Register to reply

Unbounded operators in non-relativistic QM of one spin-0 particle

Share this thread:
Fredrik
#37
Apr8-09, 04:48 AM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 9,415
The answer to my concerns in #34-35 (or at least a partial answer) is probably that there's nothing weird or unexpected about the fact that the concept of continuity depends on what topology you define on the set you're working with. The space of distributions isn't the dual space of the test functions. It's a dual space of the test functions. Or to be more accurate, the topological dual space of a set isn't defined. Only the topological dual space of a topological vector space is defined, so we have to choose a topology first, and then we get a dual space.

It still makes me wonder why their version of continuity is more desirable. It probably has something to do with the idea that derivatives of distributions should always exist, but I haven't thought that idea through yet.

Edit: I'm satisfied that I understand the definition of distributions and tempered distributions now, so we don't have to discuss them unless someone else wants to. I'm not saying that I understand every detail perfectly, but my understanding is good enough.

By they way, a thought occurred to me when I was replying to another thread. Isn't it weird to start the formulation of the simplest meaningful quantum theory (the non-relativistic QM of one spin-0 particle) by postulating that the states are rays in [itex]L^2(\mathbb R^3)[/itex] when the next postulate says that the time evolution is given by the Schrödinger equation? The Schrödinger equation only makes sense if the partial derivatives exist, and that implies continuity, but [itex]L^2(\mathbb R)[/itex] contains ridiculous functions like the one Hurkyl thought of in another thread to disprove a claim I made there:

Quote Quote by Hurkyl View Post
Consider the function:

[tex]
\psi(x) =
\begin{cases}
0 & x < 1 \\
1 & x \in [n, n + n^{-2}) \\
0 & x \in [n + n^{-2}, n+1)
\end{cases}
[/tex]

where n ranges over all positive integers. [itex]\psi(x)[/itex] does not converge to zero at [itex]+\infty[/itex]. However,

[tex]
\int_{-\infty}^{+\infty} |\psi(x)|^2 \, dx
= \sum_{n = 1}^{+\infty} \int_{n}^{n + n^{-2}} 1 \, dx
= \sum_{n = 1}^{+\infty} n^{-2} = \pi^2 / 6[/tex]
strangerep
#38
Apr9-09, 12:14 AM
Sci Advisor
P: 1,927
Quote Quote by Fredrik View Post
Maybe these two options are actually equivalent, but I would expect that they're not, mostly because I think option 1 is more natural and intuitive. I don't see why these authors would all go for option 2 if they are equivalent.
First, you can probably separate out the differentiability stuff from the topological
continuity stuff.

I think (in the absence of precise definitions of those two cases) that both notions
of convergence are equivalent. The first is just saying that you can construct a
Cauchy sequence and the second is saying that a sequence converges.
(Topologically, a sequence "converges to a point z" if, given an arbitrary open
set O containing z, all points of the sequence are eventually "in" O -- after some
integer n in the sequence.

The thing about restricting to infinitely differentiable functions is so that you
can do integral calculus, i.e., solve a DE given an initial condition, in such
a way that as to be compatible with the (rigged) Hilbert space structure.

the concept of continuity depends on what topology you define on
the set you're working with
Yes. This is a crucial point to understand in order to work with other
topologies (weak, weak*, and others). Generally speaking, convergence
is easier in weaker (coarser) topologies, but more theorems can be
proven with stronger (finer) topologies. Bit of a tradeoff depending on
exactly what one is trying to achieve.
Fredrik
#39
Apr9-09, 04:46 PM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 9,415
I think I'm going to have to study some more functional analysis before I post a bunch of new questions. I'll read some more of Conway first, and maybe I'll read Maurin's lecture notes. Strangerep and Jostpuur, thanks for the answers so far. George, feel free to bump the thread when you're less busy. I'm interested in what you have to say about all of this.
dextercioby
#40
Apr12-09, 04:55 AM
Sci Advisor
HW Helper
P: 11,927
Brilliant discussion on a very interesting topic, I must say. I was wondering, how would I build the RHS for the hydrogen atom ? The hamiltonian for the H atom has a mixed spectrum and I've never seen an application of the RHS formalism for an operator with mixed spectrum.
jostpuur
#41
Apr12-09, 10:57 AM
P: 2,066
Quote Quote by Fredrik View Post
OK, let's focus on "distributions" first, not "tempered distributions". We define the test function space (I'll call it [itex]\mathcal D[/itex]) as the set of [itex]C^\infty[/itex] functions from [itex]\mathbb R^n[/itex] into [itex]\mathbb C[/itex] that have compact support. I would like to define a "distribution" as a member of the dual space [itex]\mathcal D^*[/itex], defined as the set of continuous linear functions [itex]T:\mathcal D\rightarrow\mathbb C[/itex]. But to do that, we need to define what "continuous" means. There are at least two ways to do that.

