According to the principles of identical particles, the wavefunction of a collection of fermions must be antisymmetric and such a state is entangled. Doesn't this mean that any given electron in the universe (which is a giant system) is entangled with every other electron in the Universe? Why not?

  • 2
    $\begingroup$ Not to mention the fringe theory that there's only one electron in the universe, popping up all over in different quantum states! But I believe the more common approach is that there are so many quantum values & states that the degree of entanglement for distant pairs is unmeasurably small. $\endgroup$ Commented Oct 26, 2020 at 15:37
  • $\begingroup$ @CarlWitthoft If travelling at infinite velocity and changing between electron to quarks and others, that would make it the true "god particle" $\endgroup$
    – Michael
    Commented Oct 27, 2020 at 6:05
  • 1
    $\begingroup$ @Michael I read that theory with amusement and interest. Supposedly, the math all checks out - it travels forwards and backwards in time. Forwards as an electron, backwards as a positron. Thus, it is capable of, over a couple of uunheptaquadrillion passes represent every electron in the universe at a certain point in time. As a bonus, it can also represent every positron. It is, after all, equal in a feynman diagram whether something is an electron going forwards in time, or a positron, going backwards in time. $\endgroup$
    – Stian
    Commented Oct 27, 2020 at 8:15
  • $\begingroup$ @Michael: I was thinking of it more as being the "Walter Mitty" particle. :-) $\endgroup$ Commented Oct 27, 2020 at 12:44

4 Answers 4


Entanglement is always implicitly defined relative to a particular tensor product factorization of Hilbert space. Whether a particular state is entangled depends on which decomposition we choose. For example consider a 3 spin space $\mathcal{H}_A\otimes \mathcal{H}_B\otimes \mathcal{H}_C$ and consider the state $$ |\uparrow \downarrow \uparrow\rangle - |\downarrow \uparrow\uparrow\rangle = (|\uparrow \downarrow \rangle - |\downarrow \uparrow\rangle)|\uparrow\rangle\;. $$ If we consider the decomposition $\mathcal{H}_A$ and $(\mathcal{H}_B\otimes \mathcal{H}_C)$ then the state is entangled, whilst if we consider the decomposition $(\mathcal{H}_A\otimes \mathcal{H}_B)$ and $\mathcal{H}_C$ then it is not.

So going back to electrons, say I have 2 electrons and write the combined Hilbert space for both particles as $\mathcal{H}_A\otimes\mathcal{H}_B$. As electrons are identical fermions, any valid physical state must be antisymmetric, so if I have one electron on earth and one on Alpha Centauri the state will have the form $$ |\text{Earth}\rangle|\text{Alpha Centauri}\rangle - |\text{Alpha Centauri}\rangle|\text{Earth}\rangle $$ which is clearly entangled.

There are, however, other possible factorizations of the 2 particle Hilbert space that are possible, apart from splitting it into the states of particle $A$ and the states of particle $B$. In particular we could limit our attention to the antisymmectric subspace and use an occupation number representation (as is done in the standard Fock space construction), where we list all possible basis states for the two (single) particles (for definiteness lets choose position eigenstates) and label the states occupied by (either) one of the 2 particles with a 1 and a 0 other wise, to obtain a basis for our antisymmetrized 2 particle space.

In this representation the above state might have the form $$ |0\rangle |1\rangle |0\rangle \dots |0\rangle\text{s from here to Alpha Centauri}\dots |0\rangle |1\rangle |0\rangle $$ which is not entangled. We therefore have

Electron A is entangled with electron B but the electron in my lab is not (necessarily) entangled with the electron on Alpha Centauri.

So the question then is which factorization is physically relevant? Well the answer seems to be the second one. As electrons are identical we cannot label them as electron $A$ and $B$, so the occupation number formalism is generally more straightforward to work with (although the labeled particle formalism can be useful for few particle calculations). Quantum field theory, in particular is much closer to the occupation number approach. Therefore it is reasonable to say that the electron in my lab is probably not entangled with an electron on the other side of the universe.

