Understanding Local and Nonlocal Operators in Quantum Field Theory

[MathematicalPhysicist]

I am reading the claymath problem here:
http://claymath.org/sites/default/files/yangmills.pdf

on page 6, in the comments (section 5), they call a local operator to be an operator that satisifies:
$\mathcal{O}(\vec{x})=e^{-i\vec{P}\cdot \vec{x}}\mathcal{O}e^{i\vec{P}\cdot \vec{x}}$ where $\langle \Omega,\mathcal{O}\Omega\rangle$ where $\Omega$ is a vacuum vector.

My question is in that case how would a nonlocal operator look like?
If I remember correctly $\exp(-iP\cdot x)$ is the translation operator.
https://physics.stackexchange.com/questions/27087/a-question-from-ticcatis-red-qft-textbook

 

[Demystifier]

My question is in that case how would a nonlocal operator look like?

E.g. something like $O(x)O(y)$.

 

[MathematicalPhysicist]

E.g. something like $O(x)O(y)$.

What?
How do you define a nonlocal operator? i.e $O(\vec{x}):=?$.

 

[Demystifier]

What?
How do you define a nonlocal operator? i.e $O(\vec{x}):=?$.

You missed the point. $O(x)$ and $O(y)$ are local operators. Their product $O(x)O(y)$ isn't.

Another example. Let $O(y)$ be a local operator and let $f(x,y)$ be some function. Then
$$A(x)=\int_{-\infty}^{\infty} dy \, f(x,y) O(y)$$
is not a local operator, unless $f(x,y)=f(x-y)$.

 

[martinbn]

You missed the point. $O(x)$ and $O(y)$ are local operators. Their product $O(x)O(y)$ isn't.

I don't think that is what he is looking for, because the product is not a map from the spacetime, but from two copies.

Another example. Let $O(y)$ be a local operator and let $f(x,y)$ be some function. Then
$$A(x)=\int_{-\infty}^{\infty} dy \, f(x,y) O(y)$$
is not a local operator, unless $f(x,y)=f(x-y)$.

Leaving aside how the integral is defined, or if it even makes any sense, why is this non-local? The article says that a quantum field or a local quantum field operator is an operator valued distribution defined on the spacetime, that satisfies some axioms. They don't give the axioms, but I assume they are standard. Which axioms are not satisfied by this integral?

 

[MathematicalPhysicist]

I don't think that is what he is looking for, because the product is not a map from the spacetime, but from two copies.

Leaving aside how the integral is defined, or if it even makes any sense, why is this non-local? The article says that a quantum field or a local quantum field operator is an operator valued distribution defined on the spacetime, that satisfies some axioms. They don't give the axioms, but I assume they are standard. Which axioms are not satisfied by this integral?

I think the axioms are the ones which are described in the papers by Osterwalder and Schrader, here:
K. Osterwalder and R. Schrader,Axioms for Euclidean Green’s functions, Comm. Math.Phys.31(1973), 83–112, and Comm. Math. Phys.42(1975), 281–305.

I once asked a few questions about this paper in overflow, here:
https://physicsoverflow.org/32749/axioms-for-euclidean-greens-functionss-paper-3

 

[MathematicalPhysicist]

@Demystifier then what is the definition of a local operator?
I thought it was the one which I quoted, i.e:

$\mathcal{O}(\vec{x})=e^{-i\vec{P}\cdot \vec{x}}\mathcal{O}e^{i\vec{P}\cdot \vec{x}}$ where $\langle \Omega,\mathcal{O}\Omega\rangle=0$ where $\Omega$ is a vacuum vector.

 

[Demystifier]

@Demystifier then what is the definition of a local operator?

I gave two operators that depend on $x$ but do not satisfy the definition of local operator that you wrote. Hence they are not local operators. I though that's what you wanted, an example of an operator that is not local.

 

[Demystifier]

I don't think that is what he is looking for,

Maybe, that's why I gave the second example.

Which axioms are not satisfied by this integral?

$A(x)\neq e^{-ipx}A(0)e^{ipx}$.

 

[MathematicalPhysicist]

Wait a minute, so $O(x)=e^{-ipx}Oe^{ipx}$ and $O(y) = e^{-ipy}Oe^{ipy}$
so $O(x,y)=O(x)O(y)$ doesn't satisfy: $O(x,y)=e^{-ip(x+y)}Oe^{ip(x+y)}$ since $O(x)O(y) = e^{-ipx}Oe^{ipx}e^{-ipy}Oe^{ipy}$

Ok that's answer my question, thanks.

 

[martinbn]

I am still not sure about this. What are the axioms? Is being of the form $O(x)=e^{-ipx}Oe^{ipx}$ one of the axioms, or is it a consequence of the axioms? Also the product $O(x)O(y)$ doesn't seem to be an operator at all, let alone nonlocal.