Option 1: Define the usual inner product. The inner product gives us a norm, and the norm gives us a metric. Now we can use the definition of continuity that applies to all metric spaces.
Equipping the [itex]\mathcal{D}[/itex] with a norm

[tex]
\|\psi\|^2 = \int\limits_{\mathbb{R}^n} dx\; |\psi(x)|^2
[/tex]

and defining [itex]\mathcal{D}^*[/itex] as a topological dual

[tex]
\mathcal{D}^* = \big\{ \phi^*\in \mathbb{C}^{\mathcal{D}} \;|\; \phi^*\;\textrm{is linear},\; \sup_{\|\psi\|\leq 1} |\phi^*(\psi)| < \infty\big\}
[/tex]

is probably not what anyone wants, because for example this collection of linear forms doesn't contain delta-functions [itex]\delta_x:\mathcal{D}\to\mathbb{C}[/itex], [itex]\delta_x(\psi)=\psi(x)[/itex].

Was that what you meant by your option 1?

Quote Quote by Fredrik View Post
Option 2: This is the option that every text on the subject seems to prefer, and I don't see why. We define what it means for a sequence in [itex]\mathcal D[/itex] to be convergent by saying that [itex]\phi_n\rightarrow\phi[/itex] if there's a compact set K in [itex]\mathbb R^n[/itex] that contains the supports of all the [itex]\phi_n[/itex], and [itex]D^\alpha \phi_n[/itex] converges uniformly on K to [itex]D^\alpha\phi[/itex], for all [itex]\alpha[/itex].

Now we define "continuous" by saying that [itex]T:\mathcal D\rightarrow\mathbb C[/itex] is continuous at [itex]\phi\in\mathcal D[/itex] if

[tex]\lim_{n\rightarrow\infty}T(\phi_n)=T(\phi)[/tex]

for every sequence [itex]\{\phi_n\}[/itex] that converges to [itex]\phi[/itex]. Maybe these two options are actually equivalent, but I would expect that they're not, mostly because I think option 1 is more natural and intuitive. I don't see why these authors would all go for option 2 if they are equivalent.
I'm not 100% sure of this, but I've heard that it is possible prove the existence of such topology in [itex]\mathcal{D}[/itex], that the convergence [itex]\psi_n\to\psi[/itex] in that topology is equivalent with this definition that you described here. When [itex]\mathcal{D}[/itex] is equipped with such topology, then the standard collection of distributions [itex]\mathcal{D}^*[/itex] can be defined according to the definition of topological dual.
jostpuur
#42
Apr12-09, 11:14 AM
P: 2,066
Quote Quote by meopemuk View Post
In my understanding, the reason why standard Hilbert space formalism is not suitable for QM is rather simple. Let's say I want to define an eigenfunction of the momentum operator. In the position space such an eigenfunction has the form (I work in 1D for simplicity)

[tex]\psi(x) = N \exp(ipx)[/tex]

where [tex]N [/tex] is a normalization factor. This wavefunction must be normalized to unity, which gives


[tex] 1 =\int \limits_V |\psi(x)|^2 dx = N^2V[/tex]

where [tex]V[/tex] is the "volume of space", which is, of course, infinite. This means that the normalization factor is virtually zero

[tex] N = 1/\sqrt{V}[/tex]
While I have understood this problem, it has not yet become clear to me how this problem is supposed to be solved in a more sophisticated formalism. The only solution that I have understood so far is this:

Quote Quote by jostpuur View Post
If we ask a question that what is the probability for a momentum to be in an interval [itex][p_0-\Delta, p_0+\Delta][/itex], we get the answer from the expression

[tex]
\frac{1}{2\pi\hbar} \int\limits_{p_0-\Delta}^{p_0+\Delta} |\hat{\psi}(p)|^2 dp.
[/tex]

I'm not convinced that it is useful to insist on being able to deal with probabilities of precise eigenstates. Experimentalists cannot measure such probabilities either.
Quote Quote by Avodyne View Post
I don't see why it's necessary to go beyond Hilbert space. Rather than defining a position operator, we could define projection operators with eigenvalue 1 if the particle is in some particular volume V, and 0 otherwise; heuristically, these would be

[tex]P_{x\in V} \equiv \int_V d^3\!x\,|x\rangle\langle x|[/tex]

Similarly for a volume in momentum space. Then, instead of defining a hamiltonian whose action could take a state out of the Hilbert space, we could define a unitary time evolution operator.
I'm not convinced that the distributions are very useful. The distributions are usually defined by using rather strange topologies or norms. The Swartz norm seems to be totally different from the original idea of a Hilbert space norm. However, the Hilbert space norm is highly relevant for the probability interpretation. How are we supposed to deal with quantum mechanical probabilities if the Hilbert space norm has been replaced with some of these distribution norms?
Fredrik
#43
Apr12-09, 12:25 PM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 9,415
Quote Quote by jostpuur View Post
is probably not what anyone wants, because for example this collection of linear forms doesn't contain delta-functions [itex]\delta_x:\mathcal{D}\to\mathbb{C}[/itex], [itex]\delta_x(\psi)=\psi(x)[/itex].

Was that what you meant by your option 1?
Ahh, excellent point. Yes, that's what I meant. I get it now. (I didn't until I read your post). Delta functions wouldn't be bounded (and therefore not continous) linear functionals if we go for option 1. For delta to be bounded, there must exist M>0 such that

[tex]|\delta f|\leq M\|f\|[/tex]

for all test functions f, but the inequality is equivalent to

[tex]\frac{|f(0)|}{\|f\|}\leq M[/tex]

and we don't have to look hard to find an f that violates this. Even some constant functions will do. For example, define fr(x)=1 when |x|<r, and fr(x)=0 outside that region. Now shrink that region (i.e. choose a smaller r) until the norm of fr gets small enough to violate the inequality.
Fredrik
#44
Apr12-09, 12:47 PM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 9,415
Quote Quote by jostpuur View Post
I'm not 100% sure of this, but I've heard that it is possible prove the existence of such topology in [itex]\mathcal{D}[/itex], that the convergence [itex]\psi_n\to\psi[/itex] in that topology is equivalent with this definition that you described here. When [itex]\mathcal{D}[/itex] is equipped with such topology, then the standard collection of distributions [itex]\mathcal{D}^*[/itex] can be defined according to the definition of topological dual.
I don't think the construction of such a topology is very difficult. It would be harder to prove that the result is equivalent to what we already have, but even that looks doable. (I'm not sure I care enough to give it a try though ).

We can e.g. define a subset of [itex]\mathcal D[/itex] to be open if it can be expressed as [itex]T^{-1}(U)[/itex] where U is an open subset of the complex numbers, and T is continous in the sense defined above.

Alternatively, I think we can define a subset E of [itex]\mathcal D[/itex] to be open if for every sequence in [itex]\mathcal D[/itex] that converges to a a point in E (in the sense defined above), E contains all but a finite number of members of the sequence.

I'm guessing that these definitions are adequate (in the sense that they both define a topology on [itex]\mathcal D[/itex]), and equivalent (in the sense that those topologies are the same), but I haven't made any attempt to prove it or disprove it.
strangerep
#45
Apr12-09, 11:13 PM
Sci Advisor
P: 1,927
Quote Quote by Fredrik View Post
We can e.g. define a subset of [itex]\mathcal D[/itex] to be open if it can be expressed as [itex]T^{-1}(U)[/itex] where U is an open subset of the complex numbers, and T is continous in the sense defined above.
Yes, that defines the "T-weak" topology on [itex]\mathcal D[/itex]. (Weak topologies are
defined implicitly/indirectly by a demand that a certain set of functions be continuous
under it.)

