How does the Pauli exclusion work on fermions but not on bosons?












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How does fractional spin effect the symmetry of combined wavefunctions?



Why is Planck constant the universal anchor for determining if a particle is a Boson or a Fermion?






quantum-mechanics quantum-spin pauli-exclusion-principle







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    1












    $begingroup$


    How does fractional spin effect the symmetry of combined wavefunctions?



    Why is Planck constant the universal anchor for determining if a particle is a Boson or a Fermion?






    quantum-mechanics quantum-spin pauli-exclusion-principle







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      How does fractional spin effect the symmetry of combined wavefunctions?



      Why is Planck constant the universal anchor for determining if a particle is a Boson or a Fermion?






      quantum-mechanics quantum-spin pauli-exclusion-principle







      share|cite|improve this question











      $endgroup$




      How does fractional spin effect the symmetry of combined wavefunctions?



      Why is Planck constant the universal anchor for determining if a particle is a Boson or a Fermion?






      quantum-mechanics quantum-spin pauli-exclusion-principle




      quantum-mechanics quantum-spin pauli-exclusion-principle






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      share|cite|improve this question













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      edited Nov 24 '18 at 8:41









      Blazar

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      asked Nov 24 '18 at 8:27









      Naama-selaNaama-sela

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          The relation between spin and the exclusion principle is called the spin statistics theorem, but I don't think there is an elementary explanation for it (other than doing the math).



          Plancks constant does not determine if a particle is a boson or a fermion






          share|cite|improve this answer









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            There is the elementary handwaving that exchanging two particles is half a rotation. At least in 3D, a full rotation gives the state that one started with, but in 2D one can have anyons (fractional quantum Hall effect).
            $endgroup$
            – Pieter
            Nov 24 '18 at 12:14











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          $begingroup$

          The relation between spin and the exclusion principle is called the spin statistics theorem, but I don't think there is an elementary explanation for it (other than doing the math).



          Plancks constant does not determine if a particle is a boson or a fermion






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            There is the elementary handwaving that exchanging two particles is half a rotation. At least in 3D, a full rotation gives the state that one started with, but in 2D one can have anyons (fractional quantum Hall effect).
            $endgroup$
            – Pieter
            Nov 24 '18 at 12:14
















          3












          $begingroup$

          The relation between spin and the exclusion principle is called the spin statistics theorem, but I don't think there is an elementary explanation for it (other than doing the math).



          Plancks constant does not determine if a particle is a boson or a fermion






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            There is the elementary handwaving that exchanging two particles is half a rotation. At least in 3D, a full rotation gives the state that one started with, but in 2D one can have anyons (fractional quantum Hall effect).
            $endgroup$
            – Pieter
            Nov 24 '18 at 12:14














          3












          3








          3





          $begingroup$

          The relation between spin and the exclusion principle is called the spin statistics theorem, but I don't think there is an elementary explanation for it (other than doing the math).



          Plancks constant does not determine if a particle is a boson or a fermion






          share|cite|improve this answer









          $endgroup$



          The relation between spin and the exclusion principle is called the spin statistics theorem, but I don't think there is an elementary explanation for it (other than doing the math).



          Plancks constant does not determine if a particle is a boson or a fermion







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 '18 at 9:03









          yankylyankyl

          836




          836








          • 1




            $begingroup$
            There is the elementary handwaving that exchanging two particles is half a rotation. At least in 3D, a full rotation gives the state that one started with, but in 2D one can have anyons (fractional quantum Hall effect).
            $endgroup$
            – Pieter
            Nov 24 '18 at 12:14














          • 1




            $begingroup$
            There is the elementary handwaving that exchanging two particles is half a rotation. At least in 3D, a full rotation gives the state that one started with, but in 2D one can have anyons (fractional quantum Hall effect).
            $endgroup$
            – Pieter
            Nov 24 '18 at 12:14








          1




          1




          $begingroup$
          There is the elementary handwaving that exchanging two particles is half a rotation. At least in 3D, a full rotation gives the state that one started with, but in 2D one can have anyons (fractional quantum Hall effect).
          $endgroup$
          – Pieter
          Nov 24 '18 at 12:14




          $begingroup$
          There is the elementary handwaving that exchanging two particles is half a rotation. At least in 3D, a full rotation gives the state that one started with, but in 2D one can have anyons (fractional quantum Hall effect).
          $endgroup$
          – Pieter
          Nov 24 '18 at 12:14


















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