Hamiltonian matrix and eigenvalues

1. Jul 18, 2013

Roodles01

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.

2. Jul 18, 2013

hilbert2

It's just the normalization factor and it comes from cos(pi/4) = sin(pi/4) = 1/sqrt(2)