Understanding Local and Nonlocal Operators in Quantum Field Theory

[Demystifier]

Because there seems to be a Landau pole. I'm not an expert in axiomatic QFT, but afaik there's no complete proof for the rigorous existence of QED.

In addition, there is a lot of evidence (e.g. from lattice computations) that QED in the continuum limit is trivial.

 

[stevendaryl]

Some of the contention in this discussion follows from physicists' sloppiness about making distinctions. The symbol $x$ can mean:

1. An operator (the position operator, in QM---there is no position operator in QFT).
2. A variable
3. An unspecified constant

Similarly, an expression such as $f(x)$ can either mean a function, or the value of the function at a particular unspecified argument $x$.

I don't think that these ambiguities cause problems for physicists working alone, but they cause problems for communication.

The specific example $\delta(x) \delta(y)$ is yet another ambiguous expression. Since $\delta(x)$ is a distribution, this expression appears to be the product of distributions, which is undefined.

However, we can certainly make sense of it as a distribution on $R^2$. That is, as a functional $F$ that takes a function $f$ of type $R^2 \Rightarrow C$ and returns an element of $C$.

$F(f) = f(0,0)$

Writing this as $F(f) = \int dx dy f(x,y) \delta(x) \delta(y)$ is no more an abuse of mathematical notation than the notation $\delta(x)$ in the first place. For most purposes, the notation is useful, as long as you have a feel for when it gets you into trouble.

 

[stevendaryl]

In addition, there is a lot of evidence (e.g. from lattice computations) that QED in the continuum limit is trivial.

Professor: "Class, for your examination today, I want you to compute the magnetic moment of the electron using QED. That might seem like a lot of work to accomplish in one hour, but we have evidence that QED is actually trivial."

 

[martinbn]

Some of the contention in this discussion follows from physicists' sloppiness about making distinctions. The symbol $x$ can mean:

1. An operator (the position operator, in QM---there is no position operator in QFT).
2. A variable
3. An unspecified constant

Similarly, an expression such as $f(x)$ can either mean a function, or the value of the function at a particular unspecified argument $x$.

I don't think that these ambiguities cause problems for physicists working alone, but they cause problems for communication.

The specific example $\delta(x) \delta(y)$ is yet another ambiguous expression. Since $\delta(x)$ is a distribution, this expression appears to be the product of distributions, which is undefined.

However, we can certainly make sense of it as a distribution on $R^2$. That is, as a functional $F$ that takes a function $f$ of type $R^2 \Rightarrow C$ and returns an element of $C$.

$F(f) = f(0,0)$

Writing this as $F(f) = \int dx dy f(x,y) \delta(x) \delta(y)$ is no more an abuse of mathematical notation than the notation $\delta(x)$ in the first place. For most purposes, the notation is useful, as long as you have a feel for when it gets you into trouble.

The expected notation would be $\delta (x, y)$.

 

[Demystifier]

Professor: "Class, for your examination today, I want you to compute the magnetic moment of the electron using QED. That might seem like a lot of work to accomplish in one hour, but we have evidence that QED is actually trivial."

"Continuum limit" is the key word. In actual calculations of loop diagrams one performs a regularization which avoids dealing with the continuum limit. For more on triviality see e.g.
http://www.scholarpedia.org/article/Triviality_of_four_dimensional_phi^4_theory_on_the_lattice

 

[vanhees71]

The expected notation would be $\delta (x, y)$.

But $\delta^{(2)}(x,y)=\delta(x) \delta(y)$ (and this is not sloppy nor in any way undefined btw.).

 

[vanhees71]

Professor: "Class, for your examination today, I want you to compute the magnetic moment of the electron using QED. That might seem like a lot of work to accomplish in one hour, but we have evidence that QED is actually trivial."

"You have to do calculations in the Interaction picture. It doesn't exist, but I don't care, just shutup and calculate!" ;-)).

 

[PeterDonis]

Professor: "Class, for your examination today, I want you to compute the magnetic moment of the electron using QED. That might seem like a lot of work to accomplish in one hour, but we have evidence that QED is actually trivial."

Reminds me of the joke about the professor who starts the class by writing one equation on the blackboard; then he says "and from this, it is obvious that the following is true", and writes a second equation. Then he stops, furrows his brow, and says "wait a minute, I may be wrong". Then he spends the next hour covering the blackboard with equations, muttering to himself, and finally, just as the class is over, he steps back, sighs with relief, and says "yes, I was right in the first place; it is obvious that the second equation follows from the first."

 

[vanhees71]

One of the zillions of anecdotes about Pauli was that once in a lecture he said that something is trivial. A courageous student dared to ask, whether it's really trivial. Pauli went out of the lecture room. After 10 minutes he came back and told the students: "It's really trivial", and went on with his lecture.

 

[martinbn]

But $\delta^{(2)}(x,y)=\delta(x) \delta(y)$ (and this is not sloppy nor in any way undefined btw.).

It is at best sloppy.

