solutions_final

# solutions_final - Solutions to Final Take-home Exam Physics...

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Solutions to Final Take-home Exam Physics 623, Spring 2010, O.W. Greenberg (50 points) 1.1 (a) The degrees of freedom are space, spin and ﬂavor. With all the quarks in the s-state the space wavefunction is totally symmetric, so the wavefunction must be antisymmetric under exchange of spin and ﬂavor. Use Young diagrams and the hook rule. Combine the SU (3) flavor and SU (2) spin degrees of freedom into an SU (6) flavor - spin 6. Then we need 6 6 6. Find this in steps. 6 6 21 15, with 21 symmetric and 15 antisymmetric. 15 6 70 20, with 20 antisymmetric, so we want 20. (b) 6 (3 , 2) under SU (6) SU (3) flaovr ( SU (2) spin . Then 6 6 (3 , 2) (3 , 2) [(6 + ¯ 3 , 3 + 1)] (6 , 3) ( ¯ 3 , 1) (6 , 1) ( ¯ 3 , 3). The last two of these are antisymmetic, so we want them. Then [(6 , 1) ( ¯ 3 , 3)] (3 , 2) (10 + 8 , 2) (8 + 1 , 4 + 2) (10 , 2) (8 , 2) (8 , 4) (1 , 2) (8 , 2) (1 , 4). We want (8 , 2) (1 , 4). (c) The magnetic moment of a baryon B is μ B = h B | μ 3 | B i , where μ 3 = 2 μ 0 X q Q q S q , Q q = ( 2 3 , - 1 3 , - 1 3 ) ( u,d,s ) and S q is the z -component of the spin of quark q . is the charge of the quarks. For the magnetic moments of the proton and neutron, use the fact that the proton

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## This note was uploaded on 12/30/2011 for the course PHYSICS 623 taught by Professor Galitski during the Summer '10 term at Maryland.

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solutions_final - Solutions to Final Take-home Exam Physics...

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