Haag's Theorem, Perturbation, Existence and QFT.

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  • #26
strangerep
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[...]
In other words, [Haag's] theorem derives the non-existence of the interacting Hamiltonian from the covariant transformation law of the interacting field (1). In my formulation, I simply exchanged places of these two statements. I derived the non-existence of the covariant transformation law (1) from the existence of the interacting Hamiltonian [itex] H [/itex].

I think that both formulations are equivalent. [...]
Suppose we have a theorem which says "if A then B". We can't use it to
automatically infer "if B then A". I.e: simply exchanging the places of
the two statements is not valid without supplying a detailed proof of the
reverse direction. It's like a maths exam question that asks the student
to prove "A if and only if B". If the student only proved the forward
direction "if A then B", but didn't also prove "if B then A", then
(s)he would not receive full marks.

The reason why I chose to formulate Haag's theorem in the non-traditional form is that I have no reason to doubt the existence of interactions, however, I am very doubtful about the physical meaning and usefulness of "interacting fields". From this point of view, the theorem becomes completely harmless.
That's why I said what you wrote is one approach for "evading"
the theorem. I.e: if we don't like the consequences of a theorem, then we
must abandon one or more of its required pre-conditions. In your case,
you're advocating abandonment of the usual form of the boost transformation
law which had been phrased with a Minkowski spacetime picture in mind.
Personally, I think this is an interesting research direction to pursue, especially
in view of the well-known problems about Lorentz covariance of trajectories of
interacting particles discussed elsewhere.
 
  • #27
strangerep
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So, can we conclude that there is no covariant theory for interacting Hamiltonians?I read that the problem is circumvented by LSZ formula--how?--by dealing with the free field only?
I think all we can conclude is that the orthodox axioms for QFT lead to
logical problems when trying to tackle the physical interactions we're
most interested in.

The LSZ formalism is not entirely rigorous, so I'll just say a little bit about
its more rigorous cousin known as "Haag-Ruelle scattering theory". In the
latter, one talks about asymptotic in/out fields which are assumed to be free,
since they're what we measure in scattering experiments. One also talks
about "interpolating" fields, representing what goes on during the
interaction process. The theory discusses how the interpolating fields
approach the in/out fields in a certain precise sense. Unfortunately, all
this extra effort is not much help in practical scattering calculations.

There is a brief elementary introduction to some aspects of
non-perturbative QFT in Ch7 of Peskin & Schroeder. Other books like
Streater & Wightman, as well as Haag's "Local Quantum Physics"
are rather more difficult.
 
  • #28
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Suppose we have a theorem which says "if A then B". We can't use it to
automatically infer "if B then A". I.e: simply exchanging the places of
the two statements is not valid without supplying a detailed proof of the
reverse direction. It's like a maths exam question that asks the student
to prove "A if and only if B". If the student only proved the forward
direction "if A then B", but didn't also prove "if B then A", then
(s)he would not receive full marks.

I don't think I violated rules of logic. In a shorthand notation (the traditional form of) Haag's theorem establishes the following implication

"covariant transformations of fields" ==> "trivial interaction"

Following rules of Aristotle-Boole logic, I am allowed to take negation of both sides and invert the direction of implication. So, I obtain

"non-trivial interaction" ==> "non-covariant transformations of fields"

i.e., Haag's theorem in my formulation.

Nevertheless, I agree with you that this is a hand-waving, and that a serious approach to the theorem requires a clear formulation of all conditions and a rigorous proof. I haven't done that.



That's why I said what you wrote is one approach for "evading"
the theorem. I.e: if we don't like the consequences of a theorem, then we
must abandon one or more of its required pre-conditions. In your case,
you're advocating abandonment of the usual form of the boost transformation
law which had been phrased with a Minkowski spacetime picture in mind.
Personally, I think this is an interesting research direction to pursue, especially
in view of the well-known problems about Lorentz covariance of trajectories of
interacting particles discussed elsewhere.
Yes, I agree with you. Haag's theorem is a serious challenge to our understanding of QFT. This paradox, indeed, requires us to abandon some of the conditions. The question is: which one? There could be different opinions.

