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16.4. PROBLEM SET IV 216 16.4 Problem Set IV ...
www.tcm.phy.cam.ac.uk/~bds10/aqp/ps4.pdf22 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 -
Publications 1980 - 1989
www.tcm.phy.cam.ac.uk/~mw141/1980.html13 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. -
lectures
www.tcm.phy.cam.ac.uk/~bds10/aqp/lec16-17.pdf17 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)|. -
lectures
www.tcm.phy.cam.ac.uk/~bds10/aqp/lec2.pdf12 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! -
lectures
www.tcm.phy.cam.ac.uk/~bds10/aqp/lec2_compressed.pdf21 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! -
Chapter 2 Quantum mechanics in onedimension Following the rules ...
www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_1d.pdf8 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. -
lectures
www.tcm.phy.cam.ac.uk/~bds10/aqp/lec15_compressed.pdf9 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. -
lectures
www.tcm.phy.cam.ac.uk/~bds10/aqp/lec16-17_compressed.pdf17 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. -
lectures
www.tcm.phy.cam.ac.uk/~bds10/aqp/lec15.pdf9 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. -
Chapter 10 From molecules to solids In the previous ...
www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_molecular.pdf9 Nov 2009: Substitution confirms that this is a solution with energies(exercise),. Ek =ε 2t cos(ka)1 S cos(ka).
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