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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).
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).