One common idea is that free and interacting Hamiltonians may act in different Hilbert spaces, and even not share the same vacuum vector. This is one method to "evade" Haag's theorem. However, this method looks very suspicious to me. It contradicts almost everything I know about quantum theory.

Another idea is that the covariant transformations of interacting fields is an ill-justified statement. Maybe we can just abandon it, without any serious consequences for the theory?

Are there other ideas how Haag's theorem can be "evaded"?

Eugene.
 
  • #29
strangerep
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I don't think I violated rules of logic. In a shorthand notation (the traditional form of) Haag's theorem establishes the following implication

"covariant transformations of fields" ==> "trivial interaction"

Following rules of Aristotle-Boole logic, I am allowed to take negation of both sides and invert the direction of implication. So, I obtain

"non-trivial interaction" ==> "non-covariant transformations of fields"

i.e., Haag's theorem in my formulation.
You're ignoring the other pre-conditions of Haag's thm. Let me re-phrase my
previous illustration: given a theorem that says "if (A & B) then C", we can
do the negation as you said, to get "if not(C) then not(A&B)", the latter part
of which expands to "... not(A) or not(B)".

But let us not hijack this thread away any further from the issues
the OP was specifically interested in.
 
  • #30
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One common idea is that free and interacting Hamiltonians may act in different Hilbert spaces, and even not share the same vacuum vector. This is one method to "evade" Haag's theorem. However, this method looks very suspicious to me. It contradicts almost everything I know about quantum theory.
Please elaborate your statement--how does this contradict almost everything you know about quantum theory.
 
  • #31
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Please elaborate your statement--how does this contradict almost everything you know about quantum theory.
In quantum mechanics, every physical system is described by a Hilbert space. States of the system are described as unit vectors in this Hilbert space, and observables are described as Hermitian operators there. If we know the state vector and the operator of observable, we can calculate the probabilities of measurements of different values of this observable in this state and compare them with experiments.

In addition, we would like to know how the state vactor is transformed when the observer changes (the time dynamics is one example of such a transformation). This knowledge is provided by the interacting representation of the Poincare group acting in the Hilbert space of the system. The interacting Hamiltonian H is one (out of ten) generator of this representation. Also, one can always define the non-interacting representation of the Poincare group in the same Hilbert space with the time-translation generator H_0. Both H and H_0 act in the same Hilbert space.

The same principles work in quantum field theory. The only difference is that QFT deals with systems in which the number of particles is not conserved, so the Hilbert space is, actually, the Fock space with variable number of particles. This is how I understand quantum mechanics and QFT.


I don't know how one can justify introduction of two different Hilbert spaces (one for the interacting Hamiltonian and vacuum, and another for the non-interacting Hamiltonian and vacuum). I think these ideas (AKA "inequivalent representations of the canonical commutation relations") go against letter and spirit of quantum mechanics. These ideas are presented in many places (e.g., Umezawa "Thermo fields dynamics and condensed states") but they don't make sense to me.

Eugene.
 
  • #32
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.

Also, one can always define the non-interacting representation of the Poincare group in the same Hilbert space with the time-translation generator H_0. Both H and H_0 act in the same Hilbert space.

The same principles work in quantum field theory. The only difference is that QFT deals with systems in which the number of particles is not conserved, so the Hilbert space is, actually, the Fock space with variable number of particles. This is how I understand quantum mechanics and QFT.

Eugene.
Starting with a state of the free hamiltonian,say, we switch on the interaction--the state initially lies in Hilbert space with fixed no. (say n) of particles and later evolves into a state in the Fock space with variable number of particles.So do H and H_0 really act on the same Hilbert space?
Regarding vacuum, one can insist that vacuum should remain vacuum in the presence of interactions--but the two vacua(don't know if this is the right word!) are not really the same.
 
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  • #33
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Starting with a state of the free hamiltonian,say, we switch on the interaction--the state initially lies in Hilbert space with fixed no. (say n) of particles and later evolves into a state in the Fock space with variable number of particles.So do H and H_0 really act on the same Hilbert space?
Yes. In both cases the Hilbert space is, actually, the Fock space, which is built as a direct sum of tensor products of n-particle spaces. For example, in the world where only one particle type exists, and the 1-particle Hilbert space is H, the Fock space has the form

[tex] F = |0 \rangle \oplus H \oplus (H \otimes H) \oplus (H \otimes H \otimes H) \ldots [/tex]

where [itex] |0 \rangle [/itex] is the vacuum vector, and tensor products [itex] H \otimes H \ldots [/itex] imply symmetrization/antisymmetrization, as appropriate.

