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We know that the wavefunction of an electron transforms as Dirac spinor $(1/2,0)⊕(0,1/2)$ under the Lorentz group $SO(3,1) \sim SU(2)\times SU(2).$ Which representations can we form with two electrons?

Mozibur Ullah
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Lorentz99
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  • Well, two left-handed electrons may give you a spin 1 particle if they are antisymmetrized by spacelike L=1, like two right handed ones, and the L-R combinations yield the 3rd component of the above, OR a spinless state for L=0. Surely you don't want a turgid expression that contains everything and means nothing. – Cosmas Zachos Jan 16 '19 at 17:39
  • Ok but if I would like write as a direct sum of Lorentz representations, how could I do? – Lorentz99 Jan 16 '19 at 17:53
  • Tensor multiply and direct sum distribute. Count the multiplicities in the end to ensure you have not missed anything. Check the action of the angular momentum subgroup on it. Propose it in your question. – Cosmas Zachos Jan 16 '19 at 19:25
  • Related: https://physics.stackexchange.com/q/227857/50583, the buzzword is Young diagrams/tableaux/calculus. – ACuriousMind Jan 16 '19 at 19:30
  • Near Duplicate. – Cosmas Zachos Jan 16 '19 at 20:13
  • ...but note the reduced rep you wrote contains both an electron and a positron. So when you distribute to get (1,0)⊕(0,0)⊕(1/2,1/2)⊕(1/2,1/2)⊕(0,0)⊕(0,1), it behooves you to prune out 3/4 of the degrees of freedom, a total of 12 ! – Cosmas Zachos Jan 16 '19 at 20:46
  • sorry , why do you say that the reduced rep contains an electron and a positron? – Lorentz99 Jan 16 '19 at 21:11
  • A Dirac spinor, with 4 degrees of freedom, represents a left electron, a right electron, a left positron and a right positron. Tensoring two such 4-dim reps yields 16 degrees of freedom, as written above. – Cosmas Zachos Jan 16 '19 at 22:34
  • ok, so I suppose I should select $\left(1,0\right)\oplus\left(0,0\right)$, right? – Lorentz99 Jan 16 '19 at 23:09
  • What are you trying to do? The angular momentum subgroup is $J_L + J_R$, if you have digested the near duplicate answers... – Cosmas Zachos Jan 17 '19 at 01:30
  • See here. – Cosmas Zachos Jan 17 '19 at 01:49
  • I'm interested only in in the electrons, both left and right, so which parts of the direct sum should I keep? – Lorentz99 Jan 17 '19 at 10:44
  • It has to be worked out. Why don't you specify how the electron is sitting in the Dirac spinor you are starting from, first? The 16 operator pieces of the above distributed sum involve 4 spin triplets and 4 spin singlets. Why are you trying to do this? – Cosmas Zachos Jan 17 '19 at 15:29
  • It's an homework. I think that the request is to compose only electrons. So if I consider to compose only the parts of electrons I think that the direct sum is given by $\left(\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)\right)\otimes\left(\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)\right)$ and than the decomposition is (1,0)⊕(0,0)⊕(1/2,1/2)⊕(1/2,1/2)⊕(0,0)⊕(0,1). – Lorentz99 Jan 17 '19 at 17:17

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