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Under Schrödinger's representation, for free particles without spin, each eigenstate vector is $\delta(x-x_0)$, corresponding to the eigenvalue $x_0$, each position in the 3-dimensional configuration space of free particles would represent a quantum state.
My question is:

  1. What would be the stable ground state under the above situation?
  2. If the free particle is cooled close to 0K and becomes Boson-Einstein condensate, they should be all in the same quantum state, wouldn't this mean that all the atoms would occupy the same position?
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1 Answer 1

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The delta function $\delta(x-x_0)$ is the (general) eigenstate of the position operator.

  1. What would be the stable Ground state under the above situation?

For the computation of the ground state, you need to provide the full Hamiltonian, since the ground state is the $|\psi\rangle$ that minimises it. Most importantly, you have to specify the potential (e.g. harmonic) and the interactions.

  1. If the free particle is cooled close to 0K and becomes Boson-Einstein condensate, they should be all in the same quantum state, wouldn't this mean that all the atoms would occupy the same position?

For a discussion about the "same place" argument in a BEC, see this answer.

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