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I must confess that I'm still confused about the question of time evolution in relativistic quantum field theory (RQFT). From symmetry arguments, from the representation of the Poincare group through unitary operators on a Hilbert space, one knows that the generator of time translations is the Hamiltonian operator ($H$) of the field, and the unitary operator corresponding to a finite time translation from $t_0$ to $t_1$ is $U = \exp[-iH(t_1 - t_0)]$ (with $\hbar = 1$). So apparently, from symmetry arguments, the Schrodinger picture is live and well (i.e., it's valid and must work) in RQFT. However, I've never seen in the literature on RQFT a state vector describing the state of the quantum field at a finite time $t$, that is, I've never seen a $|\psi (t)\rangle$. Then, there is the paper by Dirac in which he takes issue with the Schrodinger time-evolution operator $U = \exp[-iH(t_1 - t_0)]$, practically showing that it doesn't make any sensible sense ("Quantum Electrodynamics without dead wood" published in Phys. Rev., http://journals.aps.org/pr/abstract/10.1103/PhysRev.139.B684).

I would very much appreciate it if you could let me know if $U = \exp[-iH(t_1 - t_0)]$ is indeed the time evolution operator in RQFT, and if the answer is yes, then why the state vector $|\psi (t)\rangle$ is never defined and calculated for a finite time $t$. This question is also important for the path-integral quantization, since in order to construct a path integral one uses the expression $U = \exp[-iH(t_1 - t_0)]$ for the time evolution operator, and also state vectors at finite time. Otherwise, the path integral approach cannot be constructed and the path-integral formulas must be postulated! Are they merely postulated?

EDIT 1: There seems to be a lot of confusion in the comments and answers between the Schrodinger picture and the Schrodinger representation (in QFT). As in all the textbooks on QM there are, by the Schrodinger picture I understand a description of a quantum system by states represented by abstract time-dependent state vectors in some Hilbert space, and whose observables are represented by time-independent operators on that Hilbert space. There is nothing unclear here, and there is no question of any ill-defined "functionals" since no representation has been introduced yet, i.e., no particular basis of the Hilbert space has been chosen in order to pass from abstract vectors to wave functions (or wave functionals) by projecting the abstract vectors on that basis.

As a matter of fact, Dirac in his paper uses abstract vectors in the formalism of second quantization. What is really puzzling, and this is the main cause of my confusion, is that the time-evolution operator should exist in the Schrodinger picture (i.e., there must be a unitary operator relating any two state vectors that have different time arguments), based on the Poincare invariance imposed on any relativistic quantum system. Yet, as Dirac shows the math just doesn't make any sense (and this is not because of some "functionals") and it seems that the Schrodinger picture must be banned from QFT, and this, in turn, comes against Poincare invariance!

EDIT 2: It's interesting to note that $U(t)$ doesn't exist even in the Heisenberg picture and not even for a free field. V. Moretti's construction is faulty since his $U(t)$ doesn't have a domain of definition. Indeed, $ \psi(x) \propto a_p + a_p^{\dagger}$. Therefore, $(\psi(x))^2 \propto (a_{p})^2 + (a_p^{\dagger})^2 + ...$, and hence $H \propto \int (\psi(x))^2 d^3x \propto \sum_p \{(a_{p})^2 + (a_p^{\dagger})^2 + ...\}.$ Hence, $U(t) \propto t^2 \sum_p (a_{p})^{2}(a_p^{\dagger})^{2}$ already at second order in $t$. Hence, acting on the vacum state $U(t)$ gives $\infty$ (for finite $t$!), as Dirac has shown in his paper. Therefore, $U(t)$ doesn't have a domain of definition and it cannot exist, even in the Heisenberg picture!

EDIT 3: @Valter Moretti You claim in your answer that you work entirely in the Heisenberg picture. Therefore, the field depends on both space $\bf{x}$ and time $t$. However, when you construct the Hamiltonian you "smear" the field with a test function, but integrate only over the space, i.e. over $\bf{x}$, but not over time $t$. Therefore, your Hamiltonian operator should depend on time $t$. You cannot simply eliminate $t$ unless you also integrate over it! There is no such construction in the literature! If it's not so, please edit your answer by showing as to how you get rid of the time dependence in the Hamiltonian.

