# Physical interpretation of entanglement in QFT vacuum

The ground state of a relativistic QFT has nonzero correlation between field operators at spatially separated points. A way to interpret this is through entanglement between different spacetime regions.

In Quantum Mechanics entanglement is between for example, two qubits which is the mathematical representation of for example, entanglement between spins of two electrons.

What is the physical interpretation of entanglement in the vacuum of relativistic QFT? I understand that the entanglement is between regions of spacetime but what are the states of the two regions which are entangled?

Update: I would like to make this questions a bit more precise using Rindler decomposition of Minkowski space. The Minkowski space can be decomposed into left and right rindler wedges. The vacuum state wavefunction can be written as (the derivation can be seen in for example Daniel Harlow's lecture notes, section 3.3) $$|\Omega\rangle=\frac{1}{\sqrt{Z}}\sum_i e^{-\pi\omega_i}|i^{*}\rangle_L|i\rangle_R$$

where $|i^{*}\rangle_L$ and $|i\rangle_R$ are $\mathrm{eigenstates^{*}}$ of the restriction of Lorentz boost operator to the right wedge ($K_r$). In this the entanglement between two pieces of the Minkowski space is evident. But, is there an operational interpretation for this entanglement? What would be a measurement procedure which would destroy this entanglement?

*Actually, $|i^{*}\rangle_L$ is not an eigenstate of $K_r$. $|i^{*}\rangle_L=\Phi |i\rangle _L$ where $|i\rangle _L$ is an eignestate of $K_r$ and $\Phi$ is a antiunitary operator that exists in all quantum field theories and is usually called CPT. (details can be found in the reference given above)

• The entanglement is attributed to the fluctuating quantum field. Even if the particle number is zero, the quantum field still has a nonzero fluctuation. Consider the ground state of a simple harmonic oscillator, if you act the number operator $a^\dagger a$ to the ground state, the number is zero, there no particle, but the ground state wave function $\exp(-x^2)$ is not trivial, the position (as well as momentum) of the oscillator is still fluctuating. This is the vacuum fluctuation, and this is where the vacuum entanglement comes from. – Everett You May 19 '17 at 1:35