The Fock space allows you to describe all kinds of time evolutions, including those in which the number of particles can change.


Regarding vacuum, one can insist that vacuum should remain vacuum in the presence of interactions--but the two vacua(don't know if this is the right word!) are not really the same.
It is true that for most Hamiltonians in QFT the original (bare) vacuum state [itex] |0 \rangle [/itex] is not an eigenstate. The same is true for bare 1-particle states [itex] |1 \rangle [/itex]. They are not eigenstates of the interacting Hamiltonian. The usual solution is to say that the vacuum and one particle states change in the presence of interaction, so that they become "physical" vacuum [itex] |vac \rangle [/itex] and "physical" particles [itex] |one \rangle [/itex], respectively, which are rather complex linear combinations of n-particle bare states. In popular books this is also described as "virtual particles", which fill vacuum and form "clouds" surrounding real particles.

In physical applications, we are not interested in properties of bare particles and bare vacuum. All real physical processes (scattering, formation of bound states, etc.) involve physical states [itex] |vac \rangle [/itex] and [itex] |one \rangle [/itex]. So, the fact that the Hamiltonian is normally expressed in terms of creation and annihilation operators of bare particles is very inconvenient.

Luckily, there is an alternative formulation of QFT in which bare particles do not appear at all. The Hamiltonian is expressed directly in terms of creation and annihilation operators of physical particles. The vacuum state is a state with no physical particles, as expected. This "dressed particle" approach was first suggested in

O. W. Greenberg, S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim., 8 (1958) 378.

It appears that traditional quantum field theories (including QED) can be recast into the "dressed particle" form without losing any of their predictive power. The characteristic feature of interacting Hamiltonians in dressed particle theories is that they yield zero when acting on physical vacuum and one-particle states. So, these states do not depend on whether the interaction is present or absent. Intuitively, this is understandable: by definition, interaction can take place only when there are two or more particles. One (or zero) particle has nothing to interact with. In other words, the "dressed particle" approach eliminates the notion of self-interaction in the vacuum and one-particle states. One can also say that even if the self-interaction is present (as in usual QFT theories), it can be always incorporated into the definition of physical vacuum and physical particles and "forgotten".

I've written a bit more about these ideas in

https://www.physicsforums.com/showpost.php?p=1391726&postcount=6

Eugene.
 
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  • #34
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Yes. In both cases the Hilbert space is, actually, the Fock space, which is built as a direct sum of tensor products of n-particle spaces. For example, in the world where only one particle type exists, and the 1-particle Hilbert space is H, the Fock space has the form

[tex] F = |0 \rangle \oplus H \oplus (H \otimes H) \oplus (H \otimes H \otimes H) \ldots [/tex]

where [itex] |0 \rangle [/itex] is the vacuum vector, and tensor products [itex] H \otimes H \ldots [/itex] imply symmetrization/antisymmetrization, as appropriate.

The Fock space allows you to describe all kinds of time evolutions, including those in which the number of particles can change.
Ok,so H_0 acts on a subspace while (operator)H acts on the complete (Fock)space.


In other words, the "dressed particle" approach eliminates the notion of self-interaction in the vacuum and one-particle states. One can also say that even if the self-interaction is present (as in usual QFT theories), it can be always incorporated into the definition of physical vacuum and physical particles and "forgotten".

I've written a bit more about these ideas in

https://www.physicsforums.com/showpost.php?p=1391726&postcount=6

Eugene.
So the virtual particles are artefacts of using bare states as our basis in QFT--we should get rid of this approach and use physical states to avoid confusion.

Coming to Haag's theorem--if you acknowledge that physical vacuum is different from bare vacuum,what implications does it have on the interpretation of Haag's theorem?
 