Alternatively, I think we can define a subset E of [itex]\mathcal D[/itex] to be open if for every sequence in [itex]\mathcal D[/itex] that converges to a a point in E (in the sense defined above), E contains all but a finite number of members of the sequence.
I don't follow this. If you haven't yet specifed a topology on [itex]\mathcal D[/itex]
then there's no meaning in the statement that a sequence of elements of
[itex]\mathcal D[/itex] "converges". But maybe I misunderstood and you
meant something else?
strangerep
#46
Apr13-09, 03:11 AM
Sci Advisor
P: 1,927
Quote Quote by bigubau View Post
I was wondering, how would I build the RHS for the hydrogen atom ? The hamiltonian for the H atom has a mixed spectrum and I've never seen an application of the RHS formalism for an operator with mixed spectrum.
The discrete part of the spectrum doesn't pose the same difficulties as encountered
with the continuous part. One has a resolution of unity as usual, consisting of an
integral over the continuous part, plus a sum over the discrete part. Rigged Hilbert
space and the nuclear spectral theorem give meaning to the continuous part.
(Not sure whether that's what you were asking, though.)

BTW, for anyone who's still a bit perplexed about the role of all this
rigged Hilbert space stuff, I noticed this pedagogical introductory paper:

Rafael de la Madrid,
"The role of the rigged Hilbert space in Quantum Mechanics",
Available as: quant-ph/0502053

It's written in a very physicist-friendly way, using the example of a rectangular
potential barrier to make everything concrete. It also explains the connections
between the bras and kets of rigged Hilbert space theory and distributions
quite clearly (while minimzing the heavy pure math that probably turns some
people off).
Fredrik
#47
Apr13-09, 05:37 AM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 9,415
Quote Quote by strangerep View Post
I don't follow this. If you haven't yet specifed a topology on [itex]\mathcal D[/itex]
then there's no meaning in the statement that a sequence of elements of
[itex]\mathcal D[/itex] "converges". But maybe I misunderstood and you
meant something else?
You understood me right. We define convergence of sequences first, and use that to define the topology. Convergence is defined as in post #35. I got that definition and the idea that it defines a topology on the test function space from the Wikipedia article on distributions. This is a direct link to the relevant section. Note in particular the sentence "It can be given a topology by defining the limit of a sequence of elements of D(U)." Unfortunately, Wikipedia doesn't say how the definition of convergence defines a topology. That's why I came up with those two guesses about how it can be done.
Fredrik
#48
Apr13-09, 07:57 AM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 9,415
Quote Quote by strangerep View Post
BTW, for anyone who's still a bit perplexed about the role of all this
rigged Hilbert space stuff, I noticed this pedagogical introductory paper:

Rafael de la Madrid,
"The role of the rigged Hilbert space in Quantum Mechanics",
Available as: quant-ph/0502053
Thanks. I have read the first 9 pages now, and I'm very pleased with it so far. I'm definitely going to read the rest later.
Fredrik
#49
Apr13-09, 12:11 PM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 9,415
I found a review of Maurin's books, and the reviewer is complaining a lot about how careless Maurin's presentations are. Link. I'm guessing it would be better to try to find the original articles than to read his books.
DarMM
#50
Apr14-09, 02:03 PM
Sci Advisor
P: 303
Quote Quote by Fredrik View Post
Quote Quote by strangerep View Post
For QFT in infinite-dimensions, even the rigged Hilbert space is not big enough.
The singular behaviours arising from the interaction terms in the Hamiltonian
become far more pathological than mere delta distributions.
Interesting...and strange. The strange part is that we can actually make predictions in spite of all of this.
I thought I'd add a bit to this since it's very interesting.

As you know the Hilbert Space for 1 particle spin-0 QM is [itex]L^2(\mathbb R^3, dx)[/itex]. Where dx is the Lesbesgue measure.

