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.

  • $\begingroup$ 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
    Commented Oct 2, 2015 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! $\endgroup$ Commented Aug 20, 2016 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. $\endgroup$ Commented Aug 20, 2016 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! $\endgroup$ Commented Aug 20, 2016 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! $\endgroup$ Commented Aug 20, 2016 at 7:56

2 Answers 2


$U=e^{-iH(t_{1}-t_{0})}$ 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. $\endgroup$ Commented Oct 3, 2015 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... $\endgroup$ Commented Oct 3, 2015 at 8:54
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    $\begingroup$ I suggest you read Berezin and Dirac. $\endgroup$
    – user7154
    Commented Oct 3, 2015 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. $\endgroup$ Commented Oct 3, 2015 at 8:57
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    $\begingroup$ OK,if you want. But all that is standard mathematical physics literature, for instance the Bratteli-Robinson textbook (two volumes) studies in great details infinite systems of identical bosons and fermions and rigorously constructs second-quantization self-adjoint Hamiltonians and so on... $\endgroup$ Commented Oct 3, 2015 at 9:07

About Dirac, taking into account the modern understanding of QFT, strictly speaking he is wrong. Dirac, like many others at his time, was frightened by «infinities», but now we understand the nature of these infinities. First, from mathematical point of view, there are trusty regularization methods that turn infinities into finite quantities having dependency on regularization parameter. The mathematical correctness of regularizations is proved. Secondly, from physical point of view, a dependency on regularization means dependence on the more low-level theory, physics at the (supposedly) Planck scale if we talk about fundamental theories.

Therefore, Dirac's prescription for Schrödinger picture should be understood in reverse way. When in a book (usually 60's or earlier) we see a statement that something must be thrown out, that something is denied to do because of an infinity, we must «translate» it into the modern language: it is allowed to do, but with the help of regularization.

Another infinities come from (optional in principle) time interval idealization. Namely, time interval $\Delta t \gg h / E $, where $E$ is process energy scale are considered as infinite $\Delta t = \infty$. This idealization together with other infinities leads to odd-looking, but understandable after some thinking, features of QFT.

Now having in mind all said above, lets see on states in QFT and their time dependency.

QFT comes with something unusual representation called physical or free particles representation. The Hilbert space $\mathcal H $ is a Fock space: $$ \mathcal H = \left\{ \sum_j c_j \prod_i (n_{ij}!)^{-1/2} (\alpha_i^+)^{n_{ij}} \vert 0 \rangle \middle\vert n_{ij} \in \mathbb N_0, \sum_i n_{ij} < \infty, c_j \in \mathbb C, \sum_j |c_j|^2 = 1 \right\}, $$ and Hamiltonian $\hat H$ matrix elements are $$ \langle a \vert \hat H \vert b \rangle = \langle a \vert \hat H_0 \vert b \rangle + \operatorname{const}, \tag{*} $$ where $ a, b \in \mathcal H $, and $ \hat H_0 $ has free Hamiltonian structure: $$ \hat H_0 = \sum_i E_i \hat \alpha_i^+ \hat \alpha_i. $$

The constant in (*) is infinite, so (as Dirac pointed out) $\hat H$ cannot be used as a time dependency generator. (But remember that «cannot» means «it can, but with proper regularization».) But equivalent (because it differs only in a meaningless constant) $\hat H_0$ does not have such problem.

But $\hat H_0$ is a free Hamiltonian! So how it can describe states evolution if, as we know, particles can react? The key is it correctly describes evolution for finite time intervals, but not for infinite one. (Remember, when infinities come into the view we need to use regularization and be accurate with limits.) Particles reaction require infinite time and they are described by $S$-matrix $$ \hat S = \exp \left( { \frac 1 {ih} \int_{-\infty}^{\infty} dt (\hat H - \hat H_0 - \operatorname{const}) } \right), $$ where proper regularization is implied (usually it includes move to a system in a finite volume box, and interaction adiabatic switching).

In the end when regularization is removed we have two well defined things describing states time evolution: free particles Hamiltonian $\hat H_0$ and $S$-matrix $\hat S$. But they do not allow to track particles reaction in time in details.

Such detailed tracking is possible using regularized true Hamiltonian $\hat H$. But the calculation is difficult technically, and in the same time the result will be completely dependent on regularization, which means physically it has heavy dependency on details of the (often unknown) more fundamental theory.

Summing up all of the above, Schrödinger picture usefulness for specific calculations is very limited. But at abstract level we always can use this picture with evolving states.

  • $\begingroup$ "which means physically it has heavy dependency on details of the (often unknown) more fundamental theory." To what extent does this dependency, though, have or not have ramifications regarding "everyday level" physics and the extent to which such embeds into RQFT? Is it that while it depends on the regularizer, that dependence can be considered "small" for the number of decimals we usually take when doing such physics? But you say the details is "heavy" on the regularizer, so that seems to contradict, no? $\endgroup$ Commented Dec 7, 2022 at 19:32
  • $\begingroup$ Lets try to do a rough estimate. Even for atomic energy scale $E \sim 1\,\text{eV}$, and $\Delta t \sim h / E \sim 10^{-15}\,\text{s}$. So using the S-matrix approach we cannot describe time evolution for intervals lesser then $10^{-15}\,\text{s}$. I'd say it's really small and does not play any role at everyday level physics. And this is for very low-energy, very smooth processes. But for time intervals much lesser then it, the time evolution details heavily depends on regularization, and there is no contradiction here. $\endgroup$
    – warlock
    Commented Dec 8, 2022 at 3:28
  • $\begingroup$ That is very interesting. The fastest chemical processes are on the order of a few tens of femtoseconds - is it possible they could be used as a probe to physics beyond what RQFT can account for by measuring them with suitably high accuracy, i.e. maybe just 3 or more decimals? $\endgroup$ Commented Dec 8, 2022 at 5:55
  • $\begingroup$ C.f.: en.wikipedia.org/wiki/Femtochemistry $\endgroup$ Commented Dec 8, 2022 at 5:56
  • $\begingroup$ No, because 1) "we cannot describe time evolution definitely" ≠ "we cannot calculate time evolution result", 2) we don't need to use S-matrix for describing chemistry processes, because we only need "full QFT" when process energy is comparable to the particles rest energy. We actually use QFT in solid state physics, that's why I took atomic energy scale as some low bound, but in principle we can do without QFT in low-energy physics. $\endgroup$
    – warlock
    Commented Dec 8, 2022 at 7:40

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