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Working with Hermitian-Adjoint - Sakurai Problem 1.4b

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Not relevant, but I have some work that reaches incorrect conclusions, and I can't see the mistake "in the middle".


    2. Relevant equations



    3. The attempt at a solution
    [tex]\begin{array}{l}
    {(XY)^\dag } = {\left( {\left\langle {a'} \right|X\left| {a''} \right\rangle \left\langle {a''} \right|Y\left| {a'''} \right\rangle } \right)^\dag } \\
    = \left( {\left\langle {a'} \right|X{{\left| {a''} \right\rangle }^\dag }} \right)\left( {\left\langle {a''} \right|Y{{\left| {a'''} \right\rangle }^\dag }} \right) \\
    = \left( {\left\langle {a''} \right|Y{{\left| {a'''} \right\rangle }^\dag }\left\langle {a'} \right|X{{\left| {a''} \right\rangle }^\dag }} \right) = {Y^\dag }{X^\dag } \\
    \end{array}[/tex]

    ...Somewhere, here, is a mistake. I distributed the adjoint operation "+" (dagger) across the two scalars, "<||>", but if I did that, this would imply:

    [tex]{\left( {XY} \right)^\dag } = \left\langle {a'} \right|X{\left| {a''} \right\rangle ^\dag }\left\langle {a''} \right|Y{\left| {a'''} \right\rangle ^\dag } = {X^\dag }{Y^\dag } = {\rm{uh - oh}}{\rm{.}}[/tex]
     
  2. jcsd
  3. Jul 19, 2010 #2
    Your first equality doesn't make sense to me.
    when you expand out a matrix product you
    should have a sum over a'',
    and when you get rid of the sum (using completeness)
    it tells you the correct order of the factors.
     
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