# 13.3: Scattering of indistinguishable particles into the same state

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Consider scattering of two indistinguishable bosons by an external potential. The wavefunction describing the bosons must be symmetric with respect to exchange.Thus the cross section for scattering of both through the same angle is: $$|2f(\theta)|^2$$: two bosons are twice as likely to be scattered into the same state as two distinguishable particles. For many bosons the effect is even more pronounced, and the probability of scattering out of the state is similarly reduced.

The tendency for bosons to clump into one state leads to superfluid behavior in He$$^4$$ and superconductivity: $$\alpha$$ particles and Cooper pairs behave as bosons. All the particles are in the same state and cannot be scattered out.

For fermions, the cross section for being scattered into the same state is $$|f(\theta ) − f(\theta )|^2 = 0$$, as we would expect from the exclusion principle.

This page titled 13.3: Scattering of indistinguishable particles into the same state is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.