In Sakurai's Modern Quantum Mechanics, second edition, $5.6.10$ is $$\begin{aligned} \dot{c}_m(t)=-\sum_nc_n(t)e^{i[\theta_n(t)-\theta_m(t)]}\langle m;t|\left[\frac\partial{\partial t}|n;t\rangle\right] \end{aligned}\tag{5.6.10}$$ where $c_m(t)$ stanstics $$\begin{aligned} |\alpha;t\rangle=\sum_nc_n(t)e^{i\theta_n(t)}|n;t\rangle \end{aligned}\tag{5.6.5}$$ But use Schrodinger Equation in $5.6.10$, it is \begin{aligned} i\hbar\frac\partial{\partial t}|\alpha;t\rangle=H(t)|\alpha;t\rangle \end{aligned} we have \begin{aligned} \langle m;t|\left[\frac\partial{\partial t}|n;t\rangle\right] = \frac{1}{i\hbar} \langle m;t|H(t)|n;t\rangle = \frac{1}{i\hbar} E_n(t) \langle m;t|n;t\rangle = \frac{1}{i\hbar} E_n(t) \delta_{mn} \end{aligned} But in the book, author don't do this thing. So are there something wrong with me?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ The point is that $H(t)\left|n;t\right>\neq E_n(t)\left|n;t\right>$ $\endgroup$– naturallyInconsistentCommented Mar 12 at 5:26
-
$\begingroup$ @naturallyInconsistent But in 5.6.3, we have this condition. $\endgroup$– liZCommented Mar 12 at 9:19
-
$\begingroup$ Sorry, you are correct. Instead, it is that $\left<m;t|\frac{\partial\ }{\partial t}|n;t\right>\neq\frac{E_n(t)}{\mathrm i\hslash}\delta_{mn}$ because Equation (5.6.3) implies a different condition. $\endgroup$– naturallyInconsistentCommented Mar 12 at 10:44
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
You are incorrectly assuming that the instantaneous eigenstates $\lvert n; t\rangle$ satisfy the Schrödinger equation. Instead, they are merely defined by the equation $$H(t)\lvert n;t\rangle = E_n(t)\lvert n;t\rangle.$$
In fact, it is a primary consequence of the adiabatic approximation (which at this point in the derivation in Sakurai has not been made yet) that an instantaneous eigenstate $\lvert n;t\rangle$ satisfies the Schrödinger equation (times dynamic and geometric phase). However, this is not generally the case.
