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Tensor Products of Su(2) Operators

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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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

A detailed explanation is given.