z_j]=0 $, $[\mathbf{Z}, H_0] + [\mathbf{K_0} , V] + [ \mathbf{Z}, V] = 0$;

Now, to verify that these relations are satisfied in the present case, please observe that due to the first given realtion expressing the transformation properties of the interaction Hamiltonian density that the action of the free Poincaré generators on the Hamiltonian density is by means of the well known differential operator realization:

[$H_0, \mathcal{H}(\mathbf{x}, 0)] = (\frac{\partial}{\partial t} \mathcal{H})(\mathbf{x}, 0)$

$[P_i , \mathcal{H}(\mathbf{x}, 0)] = i \frac{\partial}{\partial x_i} \mathcal{H}(\mathbf{x}, 0)$

$[J_i , \mathcal{H}(\mathbf{x}, 0)] = i\epsilon_{ijk} x_i \frac{\partial}{\partial x_j} \mathcal{H}(\mathbf{x}, 0)$

$[K_{0i}, \mathcal{H}(\mathbf{x}, 0)] = ((t \frac{\partial}{\partial x_i} - x_i\frac{\partial}{\partial t})\mathcal{H})(\mathbf{x}, 0) = -x_i(\frac{\partial}{\partial t} \mathcal{H})(\mathbf{x}, 0)$

What is left is to perform the substitutions. But in the addition of the given requirements, one must assume that the interaction Hamiltonian density vanishes sufficiently rapidly at infinity, and surface terms in integration by parts can be ignored. (Of course these are operators and one must specify the strength of convergence).

Here is a sample calculation of one of the required commutation relations:

$[J_i , Z_j] = \int x_j i \epsilon_{ilm} x_l \frac{\partial}{\partial x_m} (\mathcal{H}(\mathbf{x}, 0) )d^3x$

Please observe that the Poincaré generators do not act on the free $x_i$'s in the integrand, because they are only dummy integration variables. Thus after integration by parts we get:

$[J_i , Z_j] = -\int i \epsilon_{ilm} (\delta_{jm} x_l + \delta_{lm} x_j) \mathcal{H}(\mathbf{x}, 0) d^3x = + i \epsilon_{ijl} x_l \mathcal{H}(\mathbf{x}, 0) d^3x = i \epsilon_{ijk} Z_k$

Finally, please let me remark that finding an exact representation of the interacting Poincaré algebra would require knowing the exact solution (Hilbert space and operator eigenvalues) of the interacting quantum field model, which is not known outside perturbation theory, however, the interacting Poincaré generators can be deduced form the Lagrangian by means of Noether's theorem + canonical quantization. The forms of interacting Poincare algebra were already studied by Dirac in 1949.

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But one also needs to assume that $H(x,0)$ is well-defined, which is the real problem in making the proof rigorous. –  Arnold Neumaier Nov 18 '12 at 14:24
    
@David: Thanks for the proof. But you say: "Finally, please let me remark that finding an exact representation of the interacting Poincaré algebra would require knowing the exact solution of the interacting quantum field model". I do not understand: if I find a well defined $H(x)$ satisfying all the required properties, what else do I need for defining an interacting representation? –  BGal Nov 18 '12 at 18:11
    
@BGal One can easily find an interaction term, and construct the expressions of the interacting algebra in terms of the field variables. For example in a scalar field theory \mathcal{H} = \lambda \phi^4 is a possible interaction term, and it is easy to exactly construct interacting Poincaré generators in terms of the field and its derivatives. Finding a representation on the other hand would mean to find a Hilbert space on which the action of the interacting Poincaré generators is completely known. –  David Bar Moshe Nov 19 '12 at 8:15
    
cont., In 4D scalar theory, in the free case, the (free) Poincaré generators can be expressed in terms the field's creation and anihilation operators and their action on the corresponding (free) Fock space is completely known. In addition approximate representations can be constructed order by order in perturbation theory in which the interacting Poincaré generators will have higher polynomial dependence on the creation and anihilation operators (acting on the free Fock space), but an exact representation is not known. –  David Bar Moshe Nov 19 '12 at 8:16

There is no rigorous proof for this, as there is no known way to make sense of the interaction part of a local Hamiltonian at fixed time in such a way that $H$ becomes self-adjoint. The reason is that quantum fields are distributions only, and the product of distributions is ill-defined in general. There are nonrigorous recipes for renormalizing such Poincare representations, but these recipes are defined perturbatively only.

Indeed, the construction of interacting local unitary reps of the Poincare group is completely unsettled in 4 dimensions. No example is known, neither are there nogo theorems.

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But what does happen if we simply take $H(x) = \psi(x)$, where $\psi(x)$ is the usual free field? –  BGal Nov 18 '12 at 17:48
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@BGal: This is not a good Hamiltonian, as it is not bounded below. It is a reasonable interaction part, but this only leads to a free theory, as after a shift of the field (and the subtraction of an infinite constant to renormalize the Hamiltonian) we are back in the free case. –  Arnold Neumaier Nov 18 '12 at 18:00

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