Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sakurai page 181: Time evolution of ensembles

  1. May 6, 2013 #1
    From "Modern Quantum Mechanics, revised edition" by J. J. Sakurai, page 181.

    Equation (3.4.27), at some time [itex]t_0[/itex], the density operator is given by[tex]
    \rho(t_0) = \sum_i w_i \mid \alpha^{(i)} \rangle \langle \alpha^{(i)} \mid
    [/tex]Equation (3.4.28), at a later time, the state ket changes from [itex]\mid \alpha^{(i)} \rangle[/itex] to [itex]\mid \alpha^{(i)}, t_0 ; t \rangle[/itex].

    Equation (3.4.29), From the fact that [itex]\mid \alpha^{(i)}, t_0 ; t \rangle[/itex] satisfies the Schrodinger equation we obtain[tex]
    i \hbar \frac{\partial \rho}{\partial t} = \sum_i w_i \left( H \mid \alpha^{(i)}, t_0 ; t \rangle \langle \alpha^{(i)}, t_0 ; t \mid - \mid \alpha^{(i)}, t_0 ; t \rangle \langle \alpha^{(i)}, t_0 ; t \mid H \right) = - \left[\rho, H\right]
    [/tex]How does he get to (3.4.29) by applying to Schrodinger equation?

    From (2.1.27), the Schrodinger equation is given to be[tex]
    i \hbar \frac{\partial}{\partial t} \mid \alpha, t_0 ; t \rangle = H \mid \alpha, t_0 ; t \rangle
    [/tex]I can't figure out how to apply it to get (3.4.29).
  2. jcsd
  3. May 6, 2013 #2
    Both the bra and the ket are evolving in time. Formally, the time derivative of the ket-bra outer product obeys the normal product rule for derivatives:

    ##\frac{d}{dt}(|\alpha \rangle \langle \beta |) = (\frac{d}{dt} | \alpha \rangle) \langle \beta | + | \alpha \rangle (\frac{d}{dt} \langle \beta | )##.

    (Here ##|\alpha \rangle## is any ket and ##\langle \beta |## is any bra). If you apply this to your density operator you should get the desired result. Note that to evaluate the time derivative of a bra you will have to use the Hermitian conjugate of the Schrodinger equation.

    If you want to make it totally transparent why the normal derivative product rule applies in this case, you can consider the infinitesmial time evolution of a ket:

    ##|\alpha(t+dt) \rangle = |\alpha(t)\rangle + dt \frac{d}{dt}|\alpha(t)\rangle##

    If you write down the analagous formula for a bra, and then use the two formula to evaluate ##| \alpha(t+dt) \rangle \langle \beta(t+dt) |## then you should be able to derive the claimed product rule for derivatives (at the physicist level of rigor, anyway).
  4. May 6, 2013 #3


    User Avatar
    Science Advisor

    It is simply the application of the Schrödinger equation plus hermitean conjugation

    [tex]\partial_t\,(|a\rangle\langle a|) = (\partial_t \,|a\rangle)\,\langle a| + |a\rangle \, (\partial_t\,\langle a|) = (-iH|a\rangle)\langle a| + |a\rangle(\langle a|(-iH)^\dagger) = -i(H|a\rangle\langle a| - |a \rangle\langle a|H) = -i [H,\rho][/tex]
    Last edited: May 6, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook