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QM tensor product

  1. Jul 15, 2010 #1
    Hi, I am reading this article for homework about a ring in a megnetic field. It starts off by giving a hamiltonian (an adiabatic part -never mind)
    [tex]
    H_{0}= \frac{1}{2M} [ \Pi -A]^{2} -\mu B( \phi) \cdot \sigma
    [/tex]

    A- is a known operator

    where [tex] \Pi=\frac{1}{2a} \frac{d}{d \phi} -\frac{eB_{z} \pi a}{2c}[/tex] is the generalized momentum operator

    I know that the eigen states of [tex]\mu B( \phi) \cdot \sigma [/tex] (Spinors) are:
    [tex]
    | \uparrow ( \phi) > =(i \alpha e^{-i \phi}[/tex] , [tex] -\beta)^{T}
    [/tex]

    [tex]
    | \downarrow ( \phi) > =(i\beta e^{-i \phi}[/tex] , [tex] \alpha)^{T}
    [/tex]

    Now, in this article I have he sais that the eigen states of H0 can be written as
    [tex]
    | \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n}[/tex] and

    [tex]
    | \downarrow ( \phi) > \otimes \psi ^{ \downarrow}_{n}
    [/tex]

    When, [tex] \psi \^{ \uparrow}_{n}[/tex] for example is the eigenstate of a Hamiltonian:
    [tex]
    H^{up}_{0}= \frac{1}{2M} [ \Pi -const]^{2} -\mu B
    [/tex]

    How did he get it (the last Hamiltonian)???
    Someone told me to try and apply H0 on
    [tex]
    | \uparrow ( \phi) > \otimes \psi ^{ \uparrow}_{n} [/tex]
    but I got something pretty awful
    Can somone help me please???
     
    Last edited: Jul 15, 2010
  2. jcsd
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