  • #35
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Ok,so H_0 acts on a subspace while (operator)H acts on the complete (Fock)space.
No, H_0 also acts on the complete Fock space. Its expression in the entire Fock space is

[tex] H_0 = \int d^3p \sqrt{m^2c^4 + p^2c^2} a^{\dag}_p a_p [/tex]

From this formula one can obtain more explicit and physically transparent formulas in each individual sector. For example, in the 2-particle sector

[tex]H_0 = \sqrt{m^2c^4 + p_1^2c^2} + \sqrt{m^2c^4 + p_2^2c^2} [/tex]

where [itex]p_1 [/itex] and [itex]p_2 [/itex] are momenta of the two particles.



So the virtual particles are artefacts of using bare states as our basis in QFT--we should get rid of this approach and use physical states to avoid confusion.
Yes, I believe so. That's what Greenberg-Schweber "dressed particle" approach does.


Coming to Haag's theorem--if you acknowledge that physical vacuum is different from bare vacuum,what implications does it have on the interpretation of Haag's theorem?
That's an interesting question. I haven't thought it through. I am not sure whether Haag's theorem requires both interacting and non-interacting theories to have the same vacuum vector. This requirement would exclude all traditional relativistic quantum field theories known to man.

On the other hand in "dressed particle" theories the bare in interacting vacua are exactly the same, and Haag's theorem doesn't apply, because interacting fields do not transform by traditional formulas

[tex] U(\Lambda) \psi(x) U^{-1} (\Lambda) = \psi(\Lambda x) [/tex]

You can find more discussions in

M. I. Shirokov, "Dressing" and Haag's theorem, http://www.arxiv.org/math-ph/0703021 [Broken]

Eugene.
 
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  • #36
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From this formula one can obtain more explicit and physically transparent formulas in each individual sector. For example, in the 2-particle sector

[tex]H_0 = \sqrt{m^2c^4 + p_1^2c^2} + \sqrt{m^2c^4 + p_2^2c^2} [/tex]

where [itex]p_1 [/itex] and [itex]p_2 [/itex] are momenta of the two particles.
What I meant was that once an H_0 is defined thus(in a partiular sector),it continues to act only in this sector or subspace.



That's an interesting question. I haven't thought it through. I am not sure whether Haag's theorem requires both interacting and non-interacting theories to have the same vacuum vector. This requirement would exclude all traditional relativistic quantum field theories known to man.
Why would it exclude all traditional relativistic quantum field theories known to man?You may like to read the following from an article on Haag's theorem that I have downloaded.This is as the author says 'a gist of the heuristic version of Haag's original theorem'.

'This argument takes the form of a reductio. Suppose that we are trying to describe both a free scalar field and a self-interacting scalar field using the same Hilbert space [tex]H[/tex]. Suppose that we demand of the vacuum state that it be the unique (up to phase) normalized state [tex] |0> \epsilon H [/tex] that is invariant under Euclidean translations. And suppose that the vacuum state is the ground state in that it is an eigenstate of the Hamiltonian with eigenvalue 0.These suppositions are fullfilled in the case of the free scalar field with mass m > 0, the usual no-particle state [tex] |0_F> [/tex] (bare vacuum), and the free field Hamiltonian H_F .
Since the vacuum state [tex] |0_I> [/tex] of the interacting field (dressed vacuum or physical vacuum) should also be invariant under Euclidean translations,it follows from the stated assumptions that |0_I> = c|0_F>, |c| = 1, and since |0_I> is annihilated by the Hamiltonian H for the interacting field, it follows that H|0_F> = 0. But the typical Hamiltonians for interacting fields take the form H_F +H_I , where H_I describes the interaction of the field with itself, and such Hamiltonians do not annihilate |0_F> (H polarizes the vacuum)'.

If you want to read more google for 'Haag's Theorem and Its Implications for the Foundations of Quantum Field Theory'.
 
  • #37
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meopemuk said:
From this formula one can obtain more explicit and physically transparent formulas in each individual sector. For example, in the 2-particle sector

[tex]H_0 = \sqrt{m^2c^4 + p_1^2c^2} + \sqrt{m^2c^4 + p_2^2c^2} [/tex]

where [itex]p_1 [/itex] and [itex]p_2 [/itex] are momenta of the two particles.

What I meant was that once an H_0 is defined thus(in a partiular sector),it continues to act only in this sector or subspace.
Yes, we agree on this point. Each sector of the Fock space is invariant with respect to the action of the free Hamiltonian H_0. In other words, free time evolution conserves the number of particles, as expected.


