Quantum entanglement in quantum field theory How is entanglement described in quantum field theory?  For example, if a fluctuation in the electron field represents an entangled electron, is there an equal and opposite fluctuation in the positron field representing its entangled partner?
 A: There is "so much" entanglement in quantum field theory, that people don't usually make a big fuss about it. A pure bipartite state is $|\psi \rangle_{AB}$ is separable iff there exists $|\phi_1\rangle_A, |\phi_2\rangle_B$ such that$|\psi\rangle = |\phi\rangle_A \otimes |\phi\rangle_B$, or equivalently, if
$$ \langle \psi | A \otimes B |\psi \rangle = \langle \psi |A \otimes I |\psi \rangle \langle \psi | I \otimes B |\psi\rangle $$
for all operators $A, B$.
In quantum field theory, this is basically never the case. Indeed, we are often interested in quantities such as
$$ \langle 0 | \phi(x_1, t) \phi(x_2, t) |0 \rangle, $$
where $|0\rangle$ is the vacuum state. The calculation of the above expression, for example in the familiar case of a free scalar field, is done in every QFT book and it is non-zero (although exponentially decreasing with the distance), while $\langle 0 |\phi(x,t)|0 \rangle = 0$. Therefore, the vacuum state of a QFT is entangled.
In more sophisticated discussions, people often refer to the Reeh-Schlieder theorem (https://en.wikipedia.org/wiki/Reeh%E2%80%93Schlieder_theorem), which essentially states that any state of a QFT can be obtained by acting locally on the vacuum. This is reminiscent of the way in which any bipartite maximally entangled state can be obtained by starting by applying local unitaries on only one part of a Bell state.
