Question 3.8: Imagine a system in which there are just two linearly indepe...

The (time-dependent) Schrödinger equation says

$i\hbar \frac{d}{dt}|S(t)〉 =\hat{H} |S(t)〉$      (3.87)

As always, we begin by solving the time-independent Schrödinger equation:

$\hat{H}|s〉=E|s〉$       (3.88)

that is, we look for the eigenvectors and eigenvalues of $\hat{H}$ .The characteristic equation determines the eigenvalues:

$det \begin{pmatrix} h-E & g \\ g & h-E \end {pmatrix}=(h-E)^{2}-g^{2}=0\Rightarrow h-E=\mp g\Rightarrow E_{\pm }=h\pm g$

Evidently the allowed energies are  (h + g ) and (h – g) . To determine the eigenvectors, we write

$\begin{pmatrix} h & g \\ g & h\end{pmatrix} \begin {pmatrix} \alpha \\ \beta \end{pmatrix}=(h\pm g) \begin {pmatrix} \alpha \\ \beta \end{pmatrix} \Rightarrow h\alpha +g\beta =(h\pm g)\alpha \Rightarrow \beta =\pm \alpha$

so the normalized eigenvectors are

$|s_{\pm }〉 =\frac{1}{\sqrt{2} }\begin{pmatrix} 1 \\ \pm 1 \end{pmatrix}$

Next we expand the initial state as a linear combination of eigenvectors of the Hamiltonian:

$|S(0)〉= \begin{pmatrix} 1 \\0 \end {pmatrix} \frac{1}{\sqrt{2} }(|s_{+}〉 +|s_{-}〉 )$

Finally, we tack on the standard time-dependence (the wiggle factor) $exp(-iE_nt/ \hbar)$ :

$|S(t)〉= \frac{1}{\sqrt{2}}\left[e^{-i(h+g)t/\hbar }|s_{+}〉 +e^{-i(h-g)t/\hbar }|s_{-}〉\right]$
$=\frac{1}{2} e^{-iht/\hbar}\left[e^{-igt/\hbar }\begin{pmatrix}1 \\ 1 \end{pmatrix} +e^{igt/\hbar }\begin{pmatrix}1 \\ -1 \end{pmatrix}\right]$
$=\frac{1}{2} e^{-iht/\hbar}\begin{pmatrix}e^{-igt/\hbar}+e^{igt/\hbar} \\ e^{-igt/\hbar}-e^{igt/ \hbar} \end{pmatrix}$
=$e^{-iht/\hbar}\begin{pmatrix}\cos (gt/\hbar ) \\ -i\sin(gt/\hbar ) \end{pmatrix}$

If you doubt this result, by all means check it: Does it satisfy the time-dependent Schrödinger equation (Equation 3.87)? Does it match the initial state when t=0 ?