1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    Attached Files:

    • wat.png
      File size:
      29.5 KB
  5. Mar 2, 2012 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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]


    [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
    I remembered it being something simple :L

    Thanks buddy :biggrin:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook