Search

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.