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Hamiltonian matrix and eigenvalues

  1. Jul 18, 2013 #1
    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. jcsd
  3. Jul 18, 2013 #2

    hilbert2

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    Science Advisor
    Gold Member

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