Understanding Local and Nonlocal Operators in Quantum Field Theory

[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!

 

[martinbn]

https://encyclopediaofmath.org/wiki/Generalized_functions,_product_of

 

[vanhees71]

Here we have a distribution defined on test functions of two real variables (you can take $C_0^{\infty}(\mathbb{R}^2)$, $(x,y) \mapsto f(x,y)$).

Of course, $[\delta(x)]^2$, meant as a distribution on $C_0^{\infty}(\mathbb{R})$ is an undefined expression, similar to $1/0$ in the algebra of a ring or field.

Such expressions of course occur often in physics, because we physicists are pretty sloppy, and then we have to think a bit deeper. A prominent example is to use plane-wave asymptotic initial and final states with S-matrix Elements $S_{fi} \propto \delta^{(4)}(P_i-P_f)$ and then calculation probabilities by taking $|S_{fi}|^2$, which of course is an ill-defined meaningless expression. Then we "regularize" the expression and take a more careful limit to make sense of the ill-defined expressions. In this case it's simple, because one should know from classical electrodynamics and quantum mechanics that plane waves are not physically realizable descriptions of fields or states but indeed "distributions" living on the dual of the nuclear space of rigged Hilbert space. So the physical solution is obvious: You have to use true asymptotic free states, aka wave packets, and then also the scattering process being described makes sense.

Another example are infrared and collinear divergences when zero-mass fields ("particles") are involved. There the plane waves aren't even the correct "distribution-valued" (generalized) asymptotic free states, but rather "infraparticles" are. This is cured by either working with the correct asymptotic free "infraparticle" asymptotic free states (e.g., as Kulisch and Faddeev, or Kibble [*] worked out) or you do the more common approach of resumming "soft-photon ladders" (e.g., as detailed in Weinberg, QT of fields, vol. 1).

Last but not least the UV divergences in perturbative evaluation of S-matrix elements are dealed with in renormalization theory.

All this can also be formalized to avoid the mathematical night mares from the very beginning (see the above quoted book by Scharf).

Whether or not a complete mathmatically rigorous non-perturbative existence proof for any realistic QFT is possible, is not known yet.

 

[Demystifier]

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?

It's used to compute the correlation functions, which in turn are used to compute many physical quantities such as the S-matrix.

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!

It's not a field, it's a product of fields. (You do realize that here the word "field" has nothing to do with fields in abstract algebra, do you?) There is absolutely no mathematical problem with this product in classical field theory.

 

[vanhees71]

Here we have a distribution defined on test functions of two real variables (you can take $C_0^{\infty}(\mathbb{R}^2)$, $(x,y) \mapsto f(x,y)$).

Of course, $[\delta(x)]^2$, meant as a distribution on $C_0^{\infty}(\mathbb{R})$ is an undefined expression, similar to $1/0$ in the algebra of a ring or field.

Such expressions of course occur often in physics, because we physicists are pretty sloppy, and then we have to think a bit deeper. A prominent example is to use plane-wave asymptotic initial and final states with S-matrix Elements $S_{fi} \propto \delta^{(4)}(P_i-P_f)$ and then calculation probabilities by taking $|S_{fi}|^2$, which of course is an ill-defined meaningless expression. Then we "regularize" the expression and take a more careful limit to make sense of the ill-defined expressions. In this case it's simple, because one should know from classical electrodynamics and quantum mechanics that plane waves are not physically realizable descriptions of fields or states but indeed "distributions" living on the dual of the nuclear space of rigged Hilbert space. So the physical solution is obvious: You have to use true asymptotic free states, aka wave packets, and then also the scattering process being described makes sense.

Another example are infrared and collinear divergences when zero-mass fields ("particles") are involved. There the plane waves aren't even the correct "distribution-valued" (generalized) asymptotic free states, but rather "infraparticles" are. This is cured by either working with the correct asymptotic free "infraparticle" asymptotic free states (e.g., as Kulisch and Faddeev, or Kibble [*] worked out) or you do the more common approach of resumming "soft-photon ladders" (e.g., as detailed in Weinberg, QT of fields, vol. 1).

Last but not least the UV divergences in perturbative evaluation of S-matrix elements are dealed with in renormalization theory.

All this can also be formalized to avoid the mathematical night mares from the very beginning (see the above quoted book by Scharf).

Whether or not a complete mathmatically rigorous non-perturbative existence proof for any realistic QFT is possible, is not known yet.

[*]
P. Kulish and L. Faddeev, Asymptotic conditions and infrared
divergences in quantum electrodynamics, Theor. Math. Phys.
4, 745 (1970), https://doi.org/10.1007/BF01066485.