[itex]\mathbb R^3[/tex] coming from the fact that a particle can occupy any point in three dimensional space. So the set of all points [itex]\mathbb R^3[/tex] is the classical configuration space.

For Quantum Field Theory the classical configurations are the set of all possible field configurations. This set turns out to be [tex]\mathcal{D}^*(\mathbb R^3)[/tex], the space of distributions. So the Hilbert space of QFT is:
[itex]L^2(\mathcal{D}^*(\mathbb R^3), d\nu)[/itex], the space of square integrable functions over the space of distributions with respect to some measure [itex]d\nu[/itex]. A free QFT and an interacting QFT differ by their choice of [itex]d\nu[/itex].
Quantum Fields [itex]\phi(x)[/itex] are then objects that when integrated against a function [itex]f(x)[/itex] give an object [itex]\int{\phi(x)f(x)}dx = \phi(f)[/itex], which is an unbounded operator on [itex]L^2(\mathcal{D}^*(\mathbb R^3), d\nu)[/itex].

Who said rigorous QFT was hard?
strangerep
#51
Apr14-09, 09:07 PM
Sci Advisor
P: 1,927
Quote Quote by Fredrik View Post
I got that definition and the idea that it defines a topology on the test function space from the Wikipedia article on distributions. This is a direct link to the relevant section. Note in particular the sentence "It can be given a topology by defining the limit of a sequence of elements of D(U)." Unfortunately, Wikipedia doesn't say how the definition of convergence defines a topology.
OK, thanks. I think I get it now.
The key item is the use of uniform convergence. E.g., given two functions f,g it's
sensible to ask how "close" they are to each other, as follows: If we can say that
[itex]|f(x) - g(x)| < \epsilon[/itex] for all x, then [itex]\epsilon[/itex] is an expression
of the closeness of f and g (independently of x).

So one could define open sets in a manner reminiscent of open balls in a metric
topology: pick a function f and a "radius" r and then say that one particular
open set centered on f, of radius r, is all the other functions g s.t. [itex]|f(x) - g(x)| < r[/itex]
for all x. Then take unions, etc, to get a full topology.

Uniform convergence just extends this idea to sequences, the crucial bit
being how it's independent of x.

The fact that it's expressed in terms of arbitrary derivatives is just because we
want to deal with infinitely-differentiable functions only, and is a bit a red-herring
when one is focusing just on topological matters.
strangerep
#52
Apr14-09, 09:21 PM
Sci Advisor
P: 1,927
Quote Quote by Fredrik View Post
I found a review of Maurin's books, and the reviewer is complaining a lot about how careless Maurin's presentations are. Link.
Hmmm, that reviewer uses rather savage language, but I have to say I agree with
his remarks about lack of care. I noticed the same thing as soon as I started to
try and follow Maurin's proof of the nuclear spectral theorem. I had to keep flipping
backwards tediously to double-check notations, often finding things that I felt
sure were typos. Worse, even though the book came with an errata list at the
back, I had trouble matching it with what was in the text(!) (sigh).

Look's like I'll have to try harder to obtain a copy of Gelfand & Vilenkin and
see whether their presentations is any better.

TBH, I'm quite surprised and disappointed that I couldn't find a clear pedagogical
proof in a more modern text, given it's importance in modern quantum theory.

I'm guessing it would be better to try to find the original articles than
to read his books.
Dunno. Original papers tend to be written for people with a greater level
of knowledge. Books are supposed to be for lesser mortals.
strangerep
#53
Apr14-09, 09:51 PM
Sci Advisor
P: 1,927
Quote Quote by DarMM View Post
I thought I'd add a bit [...]
Hmmm, thanks for joining this thread. You just made me think about
a couple of things from a direction I hadn't explored adequately...

For Quantum Field Theory the classical configurations are the set of all possible field configurations. This set turns out to be [tex]\mathcal{D}^*(\mathbb R^3)[/tex], the space of distributions. So the Hilbert space of QFT is:
[itex]L^2(\mathcal{D}^*(\mathbb R^3), d\nu)[/itex], the space of square integrable functions over the space of distributions with respect to some measure [itex]d\nu[/itex].
I guess this gets into the problematic area of "multiplication of distributions"?
I hope you can say a bit more about that.

