What Defines a Local Operator in Position Space?

In summary: It's a tempered distribution and, hence, can be integrated against test functions. The result is then a distribution, i.e., a continuous linear functional, on the space of test functions. For the momentum space representation, you need the Fourier transform of the latter. The Fourier transform of a distribution is defined as usual, via the Fourier transform of the test functions. In this case you get$$\widehat{G}(\vec{p}\,)=\frac{1}{p^2-m^2+\mathrm{i} 0^+}.$$This is again a distribution, but a homogeneous one. Integration against a homogeneous distribution yields the value of the distribution at the origin, so that the momentum space Feynman propag
  • #1
geoduck
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Is it okay to define a local operator as an operator whose matrix elements in position space is a finite sum of delta functions and derivatives of delta functions with constant coefficients?

Suppose your operator is M, and the matrix element between two position states is <x|M|y>=M(x,y).

It seems that, at least formally, any function f(x,y) can be written as an infinite sum of deltas and derivative of deltas, with constant coefficients. So for example, the Green's function which is [itex]\langle x|(\partial^2-m^2)^{-1} |y\rangle=\text{BesselFunction}(x-y)[/itex] can be written as an infinite sum of deltas and derivatives of deltas, but not a finite sum, so it's not local.

But then what about functions of the momentum operator, such as [itex]\log[P] [/itex] or [itex]e^{iP} [/itex]? Are these local?

Is those functions aren't local, is it safe to say that only polynomials in the momentum operator can be local?

Also, the position operator squared has this matrix element: <x|X2|y>=δ(x-y)x2. If we wanted to expand this matrix element as a sum of delta functions and derivatives with constant coefficients, it would require an infinite amount. However, if we let the coefficients depend on x, then it is a finite amount.

So should the definition of a local operator be modified to as any operator whose matrix elements in position space is a finite sum of delta functions and derivatives of delta functions with coefficients depending only on position?
 
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Where did you get this terminology from? Usually "local operators" are defined in relativistic quantum field theory, using the fundamental field operators as building blocks. In a local QFT, the fundamental field operators are assumed to transform as the corresponding classical fields under proper orthochronous Poincare transformations, which are realized by unitary transformations. For a translation with four-vector ##a^{\mu}## you have, e.g., for a massive vector field
$$\hat{U}_T(a) \hat{A}^{\mu}(x) \hat{U}_T^{\dagger}(a)=\hat{A}^{\mu}(x-a)$$
and for boosts or rotations
$$\hat{U}_L(\Lambda) \hat{A}^{\mu}(x) \hat{U}_T^{\dagger}(\Lambda) = {\Lambda^{\mu}}_{\nu} \hat{A}^{\nu}(\Lambda^{-1} x),$$
where ##\Lambda \in \mathrm{SO}(1,3)^{\uparrow}##.

Such unitary representations of the proper orthochronous Poincare group are called "local realizations". A local operator is then defined as a (formal) polynom of local field operators and its space-time derivatives at the same space-time point. It's only formal, because due to the equal-time canonical commutation or anticommutation relations the field operators are operator-valued distributions that cannot be properly multiplied at the same space-time point in a strict mathematical sense.

That's why the infinities occur in loop integrals of perturbation theory, which are cured by renormalization. There's an alternative treatment, the Epstein-Glaser approach, where these artifacts are avoided by, roughly speaking, smearing out the local field-operator products over a finite space-time volume. For a good introduction into the latter approach, see

G. Scharf, Finite Quantum Electrodynamics, Springer
 
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  • #3
vanhees71 said:
That's why the infinities occur in loop integrals of perturbation theory, which are cured by renormalization. There's an alternative treatment, the Epstein-Glaser approach, where these artifacts are avoided by, roughly speaking, smearing out the local field-operator products over a finite space-time volume. For a good introduction into the latter approach, see

G. Scharf, Finite Quantum Electrodynamics, Springer

Thanks. If I have a correlation function, isn't it usually infinity even if all fields in the correlator are at different spacetime points? The exception seems to be a free field theory where everything is finite: therefore it seems infinitys come from structure of hamiltonian, and not the structure of fields.

Also, if fields are distributions, does this mean if you integrate correlation functions over space, all infinitys vanish? This does not seem to be true, because correlation functions in momentum space are infinite and also have momentum conserving delta functions, but momentum correlators are just the spacetime integrals of position ones.
 
  • #4
For correlation functions of free fields, you get well-defined functions, like the Feynman propagator in the vacuum
$$\mathrm{i} G(x)=\langle 0|\mathrm{T}_c \hat{\phi}(x) \hat{\phi}(0)|0 \rangle.$$
It's given as
$$G(x)=\int_{\mathbb{R}^4} \frac{\mathrm{d}^4 x}{(2 \pi)^4} \frac{1}{p^2-m^2+\mathrm{i} 0^+} \exp(-\mathrm{i} p \cdot x),$$
where the Minkowski product is used in the west-coast convention, i.e., ##p \cdot x=p_{\mu} x^{\mu} = p_0 x_0-\vec{p} \cdot \vec{x}##.
 

Related to What Defines a Local Operator in Position Space?

1. What are local operators and what are their functions?

Local operators are mathematical functions that operate on a specific region or neighborhood of a larger space. They are used to analyze and manipulate data within a defined area, rather than the entire data set.

2. What are some common examples of local operators?

Some common examples of local operators include convolution, Gaussian blur, and edge detection. These operators are often used in image processing and computer vision applications.

3. What is the difference between a global operator and a local operator?

The main difference between a global operator and a local operator is the area of data that they act upon. A global operator operates on the entire data set, while a local operator only operates on a specific region within that data set.

4. How are local operators used in scientific research?

Local operators are frequently used in scientific research to analyze data and extract meaningful information from large data sets. They are commonly used in fields such as physics, biology, and neuroscience to study complex systems and patterns.

5. What are some benefits of using local operators in data analysis?

Using local operators in data analysis allows for a more targeted and efficient approach to understanding complex data. They can also help in reducing noise and bias in the data, and can reveal subtle patterns and relationships that may not be apparent at a global level.

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