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I am confused about Griffith's solution (Example 10.1 pg 374) for the Hamiltonian: $$H = \dfrac{\hbar\omega_1}{2}\begin{bmatrix} \cos\alpha &e^{-i\omega t}\sin\alpha \\ e^{i\omega t}\sin\alpha & -\cos\alpha \end{bmatrix}$$ The eigenstates were straightforward to calculate: $$\chi_+ = \begin{bmatrix}\cos(\alpha/2)\\e^{i\omega t}\sin(\alpha/2) \end{bmatrix}\ \ \ \text{and}\ \ \ \chi_-=\begin{bmatrix}e^{-i\omega t}\sin(\alpha/2)\\-\cos(\alpha/2) \end{bmatrix}$$ where the states $\chi_+$ and $\chi_-$ have eigenvalues $\hbar\omega_1/2$ and $-\hbar\omega_1/2$ respectfully. Griffiths then demands the initial conditions: $$\chi(0)=\begin{bmatrix}\cos(\alpha/2)\\\sin(\alpha/2) \end{bmatrix}$$ My approach was that since we have the eigenstates, we know the general solution to be $$\chi_{\text{my guess}}(t)=c_1\chi_+e^{-i\omega_1t/2}+c_2\chi_ie^{i\omega_1t/2}$$ and so $c_1=1$ and $c_2=0$. However, this is not the solution that Griffiths states, which apparently is $$\chi_{\text{griffiths}}(t)=\begin{bmatrix}\left[\cos(\lambda t/2)-i\dfrac{\omega_1-\omega}{\lambda}\sin(\lambda t/2) \right]\cos(\alpha/2)e^{-i\omega t/2}\\\left[\cos(\lambda t/2)-i\dfrac{\omega_1+\omega}{\lambda}\sin(\lambda t/2) \right]\sin(\alpha/2)e^{i\omega t/2} \end{bmatrix}$$ where $\lambda=\sqrt{\omega^2+\omega_1^2-2\omega\omega_1\cos\alpha}$. My solution looks nothing like this and is much simpler. I am confused on where this comes from and where my train of thought went wrong.

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  • $\begingroup$ You computed an adiabatic approximation while Griffith's is the exact solution. $\endgroup$ – Jon Jun 5 '18 at 8:56
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The solution you found is not correct, because the Hamiltonian is time-dependent.

The Schrodinger equation is given by: $$i\hbar \frac{\partial \psi}{\partial t} = H \psi$$ where $\psi$ is a two-vector. Now you can diagonalise the Hamiltonian $H=SDS^{-1}$. In you case $D$ is time-independent, however the matrix $S$ is not.

If $H$ is time independent, then so is $S$ and you can write Schrodinger as $$i\hbar \frac{\partial (S^{-1}\psi)}{\partial t} = D S^{-1}\psi$$ The new basis defined by $\psi'=S^{-1}\psi$ is the eigenbasis of $H$, and your solution would hold. However in this case $S$ is time-dependent, so you can't take it inside the partial derivative.

You can do the same, but you end up with $$i\hbar S^{-1}\frac{\partial \psi}{\partial t} = D S^{-1}\psi$$ You can play with $S$ and $\psi$ to get the solution in Griffith. In particular you can write the left hand side as $$S^{-1}\frac{\partial \psi}{\partial t}=\frac{\partial (S^{-1}\psi)}{\partial t}-\frac{\partial S^{-1}}{\partial t}\psi=\frac{\partial (S^{-1}\psi)}{\partial t}+S^{-1}\frac{\partial S}{\partial t}S^{-1}\psi$$

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  • $\begingroup$ Why is $D$ always time-independent? What about $H = \text{diag}(1,t)$? $\endgroup$ – Javier Jun 5 '18 at 13:43

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