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We know that bosons donot follow Pauli exclusion principle, thus they can occupy the same state. But is it equivalent to say that bosons attract each other?

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Saying that bosons attract each other would require a careful analysis of what do we mean by attraction. In classical physics (and in everyday life), attraction between two subsystems implies the existence of some force which makes more probable to find the two subsystem close each other rather than distant. Thus, the presence of an attraction is accompanied by an increase of the pair correlation at short distances with respect to the uncorrelated case. The other way for repulsion.

Quantum systems display a peculiar behavior of pair correlations which does not originate from interactions but from the symmetry or antisymmetry of the wavefunctions. Short range correlation between bosons is always higher than what would result from the interaction only, so it is tempting to interpret it as resulting from a physical attraction, even though no attractive term is actually present in the hamiltonian.

At the lowest perturbative level, it is even possible to obtain an expression for an attractive potential (see Wasserwaage's answer) but one has to take into account that it cannot be used in a consistent way in a non-perturbative case.

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Using a statistical physics approach, you can indeed show that the properties of bosons and fermions result in an effective potential that is attractive for bosons and repulsive for fermions:

In a statistical mechanics framework, you can construct the partition functions for both classical, interacting particles in a potential $V$

$$Z_N=\frac{1}{N!} \bigg( \frac{V}{\lambda^3} \bigg)^N \int \frac{\mathrm{d}^{3N}q}{V^N}e^{ -\beta \sum_{i<j}V(\vec{q_i}-\vec{q_j}) } $$

If we now do the same for quantum gases, we get

$$Z_N \approx \frac{1}{N!} \bigg( \frac{V}{\lambda^3} \bigg)^N \int \frac{\mathrm{d}^{3N}q}{V^N}e^{ -\beta \sum_{i<j}V^S(\vec{q_i}-\vec{q_j}) } $$

Comparing these two equations, we see that the statistics of bosons and fermions express themselves as an effective potential under a configurational integral. This we call the statistical interaction $$V^S(q)= -k_BT \log (1 \pm e^{-2 \pi (q/\lambda)^2}) $$ with the thermal de-Broglie wavelength $\lambda=\sqrt{\frac{2 \pi \hbar^2}{mk_BT}}$ giving the range of the interaction.

The effective potential, attractive for Bosons ("B") and repulsive for Fermions ("F") is shown below. Note, that this "potential" follows only from the statistical properties of the particles.

For a detailed derivation, I recommend Huang's Statistical Mechanics textbook, Figure 9.3 and onwards.

enter image description here

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No. When we say that two particles "attract," we usually mean that there is some intermediary field causing the attraction, i.e. a "force carrier." You can easily make theoretical models of bosons which do not attract. (These would be called "free particles.") However, sometimes people say that bosons "like to be together." This is because it takes less energy to put bosons "near" each other than fermions. However, that doesn't mean they attract. You could make fermions attractive or bosons repulsive by tweaking the intermediary force carrier. It's just that, all else being equal, bosons take less energy to put next to each other than fermions. So while there might be a fuzzy sense in which bosons "attract," I would say that this is an abuse of language that mostly just causes extra confusion.

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