# Questions about quantum field theory

1. Nov 10, 2008

### Fredrik

Staff Emeritus
I'm not sure if this post should go here or into the quantum physics forum, but I figure this can't be a bad place to put it. I have a few questions about canonical quantization and quantum field theories with interactions that I hope someone can answer.

1. I've been told that in Klein-Gordon theory, the set of solutions of the classical field equation can be taken to be the Hilbert space of the quantum theory. Is that true, and in that case, does it contain just the one-particle states or the set of all possible states including the vaccuum?

2. Suppose we add a phi^4 term to the KG Lagrangian to get a theory with interactions. What's the Hilbert space of the quantum theory in this case? Is it the set of solutions of the classical field equation (with the phi^4 term)?

3. When we quantize the field in phi^4 theory, we do it by replacing Fourier coefficients of the solution of a classical field equation with creation and annihilation operators, but which field equation is that? The phi^4 equation or the KG equation? (I think I know the answer to this one: It's the KG equation, right?)

4. The canonical quantization procedure is a way to construct an irreducible representation of (the covering group of) the Poincaré group. Noether's theorem tells us how to express the generators in terms of creation and annihilation operators, and the theory of Lie groups and Lie algebras tells us how to use the generators to construct the symmetry operators (e.g. the translation operator). But what Lagrangian do we apply Noether's theorem to when we do this? Is it the phi^4 Lagrangian or the KG Lagrangian?

I think I should stop here because the questions I want to ask depend on the answers to these questions.

2. Nov 13, 2008

### strangerep

Yes, the quantum physics would have been a better place. But since
no one has answered, and it hasn't been moved, I'll take a stab here...

That's the general idea, although strictly speaking one takes only the
square-integrable solutions so that the states in the Hilbert space
have a well-defined (finite) norm.

1-particle. To get a multi-particle QFT, one constructs a (direct
sum of) tensor products of the 1-particle Hilbert spaces (called
a Fock space).

The standard approach is to assume that the states of the free theory
also span the Hilbert space of the interacting theory. Haag's theorem
(and some other stuff) shows that this assumption is actually invalid
(in orthodox QFT formulations). A lot of the renormalization stuff
in QFT is a way of working around this unwelcome inconvenience.

Pretty much, though the details get seriously tricked as mentioned above.

In a sense, the Lagrangian "is" the theory, so you apply the
machinery to whichever one you want to investigate. :-)

Perhaps it might be better to put any followup questions in
the quantum physics forum.

3. Nov 14, 2008

### Fredrik

Staff Emeritus
Thank you for answering. I'm probably going to have a bunch of follow-up questions about the quantization of theories with interactions, but I'm not done with Klein-Gordon theory yet. I'm going to have to start by asking about the connection between the standard way to introduce creation/annihilation operators and the Fock space construction.

(What I think of as the "standard way" is to just postulate a commutation relation that implies all the others. E.g. Mandl & Shaw used the Lagrangian to construct a momentum density, and then they postulated a commutation relation between that and the field that looked as much as possible as the familiar [x,p]=i. All other commutation relations could be derived from that).

Do you know a way to verify e.g. that a creation operator "defined" this way takes a vector in the tensor product of n one-particle Hilbert spaces to a vector in the tensor product of n+1 one-particle Hilbert spaces? How about just verifying that a creation operator takes the vacuum to vector in the Hilbert space of one-particle states? Oh, and that reminds me, how do you construct the vacuum state?

Last edited: Nov 14, 2008
4. Nov 14, 2008

### George Jones

Staff Emeritus
Fredrik, you might find

http://home.uchicago.edu/~seifert/geroch.notes/

interesting. This is a link to notes from a course that Robert Geroch gave in 1973 in which he treated with some mathematical care the quantum field theory of free particles. Near the end, he mentions interactions.

5. Nov 14, 2008

### Fredrik

Staff Emeritus
Thank you George. That text looks very good. It seems to be exactly what I need right now.

6. Nov 14, 2008

### strangerep

time yet to skim through the Geroch notes that George mentioned), so I'll just say a
couple of things to keep the thread alive. :-)

What's really happening is that one starts from classical fields, a Lagrangian, and
the (again classical) conjugate momenta of those fields wrt that Lagrangian.
Passing to a classical Hamiltonian formulation of the dynamics, and Poisson
brackets, is the motivation for the familiar [x,p]=i. I.e., it comes from the
classical dynamics. One then maps those classical fields satisfying certain
Poisson brackets over to operators on a Hilbert space satisfying corresponding
commutation relations. So an algebra of classical fields on a phase space gets
mapped over to an algebra of operators acting on a Hilbert space. (Classical
states are the points in phase space, quantum states are vectors/rays in the
Hilbert space.)

