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My question may seem quite esoteric given the title, but I think it's relatively straightforward when explained properly. Imagine a relatively simple situation of 2 hydrogen atoms (numbered 1 and 2), which we treat semi classically (first quantization QM). If we look at each atom individually and ignore the other one, we have a clear cut analytic solution that describes the electron orbital. On the other hand, if we consider the impact that proton 2 has on electron 1 (and vice versa), we need perturbation theory. Correct me if I am mistaken, but I believe we will end up with an oscillating solution whereas the electrons will trade places over time. If this is indeed the case, then these electrons' wave function will be greatly correlated at some point (in particular when it is equally likely for either of them to be orbiting proton 1 or 2). Here then are my questions :

1) Despite the approximate nature of the perturbation theory solutions, is it fair to say that such significant entanglement, or at least some reasonable measure of it, is likely ubiquitous to the exact description of such systems ? Is the back and forth "switch" from proton 1 to proton 2 for electron 1 (and vice versa) also a ubiquitous behavior ?

2) If this is the case, and if the timescale of such oscillations for typical separation size between proton 1 and 2 (say the distance between protons in the air) is much smaller than say the age of the earth's atmosphere, wouldn't it lead to all the electrons of the earth's atmosphere to be entangled in a giant mess resembling some nightmarish correlation web ? If so, shouldn't such entanglements be readily observable ? I limited the case to the earth's atmosphere almost as a gimmick to help visualize my question, though obviously the conclusion (if correct) would apply on global scales.

3) Does a QFT theoretical description substantially change the conclusions arrived at above ? My understanding of QFT is shaky at best, and breaks down completely when I try to apply it to this question.

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Does your first quantization mean quantizing particle, e.g. $[x,p]=i$, rather than quantizing field $\{a_p,a_q^{\dagger} \}= \delta(p-q)$? Are you using time-dependent perturbation or time-independent perturbation? –  user26143 Jul 10 at 8:03
    
That's exactly what I mean by quantization. And I'm using time independent perturbation theory, though I haven't done the exact derivation. I'm basically extrapolating from simpler perturbation theory problems I remember from my undergrad classes. –  ticster Jul 10 at 10:39
    
If you use time-independent perturbation theory, there is no time-evolution, such as "electrons will trade places over time" (if I am not mistaken..) –  user26143 Jul 10 at 13:23
    
Woops sorry, I mean I'm using time dependent perturbation theory (don't know how that "in" slipped in there). –  ticster Jul 10 at 15:01

1 Answer 1

up vote 2 down vote accepted

user26143 gives a correct argument about the effects of entanglement in two-atom processes, but specifically to your question, I have to reiterate the answer

1) You are not talking about entanglement

2) Yes, this effect gets effectively blocked out in the cases you consider

3) QFT does not really bring any new insight to this problem

In theory, all particles of a given kind are presumed to be entangled - globally. This is a consequence of heuristic arguments about measuring the difference between states with exchanged identical particles or the "particle-exchange symmetry" of the hamiltonian (the wikipedia article on this topic is not so great, the textbook of Ballentine provides an excellent discussion).

So, yes, at least in theory, there is an infinite web of entangled identical particles all over the place (even without interaction). This manifests macroscopically through Bose-Einstein and Fermi-Dirac statistics. The statement seems to be so fundamental we don't put any bounds on it's validity, but nobody has actually experimentally shown quantum statistics to apply over distances larger than say a few kilometers.

As to the example you have given, for now we can just say that every two-electron state is always characterized as $$|\Psi\rangle = |\psi_1\rangle |\psi_2 \rangle - |\psi_2\rangle |\psi_1 \rangle$$ Where $|\psi_{1,2}\rangle$ are one-electron states. Assuming no interaction between the two electrons (or "very weak") and fixed two protons, $|\psi_{1,2}\rangle$ can be just found as states in the two-proton potential. The exact one-particle energy eigenstates will always be eigenstates of reflection with respect to the center of mass of the two protons as this is also a symmetry of the Hamiltonian. A double reflection gives the original state, so we have only $\pm 1$ eigenvalues corresponding to symmetric and anti-symmetric one-electron states with respect to reflection (i.e. $\psi(x)=\pm \psi(-x)$ with $x$ distance from center).

Note that if you want to create a state localized more or less at one proton, you have to superpose an antisymmetric and symmetric state of very close energies because the close-shaped antisymmetric state almost cancels the peak around the second proton in the symmetric state. As a result, you get an oscillation of frequency $\Delta E/\hbar$ corresponding to the perturbation theory result. This oscillation causes the electron to jump to the second proton at times $\sim \hbar/\Delta E$ also corresponding to the result of perturbation theory.

Thus, the example you mention is actually a purely one-particle effect and just expresses the fact that the second proton (or nucleus) can "steal" the first proton's electron and this is an effect possible at any distance. However, this possibility gets cut-off at brutal rates once we consider the screening of the proton potential by his own electron and the fact that there is an ensemble of similar atoms approximately isotropically distributed all around (as is the case of gas). Even in plasma, effects such as Debye shielding make long-range "jumps" very rare. So in the atmosphere, electron stealing is more of a "contact" effect. (The situation is completely different e.g. in solids where the electrons can be "freely moving" throughout the solid and we get a delocalized mess)

But back to the entanglement as antisymmetric particles. As a consequence, we will for example never find a single spectral line in a coherent two electron state of this kind - they will be always distinct and at least slightly shifted. QFT only brings quantitative but not qualitative corrections to these results.

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Thanks, that perfectly addresses what I was asking and more. –  ticster Jul 13 at 12:25
    
Thanks for the clarification. I only thought entanglement in spin space. If entanglement is anything beyond simple product, then Slater determinant always includes it. I have a concern regarding experimental test over distance. Consider two electron in Gaussian wavepackages far from each other, $$[e^{-\alpha r_1^2-\alpha (r_2-R)^2}+ e^{-\alpha (r_1-R)^2-\alpha r_2^2} ] ( | \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle) $$. Once we measure spin of one electron at origin and got ↑, the wavefunction becomes –  user26143 Jul 13 at 13:54
    
$$ e^{-\alpha r_1^2-\alpha (r_2-R)^2} | \uparrow \downarrow \rangle - e^{-\alpha (r_1-R)^2-\alpha r_2^2} | \downarrow \uparrow \rangle$$, since the wavefunction has to be antisymmetric. The wavefunction after measurement is still "entangled" by means of not simple product. It will not have the feature that once Alice know the spin down at her place Bob immediately knows spin up at his place and Alice knows up Bob knows down, something like that, since Alice and Bob at fixed position. In this sense, the wavefunction after measurement is trivially entangled. –  user26143 Jul 13 at 13:54

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