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dan815

  • one year ago

relation between tensor product and matrices

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  1. dan815
    • one year ago
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    |dw:1443895296128:dw|

  2. dan815
    • one year ago
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    i think im just confused on the definition of the tensor product

  3. dan815
    • one year ago
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    I am assuming that is the right matrix

  4. dan815
    • one year ago
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    ok i think this is useful the book defines what these basis vectors look like

  5. Michele_Laino
    • one year ago
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    we can make tensor product between two Hilbert spaces

  6. dan815
    • one year ago
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    it said that in earlier examples \[|00> = (1,0,0,0)^T, |01> = (0,1,0,0)^T , |10> = (0,0,1,0)^T , |00> = (0,0,0,1)^T \] there are column vectors like that

  7. dan815
    • one year ago
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    so is assuming \[|0> = (1,0)^T and |1> =(0,1)^T \]right

  8. Michele_Laino
    • one year ago
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    the basis of the tensor product \[{H_1} \otimes {H_2}\] is the set: \[{e_i} \otimes {f_j}\] where \(e_i\) is the basis for \(H_1\), and \(f_j\) is the basis for \(H_2\)

  9. Michele_Laino
    • one year ago
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    of course \(e_i\) are many vectors, namely \(i=1,...,n\) if the dimension of \(H_1\) is \(n\)

  10. dan815
    • one year ago
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    are we varying i and j for ei and fj for all the basis in H1 and H2 and adding them together to get some particular H1 and H2

  11. dan815
    • one year ago
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    okay i see what u mean

  12. Michele_Laino
    • one year ago
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    yes! a generic vector of \[{H_1} \otimes {H_2}\] can be written as follows: \[\psi = \sum\limits_i {\sum\limits_j {{\psi _{ij}}\left( {{e_i} \otimes {f_j}} \right)} } \]

  13. Michele_Laino
    • one year ago
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    an observable \(A_1\) of the space \(H_1\) can be written as follows: \[{A_1} \otimes 1\] where \(1\) is the identity of \(H_2\)

  14. dan815
    • one year ago
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    okay so thsi is is like writing A1 in our new dimension space of H1 * H2

  15. Michele_Laino
    • one year ago
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    correct! Similarly for anobservable \(B_2\) of \(H_2\): \[1 \otimes {B_2}\] using Dirac notation, the tensor product between states of a particle with spin 1/2 and another particle with spin +1, can be written as follows: \[\left| { + \frac{1}{2}, + 1} \right\rangle \]

  16. Michele_Laino
    • one year ago
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    namely: \[\left| { + \frac{1}{2}} \right\rangle \oplus \left| { + 1} \right\rangle \doteq \left| { + \frac{1}{2}, + 1} \right\rangle \]

  17. Michele_Laino
    • one year ago
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    as you wrote before

  18. Michele_Laino
    • one year ago
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    oops.. \[\left| { + \frac{1}{2}} \right\rangle \otimes \left| { + 1} \right\rangle \doteq \left| { + \frac{1}{2}, + 1} \right\rangle \]

  19. dan815
    • one year ago
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    so |00> being a state is really just |0,0>

  20. Michele_Laino
    • one year ago
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    yes!

  21. Michele_Laino
    • one year ago
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    here is the action of the observable: \[{A_1} \otimes {A_2}\]: \[\left( {{A_1} \otimes {A_2}} \right)\psi \doteq \sum\limits_i {\sum\limits_j {{\psi _{ij}}\left( {{A_1}{e_i}} \right) \otimes \left( {{A_2}{f_j}} \right)} } \]

  22. Michele_Laino
    • one year ago
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    I think tat your matrix is the representative of a quantum observable, for example, which belongs to \(H1\) or to \(H2\).

  23. Michele_Laino
    • one year ago
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    for example it can be one of the Pauli matrices

  24. Michele_Laino
    • one year ago
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    when we are working with the product tensor of two Hilbert spaces, we refer to the observables of each spaces involved in such tensor product

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