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

[cristianbahena]

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

 

[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
$\alpha(r,r') =-i\hbar \partial_r \delta(r-r')$.

 

[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.

 

[cristianbahena]

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 $$A\phi (r)= \int dr´\phi(r) \alpha(r-r´) \phi(r´)$$

 

[radium]

The exchange term favors electrons of the same spin to be separated and comes directly from the Pauli exclusion principle. This comes from the anticommutation of fermions (i.e. the antisymmetry of the wavefunction). The exchange process is inherently non local since exchanging any two electrons changes the sign of the entire many body wavefunction.

More generally, you can say that fermions in general are non local because they anticommutate. In a sense they carry a string of minus signs. Another thing is that Fermion parity (even or odd) is always conserved which comes from this. There are a lot very deep consequences of this beyond this discussion.