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21 Oct 2009: i. 2γσ Bt. ]. Therefore, for initial spin configuration,. (αβ. )=. (eiϕ/2 cos(θ/2)eiϕ/2 sin(θ/2). ). With B = Bêz , Û(t) = exp[ i2 γBtσz ], |ψ(t)〉 = Û(t)| ... B = B0êz B1(êx cos(ωt) êy sin(ωt)). Paramagnetic resonance. B =

14 Jan 2009: 6]An electromagnetic wave is represented by the 4-vector potential. Aµ = (0, A cos(kz ωt), A sin(kz ωt), 0). ... a) Evaluate the electric and magnetic fields. [6]. (b) Evaluate the Lagrangian density.

17 Jul 2009: 2D d = d(sin θ, 0, cos θ) k kx [Bruun & Taylor, PRL 101, 245301 (2008)]. ... d|〉 = deff (cos ωt, sin ωt, 0)ωt r. θ. V0(r, θ; t) =d2effr3. [.

12 Oct 2009: In central well region, general solution of the form. ψ2(x) = A cos(kx) or B sin(kx),! ... leads to the condition,. k [An1 cos(ka) Bn1 sin(ka) An] =2maV0!

21 Oct 2009: B = B0êz B1(êx cos(ωt) êy sin(ωt)). Paramagnetic resonance. This result shows that spin precessionfrequency is independent of spin orientation. ... B = B0êz B1(êx cos(ωt) êy sin(ωt)). Paramagnetic resonance. B = B0êz B1(êx cos(ωt)

21 Oct 2009: In central well region, general solution of the form. ψ2(x) = A cos(kx) or B sin(kx),! ... leads to the condition,. k [An1 cos(ka) Bn1 sin(ka) An] =2maV0!

20 Jan 2009: a) The electric field is. E = Ȧ = Aω ( sin(kz ωt), cos(kz ωt), 0). ... The magnetic field is. B = A = Ak ( cos(kz ωt), sin(kz ωt), 0).

8 Oct 2009: t =2k1k2eik1a. 2k1k2 cos(k2a) i(k21 k22) sin(k2a),. Advanced Quantum Physics. 2.1. ... the form. ψ(a < x < a) = A cos(kx) B sin(kx) ,. where k =. 2m(EV0)!2. Once again we have four equations in four unknowns.

8 Oct 2009: B = B0ẑ B1(êx cos(ωt) êy sin(ωt)). The effective magnetic field in the frame rotating with the same frequency ωas the small added field is then given by. ... The EM wave has (angular) frequency ω,. Ĥ =gp2. µN Bzσz Bx cos(ωt)σx =gp2.

27 Nov 2009: r. ]P$(cos θ). If we set, ψ(r) % eikr f (θ) eikr. ... u(a0) = sin(ka0 δ0) = sin(ka0) cos δ0 cos(ka0) sin δ0. =