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So I had taken a course on BEC and Cold Atoms. I have read about the properties of non-interacting Bose gas and I was a little concerned about what we mean by two wave functions (of bosons) being the same. Does it mean that they are physically located at the same point in space? I understand that they have the same spatial probability distribution. But it is more than that. I basically want to know given two wave functions how do I say they are the same? A more physical meaning of two wave functions being the "same".

P.S. This is my first time posting. So please be kind.

pratik misra
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  • I think you mean to say that the "quantum state" is the same and not that the "wave function" is the same. I cannot think of anything that would make sense of the statement "two wave functions (of bosons) being the same"

    Quantum state being the same does have a meaning. It has to do with the symmetry property of the wavefunction describing all the bosons (there is only one wave function for the multiple bosons, which is why the 'two wavefunctoins' is not making sense to me).

     – gautam1168 May 11 '23 at 06:08
  • If you did mean to say that the queantum state is the same then you can read here for the explanation: https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Quantum_Mechanics_(Fowler)/05%3A_Interlude_-_The_Nature_of_Electrons/5.01%3A_Bosons_and_Fermions

    For fermions, two fermions cannot have the same "quantum state". I think you are looking for the Pauli Exclusion priniciple in the above link. Note that there is difference between quantum state and wave function.

     – gautam1168 May 11 '23 at 06:10

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Two wavefunctions being same is a misleading figure of speech: what is really meant is that bosons occupy the same state. When talking about multiple indistinguishable particles, we describe them by one multi-particle wavefunction, which in case of bosons should be symmetric in respect to exchanging particles.

If the particles are non-interacting (which is often the first approximation), this wave function can be indeed written as a product of identical single-particle states: $$\psi(x_1,x_2,...,x_N)=\prod_{n=1}^N\phi_0(x_n).$$ In this case the particles may indeed be found in the same point. However, if the particles are interacting, the wave function does not allow such a decomposition anymore, and the probability of finding two particles in the same point would be extremely low (or zero.) This is notably the case when we deal with a Bose-Einstein condensate of atoms - even if the atoms are electrically neutral, they cannot be considered as non-interacting, since they experience strong repulsion when brought very close together (which could be modeled, e.g., as a hard-core repulsion potential.)

Roger V.
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