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Basic Operator question

  1. Mar 1, 2012 #1
    Can someone explain to me how

    [itex]H(\sum_n w_n |a_n><a_n|) = \sum_n w_n(H|a_n><a_n|-|a_n><a_n|H)[/itex]

    I've done this before and I remember being confused about it before then finding out it was something simple.. I should really start filing my notes away for such an eventuality :frown:
    I can't seem to work out what the next step is at all.
     
  2. jcsd
  3. Mar 2, 2012 #2

    fzero

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    What you wrote is not correct. If you take an expectation value of your expression in the mth state, you'd find 0 on the RHS. We can find a correct statement if we consider the operators

    [tex] O_L = \sum_n w_n (H| a_n \rangle )\langle a_n| ,[/tex]

    [tex]O_R = \sum_n w_n | a_n \rangle (\langle a_n|H) .[/tex]

    The expectation values in an arbitrary state are equal:

    [tex] \langle a_m | O_L | a_m \rangle = \sum_n w_n \langle a_m |H| a_n \rangle \langle a_n| a_m \rangle = w_m \langle a_m |H| a_m \rangle,[/tex]

    [tex]\langle a_m | O_R | a_m \rangle = \sum_n w_n \langle a_m | a_n \rangle \langle a_n|H| a_m \rangle = w_m \langle a_m|H| a_m \rangle , [/tex]

    where in both cases we used [itex] \langle a_n| a_m \rangle = \delta_{nm}.[/itex]

    So in this sense, [itex] O_R=O_L[/itex]: we can let the operator act from the right or left side. This is the same way that expectation values work

    [tex] \langle a_n | H | a_m \rangle = \langle a_n | (H | a_m \rangle ) = (\langle a_n | H) | a_m \rangle.[/tex]


    We can therefore write

    [tex]H \left( \sum_n w_n | a_n \rangle \langle a_n| \right) = \frac{1}{2} \sum_n w_n \Bigl[ ( H | a_n \rangle ) \langle a_n| + | a_n \rangle (\langle a_n|H) \Bigr].[/tex]
     
  4. Mar 2, 2012 #3
    Yeah, I kept coming to that conclusion too yet what I wrote is what is in Sakurai's book..

    [itex]i\ \hbar \partial_t \rho = H \rho \neq -[\rho , H][/itex]

    Where [itex]\rho = \sum_n w_n |a_n \rangle \langle a_n |[/itex]

    I'll attatch an extract
     

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  5. Mar 2, 2012 #4

    fzero

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    OK, the point there is that [itex]\rho[/itex] does not satisfy Schrodinger's equation so [itex]i\ \hbar \partial_t \rho \neq H \rho [/itex]. Instead you have to use

    [tex] i \hbar \frac{\partial}{\partial t} |\alpha^{(i)},t_0;t \rangle \langle \alpha^{(i)},t_0;t | = \left( i \hbar \frac{\partial}{\partial t} |\alpha^{(i)},t_0;t \rangle \right) \langle \alpha^{(i)},t_0;t | + |\alpha^{(i)},t_0;t \rangle \left( i \hbar \frac{\partial}{\partial t} \langle \alpha^{(i)},t_0;t | \right)[/tex]

    and

    [tex] i \hbar \frac{\partial}{\partial t} \langle \alpha^{(i)},t_0;t | = - \left( i \hbar \frac{\partial}{\partial t} |\alpha^{(i)},t_0;t \rangle \right)^\dagger .[/tex]

    The derivation I gave above is valid for time-independent states and wouldn't work here.
     
  6. Mar 2, 2012 #5
    Ah!
    I remembered it being something simple :L

    Thanks buddy :biggrin:
     
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