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Bra-ket algebra

  1. Sep 22, 2007 #1
    I need to show that (XY)^(dagger)=(Y)^(dagger)(X)^(dagger) using bra-ket algebra
    where X and Y are operators

    say we started out with: if we dagger it (using *):

    XY|a> = X(Y|a>)=(X(Y|a>))=((Y|a>)X*)=<a|Y*X*

    we also know that XY|a>=<a|(XY)* by definition, so (XY)^(dagger)=(Y)^(dagger)(X)^(dagger)

    My question is on how valid this statement is:
    XY|a> = X(Y|a>)=(X(Y|a>))=((Y|a>)X*)=<a|Y*X*
     
  2. jcsd
  3. Sep 22, 2007 #2

    nrqed

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    I don't understand at all X(Y|a>)=(Y|a>)X*
     
  4. Sep 22, 2007 #3
    that's what i'm not sure about, actually, i get here and I dont know what to do:

    XY|a> = X(Y|a>) = (X<a|Y*)
     
  5. Sep 22, 2007 #4
    That's not correct. When you say X<a|Y* you are multiplying a matrix by a row vector, which is impossible.

    Basically you have two properties of the adjoint which are useful in this problem:

    |Xa> = X|a>

    and

    <Xa| = <a|X*

    And also don't forget that the product of two operators can be treated as another operator.

    EDIT: In fact only one property: <Xa| = <a|X*
     
    Last edited: Sep 22, 2007
  6. Sep 22, 2007 #5
    i'm not sure what the difference between |Xa> and X|a> is

    also, my books says you can go from X(Y|a>) to (<a|Y*)X*

    i dont know how that is possible
     
  7. Sep 22, 2007 #6
    If you translate |Xa> into words it is the column vector resulting from the multiplication of the matrix X by the a row vector having the components a0, a1, a2, ...
    X|a> would translate as the result of multiplying the matrix X by the column vector |a>
    Because |a> has the components a0, a1, a2, ... it results that |Xa> = X|a>

    On the other hand <Xa| is the conjugate transpose of the column vector resulting from the multiplication of the matrix X by the a column vector having the components a0, a1, a2, ...
    <a|X would translate as the result of multiplying the conjugate transpose of |a> by the matrix X. It is clear that they are not necessarily equal.


    Try to use the property <Xa| = <a|X*.
    Start from <XYa|. Considering that XY is also an operator, apply the property above. Then considering that Ya is a vector and X an operator, apply the property above.
     
    Last edited: Sep 22, 2007
  8. Sep 22, 2007 #7
    so I'm trying to get from X(Y|a>) to (<a|Y*)X*

    X(Y|a>)=X(<a|Y*)

    the rules that has been listed does not say what to do in this situation: X(<a|Y*)

    i could do X<Ya| but that still doesnt really help
     
  9. Sep 22, 2007 #8
    I already told you that X(<a|Y*) can't be calculated. Use the hints that I gave you.
     
  10. Sep 22, 2007 #9
    Look, I'll do the second part and you do the first part.
    I start from <XYa| and considering that Ya is a vector and X an operator, I apply the property <Xa| = <a|X* and get:

    <XYa| = <X(Ya)| = <Ya|X* then, I apply the property for <Ya| i.e. <Ya| = <a|Y* and get:
    <XYa| = <X(Ya)| = <Ya|X* = <a|Y*X*


    Now all you have to do is to start from <XYa| , consider that XY is an operator, and apply the same property <Xa| = <a|X*

    You will get <XYa| = <(XY)a| = ...
     
  11. Sep 22, 2007 #10
    using your hint of <a|XY = <X*a|Y=<Y*X*a| and <a|(XY) = <(XY)*a| to prove it, but im still not sure how my book gets from X(Y|a>) to (<a|Y*)X*
     
  12. Sep 22, 2007 #11
    That's also a solution.
     
  13. Sep 22, 2007 #12
    i know that is the solution, my question is trying to understand what my book did.

    how it went from X(Y|a>) to (<a|Y*)X*
     
  14. Sep 22, 2007 #13
    I don't think they are sayng that X(Y|a>) = (<a|Y*)X* but rather X(Y|a>) <-> (<a|Y*)X* which means that (<a|Y*)X* is the corresponding bra of the ket X(Y|a>) which is true.
     
  15. Sep 22, 2007 #14
    what does "the corresponding bra of the ket" mean?
     
  16. Sep 22, 2007 #15
    <a| is the corresponding bra of the ket |a>

    Check the wikipedia page on bra-ket notation.
     
    Last edited: Sep 22, 2007
  17. Sep 22, 2007 #16
    so, the transpose conjugate of X(Y|a>) is (<a|Y*)X* ?
     
  18. Sep 22, 2007 #17
    Exactly.
     
  19. Sep 22, 2007 #18
    thanks!

    also, i was wondering, is the kronecker delta the same thing as the dirac delta in the form delta(a-b)

    that is, [tex]\delta _{a,b} = \delta(a-b)[/tex]

    I'm really wondering if a and b were orthonomal sets, then:
    [tex]\langle a|b \rangle = \delta _{a,b} = \delta(a-b)[/tex]
     
    Last edited: Sep 22, 2007
  20. Sep 22, 2007 #19
    They are not the same but the Dirac delta has a similar role as the Kronecker delta, when the dimension of the space is infinite.
     
  21. Sep 22, 2007 #20
    so if a and b were orthonormal sets, would:

    [tex]\langle a|b \rangle = \delta(a-b)[/tex]

    or

    [tex]\langle a|b \rangle = \delta _{a,b}[/tex]
     
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