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In QFT, as I read, it appears naturally. It is connected with Poincare algebra, doesn't it?

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As explanation of the main part of the question.

Operator of relativistic orbital angular momentum 4-tensor and 4-impulse operator creates Poincare algebra. It follows that eigenvalues of vector of angular momentum operator ​​are expressed through the whole or half-integer values (in $\hbar $ units). But using only orbital momentum operator causes the possibility of having only integer values. We can artificially add operator, which implements an irreducible representation, so it doesn't connected with coordinate representation and may have half-integer values. But this method is an artificial, because without experimantal proof of existing of spin we can easily operate only with orbital angular momentum. In contrast, in quantum field theory, the spin occurs more naturally (?).

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    The short answers are yes and yes. How deep the answer can go depends on how much QM and/or QFT you have. Does the following sentence make sense to you: "Particle states are classified by unitary irreducible representations of the Poincare group." – Michael Jul 25 '13 at 11:51
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    Hi PhysiXxx you must be a bit carful with too short questions, they automatically appear in the Low Quality queue even if there is nothing wrong with them ... :-/ – Dilaton Jul 25 '13 at 12:09
  • @MichaelBrown , yes, it does. For example, elementary particle is a quantum system described by one of the irreducible rerpresentations of the Poincare group. –  Jul 25 '13 at 12:11
  • @Dilaton , and what are the consequences of this? –  Jul 25 '13 at 12:12
  • People with a certain amount of rep can review it here and decide if it looks good (what I did ;-)...), edit it, or vote for it to be deleted or closed ... – Dilaton Jul 25 '13 at 12:17
  • @Dilaton , I can "increase" the value of the question's text, but the addition won't add some useful information to the question. –  Jul 25 '13 at 12:20
  • I see, automaizing too many things can lead to stupid results :-/. If most reviewers are reasonable enough, appearing wrongly in the low quality queue does no harm to the question at the end. – Dilaton Jul 25 '13 at 12:23
  • @Dilaton , hope so. –  Jul 25 '13 at 12:25
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    Possible duplicates: http://physics.stackexchange.com/q/22449/2451 and links therein. – Qmechanic Jul 25 '13 at 13:10
  • @MichaelBrown , maybe, the answer is the next? The irreducible representations of Lie's algebra of Lorentz group can be representated as $(j_{1}, j_{2})$, where $j_{1}$ is maximum number of 3-rotation group $\mathbf {J}{3}$, $j{2}$ - maximum number of Lorentz group $\mathbf {K}{3}$. So, there are value $\Psi{\mu \nu}$, which transforms as $$ \Psi{'}{\alpha \beta} = S^{j{1}}{\mu \alpha}S^{j{2}}{\nu \beta }\Psi{\alpha \beta}, $$ where $\mathbf S^{j_{i}}$ is matrix (with a size $(2j_{i} + 1)\times (2j_{i} + 1)$) corresponding to a given irreducible representatiom. –  Jul 25 '13 at 16:40
  • So, depending on $j_{1}, j_{2}$ (half-integer or integer value), there can exist the following values: $(0, 0)$ - scalar, $(0, \frac{1}{2}); (\frac{1}{2}; 0)$ - spinor, $(0, 1); (1, 0)$ - antisymmetric tensor, $(\frac{1}{2}, \frac{1}{2})$ - 4-vector.

    Does it connected with spin of a particle?

