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I have read this question:

Does a normal torch emit entangled photons?

where Quantum Journalist says:

Although all the particles of a BEC are in a single quantum state, they are not entangled, but rather the macroscopic wavefunction can be separated as a product state

Can superconductivity be understood to be a result of quantum entanglement?

where Chiral Anomaly says:

The BCS state describing a conventional superconductor is indeed an entangled state, involving a superposition of different numbers of Cooper pairs, each of which involves a superposition of different combinations of paired momenta. An important thing about the BCS state is the specific way in which the electrons are entangled with each other. BCS superconductivity relies on the fact that many Cooper pairs can occupy the same "state," using the word "state" here like the word "orbital" in atomic physics. The particular way in which the electrons are entangled with each other is what makes this possible despite the fact that electrons individually obey the Pauli exclusion principle. But now consider the operator $$ B \equiv \sum_{k,s} B(k,s)c^\dagger(k,s)c^\dagger(-k,-s), $$ which is a rough analogue of the operator that creates a Cooper pair. The momenta and spins of the two electrons in this pair are entangled with each other (because the sum cannot be factorized). As a result of this entanglement, we have $$ B^n|0\rangle\neq 0 $$ even for large $n\gg 1$. The individual electrons still obey the Pauli exclusion principle (the operators $c^\dagger$ still anticommute with each other), but their entanglement means that plenty of terms in the product still survive even after taking this anticommutativity into account, leaving a nonzero result.

Now in a BEC, a large number of bosons occupy the same quantum state, at which point microscopic phenomena, like wavefunction interference becomes appearant macroscopically.

Now cooper pairs of electrons, attracted, can have a lower energy level then the Fermi level.

Electrons have spin-​1⁄2, so they are fermions, but the total spin of a Cooper pair is integer (0 or 1) so it is a composite boson.

This means, that if we know the spin of one of the electrons from the Cooper pair, then we have information about the spin of the other electron in the Cooper pair.

Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated, interact, or share spatial proximity in ways such that the quantum state of each particle cannot be described independently of the state of the others, even when the particles are separated by a large distance.

So basically Cooper pair of electrons should be entangled, because information of the spin of one of the electrons will give us info about the spin of the other electron in the pair.

BEC says that the wavefunctions can interfere, but it does not say they are entangled.

Question:

  1. Are Cooper pairs of electrons entangled?
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  • $\begingroup$ Maybe interessting for you: arxiv.org/pdf/1205.2455.pdf or nature.com/articles/srep07607 $\endgroup$
    – Alpha001
    Jul 22, 2019 at 12:33
  • $\begingroup$ Yes in a sense their spins are oppositively alligned. $\endgroup$
    – Jun Seo-He
    Jun 25, 2021 at 18:22
  • $\begingroup$ There are two important aspects to this question: are the two electrons that compose a Cooper pair entangled with each other? And are different Cooper pairs within a BCS ground state entangled with one another? Are you interested in both these questions? $\endgroup$
    – Rococo
    Jul 27, 2021 at 14:37
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    $\begingroup$ @Rococo I am primarily interested in the first question. $\endgroup$ Jul 27, 2021 at 15:58

2 Answers 2

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In a conventional s-wave superconductor the wavefunction of a Cooper pair looks something like: $$\psi(x_1,x_2)=\phi(x_1-x_2)\otimes\left( |\uparrow_1\downarrow_2\rangle-|\downarrow_1 \uparrow_2 \right)$$ Here $\phi$ is the relative wavefunction of the two electrons. Now, the important point is that their spins are in the antisymmetric singlet form shown above.

Does this make them entangled? Actually, that is not as trivial a question as might seem. For two different types of particle (say, a proton and an electron), if they have the joint spin state $\left( |\uparrow_1\downarrow_2\rangle-|\downarrow_1 \uparrow_2 \right)$ then they are unambigously entangled, because this cannot be factorized into two independent spin states. However, in this case we are talking about two electrons, which are identical particles, and the question of entanglement is more subtle and even controversial. I agree with the discussion and conclusion in this related question- that identical particles should be considered entangled, and in particular the two electrons in a Cooper pair are entangled in their spin degrees of freedom. However, note that this conclusion implies that all identical particles are entangled in some way, so in this sense the two electrons in a Cooper pair are not special except in that their entanglement takes a particularly simple and definite form.

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    $\begingroup$ Thank you so much! $\endgroup$ Aug 7, 2021 at 19:29
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Cooper pairs are generally seen in superconducting materials in low temperatures. These are just 2 electrons that are connected in a way to behave like a boson (in basically a Bose-Einstein Condensate state of matter). As it is a boson, its spin value should be 1 or 0 or (-1). As its spin is an integer value, the pair behaves like a boson. Also, we know that the sum of spins of any entangled pair is 0, which is an integer, thus every pair of entangled particles behave as one. So, every Cooper pair is made up of entangled supercooled electrons in the lowest energy state. Also, every entangled pair behaves like a boson because its spin is an integer.

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