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21 - 30 of 3,421 search results for KA :PC53 where 0 match all words and 3,421 match some words.
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  2. NATURAL SCIENCES TRIPOS Part II Wednesday 13 January 2016 ...

    www.tcm.phy.cam.ac.uk/~cc726/TP1/ExamFiles/exam16sol.pdf
    20 Jan 2016: 2mṙθ̇ mrθ̈ = ka sin(θ θB). d. dt. L. ṙ=L. ... r= mr̈ = mrθ̇2 kr ka cos(θ θB). d. dt. L.
  3. qcmft

    www.tcm.phy.cam.ac.uk/~bds10/tp3/secqu.pdf
    10 Oct 2012: Ĥ = JN S2 B.Z.Xk! ka†kak O(S0) (2.16). where! k = 2JS(1 cos k) = 4JS sin2(k/2) represents the dispersion relation of the spinexcitations.
  4. qcmft

    www.tcm.phy.cam.ac.uk/~bds10/tp3/ps2.pdf
    10 Oct 2012: cos ka. ——————————————–. Quantum Condensed Matter Field Theory.
  5. Lecture I 1 Lecture I: Collective Excitations: From Particles ...

    www.tcm.phy.cam.ac.uk/~bds10/tp3/lectures.pdf
    10 Nov 2013: ka†k. )] O(N. 1/20 ). cf. quantum AF in spin-wave approximation. ... 2. (akak a. †ka†k. )]As with quantum AF, Ĥ diagonalised by Bogoluibov transformation:(.
  6. phmain

    www.tcm.phy.cam.ac.uk/~bds10/phase/topol.pdf
    4 Apr 2010: The first indication of unusual behaviour in the two-dimensional XY -model (n = 2)appeared in an analysis of the high temperature series expansion by Stanley and Ka-plan (1971).
  7. 16.4. PROBLEM SET IV 216 16.4 Problem Set IV ...

    www.tcm.phy.cam.ac.uk/~bds10/aqp/ps4.pdf
    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
  8. lectures

    www.tcm.phy.cam.ac.uk/~bds10/aqp/lec2_compressed.pdf
    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!
  9. lectures

    www.tcm.phy.cam.ac.uk/~bds10/aqp/lec2.pdf
    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!
  10. lectures

    www.tcm.phy.cam.ac.uk/~bds10/aqp/lec16-17_compressed.pdf
    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.
  11. lectures

    www.tcm.phy.cam.ac.uk/~bds10/aqp/lec16-17.pdf
    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)|.

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