# What is a local operator?

I have a sort of basic question. I think an operator that acts on $$n$$-partite states is defined (up to permutation of parties) to be local if it can be written as

$$A = A_1 \otimes_{i=2}^n \mathbb{I}_i$$

where the subscript $$i$$ denotes the party on which an operator acts. This definition ensures that $$A$$ is acting only on one party (here we are assuming that each party is spatially separated from the rest). So far so good. The problem comes when I consider concatenation of operators. Consider for instance a 3-qubit system, and consider the local operators

$$\mathcal{O}_1 = Z_1 \otimes \mathbb{I}_2 \otimes \mathbb{I}_3 \, , \quad \mathcal{O}_2 = \mathbb{I}_1 \otimes Z_2 \otimes \mathbb{I}_3 \, , \quad \mathcal{O}_3 = \mathbb{I}_1 \otimes \mathbb{I}_2 \otimes Z_3 \, ,$$

i.e. $$\mathcal{O}_i$$ applies the $$Z$$ gate to party $$i$$ and does nothing to the rest. If Alice performs $$\mathcal{O}_1$$, then Bob performs $$\mathcal{O}_2$$ and then Charlie performs $$\mathcal{O}_3$$, which I understand is a protocol within the LOCC rules (actually no communication is needed at all) then the resulting gate is

$$\mathcal{O}_3 \mathcal{O}_2 \mathcal{O}_1 = Z_1 \otimes Z_2 \otimes Z_3 \, ,$$

which I would say is a non-local operator. How is it that a succession of local operators gives rise to a non local operator?

This depends on the context in which you're using the operators. You're talking about multiplying them, so I guess you're thinking of, for example, unitaries (and other circuit model elements). In this case, not only are terms $$A\otimes I$$ local operators, but $$A\otimes B$$ is also a local operator. For example, on a quantum circuit, two Hadamard gates applied on neighbouring qubits are described as $$H\otimes H$$.

It is only when you take sums of local operators that you might create a non-local operator. For example, controlled-not is written as $$I\otimes I+(I-Z)\otimes(X-I)/4$$, and cannot be written in the form $$A\otimes B$$, corresponding with the fact that it's entangling. Given that this is the setting I believe you're talking about, everything in your question is considered a local operator.

The other context in which this might arrise is a Hamiltonian. If you have a Hamiltonian that is of the form $$H=A\otimes I$$, that Hamiltonian is local. However, a Hamiltonian $$H=A\otimes B$$ is not local. There is no contradiction here because you do not combine local Hamiltonians by product, but by adding them together, so there is no expectation that $$A\otimes B$$ should be local. Ultimately, what you're really thinking about is the resultant evolution, $$e^{-iHt}$$. There, you can see that $$H=A\otimes B$$ produces unitaries that are non-local in the previous sense, while if I added local operators, $$H=A\otimes I+I\otimes B$$, that produces local operators $$e^{-iAt}\otimes e^{-iBt}$$ in the same sense as before.

There are two inequivalent definitions of "local operator" used in quantum information theory.

The first definition is used in the context of communication over a classical channel (e.g. LOCC). In this context, you have a fixed partition of the complete Hilbert space into a tensor product of $$k$$ different subsystems, and the subsystems are assumed to be so far apart that no entangling operator can ever act on both of them. In this context, a "local operator" is defined to be an operator that can be expressed as a tensor product of operators that only act on an individual subsystem, i.e. $$\bigotimes_{i=1}^k A_i$$, where the operator $$A_i$$ acts on the $$i$$th subsystem. Under this definition, the product of local operators is always a local operator, because by definition $$\left (\bigotimes_{i=1}^k A_i \right) \left( \bigotimes_{i=1}^k B_i \right) = \bigotimes_{i=1}^k (A_i B_i)$$.

The second definition is primarily used in the context of many-body quantum physics, e.g. topological quantum computing. Here, a local operator isn't defined to act nontrivially on only one qubit, but instead on a fixed number of qubits (generally up to four in practice) that does not depend on the total number of qubits in the circuit. So for example a single operator that acts on all the qubits, or a fixed fraction of the qubits, would not count as local. The idea is that as the number of qubits gets large, the operator acts on a negligible small fraction of the total.

When thinking about hardware implementations of a quantum computer, there's sometimes also a natural sense in which some qubits are "next to" each other. In hardware implementations, it's usually easier to engineer multi-qubit gates that act on qubits that are physically close together. In this case, "local" can refer to gates that only act on qubits that are a maximum physical distance away (which will depend on the geometry in which they're laid out). Under this second definition, the product of local operators will in general be non-local if they act on qubits that are sufficiently far apart.

• OK, let's say I have three qubits far apart from each other. Then I think my three operators above are local operators, while their product is not. What does this mean? Mar 20, 2020 at 15:32
• @MBolin Yes, I think that would be the standard usage if it's clear from context that the physical separation is important. It depends on the context in which the term is being used. Mar 20, 2020 at 17:55
• I think you are not answering my question. I am saying that I have the impression that the product of three local operators is not a local operators, since it implies correlations among the separated parties. Mar 20, 2020 at 18:28
• @MBolin Yes, that's correct. Sorry, but I'm not sure exactly what your question is. What do you mean by "What does this mean?"? Mar 20, 2020 at 20:40
• Is it correct? The product of three local operators (the concatenated application of three local gates) may not be local? I have slightly modified my question in order to make it clearer. Mar 21, 2020 at 0:33