A 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.

 

[Demystifier]

What is an operator in this context?

A map from a vector space $V$ to $V$. See #12.

 

[vanhees71]

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

A local operator in QFT is an operator which is a function of the space-time variables, being either a building block (like a fermionic field operator) of or directly being a representative of a local observable (like energy density, charge density,...) fulfilling the properties given in the OP #1.

Thereby the existence of an operator $P$ (the "total four-momentum") is (formally) guaranteed by the fact that the field operators and composite operators derived from them, describing local observables, are constructed such as to build a unitary local representation of the proper orthochronous Poincare group.

I always write "formally", because it's of course being far from trivial to make sense of the operator products, because indeed the operators are more like "distributions" (in the sense of "generalized functions") than usual "functions", and such distribution-valued operators cannot be so naively multiplied as we physicists just do. That's why even in the perturbative evaluation of these products we run into trouble immediately, which then is cured by the renormalization procedures. The most rigorous treatment of the perturbative QFT paradigm I know of is the Epstein-Glaser approach, aka "causal perturbation theory", nicely treated in, e.g.,

G. Scharf, Finite Quantum Electrodynamics, Springer (1995)

 

[martinbn]

A map from a vector space $V$ to $V$. See #12.

A local operator in QFT is an operator which is a function of the space-time variables, being either a building block (like a fermionic field operator) of or directly being a representative of a local observable (like energy density, charge density,...) fulfilling the properties given in the OP #1.

Well the article says

A quantum field, or local quantum field operator, is an operator-valued generalized function on space-time obeying certain axioms

So, with the notations above $O(x)$ takes as in input a test function and spits out an operator. The variable $x$ in the notation is misleading as $O(x)$ is not a function but a distribution.

 

[vanhees71]

What is confusing? Also usual distributions are written in this form, e.g., the Dirac distribution $\delta^{(4)}(x)$. Perhaps I shouldn't have been lazy in omitting the "hats" for operators to make clear that $x \in \mathbb{R}^4$ but $\hat{O}(x)$ are operator(-valued distributions).

 

[martinbn]

What is confusing? Also usual distributions are written in this form, e.g., the Dirac distribution $\delta^{(4)}(x)$. Perhaps I shouldn't have been lazy in omitting the "hats" for operators to make clear that $x \in \mathbb{R}^4$ but $\hat{O}(x)$ are operator(-valued distributions).

The notation is not confusing, i am struggling to understand what is meant. For example for the Dirac delta what does this $\delta (x)\delta (y)$ mean?

 

[PeterDonis]

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.

In the specific example you give in this thread, though, $V$ is a Hilbert space (since we're talking about QM) and $O(x)$ and $O(y)$ are supposed to be operators on this Hilbert space. What Hilbert space?

 

[Demystifier]

In the specific example you give in this thread, though, $V$ is a Hilbert space (since we're talking about QM) and $O(x)$ and $O(y)$ are supposed to be operators on this Hilbert space. What Hilbert space?

Many quantum states are in fact not in a Hilbert space. Examples are eigenstates of momentum operator and position operator. To avoid these mathematical subtleties and stay on a safe terrain, I talked simply of a vector space.

 

[PeterDonis]

Many quantum states are in fact not in a Hilbert space.

Ok, substitute a rigged Hilbert space or whatever kind of vector space is appropriate for the particular case we are discussing. My question still stands: what space is it?

 

[Demystifier]

Ok, substitute a rigged Hilbert space or whatever kind of vector space is appropriate for the particular case we are discussing. My question still stands: what space is it?

Whatever space is needed to have closeness under the action of $O(x)$. Do you think that those subtleties would really help to understand the notion of nonlocal operator?

 

[PeterDonis]

Whatever space is needed to have closeness under the action of $O(x)$.

And how do we know this is the same space as the one required to have closure under the action of $O(y)$, since $x$ and $y$ are two different spacetime points?

 

[Demystifier]

And how do we know this is the same space as the one required to have closure under the action of $O(y)$, since $x$ and $y$ are two different spacetime points?

By the requirement of closure under the action of $O(x)$ for all $x$.
(BTW, thanks for correcting my language: closeness $\to$ closure!)

 

[vanhees71]

The notation is not confusing, i am struggling to understand what is meant. For example for the Dirac delta what does this $\delta (x)\delta (y)$ mean?

It has the usual meaning, i.e., for any test function
$$\int_{\mathbb{R}} \mathrm{d} x \int_{\mathbb{R}} \mathrm{d} y \delta(x) \delta(y) f(x,y)=f(0,0).$$

 

[vanhees71]

In the specific example you give in this thread, though, $V$ is a Hilbert space (since we're talking about QM) and $O(x)$ and $O(y)$ are supposed to be operators on this Hilbert space. What Hilbert space?

For free fields it's the usual Fock space. The (rigged) Hilbert space is constructed from the observable algebra as for any QT.

 

[martinbn]

It has the usual meaning, i.e., for any test function
$$\int_{\mathbb{R}} \mathrm{d} x \int_{\mathbb{R}} \mathrm{d} y \delta(x) \delta(y) f(x,y)=f(0,0).$$

Well, no! There is no usual meaning. You cannot multiply distributions. Again you are writing symbols on paper but it's meaningless. The distribution $\delta(x)$ evaluates on test functions. In this case the test functions are $C^\infty_0(\mathbb R)$, smooth compactly supported functions in ONE variable. The distribution returns the value at the point which it is centred at, for example $\delta(5)(f)=f(5)$. There is no usual meaning of a product of two distributions, so you have to say what you mean. It cannot be what you wrote. What yo wrote doesn't have the same domain $C^\infty_0(\mathbb R)$.

 

[martinbn]

Going back to my problem. It is still not clear what is meant by $O(x)O(y)$, given that they are distributions. Even if we ignore that and think of the fields as maps, and not distributions, then it is still not clear to me what that is supposed to be. Even if you ignore the fact that they are operators and consider them scalar functions, what is meant by that product? For instance if the space-time point $x$ is here and now (with respect to me) and the space-time point $y$ tomorrow at noon in Paris, what is the meaning of the product of the two values of the field $O$ at these two events? And if you take all these products over all pairs of events, why is that a field. This is impossible to make sense even in a classical/nonquantum field theory without the mathematical subtleties!