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Two qubit state multiplication

  1. Apr 15, 2013 #1
    Hi guys having a little trouble with two-qubit state multiplication.

    Could you tell me how you work out the following? (not only the answers but the working out) I need to be able to understand these calculations before I can move on to the next step of an entanglement question. I understand single qubit bra and ket notation for example |0><0|. And I know |00> means the tensor product of |0> and |0> but I am struggling to compute the following:

    |00><00| = ?
    |00><01| = ?
    |00><10| = ?
    |00><11| = ?

    |01><00| = ?
    |01><01| = ?
    |01><10| = ?
    |01><11| = ?

    |10><00| = ?
    |10><01| = ?
    |10><10| = ?
    |10><11| = ?

    |11><00| = ?
    |11><01| = ?
    |11><10| = ?
    |11><11| = ?

    Thanks in advance. (ps i hope i posted this in the right place, please move it if not)
     
  2. jcsd
  3. Apr 15, 2013 #2

    kith

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    Science Advisor

    What do you mean by compute? I assume that you want the matrix representation in the |00>,|01>,|10>,|11>.

    You have written out all 16 possible ket-bra combinations for two quibits. Each operator corresponds to a 4x4 matrix, where one entry is a one and the others are zero. You could calculate all these matrices by writing out the kets resp. bras as column resp. row vectors. If you have done this for one or two expressions, you will immediately see how the others have to look like.

    For real calculations, it helps to know that every operator A can be written as A = Ʃi,j|ai><ai|A|aj><aj| = Ʃi,jaij|ai><aj|. So you don't have to write any matrices, you can always stick to the bra ket notation.
     
    Last edited: Apr 15, 2013
  4. Apr 15, 2013 #3
    Thanks you for your reply. Yes I will be doing calculations with it and I should keep it in bra and ket notation. Can you explain those equations you wrote? I think that is exactly what I have to use, but don't understand it. Thanks again for the reply
     
  5. Apr 15, 2013 #4

    kith

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    It's hard to answer if I don't know what exactly you don't understand.

    Let's take the simple example of a single qubit. The basis elements |ai> then are simply |0> and |1>. Now how does A = Ʃi,jaij|ai><aj| look like?
     
  6. Apr 16, 2013 #5
    A = Ʃi,jaij|ai><aj|

    To be honest from this I understand, correct me if i'm wrong:

    |ai><aj| = outer product

    if |ai> is |0> and |1> does <aj| represent <0| and <1| or have i completely misunderstood this?

    thanks again for the reply
     
  7. Apr 17, 2013 #6

    kith

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    Yes that's right, so our arbitrary operator A looks like this:
    A = a00|0><0| + a10|1><0| + a01|0><1| + a11|1><1|

    |0> and |1> are represented by column vectors (1,0)T and (0,1)T, <0| and <1| by row vectors (1,0) and (0,1). Do you know how to calculate the outer products using this? What you get is the matrix
    Code (Text):
    a[SUB]00[/SUB] a[SUB]01[/SUB]
    a[SUB]10[/SUB] a[SUB]11[/SUB]

    Here are a few examples and their matrix forms:
    σz = |0><0| - |1><1|
    Code (Text):
    1 0
    0 -1

    σx = |1><0| + |0><1|
    Code (Text):
    0 1
    1 0

    σ+ = |1><0| (if you apply this operator to |0> you get |1>, hence it is the raising operator)
    Code (Text):
    0 0
    1 0
     
    Last edited: Apr 17, 2013
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