Larmor Precession e = S B = B 0 z m 1 B = B 0 S - - PowerPoint PPT Presentation
Larmor Precession e = S B = B 0 z m 1 B = B 0 S z = 0 H = 0 1 2 i t ( t ) = H ( t ) cos( / 2) e i t/ 2 ( t ) = sin( / 2) e
i ∂ ∂tψ(t) = Hψ(t) ψ(t) = cos(α/2)e−iωt/2 sin(α/2)e+iωt/2
z H = − µ · B = −γB0Sz = ω 2 1 −1
m
S
ψ|Sz|ψ =
2 1 −1
cos(α/2)e−iωt/2 sin(α/2)eiωt/2
ψ|Sx|ψ =
2 1 1
cos(α/2)e−iωt/2 sin(α/2)eiωt/2
=
ψ|Sy|ψ =
2 −i i
cos(α/2)e−iωt/2 sin(α/2)eiωt/2
=
ψ|Sx|ψ =
ψ|Sy|ψ =
ψ|Sz|ψ =
ω = −γB0
H = −γ B · S
x + (B0 + αz)ˆ z
B = 0 H = t < 0 −γ(B0 + αz)Sz 0 ≤ t ≤ T T < t
χ(t < 0) = aχ↑ + bχ↓ ω = −γB0 χ(0 < t < T) = aχ↑e−iωt/2+iγαzt/2 + bχ↓eiωt/2−iγαzt/2
χ(t > T) = aχ↑e−iωT/2+iγαzT/2 + bχ↓eiωT/2−iγαzT/2 pz = ±αγT 2
Z protons ignoring repulsion, (n, l, m) orbitals 2 electrons per orbital ↑↓ (spin singlet) n2-fold degeneracy
n = 1 2 electrons n = 2 22 · 2 = 8 electrons n = 3 32 · 2 = 18 electrons n = 4 42 · 2 = 32 electrons n = 5 52 · 2 = 50 electrons . . . . . .
Z Element
1 Hydrogen 1 in (1,0,0) 2 Helium 2 in (1,0,0) 3 Lithium 1 in (2,0,0) 4 Beryllium 2 in (2,0,0) 5 Boron 1 in (2,1,m) 6 Carbon 2 in (2,1,m) 7 Nitrogen 3 in (2,1,m) 8 Oxygen 4 in (2,1,m) 9 Fluorine 5 in (2,1,m) 10 Neon 6 in (2,1,m) 11 Sodium 1 in (3,0,0) 12 Magnesium 2 in (3,0,0) 13 Aluminum 1 in (3,1,m) 14 Silicon 2 in (3,1,m) 15 Phosphorous 3 in (3,1,m) 16 Sulfur 4 in (3,1,m) 17 Chlorine 5 in (3,1,m) 18 Argon 6 in (3,1,m) 19 Potassium 1 in (4,0,0) 20 Calcium 2 in (4,0,0) 21 Scandium 1 in (3,2,m) . . . . . . . . .
2s+1LJ