# Evolution operator in QFT - why Schrodinger?

• A
Gold Member
I'm reading through a couple of books (Lahiri & Pal's "A First Book of Quantum Field Theory" and Greiner & Reinhardt's "Field Quantization" and have come to the derivation of the evolution operator which leads to the S-matrix. In both books, the derivation starts with the Schrodinger equation in the form $$i\frac{d}{dt}\left|\Psi\left(t\right)\right\rangle =\left(H_{0}+H_{I}\right)\left|\Psi\left(t\right)\right\rangle$$ where ##H_0## is the free-field Hamiltonian and ##H_I## is the interaction Hamiltonian. The derivation from this point on is fairly clear in both books, and leads to the differential equation for the evolution operator ##U(t)##
\begin{align} i\frac{dU\left(t\right)}{dt} & =U_{0}^{\dagger}\left(t\right)H_{I}U_{0}\left(t\right)U\left(t\right)\\ & =H_{I}\left(t\right)U\left(t\right) \end{align}
which leads to the time-ordered product and Wick's theorem.

My question is: is it valid to use the Schrodinger equation (which I always assumed was for nonrelativistic QM) to derive things in a relativistic quantum field theory? Neither book makes any comment on this point. I would have thought you'd need to use the Klein-Gordon or Dirac equation, depending on the particles concerned.

Thanks for any insight.

hilbert2
Gold Member
The choice of field equation affects the Hamiltonian, not the way how time propagation of the state vector happens.

samalkhaiat
My question is: is it valid to use the Schrodinger equation (which I always assumed was for nonrelativistic QM) to derive things in a relativistic quantum field theory?
States of a quantum system (relativistic or not) evolve according to the Schrödinger equation $i \partial_{t}| \Psi \rangle = H | \Psi \rangle$.

I would have thought you'd need to use the Klein-Gordon or Dirac equation, depending on the particles concerned.
These are relativistic field equations not the Schrödinger equations of the relativistic QFT's.

nrqed, dextercioby and bhobba
haushofer
It's nomenclature. E.g. for a scalar field the Klein Gordon eqn tells you how the quantum field evolves. The corresponding Schrodinger equation (as Sam writes it down above me) tells you how the wave function evolves. The corresponding Hamiltonian is relativistic.

Demystifier
Gold Member
There are two different meanings of the expression "Schrodinger equation". The original narrow meaning, that can really be attributed to Schrodinger, is
$$H|\psi(t)\rangle=i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle \;\;\;\; (1)$$
where ##H## is the Hamiltonian of ##n## non-relativistic particles of the form
$$H=\sum_{a=1}^n \frac{{\bf p}_a^2}{2m_a} +V( {\bf x}_1,\cdots,{\bf x}_n) \;\;\;\; (2)$$
The more general meaning is any equation of the form (1), where ##H## does not need to have a form (2). In QFT, the Schrodinger equation is satisfied in this generalized sense, not in the original narrow sense.

hilbert2
Gold Member
The most important feature of this is that the evolution equation of the state vector is always a first order DE in time, so you only need to know the initial state to be able to deduce the later states. No rate of change initial conditions needed like in classical mechanics.

Demystifier
Gold Member
The most important feature of this is that the evolution equation of the state vector is always a first order DE in time, so you only need to know the initial state to be able to deduce the later states. No rate of change initial conditions needed like in classical mechanics.
That's closely related to the fact that QM provides less information than classical mechanics. For instance, if position is given then QM tells you nothing about the momentum. The easiest way to see the relation is to note that Schrodinger equation in QM is very similar to Hamilton-Jacobi equation in classical mechanics, which is also first order in time, and does not, by itself, define particle velocities.

Demystifier
Gold Member
Interestingly, there is a theorem, that such a formulation of interacting dynamics from free dynamics is not possible:
https://en.wikipedia.org/wiki/Haag's_theorem
That's just one of many manifestations of the fact that QFT is not well defined due to the infinite number of degrees of freedom. The Haag's theorem can be avoided by taking an IR cutoff (which in calculations of Feynman diagrams must be done anyway), which however violates the Poincare invariance.

