Understanding Local and Nonlocal Operators in Quantum Field Theory

[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 positive 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.

 

[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 $$\delta^{3}(\vec{x}) = \delta^{2} (x_{1} , x_{2}) \ \delta (x_{3}) = \delta (x_{1}) \ \delta^{2}(x_{2} ,x_{3})$$.