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13 Dec 2009: Phys. A18 3007-3026 (1985) (23) Frequency Dependence of NMR Spin-Lattice Relaxation in Bilayer Membranes JA Marqusee, M Warner and KA Dill, J.

22 Nov 2009: and verify that the scattering is isotropic when the energy of the incidentparticle or the size of the scatterer is sufficiently low, so that Ka 1.Obtain an expression for the

21 Oct 2009: κa =. {ka tan(ka) evenka cot(ka) odd. κa =. (2ma2V0! 2 (ka)2. ... leads to the condition,. k [An1 cos(ka) Bn1 sin(ka) An] =2maV0!

12 Oct 2009: κa =. {ka tan(ka) evenka cot(ka) odd. κa =. (2ma2V0! 2 (ka)2. ... leads to the condition,. k [An1 cos(ka) Bn1 sin(ka) An] =2maV0!

17 Nov 2009: eikna = ks (eika eika). 1N. eikna2ks cos(ka). We therefore find that. ... Classical chain: normal modes. ωk = 2. ksm| sin(ka/2)|. At low energies, k 0, (i.e.

17 Nov 2009: Substituted into the equations of motion, we obtain. (mω2 2ks ) = 2ks cos(ka). ... ksm| sin(ka/2)|. Classical chain: normal modes. ωk = 2. ksm| sin(ka/2)|.

9 Nov 2009: For N lattice sites and periodic boundary condition, αnN = αn,solution given by αn =. 1N. eikna,. E = Ek =ε 2t cos(ka)1 S cos(ka). i.e.

9 Nov 2009: For N lattice sites and periodic boundary condition, αnN = αn,solution given by αn =. 1N. eikna,. E = Ek =ε 2t cos(ka)1 S cos(ka). i.e.

9 Nov 2009: Substitution confirms that this is a solution with energies(exercise),. Ek =ε 2t cos(ka)1 S cos(ka).

8 Oct 2009: Thefigure (right) compares κa = ( 2mV0a. 2! 2 (ka)2)1/2 with ka tan(ka) for the even. ... Bn1 =2maV0! 2k Bn sin(ka) Bn cos(ka) An sin(ka). Advanced Quantum Physics.