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Density Operators

  1. May 6, 2015 #1
    I have difficulty in understanding the Density Operator. Please see attached file. (From the Book " Quantum Mechanics Demystified Page 250)

    Most grateful if someone could help!!

    Peter Yu
     
  2. jcsd
  3. May 6, 2015 #2
    I have difficulty in understanding the Density Operator. Please see attached file. (From the Book " Quantum Mechanics Demystified Page 250)

    Most grateful if someone could help!! Here is the attached file!!

    Peter Yu
     

    Attached Files:

  4. May 6, 2015 #3

    aleazk

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    Gold Member

    The duality operation transforms those ##c_i## into their conjugates. Do the outer product between the vector and its dual and you will get the result (remember that you have two sums, the expansion of the vector and the one of the dual, you have two sums and that's why you have two indices; when the values of the indices coincide, you get ##c_i## multiplied by its conjugate, i.e., the square modulus)
     
  5. May 6, 2015 #4

    bhobba

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    Science Advisor
    Gold Member

    Its best understood the other way around. Forget what you have been told about states - they are not in general elements of a vector space - they are positive operators of unit trace. Operators of the form |u><u| are by definition called pure. The |u> can be mapped to a vector space so pure operators are the ones usually talked about. Operators of the form ∑pi |bi><bi| are called mixed. It can be shown all states are pure or mixed.

    The Born rule using this definition of a state is, given an observable O, the expected value of an observation using O, E(O), when a system is in state P, is E(O) = Trace (PO).

    Thanks
    Bill
     
  6. May 6, 2015 #5

    DrClaude

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    Staff: Mentor

    To complement aleazk's answer, try it for yourself. Consider the simple case
    $$
    | \psi \rangle = c_1 | u_1 \rangle + c_2 | u_2 \rangle
    $$
    Find ##\langle \psi |##, then calculate ##\rho = | \psi \rangle \langle \psi |##.
     
  7. May 6, 2015 #6
    Hi All,
    Thank you very much your all your help!
    Now I understand the first part on the first page. But I still have difficulty on the second part on the second page of my attached file.
    Thank again!!
    Peter Yu
     
  8. May 6, 2015 #7

    DrClaude

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    Staff: Mentor

    Which part of that equation causes you problems? The first or the second equality, or both? Or is it the left-hand side you don't understand?
     
  9. May 6, 2015 #8
    Hi DrClaude,
    First of all many thanks for your kind assistance. I do not understand both side.
    Regards,
    Peter Yu
     
  10. May 6, 2015 #9

    DrClaude

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    Staff: Mentor

    The first equality you can find by replacing ρ using the first equation you didn't understand. When first calculating ##\rho | u_j \rangle##, you will get a bunch of ##\langle u_i | u_j \rangle##, which by orthogonality of the basis states ##| u_i \rangle## will result in Kronecker deltas ##\delta_{ij}##. You will thus get ##c_i^* \langle u_i |## from ##\rho## and ##| u_j \rangle## on the left-hand side, resulting in
    $$
    c_i^* \langle u_i | u_j \rangle = c_i^* \delta_{ij} = c_j^*
    $$
    You then do the same with the bra ##\langle u_i |## applied to ##\rho##, giving you ##c_i##.

    Tjhe second equality you get by rewriting the complex numbers ##c_i## and ##c_j^*## in polar form, as explained at the bottom of page 251.
     
  11. May 6, 2015 #10
    Hi All,
    I got it now!!
    Thank you very much for your help!
    Regards!
     
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