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  Time translation vs time evolution in QM and QFT

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2434 views

I would be very grateful if you could help me with some insight into the concepts of time translation and dynamic evolution in QM and QFT.

Any relativistic quantum system must be Poincare covariant. Cf. Wigner, the Poincare transformations are represented by unitary operators in the system's Hilbert space. The infinitesimal generators of the unitary operators are Hermitian operators corresponding to basic observables. The very definition of the way the time translation operator acts in either Schroedinger or Heisenberg picture is formally equivalent with the Schroedinger and Heisenberg equations of motion (in integrated form), respectively. For example, $\psi (t+a) = U(a)\psi (t)$, etc.

Question 1: Is it sufficient to postulate Poincare covariance instead of postulating the Schroedinger or Heisenberg eqs. of motion? That is, are the Schroedinger or Heisenberg eqs. of motion really consequences (byproducts) of postulating Poincare invariance?

Question 2: I've read a bit axiomatic qft (AQFT) and saw that there is no axiom (postulate) for dynamics among Wightman's postulates. Why is it so? Is it not needed due to an yes answer to Question 1?

I understand that it's hard to come up with a nice Hamiltonian $H$ for a realistic system in AQFT, but the equations of motions should exist at least formally, with an yet to be found $H$. Am I wrong? What are the dynamical equations for a general system (field or otherwise)?

Question 3: Is there a difference between time translation and dynamical evolution? If yes, which is that from the math and physics point of view? Are they described by different operators? Could you provide a simple example for illustration?

Thank you!

asked Aug 26, 2016 in Theoretical Physics by mhrt (55 points) [ revision history ]
recategorized Aug 26, 2016 by Dilaton

1: yes, yes. - 2. yes, yes. - 3. no,no. -- It is too late now to give a detailed answer.

@ArnoldNeumaier In 1970, while presenting Wightman's axioms at a physics school (see Haag's lecture, p.14 in Stanley Deser, Marc Grisaru, and Hugh Pendleton, "Lectures on Elementary Particles and Quantum Field Theory", vol.2, 1970), Haag states: "It has, however, so far not been possible to incorporate a formulation of a specific dynamical law into the framework of axiomatic field theory." How does this square with an "yes, yes" answer to my question 2)?

1 Answer

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1. The dynamical law (Schroedinger resp. Heisenberg equation) is simply the statement that time translations form a unitary representation of the 1-dimensional Lie group $U(1)$. Its infinitesimal generator $H$ automatically has all the required properties (densely defined action on states satisfying the Schroedinger equation, self-adjointness, and adjoint action on observables satisfying the Heisenberg equation of motion). Thus restricting a unitary representation of the Poincare group to time translations provides the dynamics.

2. In case of a QFT satisfying the Wightman axioms, the states are the linear combinations of $N$-point states; the latter are obtained from the distributional (improper) states $|x_1,\ldots,x_N\rangle:=\Phi(x_1)\cdots\Phi(x_N)|vac\rangle$ by integration with suitable weight functions. Spacetime translations (and in particular the dynamics) are now easily seen to be given by $U(x)|x_1,\ldots,x_N\rangle:=|x_1-x,\ldots,x_N-x\rangle$. In particular, once one has a realization of the Wightman axioms, the Hamiltonian is very explicitly given by $H=i\hbar \sum_j \partial/\partial x_j$.

Haag's comment refers not to the lack of an explicit dynamics but to a lack of an explicit construction of an interacting 4D relativistic QFT in which given nonlinear field equations (e.g., equations related to the Lagrangian of QED) are realized. The latter constitute the ''specific dynamical law'' in his quote. In modern terminology, what is missing is a rigorous construction in which a nontrivial operator product expansion (Wilson's replacement of the notion of nonlinear field equations) can be proved.

3. The answer is already in the formula for the Hamiltonian given above.

answered Aug 27, 2016 by Arnold Neumaier (15,787 points) [ revision history ]
edited Aug 27, 2016 by Arnold Neumaier

@ArnoldNeumaier Thank you so much for your very illuminating answer! Now it's clear to me what Haag meant by his statement.

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