 

[vanhees71]

Why? The definition is
$$\int_{\mathbb{R}} \mathrm{d} x \int_{\mathbb{R}} \mathrm{d} y f(x,y) \delta(x) \delta(y)=f(0,0).$$
That's well-defined and standard in all the literature I know.

 

[martinbn]

Why? The definition is
$$\int_{\mathbb{R}} \mathrm{d} x \int_{\mathbb{R}} \mathrm{d} y f(x,y) \delta(x) \delta(y)=f(0,0).$$
That's well-defined and standard in all the literature I know.

Can you give me a title and a page.

 

[vanhees71]

This should be in any textbook dealing with $\delta$ distributions. One concrete example is

M. L. Boas, Mathematical Methods in the Physical Sciences

 

[martinbn]

This should be in any textbook dealing with $\delta$ distributions. One concrete example is

M. L. Boas, Mathematical Methods in the Physical Sciences

I found it. It is sloppy there too.

 

[vanhees71]

Perhaps we should open another thread somewhere in the Math forum, but why is this sloppy? I'm not a mathematician, but I cannot imagine that there's something problematic with this formula. One way to prove it is to use some weak limit, e.g., of Gaussians in the limit of vanishing standard deviations. There the functions factorize, and I don't see why there is a problem with taking the weak limit leading to the factorization of the multi-dimensional $\delta$-distribution, i.e., like
$$\delta^{(3)}(\vec{x}-\vec{x}_0)=\delta(x-x_0) \delta(y-y_0) \delta(z-z_0).$$

 

[martinbn]

Perhaps we should open another thread somewhere in the Math forum, but why is this sloppy? I'm not a mathematician, but I cannot imagine that there's something problematic with this formula. One way to prove it is to use some weak limit, e.g., of Gaussians in the limit of vanishing standard deviations. There the functions factorize, and I don't see why there is a problem with taking the weak limit leading to the factorization of the multi-dimensional $\delta$-distribution, i.e., like
$$\delta^{(3)}(\vec{x}-\vec{x}_0)=\delta(x-x_0) \delta(y-y_0) \delta(z-z_0).$$

It looks like a product, but it is something else.

 

[Demystifier]

It looks like a product, but it is something else.

What?

 

[Demystifier]

"You have to do calculations in the Interaction picture. It doesn't exist, but I don't care, just shutup and calculate!" ;-)).

Perhaps it doesn't exist by a mathematician's definition of "exists", but it exists by a physicist's definition. If it gives good results, then it exists. And if its existence contradicts certain axioms, then so much worse for the axioms.

 

[martinbn]

What?

The Dirac delta on a different space of test funtions.

 

[samalkhaiat]

It looks like a product, but it is something else.

It is not the undefined (point-wise) product of distributions. It is the well-defined direct product of distributions which is commutative and associative: $$\delta^{3}(\vec{x}) = \left( \delta (x_{1}) \delta (x_{2})\right) \ \delta (x_{3}) = \delta (x_{1}) \ \left( \delta (x_{2}) \delta (x_{3})\right),$$ or $$\delta^{3}(\vec{x}) = \delta^{2} (x_{1} , x_{2}) \ \delta (x_{3}) = \delta (x_{1}) \ \delta^{2}(x_{2} ,x_{3})$$.

 

[vanhees71]

Perhaps it doesn't exist by a mathematician's definition of "exists", but it exists by a physicist's definition. If it gives good results, then it exists. And if its existence contradicts certain axioms, then so much worse for the axioms.

Again, it exists after doing some regularization and taking the appropriate weak limit at the end of the calcultion. One most simple way is to introduce a finite spatial "quantization volume" and imposing periodic spatial boundary conditions for the fields. For a through treatment about the "irrelevance" of Haag's theorem FAPP, see

A. Duncan, The conceptual framework of quantum field
theory, Oxford University Press, Oxford (2012).

 

[vanhees71]

The Dirac delta on a different space of test funtions.

What do you mean by that. Of course a "3D $\delta$ distribution" is a linear form on some corresponding test-function space, e.g,. $C_0^{\infty}(\mathbb{R}^3)$. I really don't see any problem with this kind of "product". At least it works FAPP. E.g., you need it to express it in terms of non-Cartesian coordinates, like
$$\delta^{(3)}(\vec{x}-\vec{x}')=\frac{1}{r^2 \sin \vartheta} \delta(r-r') \delta(\vartheta-\vartheta') \delta(\varphi-\varphi')$$
for spherical coordinates. Here of course you have to be careful with the coordinate singularity along the polar axis.

 

[martinbn]

It is not the undefined (point-wise) product of distributions. It is the well-defined direct product of distributions which is commutative and associative: $$\delta^{3}(\vec{x}) = \left( \delta (x_{1}) \delta (x_{2})\right) \ \delta (x_{3}) = \delta (x_{1}) \ \left( \delta (x_{2}) \delta (x_{3})\right),$$ or $$\delta^{3}(\vec{x}) = \delta^{2} (x_{1} , x_{2}) \ \delta (x_{3}) = \delta (x_{1}) \ \delta^{2}(x_{2} ,x_{3})$$.