rubi
That's just one of many manifestations of the fact that QFT is not well defined due to the infinite number of degrees of freedom. The Haag's theorem can be avoided by taking an IR cutoff (which in calculations of Feynman diagrams must be done anyway), which however violates the Poincare invariance.
That's not what Haag's theorem says. Free QFT's are well defined in any dimension and well defined QFT's have been shown to exist in lower dimensions (and Haag's theorem still applies there). There's no indication that interacting QFT's can't exist in higher dimensions, but their construction is a difficult open problem. Haag's theorem just says that an interacting QFT is never unitarily equivalent to a free QFT. And that really shouldn't be so surprising. In fact, the opposite would be much more surprising, because there is no physical reason for such a unitary equivalence to exist. It would rather be a mathematical coincidence.

dextercioby
Demystifier
Gold Member
That's not what Haag's theorem says.
True, but that's what some later analysis of the Haag's theorem revealed. See e.g. A. Duncan, The Conceptual Framework of Quantum Field Theory (Oxford 2012).

Demystifier
Gold Member
In fact, the opposite would be much more surprising, because there is no physical reason for such a unitary equivalence to exist.
The unitary equivalence exists in all quantum theories in which Hamiltonian has a finite number of degrees of freedom. I think it's quite a good physical reason to expect that QFT should be unitary as well.

rubi
True, but that's what some later analysis of the Haag's theorem revealed. See e.g. A. Duncan, The Conceptual Framework of Quantum Field Theory (Oxford 2012).
The statement is still wrong. The infinitely many degrees of freedom don't spoil the possibility for well defined QFT's. As I said, counterexamples exist (for example the ##P(\phi)_2## class of theories, to be concrete). Duncan's book doesn't contain any new analysis in this respect. The fact that finite dimensional systems (and thus cut-off systems) are unitarily equivalent (under the assumption of weak continuity), is known as the Stone-von-Neumann theorem. If you place cut-offs and manage to take the continuum limit, then you will still necessarily get a unitarily inequivalent QFT. By the way, that's exactly how interacting QFT's are actually constructed.

The unitary equivalence exists in all quantum theories in which Hamiltonian has a finite number of degrees of freedom. I think it's quite a good physical reason to expect that QFT should be unitary as well.
First of all, that's not a physical argument. But essentially you're saying that if some theorem (the Stone-von-Neumann theorem in this case) holds under certain assumptions (finitely many degrees of freedom), then we should also expect it to hold if we drop the assumptions. That makes no sense to me.

dextercioby
The Schrodinger equation can be derived from the Heisenberg uncertainty principle, the Born rule, and Bohr's correspondence principle [assuming classical mechanics (both relativistic and non-relativistic) in a 'quasi-classical limit'] alone, as in Landau's QM Volume 3.

The actual explicit form of the Hamiltonian, as in post 5, is then derived explicitly from the assumption of Galilean relativity/invariance, and you get non-relativistic quantum mechanics from this, however this is not intrinsic, requiring instead Lorentz invariance one can obtain other Hamiltonian's, e.g. the Dirac equation is a Schrodinger equation, there's nothing wrong with it.

Once you have the Schrodinger equation, you can then transition to the Heisenberg picture, and change basis to the occupation number basis - both non-relativistic and relativitic quantum field operators fall out of this very naturally depending on what form of relativity you assume, c.f. Landau Vol 3 ch. 9.

From this perspective, what is called quantum field theory is simply a re-expression of normal quantum mechanics in the occupation number formalism assuming Einsteinian relativity.

A book like Lahiri-Pal will basically set up quantum field operators by postulating the commutation relations on fields and their canonical momenta directly motivated by the relation between position and momentum operators, as they do in chapter 3 pretty early, which avoids 'deriving' the necessity of field operators and their commutation relations as is done in ch. 9 above, maybe making it seem like QFT is disjoint from non-relativistic quantum mechanics.

In assuming the commutation relations, as Lahiri-Pal do, they have assumed the same material needed to derive the Schrodinger equation, which is why they can start from it in chapter 5.

Whatever about Haag's theorem, one can understand how useless the idea of describing interacting theories is from the statement contained and justified in the first 3 pages of the reference below (the first 3 pages which are important should be previewable on amazon):

• "the description of such a process as occurring in the course of time is therefore just as unreal as the classical paths are in non-relativistic quantum mechanics. The only observable quantities are the properties (momenta, polarisations) of free particles: the initial particles which come into interaction, and the final particles which result from the process"

https://www.amazon.com/dp/0750633719/?tag=pfamazon01-20&tag=pfamazon01-20

Demystifier
Gold Member
But essentially you're saying that if some theorem ... holds under certain assumptions ..., then we should also expect it to hold if we drop the assumptions. That makes no sense to me.
That's called a generalization. Many mathematical theorems have their generalizations.

