atyy said:
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Indeed, (1) is an abuse of notation, and it's not the same (at least not in the usual notation) as (2). How can this go through in highly respected journals? As I referee, I never let through something like this. What is reviewing good for if not to make the content of the paper clear (provided it's not flawed; then you have to reject it anyway)?
Obviously (1) doesn't make sense, because the vacuum has no position information. If I understand the author right, what he labels with subscripts ##A## and ##B## are locations of observers Alice and Bob, but how can that make sense for the vacuum state, which is Poincare (or Galilei for non-relativistic QT) invariant? So again, what's the meaning of such notation?
A single particle state is always created from the vacuum with a creation operator acting on the vacuum. If you want to describe a single-particle state which is peaked at different locations, e.g., prepared by sending a particle through a potential well, where the "matter wave" gets partial reflected and transmitted, the correspoding annihilation operator reads (in the Heisenberg picture)
$$\hat{\psi}^{(+)}(t,x)=\int_{\mathbb{R}} \mathrm{d}^3 k \hat{a}(k) [A(k) \exp(-\mathrm{i} E_k +\mathrm{i} k x + B(k) \exp(-\mathrm{i} E_k-\mathrm{i} k x)],$$
and the corresponding state (time-independent) state is given by
$$|\Psi \rangle=\hat{\psi}^{(+) \dagger}(0,x)|\Omega \rangle,$$
where ##|\Omega \rangle## is the vacuum. If you choose ##A(k)## and ##B(k)## both to peak around ##k=k_0## at long times you have a single particle operator which is peaked at opposite positions.
If you put A somewhere far to the "right" and B somewhere far to the left, indeed only one of them will detect a particle, i.e., if A detects a particle, then necessarily B won't detect it and vice versa. I guess, that's what the author wanted to express with his Eq. (1), but to get the probabilities for finding the particles by A and B you must use simply a single-particle position eigenstate ##|x,t \rangle## (note that in the Heisenberg picture the eigenstates of operators are time dependent). Then the probability distribution to find the particle around ##x## is simply given by
$$\psi(t,x) = \langle x,t|\Psi \rangle.$$
You can of course as well use the Schrödinger picture. Then
$$|\Psi,t \rangle=\hat{\psi}^{(+)\dagger}(t,x)|\Omega \rangle$$
and
$$\psi(t,x)=\langle x|\Psi,t \rangle,$$
where now ##|x \rangle## is the time-independent Schrödinger-picture position eigenstate and ##|\Psi,t \rangle## the time-dependent Schrödinger-picture state. Now ##\psi(t,x)## for large times is peaked on opposite positions (with little overlap of the two parts, if ##A(k)## and ##B(k)## are appropriately chosen).
Of course, what's said in the introduction is true. You don't need two- or more-particle states to have entanglement. You can also have entanglement between different observables of a single particle. A well-known example is the preparation of position-spin entangled states by a Stern-Gerlach apparatus: After the beam rund through the magnetic field it splits in partial beams, each with a (nearly perfectly) prepared spin component in direction of the magnetic field. So you created a state with position-spin entanglement.
Recently they even created states where position and magnetic moment of a neutron seem to be transported on separate paths through a beam splitter:
T. Denkmayr et al, Observation of a quantum Cheshire Cat in a matter-wave interferometer experiment, Nature Communications 5, Article number: 4492
http://dx.doi.org/10.1038/ncomms5492
The article is open access!So maybe the author in the above cited PRA article had in mind the following (take a spin-1/2 particle like a neutron as an example) the state defined in the above explained way by using the annihilation operator
$$\hat{\psi}(t,x) = \int_{\mathrm{R}} \mathrm{d} k [A(k) \hat{a}(k,\sigma=+1/2) \exp(-\mathrm{i} E_k t +\mathrm{i} k x) + B(k) \hat{a}(k,\sigma=-1/2) \exp(-\mathrm{i} E_k t-\mathrm{i} k x)].$$
Then spin up and down are the properties labeled with A and B in the author's notation. Suppose ##A(k)## and ##B(k)## are both peaked around ##k=k_0## and we look at times, were the partial waves are well separated in opposite directions, then A will measure with probabilities very close to
$$P_{\sigma=1/2,A}=\int_{\mathbb{R}} \mathrm{d} k |A(k)|^2, \quad P_{\sigma=-1/2,A}=0,$$
i.e., either a particle with spin up with the given probability but with certainty never a particle with spin down, while for B it's
$$P_{\sigma=-1/2,B}=\int_{\mathbb{R}} \mathrm{d} k |B(k)|^2, \quad P_{\sigma=1/2,B}=0.$$
Of course the single-particle state is position-spin entangled, and indeed, if A detects a particle with spin 1/2, she knows that B can't find one with spin -1/2, because she knows that a single-particle state was prepared, but I still don't understand what (1) means.
Of course you could also use the occupation-number basis. Then the tensor product makes some sense again, because then you define
$$|\{N_i \}_{i} \rangle = \prod_i \frac{1}{\sqrt{N_i!}} (\hat{a}^{\dagger}_i)^{N_i} |\Omega \rangle.$$
This is indeed a (anti-)symmetrized product state for (fermions) bosons. Then with (1) the authors means a state
$$|\psi \rangle_{A,B} = \frac{1}{\sqrt{2}} (|\{N_B=1,N_{i \neq B}=0 \} \rangle + |\{N_A=1,N_{i \neq A}=0 \} \rangle,$$
which indeed is precisely what he wrote in Eq. (2). If you interpret (1) in this way, it makes sense and is consistent, while in the Nature article it's not even clear what's meant with their Eq. (1) (at least not for me).