What do you mean by that. Of course a "3D $\delta$ distribution" is a linear form on some corresponding test-function space, e.g,. $C_0^{\infty}(\mathbb{R}^3)$. I really don't see any problem with this kind of "product". At least it works FAPP. E.g., you need it to express it in terms of non-Cartesian coordinates, like
$$\delta^{(3)}(\vec{x}-\vec{x}')=\frac{1}{r^2 \sin \vartheta} \delta(r-r') \delta(\vartheta-\vartheta') \delta(\varphi-\varphi')$$
for spherical coordinates. Here of course you have to be careful with the coordinate singularity along the polar axis.

Of course you can do that, but it is sloppy and produces a distribution on a different space. The question in this thread was to give an example of a nonlocal quantum field, which is supposed to be a function/distribution on space-time, not on the product of two copies of space-time.

 

[Demystifier]

nonlocal quantum field ... is supposed to be a function/distribution on space-time, not on the product of two copies of space-time.

Who said that?

 

[vanhees71]

What do you mean by "it produces a distribution on a different space". There is a space of test functions, e.g., $C_0^{\infty}(\mathbb{R}^3)$, and there the distribution $\delta(x)\delta(y)\delta(z)$ is defined by its action on a member of this test-function space. Of course, it's not defined as a "point-wise product". A "point-wise" interpretation of a distribution doesn't make sense to begin with. As @samalkhaiat said, it's a "direct" (or tensor) product.

 

[martinbn]

Who said that?

That's what a local quantum field is. A distribution on spacetime satisfying some axioms. If you ask for a nonlocal one, it has to be the same type of object, that doesn't satisfy these axioms. Otherwise you can literary give anything as an example. Say here is the equations of a parabola $y=x^2$, this is not a local quantum field, thus it is what you are looking for!

 

[vanhees71]

A definition is a definition. There's not much to argue about it, and @Demystifier gave examples of "non-loacal operators" early on in this thread. I'm a bit lost, about what we are discussing, agreeing, or disagreeing.

 

[Demystifier]

That's what a local quantum field is. A distribution on spacetime satisfying some axioms. If you ask for a nonlocal one, it has to be the same type of object, that doesn't satisfy these axioms. Otherwise you can literary give anything as an example. Say here is the equations of a parabola $y=x^2$, this is not a local quantum field, thus it is what you are looking for!

Intuitively, a nonlocal field is something that should have the following properties:
(i) It is an operator that acts on the same vector space as the local field $O(x)$.
(ii) It cannot be associated with one point $x$.
Perhaps the best example of a nonlocal field (best in the sense that mathematical physicists frequently use it) is a smeared operator
$$O(f)=\int dx\, O(x)f(x)$$
where $f(x)$ is a smearing function.

 

[martinbn]

Intuitively, a nonlocal field is something that should have the following properties:
(i) It is an operator that acts on the same vector space as the local field $O(x)$.
(ii) It cannot be associated with one point $x$.
Perhaps the best example of a nonlocal field (best in the sense that mathematical physicists frequently use it) is a smeared operator
$$O(f)=\int dx\, O(x)f(x)$$
where $f(x)$ is a smearing function.

This is also true for the local ones. That is why they are distributions, not functions.

 

[martinbn]

A definition is a definition. There's not much to argue about it, and @Demystifier gave examples of "non-loacal operators" early on in this thread. I'm a bit lost, about what we are discussing, agreeing, or disagreeing.

And what is the definition of a local field? So far no one answered that. An example was given, then i asked is every local field of that type, no one answered. In the article nothing along these lines was said. So, can anyone give the definition and a reference?

 

[Demystifier]

"direct" (or tensor) product

Since this thread turns out to be about nitpicking on sloppy (physicist's?) and precise (mathematician's?) definitions, it should be mentioned that in mathematics direct product and tensor product are different things, but physicists often say "direct" product when in fact they mean tensor product. In the example above, I think it was really the direct (not the tensor) product.

 

[Demystifier]

This is also true for the local ones.

WHAT is also true for the local ones?

 

[Demystifier]

And what is the definition of a local field? So far no one answered that.

I said it at least two times, so here is the third. According to the reference in the first post, a field operator $O(x)$ is local iff $O(x)=e^{-ipx}O(0)e^{ipx}$.

 

[vanhees71]

The definition has been given in #1, it's an operator with
$$\hat{O}(x)=\exp(-\mathrm{i} \hat{P} \cdot x) \hat{O}(0) \exp(\mathrm{i} \hat{P} \cdot x),$$
where $\hat{P}$ is the total-four-momentum operator.

 

[martinbn]

I said it at least two times, so here is the third. According to the reference in the first post, a field operator $O(x)$ is local iff $O(x)=e^{-ipx}O(0)e^{ipx}$.

That is not how i read it. Where is that definition? To me this looked like an example of a local operator. Similar to saying that a polynomial is a continuous function. That is not what the definition of a continuous function.