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8 Nov 2019:   cos φ sin φ 0sin φ cos φ 00 0 1. ... ψ̇ φ̇ cos θ. . (b) [optional extra] Calculate the matrix B = R>Ṙ and verify that it is antisymmetric.

13 Mar 2019: What. motivated your choice? [Suggestions: You could try ψ = cos(πx/2α) or ψ = (α2x2) for |x| < α and vanishingoutside this interval.]. ... 8. Consider the scaled Schrödinger equation. d2ψ. dx2+ λ cos 2xψ = Eψ.

18 Nov 2019: q1 = Q1 cos λ P2 sin λ, q2 = Q2 cos λ P1 sin λ,. ... p1 = Q2 sin λ P1 cos λ, p2 = Q1 sin λ P2 cos λ.

23 Nov 2019: The. Hamiltonian felt by the protons will then be. H = µ (B b cos(ωt) ) = 2µ. ... The TDSE for this system is thus(ȧ. ḃ. )= i. (B b cos(ωt).

16 Jul 2019: Compute the Moore-Penrose inverse of the operator K : L2([a,b]) R with. ... b) Show that the functions ϕpq C() with. ϕpq(x1,x2) := vpwq cos (πqx1) cos (πpx2).

27 May 2019: 1, 2), B = B(r){cos(ot),sin(ot),0}is the background magnetic field, ẑ is the normal to the air–water interface, B(r) is the strength of the ... 4h2sintyðtÞ; (18). where n̂ = {sin[c(t)],cos[c(t)],0}, b 3mB0ðC

6 Dec 2019: b〉 = cos(θ. 2. )eiφ/2|〉 sin. (θ. 2. )eiφ/2|〉 (10.47). as we found in the second problem sheet, where θ = cos1(a b) and φ is the azimuthalangle ... In particular, suppose a and b are unit vectors, with a b = 0 and b′ = b cos α a sin α.Then

6 Feb 2019: This implies a = b,so using the formula cos θ 1 = 2 cos2(θ/2), we obtain,. ... 4.11). The antiderivative of cos(2 arctan(2x)) is arctan(2x) x, so there exist (new) realconstants a, b and c such that.

19 Jul 2019: x(t) = X R sin(!B(t t0)) and y(t) = Y R cos(!B(t t0)). ... 4r. 1 cos. sin(1.27). This again gives rise to the magnetic field (1.26).

19 Jul 2019: 161 –. by a sign, since this can be compensated by the cos function. ... Insisting that the two. cos functions agree, up to sign, gives us the condition.

22 Jan 2019: ζ,fibLfib (sin2 α ρfib cos2 α) 6π μah, (26)B = [ζ,t Lt (1 ρt )ty tz hζ,fibLfib (1 ρfib ) sin α cos α]Rf l, (27). ... where. A = {ζ,t Lt (cos α[1 (1 ρt )t 2z ] h sin α(1 ρt )ty tz ) μ̃Lfib cos α 6π μah cos α}, (53)B = Lfib sin α,

22 Jul 2019: u =e cos(2py). I0() 1, v = 0. (3.2a,b). Given this fixed, known velocity the concentration simply evolves via the advection–diffusion equation (2.7). ... u(x, y, t) =e cos(2py)Rc0(x,y,t). Z 1. 0e cos(2ps)Rc. 0(x,s,t) ds 1.

27 May 2019: CðtÞ uð0Þ uðtÞh i Ue2X1q0. Pq û0 ûq. y;f. Ue2X1q0. Pq cos yh iqy;. ... CðtÞ Ue2X1q0. t=Tð Þq. q!et=T cos yh iqy. Ue2 exp 1 cos yh iy.

19 Jul 2019: The diagonal matrixelements are particularly simple. h i|H| ii = eE0 h i|x| ii cos(!t) = 0. ... A cos. p2 2. 2t! B sin. p2 2. 2t! #. We’ll again require that all the particles sit in the ground state | 1i at time t = 0.This fixes A =

6 May 2019: h = h0[1 a cos cosh b sin sinh ], (4.1). ... where =. p. 2/P(4.2). and. a = cosh sin cos sinh cos sin cosh sinh. ,

19 Jul 2019: along the a = (sin, 0, cos) axis or along the b = (sin, 0, cos) axis. ... hEPR|Az B |EPRi = sin2. 2 cos2. 2= cos. This means that if we take the four operators A2, B1, A1 and B2 to be spin operators,.

19 Jul 2019: b = R sin. 2. 2. = R cos. 2. If b > R, clearly there is no scattering. ... incident(r) = eikz = eikr cos. is also a solution to the Schrödinger equation.

27 May 2019: d(0)3 =(. sin α0 sin(. 2π s. 0. ), h sin α0 cos. ... A = [ζL(sin2 α0 ρ cos2 α0 ) 6π μah], (100)B = ζhR0L(1 ρ ) sin α0 cos α0, (101).

22 Jan 2019: cosh τ µer, hτ =. kcosh τ µ. , (2.6a,b). eµ =sinh τ. ... cosh τ µ). 1µ2, (2.7a,b). eφ =cos φ ex sin φ ey, hφ =k.

22 Aug 2019: From (1.30a) and (1.31). u = ϕ =Amn(ik cos. mπy. hcos. ... Then. ϕ [eiωc0. (x cos ψy sin ψc0t) eiωc0. (x cos ψy sin ψc0t)].

8 Dec 2019: 2Ey+. n=1. {en[cosh(kny)coth(kn)y sinh(kny)]. dn[sinh(kny) tanh(kn)y cosh(kny)]}cos(knx), (4.3). where kn =2πn/L, and B, E, en ... Hence, we obtain the generalcondition. 0=uy. I. 1E. n=1. dnβn cos(knx) (4.9).

18 Oct 2019: 0:. 0 = AB cos ωτω = B sin ωτ. Which rearranges to:. ... cos 2θ cos θ αβ = 0 (12)sin 2θ sin θ = 0 (13).

6 Nov 2019: b). Its quantum vortex counterpart228is the Iordanskii force, with the Stokes drift corresponding to the phonon current. ... p. t+ H B = (HuL)p FFF , (3.4). with vertical averaging understood in problem (iii).