# What are local and non-local operators in QM?

## Main Question or Discussion Point

In Hartree-Fock method, I saw the Fock operator has two integrals: Coulomb integral and exchange integral. One can define two operator. "The exchange operator is no local operator" why? Whats de diference: local and no local operator?

And why do the operators have singularities?

thanks

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DrDu
A general one-particle operator ##A## can be defined via it's action on a wavefunction in position representation
##\{A\psi\}(r)=\int d^3r' \alpha(r,r') \psi(r')##.
The function ##\alpha(r,r')## is called the kernel of the operator.
When ##\alpha(r,r')=f(r)\delta^3(r-r')##, with Dirac's delta function, we say that the operator is local.
Obviously, the position operator is local with f(r)=r, while for example, the momentum operator is not local as
##\alpha(r,r') =-i\hbar \partial_r \delta(r-r')##.

• bhobba, Tio Barnabe, dextercioby and 1 other person
Tio Barnabe
A general one-particle operator ##A## can be defined via it's action on a wavefunction in position representation
##\{A\psi\}(r)=\int d^3r' \alpha(r,r') \psi(r')##.
The function ##\alpha(r,r')## is called the kernel of the operator.
When ##\alpha(r,r')=f(r)\delta^3(r-r')##, with Dirac's delta function, we say that the operator is local.
Obviously, the position operator is local with f(r)=r, while for example, the momentum operator is not local as
##\alpha(r,r') =-i\hbar \partial_r \delta(r-r')##.
Sorry my ignorance, but what should result application of the position operator into the wavefunction as given above?

DrDu
Sorry my ignorance, but what should result application of the position operator into the wavefunction as given above?
The wavefunction gets multiplied at each point with the respective value of r.

• Tio Barnabe
In Hartree-Fock method, I saw the Fock operator has two integrals: Coulomb integral and exchange integral. One can define two operator. "The exchange operator is no local operator" why? Whats de diference: local and no local operator?

And why do the operators have singularities?

thanks
When F(r)= r
One get:

$${A \phi}(r)= r \phi{r}(r)$$ its a eigenvalue equation
when $$f(r)= -i \hbar \partial_r$$

One get:

$${A \phi}(r)= -i \hbar \partial_r \phi{r}(r)$$ it has no sense.
is the why is local or no local
am i right?

Note: i´m using $$A\phi (r)= \int dr´\phi(r) \alpha(r-r´) \phi(r´)$$

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