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Is the eigenstates of p right?

  1. Aug 18, 2008 #1
    Quantum Arnold’s cat is a special system.
    The Hamiltonian is H=p2+Kq2[tex]\delta[/tex]1(t)/2, where p[tex]\in[/tex](0,1],q[tex]\in[/tex](0,1].
    The system is in an N-dimensional Hilbert space, where N=1/h.
    Thus we can define : The eigenstates of [tex]\widehat{q}[/tex] are |j>, j=1,….,N, and the eigenstates of [tex]\widehat{p}[/tex] are |L>, L=1,…,N.
    So [tex]\hat{q}[/tex]|j>=[tex]\frac{j}{n}[/tex]|j>, [tex]\hat{p}[/tex]|L>=[tex]\frac{L}{N}[/tex] |L>.

    Now let’s obtain the eigenstates of [tex]\hat{p}[/tex].
    Because [tex]\hat{p}[/tex]|L>=[tex]\frac{L}{N}[/tex] |L>, -ih[tex]\frac{d\psi(q)}{dq}[/tex]=L/N[tex]\psi(q)[/tex].
    Therefor the eigenstates of [tex]\hat{p}[/tex] is [tex]\psi(q)[/tex]=exp(i2[tex]\pi[/tex]Lq).

    Is it right?
  2. jcsd
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