It's kinda "by construction". From a basic Lie algebra of creation/annihilation operators
(ie a Heisenberg algebra), one can form a larger enveloping algebra by polynomial
expressions in the basic operators. The latter is pretty much defined as a (direct sum of)
tensor products of copies of the basic algebra, modulo the commutation relations.
Wiki has a bit more: http://en.wikipedia.org/wiki/Universal_enveloping_algebra

Again, that's by construction. Having formed the enveloping algebra I mentioned above,
there's something called a "GNS construction" which lets you make a Hilbert space
from that algebra (roughly speaking). Basically, you just postulate a special state
(called "vacuum", or "cyclic vector") and act on it with all the members of the algebra,
postulating that the annihilation operator takes the vacuum to zero. Thus, the Hilbert
space is constructed (or generated) from a single vector and the algebra, and is
thus a representation of the enveloping algebra by construction.

For 1-particle QM, it's the lowest-energy eigenstate of the Hamiltonian. In
the QFT construction above, it's postulated (together with the assumption that
the annihilation operator takes the vacuum to zero).

7. Nov 15, 2008

### Fredrik

Staff Emeritus
I'm going to read the Geroch text from the start, and I tend to get stuck on a lot of details, so it's going going to take me a while. I haven't gotten to the relevant parts yet. (I think it will eventually answer most of my questions. The topics it covers are exactly those I have always felt are missing from my QFT books (M&S, P&S and W)).

Yes, but which Hilbert space? OK, you have already told me the answer to that. It's the Fock space "H" constructed from the Hilbert space "S" of square-integrable solutions of the field equation. But is the commutation relation $[a(\vec p),a(\vec p')^\dagger]=\delta^{(3)}(\vec p-\vec p')$ really sufficient to tell us that if $a(\vec p)^\dagger$ acts on a member of S, the result is in $S\otimes S$? (I don't fully understand all the details of the Fock space construction, but I assume that it makes sense to think of S and $S\otimes S$ as subspaces of H).

Now I'm not sure if I misunderstood the universal enveloping algebra. It appears to be a "Fock space" constructed from the vector space of operators acting on the Hilbert space of one-particle states. So it's a vector space of operators. Do we really need a Hilbert space structure on that?

Hm, I'm guessing that you're talking about the algebraic approach to QFT here. You start with an algebra that's defined abstractly (and not as operators acting on a Hilbert space) and then construct everything from that. I'm actually very interested in that, and I'm looking forward to learning more about it from the Geroch text. I'm even thinking about buying Haag's book, but I'll probably wait until I'm done with Geroch. But do we really need to consider that approach just to understand the vacuum state in Klein-Gordon theory, or to see why $a(\vec p)^\dagger$ takes vectors in S to vectors in $S\otimes S$? (I'm guessing the answer is no).

But how does it fit into the Fock space construction? The vacuum isn't a vector in the Hilbert space of one-particle states, is it? Do we include a one-dimensional Hilbert space into the direct sum and postulate that $a(\vec p)$ takes any vector in it to zero? Can we really postulate that? I mean, aren't we supposed to use the commutation relations to prove that there's a $|0\rangle$ that $a(\vec p)$ takes to 0?

8. Nov 15, 2008

### DarMM

Yes, the commutation relations are sufficient to derive this structure. Provdied your using a Fock representation of those commutation relations.
However to prove this it is better to work with their smeared form.
$$[a(f),a(g)^\dagger]\psi = (f,g)\psi$$
Where $$a(f)$$ is the annhilation operator integrated against some function $$f(p)$$.

Find the axioms for the canonical commutation relations* and then you should see the additional axioms for a Fock representation give you enough structure to prove exactly what you are asking.

*I'm sorry to say I can't find a website which lists the axioms in a decent way. I'll post a link if I find one. They are given in almost any book on rigorous QFT that dates from after the 1970s.

9. Nov 15, 2008

### strangerep

OK, I think I should back off until you have time to get further into
Geroch. I'm probably pitching my answers at the wrong level, causing
confusion rather than helping.

It's a Hilbert space constructed such that the creation/annihilation
operators act "irreducibly" -- meaning _every_ operator on the Hilbert
space can be expressed as a polynomial in the c/a ops. (Recall that the
definition of an "operator" on a vector space is just a linear mapping
between vectors in that space.)

(Probably less confusing if I don't answer the rest of that part of
your question until you're further into Geroch.)

In any theory purporting to describe some class of physical
systems we need two ingredients: a set of (abstract) observable
properties characterizing that class of systems (eg momentum, charge, etc),
and a set of mappings from those observable properties to ordinary
numbers. Each such individual mapping is a possible "state" of
the system. Ie, it's more general and powerful to think of
"state" as such a mapping.

In general, the set of abstract observable properties form an algebra
"g" (eg the Poincare algebra, or maybe its enveloping algebra) and the
states are elements of the dual space "g*" (the space of linear
mappings from elements of g to ordinary complex numbers C). It can be
shown that g* is a vector space, and the usual notion of Hilbert space
is just one way to realize g*. Suppose $P_i$ is an element
of g, ie a particular abstract observable property (which I'll call the
i'th component of linear momentum). When we write stuff like
$<\psi|P_i|\psi>$ (the expectation value for
$P_i$ in the state $\psi$), this is just the
concrete Hilbert space way of expressing the underlying idea that
$\psi$ determines one particular mapping from g to C.
Sometimes, it's better to write expectation value as
$\psi(P_i)$ to make this more explicit.

When this framework of observable algebra and its dual space are first
encountered, it seems weird and backwards. Usually textbooks talk about
state space first, and then about operators acting on that space. But
the above algebraic way of thinking about it is more powerful: to
construct a physical theory we first specify an algebra of observable
properties (which characterize the class of systems we're interested
in), and the dual space over that algebra (mappings from the algebra to
numbers) is then the state space for any system in that system class.
So the answer to the question "Do we really need a Hilbert space
structure on that?" is that what we really need is a dual space over
the observable algebra which also satisfies a few other axioms that
make the numbers suitable for interpretation as probabilities (ie the
mappings should be positive-definite, non-degenerate, and a few other
things). The axioms of a Hilbert space fit the bill quite well, so we
use it.

One needs to understand enough of it to see that we're just
constructing the Fock space in such a way that it's true and we do so
because we're interested in physical systems with variable numbers of
particles.

One usually expresses the Fock space F as:

$$F ~:=~ C \oplus S \oplus (S\otimes S) \oplus (S\otimes S\otimes S) \oplus ...$$

The vacuum "sector" corresponds to the "C" (ordinary complex numbers),
which all represent the vacuum state $|0\rangle$ since states here
are rays.

Again, I'll defer further answers for now since I'm sure I'd just
deepen the confusion.

10. Nov 16, 2008

### Fredrik

Staff Emeritus
I'm still confused, but at least I'm confused on a higher level. Seriously, I appreciate the efforts, and I'm a little less confused than before thanks to you. I'll save my next set of follow-up questions until I have read more of the Geroch text.

11. Nov 16, 2008

### tim_lou

What insight!!! i never thought about how we are basically taking the universal enveloping algebra by imposing one single commutation relations (parametrized by p of course)! I don't want to steal this thread, but what are the physical interpretations that there is a Hopf algebra structure associated with it? What if we deform this hopf algebra? These questions may sound trivial but I'm just beginning to learn about Hopf algebras and I'm starting to make some connections here and there--this is one of them.

12. Nov 16, 2008

### strangerep

A UEA is pretty much a Hopf algebra already. Some theorists look for insights in how
to obtain better algebras (ie describing the physics better) by deforming known algebras.
A simple case is "Lie deformation" wherein one takes a Lie algebra, deforms its
structure constants infinitesimally, and then investigates whether the new algebra is
isomorphic to the original. If the answer is "yes", the algebra is said to be "rigid" or
"stable". Rigid/stable algebras are regarded as better candidates for physical theories
since small errors in the structure constants do not change the gross qualitative behaviour
of the theory in any deep or fundamental way (and thus the theory is more robust in the
face of experimental errors). Semisimple Lie algebras are known to be rigid, and
hence deforming them is unlikely to bear physically interesting fruit. However,
deforming their UEAs can produce non-isomorphic structures. Some people hope to
construct better quantum theories of interacting fields this way, though (afaik) this
is still just a hope.

Mathematicians have taken to hijacking the adjective "quantum" for lots of things
(eg "quantum groups"), etc, but I find this a bit confusing and unhelpful.
"Deformed" might have been a better adjective. Whether such mathematical
research can produce physically useful theories is unclear. (Perhaps you
should ask the question again over on sci.physics.research.)