Why would it exclude all traditional relativistic quantum field theories known to man?You may like to read the following from an article on Haag's theorem that I have downloaded.This is as the author says 'a gist of the heuristic version of Haag's original theorem'.

'This argument takes the form of a reductio. Suppose that we are trying to describe both a free scalar field and a self-interacting scalar field using the same Hilbert space [tex]H[/tex]. Suppose that we demand of the vacuum state that it be the unique (up to phase) normalized state [tex] |0> \in H [/tex] that is invariant under Euclidean translations. And suppose that the vacuum state is the ground state in that it is an eigenstate of the Hamiltonian with eigenvalue 0.These suppositions are fullfilled in the case of the free scalar field with mass m > 0, the usual no-particle state [tex] |0_F> [/tex] (bare vacuum), and the free field Hamiltonian H_F .
Since the vacuum state [tex] |0_I> [/tex] of the interacting field (dressed vacuum or physical vacuum) should also be invariant under Euclidean translations,it follows from the stated assumptions that |0_I> = c|0_F>, |c| = 1, and since |0_I> is annihilated by the Hamiltonian H for the interacting field, it follows that H|0_F> = 0. But the typical Hamiltonians for interacting fields take the form H_F +H_I , where H_I describes the interaction of the field with itself, and such Hamiltonians do not annihilate |0_F> (H polarizes the vacuum)'.

If you want to read more google for 'Haag's Theorem and Its Implications for the Foundations of Quantum Field Theory'.
Thank you for the reference. I downloaded this article, and started to read it. It looks interesting.

This "heuristic" version of Haag's theorem uses the fact that all relativistic quantum field theories "known to man" polarize vacuum. This means that the lowest energy eigenstate [itex] |vac \rangle [/itex] of the full interacting Hamiltonian H = H_0 + V (or H = H_F + H_I in your notation) is different from the lowest energy eigenstate [itex] |0 \rangle [/itex] of the free Hamiltonian H_0. In QED, this fact is evident from the presence of tri-linear terms in the interaction Hamiltonian V. For example, there are terms with three creation operators

[tex] a^{\dag}b^{\dag}c^{\dag} [/tex]........(1)

with a non-trivial action on the vector [itex] |0 \rangle [/itex] (here [itex] a^{\dag}b^{\dag}c^{\dag} [/itex] are creation operators for electrons, positrons, and photons, respectively). The presence of these terms implies that [itex] |0 \rangle [/itex] is not an eigenvector of H.

This version of Haag's theorem does not apply to "dressed particle" theories (including the "dressed particle" version of QED), because in such theories terms of the type (1) are explicitly absent in interaction Hamiltonians and the vacuum is not "polarized".

In my previous posts I discussed the version of Haag's theorem, which is marked as "HWW theorem, Part II" in the paper you referred to. My major objection was related to the use of the covariant transformation law for interacting fields

[tex] U(\Lambda) \phi(x) U^{-1}(\Lambda) = \phi(\Lambda x) [/tex]

(which, in a different notation, is given in eq. (11) for j=2). My point was that there is no empirical or theoretical evidence for the validity of this transformation law. For example, I am pretty sure that interacting fields in QED do not obey such transformations. However, this doesn't mean that QED is inaccurate or relativistically non-invariant or inappropriate for any other reason.

Eugene.
 
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  • #38
strangerep
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I think I can now clarify a few more things...

Perhaps the most important thing to know about Haag's thm
is that it's formulated within the framework of (orthodox)
axiomatic QFT. This point is critical because if one
doesn't have solid axioms, one doesn't have a solid
mathematical framework, and therefore cannot prove any
worthwhile theorems.

First, here's a summary of the (orthodox) Axioms...

Axiom-1 postulates the physical existence of Minkowski
spacetime, asserting the "events happen in a 4D Minkowski
spacetime, an event being something about which it makes
sense to say that it does or does not occur".

Axiom-1 also postulates that a quantum state of a physical
system is described by a ray in a separable Hilbert
space "H". To every measurable physical quantity "a"
there corresponds a self-adjoint, generally unbounded,
operator "A" in the separable Hilbert space.