T. W. B. Kibble, Coherent Soft-Photon States and Infrared
Divergences. I. Classical Currents, Jour. Math. Phys. 9, 315
(1968), https://doi.org/10.1063/1.1664582.

T. W. B. Kibble, Coherent Soft-Photon States and Infrared
Divergences. II. Mass-Shell Singularities of Green’s Functions,
Phys. Rev. 173, 1527 (1968),
https://doi.org/10.1103/PhysRev.173.1527.

T. W. B. Kibble, Coherent Soft-Photon States and Infrared
Divergences. III. Asymptotic States and Reduction Formulas,
Phys. Rev. 174, 1882 (1968),
https://doi.org/10.1103/PhysRev.174.1882.

T. W. B. Kibble, Coherent Soft-Photon States and Infrared
Divergences. IV. The Scattering Operator, Phys. Rev. 175,
1624 (1968), https://doi.org/10.1103/PhysRev.175.1624.

 

[martinbn]

It's not a field, it's a product of fields. (You do realize that here the word "field" has nothing to do with fields in abstract algebra, do you?) There is absolutely no mathematical problem with this product in classical field theory.

Of course, but that's my point. The question in this thread was to give an example of a non-local FIELD. Now you say that your example is NOT A FIELD!

Here we have a distribution defined on test functions of two real variables (you can take $C_0^{\infty}(\mathbb{R}^2)$, $(x,y) \mapsto f(x,y)$).

Same as above. Of course you can do that. How is that relevant to the question. You may as well have given an example of some completely different mathematical object.

 

[Demystifier]

Of course, but that's my point. The question in this thread was to give an example of a non-local FIELD. Now you say that your example is NOT A FIELD!

Fair enough! But as I said several times, it can still be viewed as a field if $y$ is treated as a constant. Just like $f(x)f(y)$ can be viewed as a function of one variable if $y$ is treated as a constant.

 

[martinbn]

Fair enough! But as I said several times, it can still be viewed as a field if $y$ is treated as a constant. Just like $f(x)f(y)$ can be viewed as a function of one variable if $y$ is treated as a constant.

But if you multiply something by a constant, then it is not much different than the original. How is $5O$ nonlocal if $O$ was local, whatever local means in this context? Is this related to the fact that here we don't have numbers but operators and they don't commute, and we are not multiplying by a constant number but a constant operator? How? Is it ok to multiply like that?

 

[Demystifier]

Is this related to the fact that here we don't have numbers but operators and they don't commute, and we are not multiplying by a constant number but a constant operator?

Of course. I thought it was obvious all along. Let $O(x)=e^{-ipx}O(0)e^{ipx}$ where $p$ is the translation operator. By definition, this equality means that $O(x)$ is a local operator. Now consider the operator
$$A(x)=O(x)O(7)$$
It follows that
$$A(x) \neq e^{-ipx} A(0) e^{ipx}$$
Hence, by definition, $A(x)$ is not a local operator.

 

[vanhees71]

Of course, but that's my point. The question in this thread was to give an example of a non-local FIELD. Now you say that your example is NOT A FIELD!

Same as above. Of course you can do that. How is that relevant to the question. You may as well have given an example of some completely different mathematical object.

You asked for the definition, and I gave it. I don't know, what's so difficult to accept that a non-local operator depends not only on one space-time argument. There's nothing to understand, because it's a definition.

 

[vanhees71]

$O(7)$ doesn't make sense to begin with. $x$ and $y$ are four-vector arguments.

 

[PeterDonis]

By the requirement of closure under the action of $O(x)$ for all $x$.

I'm sorry, but this doesn't help. The action of $O(x)$ for any given $x$ doesn't change $x$. So this still doesn't help me to understand how $O(x)$ and $O(y)$, with $y \neq x$, can both be operators on the same vector space. Why is there not a separate vector space for each individual $x$?

 

[PeterDonis]

For free fields it's the usual Fock space.

I understand how to make sense of a product of operators for different momenta in the momentum representation of Fock space. But how do I make sense of a product of operators for different positions? The basis states in the Fock space are, as you said in another post, plane waves, which means they aren't localized to any position at all.