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  • $\begingroup$ Schroedinger functionals $\Psi[\phi,t]$ (here for a scalar field $\phi$) are somewhat impractical, but they even have a wikipedia entry en.wikipedia.org/wiki/Schr%C3%B6dinger_functional , and yes, the time evolution operator is the ``same as it ever was'' (to paraphrase David Byrne). $\endgroup$ – Thomas Oct 2 '15 at 17:45
  • $\begingroup$ @ValterMoretti I'm not trying to offend you in any way. I just want to know your opinion. Am I right or wrong in my statements? An yes or no would be sufficient. Thank you! $\endgroup$ – Andrea Becker Aug 20 '16 at 7:04
  • $\begingroup$ Regarding Edit 2. Performing rigorous computations avoiding formal objects like $a_k$ or performing these formal computations with the due care, you see that the problem does not exist (you have to use the so-called normal order of operators). Regarding Edit 3. The problem does not exist actually: H is a constant of motion exactly as in classical mechanics and classical field theory even if it is function of the classical fields which depend on time. However inside constants of motion all time dependences cancel each other as is well known. $\endgroup$ – Valter Moretti Aug 20 '16 at 7:05
  • $\begingroup$ @ValterMoretti Thank you for your opinion! I'm a very hard to convince person and I don't believe anything, unless I see a rigorous proof. Sorry about that. SE is also a site for learning, and I would be extremely grateful if you or somebody else could rigorously disprove, by providing a rigorous math demonstration, the statements I've made in EDIT 2 and EDIT 3. Thank you very much! $\endgroup$ – Andrea Becker Aug 20 '16 at 7:23
  • $\begingroup$ @ValterMoretti There is no dynamical principle in AQFT, i.e. no Heisenberg equation of motion is being postulated! How do you know which are the constants of motion, then? Thank you! $\endgroup$ – Andrea Becker Aug 20 '16 at 7:56
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Consider for the shake of simplicity a free neutral scalar field $\phi$. Passing to the second quantization picture, it is a operator valued distribution $$C_0^\infty(M;\mathbb R) \ni f \mapsto \phi(f)$$ where $M$ is Minkowski spacetime and $\phi(f)$ is a densely defined symmetric operator on the Hilbert space $$F_+(\cal H) = \mathbb C \oplus \cal H \oplus(\cal H \otimes \cal H)_S \oplus (\cal H \otimes \cal H \otimes \cal H)_S \oplus \cdots $$ Obviously $F_+(\cal H)$ is the symmetrized ($(\cdots)_S$) Fock space generated by the one-particle Hilbert space $\cal H$. And $$\phi(f) = a_{Pf} + a^\dagger_{Pf} \tag{1}$$ Above $P$ extract the positive frequency part of the $4D$ Fourier transform of $f$ and $a^\dagger$ and $a$ are the standard creation and annihilation operators. Formally $$\phi(f) = \int_{\mathbb R^4} f(x) \phi(x) d^4x\:,\tag{2}$$ The domain of definition of $\phi(f)$ is the dense invariant span $D$ of vectors with finite number of particles associated with all possible $f$ as above. With this definition it is possible to give a rigorous meaning to objects (densely defined symmetric operators) of the form $$\int_{\mathbb R^3} f(\vec{x}) \phi(t,\vec{x})^2 d^3x \:\quad \mbox{and}\quad \int_{\mathbb R^3} g(\vec{x}) \partial_a \phi(t,\vec{x}) \partial_b \phi(t,\vec{x}) d^3x$$ where $\mathbb R^3$ is the rest space of an inertial reference frame referred to Minkowskian coordinates. The procedure to do it is usually called normal ordering. Also the limit for $f \to 1$ constantly on $M$ is well defined producing essentially self adjoint operators. In particular integrating the component $T_{00}$ of the stress energy tensor over the rest space $\mathbb R^3$ you obtain the Hamiltonian operator, $H$, of the field as the unique self-adjont extension of the integrated operator. It is finally possible to prove that $$H = \oplus_{k=0}^{+\infty} H_k\:,$$ where $H_0 =0$, $H_1 = \sqrt{m^2+ \vec{P}^2} : \cal H \to \cal H$ is the standard relativistic Hamiltonian operator of one particle associated with the quantumfield $\phi$, $P_\mu$ being the standard four momentum operator of one particle in $\cal H = L^2(\mathbb R^3, d\vec{p}/p^0)$, and $$H_n = H+ \cdots + H \quad \mbox{(n times)}\quad : \cal H \oplus \cdots \oplus \cal H \to \cal H \oplus \cdots \oplus \cal H$$ is the Hamiltonian of a system of $n$ particles. The time evolutor is finally well defined, as $H$ is self-adjoint and reads $$U_t := e^{-itH}$$ With this definition, it is possible to prove that with an interpretation of $\phi$ as the one given in (2), $$U_\tau \phi(t,x) U_\tau^{-1} = \phi(t+\tau, x)$$ which is the integrated version of Heisenberg equation of motion. This identity can be written in a rigorous form using the rigorous interpretation of $\phi$ provided in (1).