rubi
That's called a generalization. Many mathematical theorems have their generalizations.
But this one doesn't and Haag's theorem proves that. We can't just pretend that Haag's theorem isn't there. Also, we would expect generalizations, if they did exist, to have weaker conclusions, especially in the case of the Stone-von-Neumann theorem, which is in fact a rather strong and surprising result.

dextercioby
Demystifier
Gold Member
Stone-von-Neumann theorem, which is in fact a rather strong and surprising result.
Why is it surprising?

rubi
Why is it surprising?
Because usually, algebras have plenty of (weakly continuous) representations and not just one (up to unitary equivalence). The Weyl algebra is very exceptional in this respect.

Demystifier
Demystifier
Gold Member
Because usually, algebras have plenty of (weakly continuous) representations and not just one (up to unitary equivalence). The Weyl algebra is very exceptional in this respect.
If I understood it correctly, it means the following. The algebra
$$[x_k,p_{k'}]=i\delta_{kk'}$$
has only one representation (up to unitary equivalence), but the continuous version
$$[\phi(x),\pi(x')]=i\delta(x-x')$$
has many representations. Is that correct?

rubi
If I understood it correctly, it means the following. The algebra
$$[x_k,p_{k'}]=i\delta_{kk'}$$
has only one representation (up to unitary equivalence), but the continuous version
$$[\phi(x),\pi(x')]=i\delta(x-x')$$
has many representations. Is that correct?
Yes, that's correct up to subtleties. (In order to get the uniqueness result for the finite dimensional case, one has to use the Weyl version, i.e. the exponentiated version of the algebra. Otherwise, there are counterexamples even in the finite dimensional case, see Reed & Simon Vol. 1)

Demystifier
Demystifier
Gold Member
Yes, that's correct up to subtleties. (In order to get the uniqueness result for the finite dimensional case, one has to use the Weyl version, i.e. the exponentiated version of the algebra. Otherwise, there are counterexamples even in the finite dimensional case, see Reed & Simon Vol. 1)
So, if one of the representations of the continuous version of the algebra is
$$\pi(x)=i\frac{\delta}{\delta\phi(x)}$$
then what would be another representation? Or is it that the functional derivative above itself is ambiguous?

Demystifier
Gold Member
Because usually, algebras have plenty of (weakly continuous) representations and not just one (up to unitary equivalence). The Weyl algebra is very exceptional in this respect.
Do you perhaps know what is that status of Virasoro algebra, relevant to 2D conformal fields and string theory? Is it also exceptional in that sense? Or if not, is it a problem for unitarity of string theory?

rubi
So, if one of the representations of the continuous version of the algebra is
$$\pi(x)=i\frac{\delta}{\delta\phi(x)}$$
then what would be another representation? Or is it that the functional derivative above itself is ambiguous?
The simplest example I can think of is the case of a free scalar field with mass ##m_1## and another one with mass ##m_2##. If you write down the respective Fock spaces and the corresponding action of the field operators, for example by expressing them in terms of creation and annihilation operators, then these two representations won't be unitarily equivalent unless ##m_1 = m_2##. A calculation can be found in Reed & Simon Vol. II, if I remember correctly.

Do you perhaps know what is that status of Virasoro algebra, relevant to 2D conformal fields and string theory? Is it also exceptional in that sense? Or if not, is it a problem for unitarity of string theory?
The Virasoro algebra too has infinitely many representations, and string theorists do study them. The fact that there are many representations is not a problem for the unitarity of a theory, because you just pick one representation and build the theory within that representation. A problem only occurs if you want to relate two theories built on inequivalent representations with each other, for example if you want to relate the Hilbert space of a scalar field with the Hilbert space of a spin-1 field or if you want to relate the Hilbert spaces of fields with different masses or if you want to relate the Hilbert space of a free QFT to the Hilbert space of an interacting QFT, and so on. The theories may just be too different for such a simple relation to exist.

Demystifier
Demystifier
Gold Member
A problem only occurs if you want to relate two theories built on inequivalent representations with each other, for example ... if you want to relate the Hilbert space of a free QFT to the Hilbert space of an interacting QFT,
So the Haag's theorem does not say that interacting QFT in the Heisenberg or Schrodinger picture doesn't exist. It only says that the intermediate interaction picture doesn't exist. Is that correct?

Last edited:
rubi