A Understanding Local and Nonlocal Operators in Quantum Field Theory

[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.

 

[martinbn]

WHAT is also true for the local ones?

What you wrote in your post.

 

[Demystifier]

What you wrote in your post.

So you are saying that $O(f)$ is a local operator?

 

[vanhees71]

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.

Do we have to define all well-known terminology on PF? It all started with "reference frame". Now we get lost in discussions about standard terminology in quantum (field) theory. It's getting somewhat annoying!

A direct, Kronecker, or tensor product in quantum theory is a construct like $|x \rangle \otimes |y \rangle \otimes |z \rangle=|\vec{x} \rangle$ for vectors or, in this case, generalized eigenvectors and the corresponding definitions for operators $\hat{O}_1 \otimes \hat{O}_2 \otimes \hat{O}_3$ etc.

Of course you can define all kinds of "direct products" between all kinds of mathematical objects:

https://en.wikipedia.org/wiki/Direct_product

 

[Demystifier]

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.

The paper says: "... for any local quantum field operator ${\cal O}(\vec{x}) = e^{-i{\vec P}\cdot{\vec x}}{\cal O}e^{i{\vec P}\cdot{\vec x}}$ ..." How could the word "any" be interpreted as an example?

 

[Demystifier]

Do we have to define all well-known terminology on PF.

For @martinbn , yes we do.

 

[martinbn]

So you are saying that $O(f)$ is a local operator?

No, I said that local operators are also of that type, they evaluate on test function. Are you saying they are not?

The paper says: "... for any local quantum field operator ${\cal O}(\vec{x}) = e^{-i{\vec P}\cdot{\vec x}}{\cal O}e^{i{\vec P}\cdot{\vec x}}$ ..." How could the word "any" be interpreted as an example?

Because it means any of these. And they have a definition for the general case early on, which I already quoted. How would you understand the statement "For any polynomial $a_nx^n+\cdots+a_1x+a_0$, we have..."? As a definition of polynomial? That all polynomials are of that form? Then is $x^2+y^2$ a polynomial?

 

[martinbn]

Do we have to define all well-known terminology on PF?

You don't have to do anything, but if asked, why would you argue about nonrelevant things for two pages instead of give the definition? Have you never asked about a definition here? Is it not allowed to ask about definitions here?

 

[vanhees71]

The definition was given in #1. I don't undestand, why we have to discuss standard definitions. I stop from watching this thread now, because obviously I've nothing to add to clarify the confusion. I don't understand, how you can come to the conclusion that @Demystifier 's $O[f]$ might be a local operator. It hasn't even a space-time argument and is an integral of a local field operator over all space!

 

[Demystifier]

Because it means any of these. And they have a definition for the general case early on, which I already quoted. How would you understand the statement "For any polynomial $a_nx^n+\cdots+a_1x+a_0$, we have..."? As a definition of polynomial? That all polynomials are of that form? Then is $x^2+y^2$ a polynomial?

The latter is a polynomial in $x$ with $a_0=y^2$.

 

[Demystifier]

No, I said that local operators are also of that type, they evaluate on test function. Are you saying they are not?

Local operator $O(x)$ in QFT does not evaluate on test function. For instance, for the free Klein-Gordon field operator $\phi(x)$ you can calculate $\phi(x)|0\rangle$ without using any test function. If you smear the local operator $O(x)$ with a test function $f(x)$, the resulting operator $O(f)=\int dx \, O(x)f(x)$ is no longer a local operator.

 

[Nullstein]

What if I define $O(0) := \int \phi(x) f(x) dx$ and $O(x) := e^{-i P x} O(0) e^{i P x}$? Is $O(x)$ a local operator then, because it agrees with the definition?

 

[Demystifier]

What if I define $O(0) := \int \phi(x) f(x) dx$ and $O(x) := e^{-i P x} O(0) e^{i P x}$? Is $O(x)$ a local operator then, because it agrees with the definition?

Good example! Yes, it is a local operator.

 

[Nullstein]

But if we expand it, we get $O(x) = e^{-i P x} \int \phi(y) f(y) dy e^{i P x} = \int e^{-i P x} \phi(y) e^{i P x} f(y) dy = \int \phi(y + x) f(y) dy = \int \phi(y) f(y - x) dy$, so $O(x) = \int \phi(y) g_x(y) dy$ with the new function $g_x(y) := f(y-x)$ and so $O(x)$ is again just $\phi(x)$ smeared with some non-local function $g_x$ and should really be a non-local operator according to your earlier post? I don't see what I'm missing.

 

[MathematicalPhysicist]

But if we expand it, we get $O(x) = e^{-i P x} \int \phi(y) f(y) dy e^{i P x} = \int e^{-i P x} \phi(y) e^{i P x} f(y) dy = \int \phi(y + x) f(y) dy = \int \phi(y) f(y - x) dy$, so $O(x) = \int \phi(y) g_x(y) dy$ with the new function $g_x(y) := f(y-x)$ and so $O(x)$ is again just $\phi(x)$ smeared with some non-local function $g_x$ and should really be a non-local operator according to your earlier post? I don't see what I'm missing.

I think in the integral it should be $e^{-iPx}\phi(y)f(y)e^{iPx}=\phi(x+y)f(x+y)$, am I wrong here?

 

[Nullstein]

For the translation operator, $f(x)$ is just a constant real number and can be pulled out. But even if it were true, we would just get $O(x) = \phi[f]$ instead of $O(x) = \phi[g_x]$, which would still not solve the problem.