SpinMagMom - S-1 Spin Recall that in the H-atom solution,...

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S-1 Spin ½ Recall that in the H-atom solution, we showed that the fact that the wavefunction Ψ (r) is single-valued requires that the angular momentum quantum nbr be integer: l = 0, 1, 2. . However, operator algebra allowed solutions l = 0, 1/2, 1, 3/2, 2… Experiment shows that the electron possesses an intrinsic angular momentum called spin with l = ½. By convention, we use the letter s instead of l for the spin angular momentum quantum number : s = ½. The existence of spin is not derivable from non- relativistic QM. It is not a form of orbital angular momentum; it cannot be derived from . (The electron is a point particle with radius r = 0.) Lrp GG G Electrons, protons, neutrons, and quarks all possess spin s = ½. Electrons and quarks are elementary point particles (as far as we can tell) and have no internal structure. However, protons and neutrons are made of 3 quarks each. The 3 half-spins of the quarks add to produce a total spin of ½ for the composite particle (in a sense, ↑↑↓ makes a single ). Photons have spin 1, mesons have spin 0, the delta-particle has spin 3/2. The graviton has spin 2. (Gravitons have not been detected experimentally, so this last statement is a theoretical prediction.) Spin and Magnetic Moment We can detect and measure spin experimentally because the spin of a charged particle is always associated with a magnetic moment. Classically, a magnetic moment is defined as a vector µ associated with a loop of current. The direction of µ is perpendicular to the plane of the current loop (right-hand-rule), and the magnitude is 2 iA i r µ == π . The connection between orbital angular momentum (not spin) and magnetic moment can be seen in the following classical model: Consider a particle with mass m, charge q in circular orbit of radius r, speed v, period T. µ () 2 q2 r q v q v i,v i i A r TT 2 r 2 r ⎛⎞ π = = π = ⎜⎟ ππ ⎝⎠ q v r 2 r i i r m, q 4/26/2008 Dubson, Phys3220
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S-2 | angular momentum | = L = p r = m v r , so v r = L/m , and qvr q L 22 m µ= = . So for a classical system, the magnetic moment is proportional to the orbital angular momentum: q L (orbital) 2m G G . The same relation holds in a quantum system. In a magnetic field B, the energy of a magnetic moment is given by (assuming z EB =−µ⋅ = −µ G B G ˆ BB z = G ). In QM, z Lm = = . Writing electron mass as m e (to avoid confusion with the magnetic quantum number m) and q = –e we have z e e m µ=− = , where m = l .. + l . The quantity B e e µ≡ = is called the Bohr magneton. The possible energies of the magnetic moment in ˆ z = G is given by orb z B =− B m µ =−µ . For spin angular momentum, it is found experimentally that the associated magnetic moment is twice as big as for the orbital case: q S( s p i n ) m G G (We use S instead of L when referring to spin angular momentum.) This can be written z e e m2 m = −µ = B m m . The energy of a spin in a field is spin B E2 B (m = ± 1/2) a fact which has been verified experimentally.
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This note was uploaded on 02/27/2012 for the course PHYSICS 3220 taught by Professor Stevepollock during the Fall '08 term at Colorado.

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SpinMagMom - S-1 Spin Recall that in the H-atom solution,...

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