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?


2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – MBolin
    Mar 20, 2020 at 15:32
  • $\begingroup$ @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. $\endgroup$
    – tparker
    Mar 20, 2020 at 17:55
  • $\begingroup$ 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. $\endgroup$
    – MBolin
    Mar 20, 2020 at 18:28
  • $\begingroup$ @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?"? $\endgroup$
    – tparker
    Mar 20, 2020 at 20:40
  • $\begingroup$ 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. $\endgroup$
    – MBolin
    Mar 21, 2020 at 0:33

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