# What is a local operator in quantum mechanics?

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?

From a quantum informational perspective, given a multipartite state $$\lvert\Psi\rangle$$, a local operation is one that acts independently on each subspace. 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 simple example of local operation is some $$A\equiv (A'\otimes I)$$. This is an operation that only acts nontrivially on one part of the system. More explicitly, its action would then read: $$A\lvert\Psi\rangle=\sum_{ij} c_{ij}(A'\lvert i\rangle)\otimes\lvert j\rangle.$$ As another example, an operation of the form $$A'\otimes B'$$ is also local, and its action on a product state $$\lvert\psi\rangle\otimes\lvert\phi\rangle$$ would read $$\lvert\psi\rangle\otimes\lvert\phi\rangle\to (A'\lvert\psi\rangle)\otimes(B'\lvert\phi\rangle).$$ One notable property of local (unitary) operations is that they do not affect the entanglement (in a way that can be made precise).

Global operations are operations that are not local. Albeit in some contexts it's possible that "global" and "non-local" refer to different properties (for example an operation on a three-qubit system that only acts nontrivially on two of the three qubits might be referred to as a non-local, but also not global).

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

• Thank you for the quantum information perspective! Commented Jun 10, 2019 at 13:22
• Easily explained! Just one thing more if you could mention. The 1st eqn says that you have written the state in tensor product space as a product state ( as it's not a schmidt decomposition). On product states A acts on the 1st part and Identity acts on the 2nd part ( if there is no operator B) correct? But same thing happens for entangled states too. A acts on 1st part and identity or B acts on other. So is there no difference between the action of a local operator on products states and entangled states? Does a local operator acts in the same way on a product state and entangled state Commented Nov 25, 2019 at 14:12
• @Shashaank the action of an operator is defined regardless of the states on which it is made to act
– glS
Commented Nov 25, 2019 at 15:14
• @glS thanks for the clarufication Commented Nov 25, 2019 at 15:32

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.

• Okay, so a non-local operator may not mean it is a global operator, right? Commented Jun 5, 2019 at 20:39
• @JoãoBravo Yes, the opposite of "local" in this context is "nonlocal," not "global."
– Buzz
Commented Jun 5, 2019 at 20:41
• 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? Commented Jun 5, 2019 at 20:42
• 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? Commented Jun 6, 2019 at 15:40
• 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. Commented Jun 9, 2019 at 16:50