 

[Demystifier]

Also the product $O(x)O(y)$ doesn't seem to be an operator at all

Why?

Let $|v\rangle$ be a vector in a vector space (not necessarily Hilbert space) $V$. If $O(y)$ is an operator, then $|v'\rangle=O(y)|v\rangle$ is in $V$. Hence $|v''\rangle=O(x)|v'\rangle=O(x)O(y)|v\rangle$ is also in $V$. As far I can see, this shows that $O(x)O(y)$ is an operator.

 

[martinbn]

Why?

I am following the article. They define a local field operator to be an operator valued distribution on spacetime, that satisfies some axioms. In other words for each spacetime point $x$ you have an operator valued distribution $O(x)$, that satisfies some axioms. To be a nonlocal operator would mean to not satisfy some of the axioms. The object $O(x)O(y)$ is not a map from spacetime to operator valued distributions. The proper multiplication is $O_1(x)O_2(x)$. Here is an analogy. A continuous function is a map from real numbers to real numbers usually denoted by $f(x)$, that satisfies some conditions. An example of noncontinuous function wouldn't be $f(x)f(y)$, it is not even a function of one variable.

 

[Demystifier]

An example of noncontinuous function wouldn't be $f(x)f(y)$, it is not even a function of one variable.

What if $y$ is treated as a constant, say $y=7$? Is $f(x)f(7)$ a function of one variable?

 

[martinbn]

What if $y$ is treated as a constant, say $y=7$? Is $f(x)f(7)$ a function of one variable?

Yes, and?

 

[vanhees71]

But $O(x) O(y)$ is an operator product, which is (only formally though) well defined, but it's not local, because the translation operator doesn't act on it in the way as defined above.

 

[Demystifier]

Yes, and?

Then $f(x)f(y)$ can be interpreted as a function of one variable $x$.

 

[vanhees71]

But it's still not a local operator of course.

 

[martinbn]

Then $f(x)f(y)$ can be interpreted as a function of one variable $x$.

How?!

 

[martinbn]

But $O(x) O(y)$ is an operator product, which is (only formally though) well defined, but it's not local, because the translation operator doesn't act on it in the way as defined above.

But the result is not a map from spacetime to...

 

[Demystifier]

How?!

By using $y=7$.

 

[martinbn]

By using $y=7$.

Then it is not $f(x)f(y)$ it is $cf(x)$, where $c$ is the number $f(7)$.

 

[Demystifier]

Then it is not $f(x)f(y)$ it is $cf(x)$, where $c$ is the number $f(7)$.

What is the difference between
(i) $f(7)$
and
(ii) $f(y)$ for $y=7$?

 

[martinbn]

What is the difference between
(i) $f(7)$
and
(ii) $f(y)$ for $y=7$?

But it is not what you are writing, you are writing $f(y)$, here $y$ is a variable.

 

[Demystifier]

here $y$ is a variable.

Who said that?

 

[martinbn]

Who said that?

Then what is it?!

 

[Demystifier]

Then what is it?!

An unspecified constant.

 

[martinbn]

An unspecified constant.

So, what is the example then? And why doesn't it satisfy the axioms? By the way what are they?

 

[Demystifier]

So, what is the example then?

$y=7$

 

[martinbn]

$y=7$

What is the example of a nonlocal operator?

 

[Demystifier]

What is the example of a nonlocal operator?

You give me an example of a local operator and then I will give you an example of a nonlocal one.

 

[martinbn]

You give me an example of a local operator and then I will give you an example of a nonlocal one.

There is an example in the first post!

 

[vanhees71]

I'm a bit confused by this discussion. It's obvious that $O(x) O(y)$ is a formally valid operator product, but it's obviously non-local, because it doesn't depend on only one space-time argument.

Indeed, if $O$ is a local operator, by definition
$$O(x)=T(x) O(0) T^{\dagger}(x), \quad O(y)=T(y) O(0) T^{\dagger}(y)$$
with the translation operator (using the west-coast convention and the Heisenberg picture of time evolution)
$$T(x)=\exp(\mathrm{i} x \cdot P)$$
and thus
$$O(x) O(y) = T(x) O(0) T^{\dagger}(x+y) O(0) T(y),$$
which obviously cannot be written in terms of some translation operation with a single space-time argument.

Of course any power of $O$ at the SAME space-time argument is local, because, e.g.,
$$O^2(x)=[T(x) O(0) T^{\dagger}(x)]^2=T(x) O^2(0) T^{\dagger}(x).$$

 

[martinbn]

I'm a bit confused by this discussion.

Most likely I am missing something because I haven't seen the definitions. What is an operator in this context?

 

[Demystifier]

There is an example in the first post!

Then the second post with $y=7$ is the example of nonlocal operator.