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Vectors in Hilbert space

  1. Jul 2, 2008 #1
    In QM we require that an operator acting on a state vector gives the corresponding observable multiplied by the vector.

    Spin up can be represented by the state vector [tex] \left( \begin{array}{c} 1 \\ 0 \end{array} \right) [/tex], while spin down can be represented by [tex] \left( \begin{array}{c} 0 \\ 1 \end{array} \right) [/tex]

    As I understand the Hamiltonian is represented by an infinite dimensional matrix, because there is an infinite number of energy eigenstates. My question is, how can we satisfy both

    [tex] \hat{H} \left | \psi \right \rangle = E \left | \psi \right \rangle [/tex]


    [tex] \hat{L_{z}} \left | \psi \right \rangle = m\hbar \left | \psi \right \rangle[/tex]

    when in one case [tex] \left | \psi \right \rangle [/tex] is a 2 dimensional vector, and in the other it is an infinite dimensional vector.
  2. jcsd
  3. Jul 2, 2008 #2


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    different bits!

    Hi Bobhawke! :smile:

    I think the answer is that an electron say could be

    [tex]\left( \begin{array}{c} 1 \\ 0 \end{array} \right) \sum a_n\psi_n[/tex] ,

    and the rank-two L acts on the left bit, while the infinite-rank H acts on the right bits. :smile:
  4. Jul 2, 2008 #3
    Ah of course I should have seen that.

    Thanks tiny tim!
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