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In quantum mechanics, what exactly is meant by "local" operator?

What about a "global" or a "non-local" operator? Are these the same?

Can you also also help me understand what exactly is a local perturbation and a local symmetry?

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A local operator is one whose action only depends on the value of the wave function (and its derivatives) at a single point. Almost all the ordinary operators one encounters are local in this sense, including $\hat{x}$, $\hat{p}$, $\hat{L}_{z}$, etc. The opposite of "local" in this context is not "global," but rather "nonlocal."

A nonlocal operator would act in something like the following way: $$\hat{W}\psi\left(\vec{x}\right)=\int d^{3}x'\, W\left(\vec{x},\vec{x}'\right)\psi\left(\vec{x}'\right).$$ The nonlocality comes from the fact that the value of $\hat{W}\psi$ at a point $\vec{x}$ depends on the value of $\psi$ at other points. The condition for $\hat{W}$ to be Hermitian is $W\left(\vec{x},\vec{x}'\right)=W\left(\vec{x}',\vec{x}\right)^{*}$.

The use of the terms "local perturbation" and "local symmetry" are less clearcut. A local perturbation might be one of two things. It might be a perturbation (added to the Hamiltonian) that is represented by a local operator. Or it might mean a position-dependent perturbation that goes to zero at spatial infinity. (Without further context, it is not possible to know which is meant. I requested a monograph from the library once, since it was supposed to be about "nonlocal solitons," thinking that it would cover solitary waves with nonlocal interactions like $\hat{W}$ above. In fact, it turned out to be a book about non-localized solitonic phenomena.)

Having a local symmetry means that the symmetry transformation will be allowed to depend on position, but there may be other conditions applied. Some people restrict the term "local symmetry" to mean a gauge symmetry of the second kind; other people use it to mean something different.

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  • $\begingroup$ Okay, so a non-local operator may not mean it is a global operator, right? $\endgroup$ – João Bravo Jun 5 at 20:39
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    $\begingroup$ @JoãoBravo Yes, the opposite of "local" in this context is "nonlocal," not "global." $\endgroup$ – Buzz Jun 5 at 20:41
  • $\begingroup$ Thank you! In regards to the last part of my post, can I say that a local symmetry is an invariance of the system to the application of a local operator? If that is the case, are global symmetries a subset of local symmetries? $\endgroup$ – João Bravo Jun 5 at 20:42
  • $\begingroup$ But isn't "a position-dependent perturbation that goes to zero at spatial infinity" also described by a local operator that you add to the hamiltonian? $\endgroup$ – João Bravo Jun 6 at 15:40
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    $\begingroup$ Just to abuse terminology, the opposite of a local operator is not a global operator in the same sense that a the opposite of a local government (such as a city government) is any larger government (such as a state government or even a few nations signing treaties)... a "global government" is a much larger concept indeed. $\endgroup$ – Cort Ammon Jun 9 at 16:50
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If you are dealing with a multipartite state $\lvert\Psi\rangle$, then the distinction between local and non-local operations is an important one for example for the study of entanglement.

Consider for simplicity a bipartite state $\lvert\Psi\rangle$, that is, a state that can be written as $\lvert\Psi\rangle=\sum_{ij}c_{ij}\lvert i\rangle\otimes\lvert j\rangle$. A local operation $A$ is one that only acts on one part of the system. For example, an operation of the form $$A\lvert\Psi\rangle=\sum_{ij} c_{ij}(A'\lvert i\rangle)\otimes\lvert j\rangle$$ is local in that it only affects the first part of the system. In the simple case of a product state $\lvert\psi\rangle\otimes\lvert\phi\rangle$ you would have an evolution such as $$\lvert\psi\rangle\otimes\lvert\phi\rangle\to (A'\lvert\psi\rangle)\otimes(B'\lvert\phi\rangle).$$ One notable property of local operations is that they cannot affect the entanglement of the state (in a way that can be made precise).

Global operations are operations that are not local.

For a two-qubit system, an example of a global operation can be a CNOT, while a local operation is any single-qubit gate.

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  • $\begingroup$ Thank you for the quantum information perspective! $\endgroup$ – João Bravo Jun 10 at 13:22

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