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?


2 Answers 2


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.

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    $\begingroup$ Thank you for the quantum information perspective! $\endgroup$ Commented Jun 10, 2019 at 13:22
  • $\begingroup$ 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 $\endgroup$
    – Shashaank
    Commented Nov 25, 2019 at 14:12
  • $\begingroup$ @Shashaank the action of an operator is defined regardless of the states on which it is made to act $\endgroup$
    – glS
    Commented Nov 25, 2019 at 15:14
  • $\begingroup$ @glS thanks for the clarufication $\endgroup$
    – Shashaank
    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.

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

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