     –  Jul 25 '13 at 16:46
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    You're almost there. $j_1$ and $j_2$ are not the weights of rotations and boosts resp. Rather the Lorentz group breaks up into $SU(2)\otimes SU(2)$ with the two groups generated by $J_i \pm i K_i$ resp. So the angular momentum is related to $j_1 + j_2$. So the $(0,1/2);(1/2;0);(0,1);(1,0)$ reps work out the way they should and notice that $(1/2,1/2)$ has spin 1/2+1/2=1 and 1/2-1/2=0 components (just the usual addition of angular momentum rules). The spin 1 part is the spatial vector and the spin 0 part is the time component, which is a scalar under spatial rotations. – Michael Jul 25 '13 at 22:58
  • @MichaelBrown , thank you! But your answer created a few new questions. Can you help me, if you please? 1. I don't understand how to identify the $j_{1} + j_{2}$ value with the experimentally observed spin. 2. We introduce operators $\mathbf J_{i} + i\mathbf K_{i}, \mathbf J_{i} - i\mathbf K_{i} $ for transformation of algebra of Lorentz group generators (to algebra of SU(2) group generators). Thus, spin is connected with rotations and Lorentz transformations. But it doesn't connected with space-time shift. So why we use Poincare algebra, not homogeneous Lorentz algebra, for introducing spin? –  Jul 25 '13 at 23:18
  • "...Thus, spin is connected with rotations and Lorentz transformations..." I thought that it is connected only with rotations. This is strange. –  Jul 26 '13 at 00:13
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    Just to be sure we are on the same page, you do realize $j_1+j_2$ corresponds to some representation of $\frac{1}{2}(J_i+K_i)+\frac{1}{2}(J_i-K_i)=J_i$, which are the generators of rotations, right? And I assume you have learnt in QM that spin operators give a representation of generator of rotations? The full story is much longer than these but if you know these already it will be a good start. – Jia Yiyang Jul 26 '13 at 14:05
  • @JiaYiyang : I don't understand how did you come to a conclusion about corresponding of $j_{1} + j_{2}$ to the representation of $\frac{1}{2}(J_{i} + iK_{i}) + \frac{1}{2}(J_{i} - iK_{i})$ generator. –  Aug 04 '13 at 13:24
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    @PhysiXxx: For notational convenience let's define $A_i=\frac{1}{2}(J_{i} + iK_{i})$, $B_i= \frac{1}{2}(J_{i} - iK_{i})$. Now by a direct computation you can show the three $A_i$'s form a SU(2) Lie algebra, so are $B_i$'s, in addition , $[A_i, B_j]=0$. Thus we really have two independent sets of SU(2) generators, so to see the representation of $J_i=A_i+B_i$, you just need exactly the same kind of math as when you add the spins of two particles in elementary QM(usually covered in a 2nd course of QM), so if you are already familiar with that, you know the possible spin quantum numbers are... – Jia Yiyang Aug 04 '13 at 13:52
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    @PhysiXxx: $j_1+j_2$,$j_1+j_2-1$,..., $|j_1-j_2|$, these quantum numbers are just the biggest possible values of $J_3$ – Jia Yiyang Aug 04 '13 at 13:54
  • @JiaYiyang : thank you, I understand this now! So, at the beginning we talk about a classification of values $\psi_{\alpha \beta}$ transforming as $ \psi_{\alpha \beta}{'} = S^{j_{1}}{\alpha \mu}S^{j{2}}{\beta \nu}\psi{\mu \nu}$ with matrices of irreducible representations (and also about the number of the degrees of freedom of $\psi_{\alpha \beta}$ via $(2j_{1} + 1)(2j_{2} + 1)$), then we talk about value $j_{1} + j_{2}$, which is one of the eigenvalues of irreducible representation of the generator of rotations (so corresponds to angular momentum). Right? –  Aug 04 '13 at 14:28
  • And $\hat {\mathbf S}^{j_{1}}{i}$ corrseponds to irreducible representation of $\hat J{i} + i\hat K_{i}$, $\hat {\mathbf S}^{j_{1}}{i}$ - $\hat J{i} - i\hat K_{i}$? –  Aug 04 '13 at 14:34
  • @PhysiXxx: Yes,in short, $(j_1,j_2)$ classifies an irreducible rep of the full lorentz group, but if you restrict this rep on the rotational subgroup, it's actually reducible. – Jia Yiyang Aug 04 '13 at 15:04
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    @PhysiXxx: But I must make you aware of a subtlety, we know lorentz group is non-compact and it's a standard theorem that it cannot have an unitary, faithful, and finite dimensional irreducible rep, but $(j_1,j_2)$ is clearly a faithful finite-dim irr rep of lorentz group, so it can't be unitary, but particle must live in Hilbert space so it requires unitary reps(Wigner's theorem), so $(j_1,j_2)$ cannot directly describe a particle state, then it becomes unclear why $j_1+j_2$ still actually describes the spin, because although rotational subgroup is represented unitarily, but it just a part.. – Jia Yiyang Aug 04 '13 at 15:06
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    ..of non-unitarily represented Lorentz group. This is why I said the full story is much longer. A true satifactory answer can be obtained from Weinberg's QFT Vol1, chap2 and chap5 combined. This is a lot of reading. – Jia Yiyang Aug 04 '13 at 15:10
  • Thank you, I'll read. But I don't understand few aspects:
    1. Operator of irreducible representation of $J_{i}$ with eigenvalue of $j_{1} + j_{2}$ corresponds to Hermitian operator and characterizes spin, doesn't it?
    2. According to your words, why should we classify objects according to their transformation by non-unitary representations of Lorentz group?
     –  Aug 04 '13 at 21:55
  • Yes, that's why I said "because although rotational subgroup is represented unitarily...", however "it just a part of non-unitarily represented Lorentz group". My point is since the full group rep is non-unitary, it can't be a rep on Hilbert space, thus not a direct description of particle state, so although the rotational subgroup is represented unitarily, it's not clear why it really describes the physical spin(because spin is part of particle state so it should live in part of Hilbert space. @PhysiXxx
    • – Jia Yiyang Aug 05 '13 at 02:14
  • 2.No,I did not mean that. It makes much more sense to classify particles according to unitary rep, because we know particle states shall constitute a Hilbert space, and symmetry transformations(e.g. Lorentz transformations) must be represented unitarily(Wigner). However as we've discussed,$(j_1,j_2)$ is not a unitary rep, so exactly how is it related to particle state? This is what I mean by "the full story is much longer".@PhysiXxx – Jia Yiyang Aug 05 '13 at 02:20
  • @JiaYiyang . I don't read anywhere about the classification of fields representation (scalar, vector, tensor etc.) via unitary rep, only via Lorentz group. In Weinberg's book I also didn't find this. Can I find in this book the information? –  Aug 05 '13 at 09:52
  • Not classification of fields, but classification of particles, I think Weinberg has it under the title "one-particle state", somewhere in chap2. I mentioned it since you were asking how the spin of a particle arises in field theory. The subtlety I tried to emphasize can be mathematically put as "what's the relation between classification of particles and classification of fields?"@PhysiXxx – Jia Yiyang Aug 05 '13 at 10:53