Axiom-2 postulates that we are given a finite number of
operator-valued distributions phi_i(x) over Minkowski
spacetime, called "fields". These fields, smeared out
with test functions and then denoted phi(f), define an
(in general unbounded) operators acting in the Hilbert space.

It is not until Axiom-3 that we come to the postulate that
there exists in H a unitary continuous representation of
the Poincare group. It is here that the usual expression
of the transformation rules for fields enters, expressed
in terms of how these transformations work for Minkowski
spacetime points. (This is the item about which Eugene has
frequently been asking whether a "proof" exists.
Answer: "no, it's an axiom".)

Axiom-4 deals with the spectral conditions of the mass^2
and energy operators.

Axiom-5 deals with causality, using the notion of test
functions with compact support in Minkowski space to
define "local" operator fields which either commute
or anti-commute with each other if the supports of the
respective test functions are spacelike-separated.

Axiom-6 postulates that the vacuum vector (being
the eigenstate of the energy operator with lowest
eigenvalue) is "cyclic" with respect to the field
operators phi(f). This means (loosely) that the whole
Hilbert space H can be generated by the action of the
phi(f) operators on the vacuum. It also involves the
operators being "irreducible" in H, meaning that every
operator in H can be approximated arbitrarily closely
by functions of the phi(f).

Then there's also an "Axiom-0" postulating that the
Hilbert space decomposes in coherent Hilbert spaces
corresponding to the various physically-meaningful
quantum numbers (e.g: spin, charges, etc).

----- (End of Axioms) -----

You don't have to read very far into these axioms to see
that Eugene's approach to QFT is outside this framework.
E.g: the use of Minkowski spacetime points as physical
events, and "attaching" field operators to these points,
is distinctly different from an approach which insists strictly
that physically-measurable positions must correspond
to the eigenvalues of a self-adjoint operator.

Now, about Haag's thm... It doesn't talk specifically
about "interacting" and "non-interacting" representations.
Rather it takes the 1st representation as free and assumes
the existence of a 2nd Poincare representation in the same
Hilbert space described by the above axioms (and with
the same kind of expression for Lorentz transformations
of the fields). It finds that this 2nd representation
must also be free.

Discussions then turn to what should be "done" about
Haag's thm, i.e: which pre-condition(s) or axiom(s)
should be abandoned or modified. Eugene (and Shirokov,
afaict) advocate that the problem lies in the form
of Lorentz boosts acting on the fields (which was
originally motivated by the way the fields are "attached"
to points of Minkowski spacetime). The Earman/Fraser
paper mentioned earlier (and also other authors,
e.g: Barton, as well as some others who start from
an algebraic approach to QFT) advocate the importance
of unitarily inequivalent representations (with their
different vacuum vectors and disjoint Hilbert spaces).
Ironically, this is closely related to the way Haag
originally discussed his theorem in the paper
mentioned earlier in this thread.

Also earlier in this thread, the question arose about whether
Haag's thm excluded "all traditional relativistic QFTs
known to man". I think the key point here is that the
orthodox axiomatic QFT framework does not work at all
well with modern gauge field theories such as the Standard
Model. Hence, Haag's thm is pretty much irrelevant there.
The modern approach is to use Path Integral techniques,
but even this has not yet been rigorously successful
wrt proving existence of a 4D interacting QFT.

[And BTW, the irrelevance of Haag's thm seems also to
apply to the famous Coleman-Mandula thm which motivated
supersymmetry - since the C-M thm is also formulated
under the auspices of axiomatic QFT.]

I'm not sure whether the above sufficiently addresses all the issues
previously raised in this thread, so I'll leave it at that for now and
wait to see if anyone wants to talk further.
 
  • #39
Haelfix
Science Advisor
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Coleman Mandula is considerably stronger as it requires far less axioms than Haags theorem.. AFAICR, you basically only need finite particle species below some mass scale and analytic cross sections for elastic 2 body scattering.

Strictly speaking I seem to recall it hasn't been proved for all QFTs of interest, and the cases where you can evade the axioms are all thoroughly studied and interesting in their own right.
 
  • #40
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Thank you strangerep,

your summary is very good. I agree completely.

For me your analysis means that AQFT (with its present axioms) is a purely formal exercise disconnected from reality.

Eugene.
 

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