  • $\begingroup$ Dear @BySymmetry, thanks for your answer. In your example, the state that you've chosen is not antisymmetric with respect to the interchange of the 2nd and 3rd spins. It becomes $|\uparrow\uparrow\downarrow\rangle-|\downarrow\uparrow\uparrow\rangle$ which is not minus the state you've chosen. $\endgroup$ Commented Oct 28, 2020 at 14:08
  • $\begingroup$ Yes. I was trying to illustrate that whether or not a state is entangled depends on the factorization of Hilbert space being considered. I used distinguishable spins as it allows for much simpler examples. The later paragraphs go on to apply the idea to fermions $\endgroup$ Commented Oct 28, 2020 at 14:16
  • $\begingroup$ Can you write an antisymmetric 3-particle state that obeys the property of your opening paragraph? $\endgroup$ Commented Oct 31, 2020 at 5:34
  • $\begingroup$ A totally antisymmetric N-particle wavefunction will, I believe, always be fully entangled, with respect to a decomposition of Hilbert space into (labeled) single particle states $\endgroup$ Commented Oct 31, 2020 at 11:50
  • $\begingroup$ Having said that the second half of the answer is essentially saying that I can find a decomposition (in a rather more complicated way) such that the state is not inherently entangled (and that this decomposition is arguably more physically meaningful) $\endgroup$ Commented Oct 31, 2020 at 11:53

I would say that the answer to this question is a matter of perspective. In principle, a small, closed quantum system interacting with itself evolves following a unitary time-evolution (i.e., it evolves in a deterministic and reversible manner). Important here is that it stays quantum until a measurement takes place, in that case decoherence happens and the system is not a (possibly entangled) quantum system anymore.

Now, in principle, the whole universe is a closed quantum system that evolved from the Big Bang up to now and therefore should remain quantum. So why don't we see the universe as the whole as a quantum system?

Here is where the perspective comes in: When we go back to our small, closed quantum system and make a measurement with some detector, we let the (up to know) closed system interact with an external device. This detector is so large that we cannot keep track of all its degrees of freedom. Therefore, we loose information (namely, what changed in our detector) but in turn measure the small quantum system. This measurement "destroys" quantum behavior (like entanglement). What was a probability distribution before (the wave-function of the whole quantum system) is now in a definite state. But if we would look at both the small system and the detector as a new quantum system, no information would be lost and the whole system would still be quantum (and possibly entangled).

When we again consider the universe, the very same happens. As a whole, we have an incredibly complicated quantum system which is interacting with itself. But we cannot even begin to understand the whole process, so our lack of information is what makes it impossible for us to see the quantum behavior. So, everything is connected, but if we don't know, what everything is doing, we do not know how it is connected.

One point I glossed over is that this decoherence, the transition from quantum to classical, is actually really hard to understand and (Imho) one of the most important open questions in physics. That is the so-called measurement problem.


Everything is entangled with everything in the universe, which is described by one epic wavefunction in a likewise epic state space. Fortunately, that space can be factorized which is what makes it possible to perform calculations and have them closely match experimental data.

Entanglement is the norm and coherence is the exception. That we can neglect the Alpha Centauri electrons and still get meaningful data does not mean they aren’t entangled.


There is a trick in saying "it stays quantum until a measurement takes place." This depends on the measurement! According to me, a measurement is an interaction between objects, especially one in which information is exchanged (I don't believe information can go only one way, so that one object is altered while the other is not; see: conservation laws). The interesting and puzzling thing -- "the only mystery," in Feynman's words -- happens when two incompatible objects interact.

So if, say, the object is a state of definite momentum, and you measure its position, you will get the puzzling phenomenon.

But if the object is in a state of definite momentum, and you measure its momentum, you don't get the puzzling phenomenon. ("decoherence")

  • $\begingroup$ "it stays quantum until a measurement takes place." After a measurement the outcome of the measurement is entangled as well. Else Schrödinger's cat can't be both alive and dead. The cat exists in both states. From the cat's perspective it thinks there is only one state and doesn't think it is entangled. This can be escalated further and further leading to OP's question. $\endgroup$
    – Tin Nguyen
    Commented Oct 27, 2020 at 8:36
  • $\begingroup$ I don't think the voters liked the "trick" description, rather than it (measurement) being a shift in perspective. Beware of mathematics bearing gifts;-) $\endgroup$ Commented Nov 11, 2020 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.