 

[vanhees71]

For a free field you have an expansion in terms of creation and annihilation operators wrt. some energy eigenbasis. In relativistic QFT it's convenient to choose the momentum eigenstates which are also energy eigenstates with energy eigenvalues $E_{\vec{p}}=\sqrt{\vec{p}^2+m^2}$ (using the usual natural units with $\hbar=c=1$). E.g., for a scalar field
$$\hat{\phi}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 p \frac{1}{(2 \pi)^3 \sqrt{2 E_{\vec{p}}}} \left [\hat{a}(\vec{p}) \exp(-\mathrm{i} p \cdot x) + \hat{b}^{\dagger}(\vec{p}) \exp(+\mathrm{i} p \cdot x) \right]_{p^0=+E_\vec{p}}.$$
Here $p \cdot x=p_{\mu} p^{\mu}=(p^0)^2-\vec{p}^2$ is the usual Minkowski product with signature (+---).

The field operators are by construction operator-valued functions of the space-time coordinates, obeying the canonical equal-time commutation relations (spin 0 implies the quantize as bosons to fulfill the microcausality condition as well as the postive semidefinitenes of $\hat{H}$), resulting in the usual bosonic commutators for the annihilation and creation operators, which are normalized such that
$$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{p}').$$
Now you build with field operators local observables, e.g., the current
$$\hat{J}_{\mu}(x)=\mathrm{i} :\hat{\phi}^{\dagger}(x) \overleftrightarrow{\partial_{\mu}} \hat{\phi}(x):$$
Here the colons denote normal ordering and
$$A(x) \overleftrightarrow{\partial_{\mu}} B(x)=A(x) \partial_{\mu} B(x)-[\partial_{\mu} A(x)] B(x).$$
The four-momentum operator is given by
$$\hat{P}^{\mu} = \int_{\mathbb{R}^3} \mathrm{d}^3 p \frac{1}{(2 \pi)^3} p^{\mu} [\hat{a}^{\dagger}(\vec{p}) \hat{a}(p) + \hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})], \quad p^0=E_{\vec{p}}.$$
With that you get the said property of a local operator for the fields as well as for local composite operators.

Operator products of the kind $\hat{\phi}(x) \hat{\phi}(y)$ are not local operators in this sense, because they do not fulfill this property.

 

[Demystifier]

$O(7)$ doesn't make sense to begin with. $x$ and $y$ are four-vector arguments.

I was working in one dimension. Some still cannot understand what is local operator in one dimension, so four dimensions would be an overkill.

 

[Demystifier]

I'm sorry, but this doesn't help. The action of $O(x)$ for any given $x$ doesn't change $x$. So this still doesn't help me to understand how $O(x)$ and $O(y)$, with $y \neq x$, can both be operators on the same vector space. Why is there not a separate vector space for each individual $x$?

Take, for example, free scalar (Klein-Gordon) field $\phi(x)$. Aren't $\phi(x)$ and $\phi(y)$ operators on the same vector space?

 

[Demystifier]

The basis states in the Fock space are, as you said in another post, plane waves, which means they aren't localized to any position at all.

So what? Have you ever calculated something like $\phi(x)|0\rangle$ for the free scalar field?

 

[PeterDonis]

For a free field you have an expansion in terms of creation and annihilation operators wrt. some energy eigenbasis.

Ok, so for a product like $\hat{\phi}(x) \hat{\phi}(y)$, we would multiply two of these integrals, one with $\exp(i p \cdot x)$ and one with $\exp(i p^\prime \cdot y)$. (The $p$ in each integral would be a different dummy integration variable, which is why I put a prime on the second one.) Then we're back to the issue @martinbn has already raised regarding the product of distributions, since each integral by itself gives a distribution.

 

[vanhees71]

Yes. We all know that the physicists' reckless manipulations with distributions leads to trouble since around 1926. As I said above, the physicists have also found ways to cure it (normal ordering, renormalization) and mathematicians have made it rigorous as far as they could get it rigorous (e.g., the Epstein-Glaser approach to perturbation theory, where the product of local operators and distributions are made rigorous). AFAIK the problem, whether a realistic interacting QFT exists in a rigorous sense, is not solved yet. For QED it's likely not to exist, while for QCD (asymptotic free gauge theory) it may well be, but nobody knows it for sure yet.

 

[MathematicalPhysicist]

Yes. We all know that the physicists' reckless manipulations with distributions leads to trouble since around 1926. As I said above, the physicists have also found ways to cure it (normal ordering, renormalization) and mathematicians have made it rigorous as far as they could get it rigorous (e.g., the Epstein-Glaser approach to perturbation theory, where the product of local operators and distributions are made rigorous). AFAIK the problem, whether a realistic interacting QFT exists in a rigorous sense, is not solved yet. For QED it's likely not to exist, while for QCD (asymptotic free gauge theory) it may well be, but nobody knows it for sure yet.

Why is it likely not to exist for QED?

 

[vanhees71]

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.