# How does spin appear in QFT?

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 Brown Jul 25 '13 at 11:51
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
Possible duplicates: physics.stackexchange.com/q/22449/2451 and links therein. – Qmechanic Jul 25 '13 at 13:10
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 Brown Jul 25 '13 at 22:58
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