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    Tensor Products of Su(2) Operators

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    In the Hilbert space C^2 x C^2
    (where x is used in all of my notation to mean the cross product)

    Consider the state vector:
    Psi = 1/sqrt(2) ( e_1 X e_2) + 1/sqrt(2) ( e_2 X e_1 )

    (Part a)
    What is the probability that the measurement of q_3 X I gives the value -1 and how does the state vector change in this case?

    (NOTE: where q_3 is sigm_3, X is tensor product, I is identity matrix, and we are in C^2 X C^2)

    (Part b)
    The same question for the measurement of I x q_3 and the value +1

    (Part c)
    Find the state vector psi(t) at time t if the Hamiltonian is: q_3 x I + I x q_2

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    Solution Preview

    Part a)

    I assume that e1 = [1,0] and e2 = [0,1].

    We first write down the eigenstates of q_3 X I. Clearly these are given by the tensor products of the eigenstates of q_3 and I. We can thus take these eigenstates to be:

    |v_{i,j}> = |e_i> |e_j>

    Note that you can write two kets next to each other to indicate a tensor product. If you apply an operator A X B to this then the A will act on the first ket and B on the second ket. We have:

    q_3 X I |e_i> |e_j> = (-1)^(i+1) |e_i> |e_j>

    To find the probability you expand |Psi> in these eigenvectors. But this expansion is already given:

    |psi> = 1/sqrt[2] [|e_1>|e_2> + |e_2>|e_1>]

    So, all you have to do is sum the ...

    Solution Summary

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