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The generator of spin corresponds to the 90 degree rotation 2x2 matrix, right? A block-diagonal matrix composed of these 2x2 matrices will generate the rotations of the vectors in a wave function of vector components

But this 2x2 matrix has eigenvalues $+i$ and $-i$. Or maybe $+1$ and $-1$ after Hermiticising. Then how come we have $+\frac{1}{2}$ and integral eigenvalues of spin?

Qmechanic
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Ryder Rude
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  • Possible duplicate: https://physics.stackexchange.com/q/20581/2451 – Qmechanic Jan 24 '22 at 07:11
  • @Qmechanic I saw the same proof in my book and I agree with it. This post is asking where does the "proof" in this post go wrong? Why aren't the eigenvalues are +1 and -1 like that of the 2x2 rotation matrix? – Ryder Rude Jan 24 '22 at 07:42
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    Compare the Lie algebra of the spin operators with the Lie algebra of Pauli matrices: What relationship do you see? – Qmechanic Jan 24 '22 at 07:46
  • @Qmechanic I haven't studied lie algebras or Pauli matrices yet. Shankar's book introduces spin before either of those. I only know that the spin matrix generates rotations of a vector valued wave-function. As in, the angular momentum generates rotations of the co-ordinate system and the spin generates rotations of the vectors themselves. So if we ignore the Angular momentum operator, spin alone just generates rotations of vectors, right?. The generator of vector rotations is the 2x2 matrix with eigenvalues +i and -i – Ryder Rude Jan 24 '22 at 07:59
  • Do you know that $S_k=\frac{\hbar}{2}\sigma_k$ in that representation? – Qmechanic Jan 24 '22 at 08:50
  • @Qmechanic I think I got it. The Spin matrices are NOT actually 2x2 rotation matrices, right? That was maybe just an analogy in the book. The wave-function is not actually vector valued. It's spinor-valued, and spinor matrices are different from 2x2 rotation matrices. – Ryder Rude Jan 24 '22 at 10:31
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    It seems you got it. – Qmechanic Jan 24 '22 at 12:45

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