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I understand that Bose Einstein condensate is:

A Bose–Einstein condensate (BEC) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero. Under such conditions, a large fraction of bosons occupy the lowest quantum state, at which point microscopic quantum phenomena, particularly wavefunction interference, become apparent.

I understand that all the particles in the condensate will be described by the same wavefunction.

Now entanglement is:

Quantum entanglement is a physical phenomenon which 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 other(s), even when the particles are separated by a large distance—instead, a quantum state must be described for the system as a whole.

I understand that entangled particles can be described by the same wavefunction too.

So in both cases the particles can be described by the same wavefunction.

Question:

  1. Is BEC the same as entanglement for a larger (macroscale) number of particles?

  2. What is the real difference between BEC and entanglement, other then the number of particles?

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The key difference is the entanglement. The particles in a BEC indeed overlap and therefore when in the BEC can be described by the same wave function. However, when a BEC gets released, the constituent parts will move away from each other and can be then described on their own. If we were to measure on of those particles it would not tell us anything about the other particles.

However, when you have entangled particles, their quantum states depend on one another. So if we have two entangled particles, a measurement of one will tell me the exact state of the other.

In the end, just because particles overlap in a BEC does not entangle them.

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Entanglement and BECs are not the same at all: if you consider the idealized situation of a pure BEC of $N$ bosons, you can write the whole N-boson wave function as \begin{equation} \Psi(x_1,\dots,x_n) = \phi(x_1)\phi(x_2)\times\dots\times\phi(x_N) \end{equation} with a single function $\phi(x)$. Then the probability density to find particles at $x_1,\dots,x_n$ factorizes entirely: \begin{equation} P(x_1,\dots,x_n)=p(x_1)p(x_2)\times\dots\times p(x_N) \end{equation} with $p(x)=\vert\phi(x)\vert^2$. So all detections are conditionally independent of each other in a pure BEC. Measuring one particle has no effect on where the others will be detected. That's the complete opposite of the EPR thought experiment, where two particles are entangled and measuring the state of one particle determines the state of the other.

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    $\begingroup$ To put this in more technical language: the BEC state is a separable state, because, as you show, it is a single explicit product of single-particle states. The correct definition of an entangled state is 'a state that is not separable'. BECs are as far from entanglement as it gets. $\endgroup$ – Emilio Pisanty Sep 6 '18 at 9:46
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    $\begingroup$ @emilio-pisanty "BECs are as far from entanglement as it gets." correct, but only for pure BECs.; there are also depleted and fragmented BEC states that cannot be written in separable form. There are also issues with defining entanglement for more than 2 particles consistently. So it's not an exact antipode. $\endgroup$ – jkds Sep 6 '18 at 10:15
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Here are my sketched out answers. Hopefully others add more rigor.

  1. A characteristic of entangled system is that if you can only look at parts of the system, you don't gain full information. It looks thermal. The full information and the purity of the system is ensured when you look at the whole system. There is an interesting experiment on this. See Science Vol. 353, Issue 6301, pp. 794-800.

  2. BEC is not an entangled system. For simplicity, if we consider the non-interacting case, the wavefunction is factorizable. It looks like a coherent state: $ \frac{(\hat{a}^\dagger)^{N}}{\sqrt{N!}}|vacuum\rangle$. Or to allow number fluctuation (work in the grand canonical ensemble), $\Pi_{i} (u_i + v_i \hat{a_i}^{\dagger})|vacuum \rangle$. An entangled wavefunction cannot be factorized.

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