A free QFT and an interacting QFT differ by their choice of [itex]d\nu[/itex].
Quantum Fields [itex]\phi(x)[/itex] are then objects that when integrated against a function [itex]f(x)[/itex] give an object [itex]\int{\phi(x)f(x)}dx = \phi(f)[/itex], which is an unbounded operator on [itex]L^2(\mathcal{D}^*(\mathbb R^3), d\nu)[/itex].
Let me see... in the simpler case of rigged Hilbert space, translating into
Dirac bra-ket notation, we have
[tex]
\phi(f) ~=~ \int{\phi(x)f(x)}dx ~=~ \langle\phi| \int{|x\rangle\langle x|}dx ~ |f\rangle
~=~ \langle \phi | f\rangle
[/tex]
with
[tex]
\langle x|x'\rangle ~=~ \delta(x - x') ~,~~~\mbox{etc.}
[/tex]

So in the more general case we're dealing with functions over distributions (which
involving products of distributions in general), and then trying to figure out
how to construct (higher-order?) functionals over these functions?
DarMM
#54
Apr15-09, 04:29 AM
Sci Advisor
P: 303
Quote Quote by strangerep View Post
I guess this gets into the problematic area of "multiplication of distributions"?
I hope you can say a bit more about that.
Yes, in fact it's one of the big issues in rigorous QFT. Your question below will naturally lead to it.

Quote Quote by strangerep View Post
So in the more general case we're dealing with functions over distributions (which
involving products of distributions in general), and then trying to figure out
how to construct (higher-order?) functionals over these functions?
Let me give a simple example of a function over distributions.
As you know a distribution [itex]T(x)[/itex] is a continuous linear functional over the space of test functions [itex]\mathcal{D}[/itex]. However one can reverse this and consider a test function [itex]f[/itex] as a function on distributions. A given test function [itex]f[/itex] maps any distribution [itex]T[/itex] to [itex]T(f)[/itex], resulting in a map over the whole space of distributions.
[itex]f:\mathcal{D}^*(\mathbb R^3) \rightarrow \mathbb R[/itex].

So one writes this function over distributions as [itex]f(T)[/itex]. One can then immediately arrive at more general functions for instance:
[itex]f(T)g(T)[/itex]
[itex]f(T)g(T)h(T)[/itex]
or [itex]e^{f(T)}[/itex]

Since one can choose a any test function, this results in an enormous space of functions over Distributions.

In the case of a free quantum field, the quantum field operator acts as a "multiplication operator". For example:
[itex]\phi(g) = \int{\phi(x)g(x)}dx[/itex]
Then,
[itex]\phi(g)f(T) = g(T)f(T)[/itex]

The conjuagate momentum field [tex]\pi(x)[/tex] acts like the momentum operator from regular QM by being a derivative. In rough language:
[itex]\pi(x)f(T) = -i\frac{\delta}{\delta T}f(T)[/itex]

The Hamiltonian is composed of [itex]\pi(x)[/itex] and [itex]\phi(x)[/itex] and so it acts in a natural way.

Like all quantum mechanical theories though, we don't allow just any functions, but only square integrable ones. So one requires a notion of integration for this functions, formally something like:
[itex]\int{f(T)}dT[/itex]. This can be done, although the details are a bit involved. For the purpose of the remainder of this post let's assume we have done this for the free theory. We have the correct measure [itex]dT_{free}[/itex], the Hilbert space of square integrable functions with respect to it and the Hamiltonian, field and conjugate momentum operators acting in the way I described above. This space is basically Fock space.



However when one comes to [itex]\phi(x)^4[/tex] theory for example, the Hamiltonian gains an additional term:
[itex]\int{\phi(x)^4}dx[/itex]

It is very difficult to know exactly how one should interpret this.
[itex](\int{\phi(x)^4}dx)f(T) = ?[/itex]

There are two problems here. [itex]\phi(x)[/itex] is a operator valued distribution. When integrated it gives operators on [itex]L^2(\mathcal{D}^*(\mathbb R^3), dT_{free})[/itex]. Due to its distributional nature no sense can be made of [tex]\phi(x)^4[/itex]. One could attempt to come up with some way of powering distributions, however this powering system must respect the fact that [itex](\int{\phi(x)^4}dx)[/itex] is meant to be a linear operator and respect certain commutation relations.

