- #1
Roodles01
- 128
- 0
OK. An example I have has me stumped temporarily. I'm tired.
General spin matrix can be written as
Sn(hat) = hbar/2 [cosθ e-i∅sinθ]
...... [[ei∅sinθ cosθ]
giving 2 eigenvectors (note these are column matrices)
I up arrow > = [cos (θ/2)]
.....[ei∅sin(θ/2)]
Idown arrow> = [-e-i∅sin(θ/2)]
......[cos (θ/2)]
Using these eigenvectors and assuming that the magnetic field in question (from Hamiltonian matrix generalised to deal with any angle) is along the z-axis, then θ= ∏/2 and ∅= 0 then eigenvectors are;
Ispin up> = 1/√2 [1]
.......[1]
and
Ispin down> = 1/√2 [-1]
......[ 1]
Could someone help with where the 1/√2 bit came in.
I'm liable to not see the wood for the trees sometimes.
General spin matrix can be written as
Sn(hat) = hbar/2 [cosθ e-i∅sinθ]
...... [[ei∅sinθ cosθ]
giving 2 eigenvectors (note these are column matrices)
I up arrow > = [cos (θ/2)]
.....[ei∅sin(θ/2)]
Idown arrow> = [-e-i∅sin(θ/2)]
......[cos (θ/2)]
Using these eigenvectors and assuming that the magnetic field in question (from Hamiltonian matrix generalised to deal with any angle) is along the z-axis, then θ= ∏/2 and ∅= 0 then eigenvectors are;
Ispin up> = 1/√2 [1]
.......[1]
and
Ispin down> = 1/√2 [-1]
......[ 1]
Could someone help with where the 1/√2 bit came in.
I'm liable to not see the wood for the trees sometimes.