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    $\begingroup$ You've just integrated the Heisenberg equations of motion for a system with an infinite number of degrees of freedom; everyone knows this works. However, the question asked if $U=e^{itH}$ makes sense in the Schrodinger picture where $U$ is carried on the space of wave functionals $|\psi\rangle$. All you've done is show that the operators of the defining rep of the infinite-dimensional symplectic group Sp($\infty$,R) are well behaved. Now, in order to answer the question, you need to construct the unitary rep of Sp($\infty$,R) carried on the space of wave functionals. $\endgroup$ – Stephen Blake Oct 3 '15 at 14:34
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    $\begingroup$ No, what I wrote is completely sufficient. Perhaps you have a different opinion on what the Schroedinger representation is! $\endgroup$ – Valter Moretti Oct 3 '15 at 15:36
  • $\begingroup$ "wave functional" who needs them? Could you, please, be more explicit on your definition of Schroedinger representation in this context. I guess that here is the point of our misunderstanding. $\endgroup$ – Valter Moretti Oct 3 '15 at 15:38
  • $\begingroup$ As far as I am concerned the Schroedinger picture is the one where states evolve and observables do not. In the Heisenberg picture, instead, states do not evolve and observable do. Usually QFT is described by means of the latter. $\endgroup$ – Valter Moretti Oct 3 '15 at 15:53
  • $\begingroup$ I think I understand. It seems to me that you are saying that in the Schroedinger representation the states are functionals over a space of field configurations forming a $L^2$ Hilbert space with respect to a suitable measure. It is however known, as you correctly say, that it is just an heuristic viewpoint (similarly, the projective infinite Hilbert space does not admit an unitary invariant measure)...OK this is the point. I am not sure on the interpretation adopted by the OP. If you edit your answer precisely stating your definition of Schroedinger picture I will remove my -1. $\endgroup$ – Valter Moretti Oct 3 '15 at 16:02
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$U=e^{-iH(t_{1}-t_{0})}$ is indeed the formal time evolution operator in RQFT. However, typically, the $U$ does not exist for a system with an infinite number of degrees of freedom and that is why one never sees a $\psi(t)$ calculated explicitly.

In Dirac's 1966 book "Lectures on Quantum Field Theory", section 7 on page 31 studies a model Hamiltonian for an infinite-dimensional fermion system. Dirac shows that one can integrate the Heisenberg equations of motion for the model Hamiltonian and nothing bad happens. He then shows that the attempt to get $U$ by integrating the Schrodinger equation fails because $U$ blows up.

F. A. Berezin's book "The Method of Second Quantization" gives explicit formulas for $U$ for infinite-dimensional bosonic and fermionic systems where the generator (e.g. Hamiltonian) is quadratic in the creation and annihilation operators (Berezin shows that $U$ is a unitary rep of the infinite-dimensional symplectic group Sp($\infty$,R)). In order to get a better understanding of Dirac's lectures, I used Berezin's $U$ for Dirac's model Hamiltonian - but changed to a bosonic system - and found that $U$ blows up and does not exist.

This means that one has to read QFT books written after the 1960s (e.g. Itzykson and Zuber, Weinberg, Peskin and Schroeder) as works of fiction because $U$ appears and the books don't mention that $U$ does not exist. The success of the practical calculations in QFT seems to result from effectively working with a finite number of degrees of freedom (when $U$ does exist) by putting the system on a lattice in a box.

My own view is that the non-existence of $U$ is saying that QFT cannot say anything about the time evolution of $\psi(t)$ so that only a description at infinity makes sense.

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    $\begingroup$ I do not understand what "$U$ does not exist" means! For a free field it exists. If you include interactions there are problems to deal with local interactions (products of at least three fields at the same point), otherwise $U$ is well defined. $\endgroup$ – Valter Moretti Oct 3 '15 at 8:51
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    $\begingroup$ You are maybe referring to the so called Haag's theorem, which states that the interaction picture does not exist and this is another issue... $\endgroup$ – Valter Moretti Oct 3 '15 at 8:54
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    $\begingroup$ I suggest you read Berezin and Dirac. $\endgroup$ – Stephen Blake Oct 3 '15 at 8:54
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    $\begingroup$ Second quantization is well defined the Hamiltonian of a bosonic field is the direct sum $H = \oplus_{k=0}^{+\infty} H_k$ where $H_k$ is the well defined Hamiltonian for a system of identical $k$ bosons. $H$ is selfadjoint because it admits a dense set of analytic vectors. Therefore, by elementary spectral theory $U_t=e^{-itH}$ is a well defined strongly continuous one-parameter group of unitary operators. $\endgroup$ – Valter Moretti Oct 3 '15 at 8:57
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    $\begingroup$ Since you insisted on this wrong statement I voted -1, sorry. $\endgroup$ – Valter Moretti Oct 3 '15 at 8:58

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