Let's approach this problem first in two dimensions then in three dimensions.
Two dimensions
It turned out (again avoiding details) that there was one unique notion of powering which respects the structure of [itex](\int{\phi(x)^4}dx)[/itex] as a quantum operator. This notion was actually already known to physicists as Wick ordering. So we replace:
[itex](\int{\phi(x)^4}dx)[/itex] with [itex](\int{:\phi(x)^4:}dx)[/itex].
One can then prove that the full Hamiltonian is self-adjoint and semi-bounded (stable, no negative energy), the time evolution operator is unitary, e.t.c. Everything one wants from a QFT basically. This was done by James Glimm and Arthur Jaffe in the period 1968-1973.

Three dimensions
Here things get significantly harder. Where as one can put the proof of the two dimensional case into a semester long graduate course, the proof of the three dimensional case is a monster. However let me paint a picture.

Basically we try what we did before and replace [itex](\int{\phi(x)^4}dx)[/itex] with [itex](\int{:\phi(x)^4:}dx)[/itex]. However one finds that this does not result in the full Hamiltonian being a self-adjoint operator. If you are very clever you can prove that this problem can not be overcome, unless you leave [itex]L^2(\mathcal{D}^*(\mathbb R^3), dT_{free})[/itex], that is leave Fock space and move to another space of square integrable functions over distributions. Let me call it:
[itex]L^2(\mathcal{D}^*(\mathbb R^3), dT_{Glimm})[/itex].

However finding this space is incredibly diffcult, so one uses the condition that on this correct space the physical mass of a particle should be finite. This condition show you how to get to this new space. Basically, cutoff the fields so that you do the analysis in a more clear cut way, then use the condition of finite mass. This condition tells you that you should add the term:
[itex]\delta m^2(\int{:\phi(x)^2:}dx)[/itex]
to the Hamiltonian. The [itex]\delta m^2[/itex] is a constant that will go infinite when the cutoff is removed, whose explicit form is given by the finite mass condition. As the cutoff is removed, this extra term does the job of "pushing" us into the correct space [itex]L^2(\mathcal{D}^*(\mathbb R^3), dT_{Glimm})[/itex], which is alot easier than finding it abstractly. Physicists will know this procedure as mass renormalization.

We then take the remove the cutoff. James Glimm proved that this moves us into the correct space and that the full Hamiltonian is self-adjoint there. Glimm and Jaffe later proved that the Hamiltonian is also semibounded (in what is considered the most difficult paper in mathematical physics, lots of heavy analysis). Shortly after came all the the other properties one wants, Lorentz invariance, e.t.c.

Thus proving the rigorous existence of [itex]\phi^4[/itex] in three dimensions.

As you can imagine the four dimensional case is extremely difficult. For five dimensions and greater, Jürg Fröhlich has shown that there is no space where the [itex](\int{\phi(x)^4}dx)[/itex] term makes sense, unless you set the interaction to zero, that is make it free.

People have also shown the existence of Yukawa theories in two and three dimensions, as well as gauge and Higgs theories in two and three dimensions. Tadeusz Balaban has gotten pretty far although with Yang-Mills theory in two and three dimensions. He has even gotten some results for the four dimensional case.


One final thing I wish to emphasize is that renormalization is not the difficulty in rigorous quantum field theory. Renormalization is a necessary technique, as it has been known for a long time the most interacting theories simply cannot live in the same space as the free theory (Haag's theorem).
The actual difficulty is with standard analytic stuff, proving convergence and obtaining estimates which tell you something about a limit besides the fact that it exists.

I hope this helps.


Register to reply

Related Discussions
Raising and lowering operators for spin Quantum Physics 3
Spin angular momentum operators Advanced Physics Homework 1
Spin angular momentum operators Quantum Physics 5
Spin operators Advanced Physics Homework 2
Help me (Spin Operators) Introductory Physics Homework 6