dEdt
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To cut to the chase, I have to solve for the evolution of a two-state system where the system's state at time t satisfies the equation
\mathrm{i}\hbar\left( <br /> \begin{array}{cc} <br /> \dot{c}_1(t)\\ <br /> \dot{c}_2(t)<br /> \end{array} <br /> \right)=\left(<br /> \begin{array}{cc} <br /> 0 & \gamma \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\ <br /> \gamma \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 0 <br /> \end{array} <br /> \right)\left(<br /> \begin{array}{cc} <br /> c_1(t)\\ <br /> c_2(t)<br /> \end{array} <br /> \right)
with initial conditions c_1(0)=1 and c_2(0)=0.
I decided to solve this by diagonalization. The eigenvectors of the Hamiltonian are
\frac{1}{\sqrt{2}}\left(<br /> \begin{array}{cc} <br /> \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\ <br /> 1<br /> \end{array} <br /> \right) and \frac{1}{\sqrt{2}}\left(<br /> \begin{array}{cc} <br /> \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\ <br /> -1<br /> \end{array} <br /> \right), with eigenvalues \gamma and -\gamma, respectively.
The change-of-basis matrix then is just
U^\dagger = \frac{1}{\sqrt{2}}\left( <br /> \begin{array}{cc} <br /> \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 1\\ <br /> \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & -1<br /> \end{array} <br /> \right). So in the new basis, the state vector becomes
\frac{1}{\sqrt{2}}\left( <br /> \begin{array}{cc} <br /> \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 1\\ <br /> \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & -1<br /> \end{array} <br /> \right)\left(<br /> \begin{array}{cc} <br /> c_1(t)\\ <br /> c_2(t)<br /> \end{array} <br /> \right):=\left(<br /> \begin{array}{cc} <br /> d_1(t)\\ <br /> d_2(t)<br /> \end{array} <br /> \right).
Therefore, we have
\mathrm{i}\hbar\left(<br /> \begin{array}{cc} <br /> \dot{d}_1(t)\\ <br /> \dot{d}_2(t)<br /> \end{array} <br /> \right)=\left(<br /> \begin{array}{cc} <br /> \gamma & 0\\ <br /> 0 & -\gamma<br /> \end{array} <br /> \right)\left(<br /> \begin{array}{cc} <br /> d_1(t)\\ <br /> d_2(t)<br /> \end{array} <br /> \right)
which can be solved easily. Once the d's are found, the c's are easily computed.
Problem is, this doesn't give the right answer. I suspect it has something to do with using a time-dependent change of basis matrix, but I'm not sure.
\mathrm{i}\hbar\left( <br /> \begin{array}{cc} <br /> \dot{c}_1(t)\\ <br /> \dot{c}_2(t)<br /> \end{array} <br /> \right)=\left(<br /> \begin{array}{cc} <br /> 0 & \gamma \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\ <br /> \gamma \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 0 <br /> \end{array} <br /> \right)\left(<br /> \begin{array}{cc} <br /> c_1(t)\\ <br /> c_2(t)<br /> \end{array} <br /> \right)
with initial conditions c_1(0)=1 and c_2(0)=0.
I decided to solve this by diagonalization. The eigenvectors of the Hamiltonian are
\frac{1}{\sqrt{2}}\left(<br /> \begin{array}{cc} <br /> \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\ <br /> 1<br /> \end{array} <br /> \right) and \frac{1}{\sqrt{2}}\left(<br /> \begin{array}{cc} <br /> \mathrm{e}^{\mathrm{i}(\omega - \omega_{21})t}\\ <br /> -1<br /> \end{array} <br /> \right), with eigenvalues \gamma and -\gamma, respectively.
The change-of-basis matrix then is just
U^\dagger = \frac{1}{\sqrt{2}}\left( <br /> \begin{array}{cc} <br /> \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 1\\ <br /> \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & -1<br /> \end{array} <br /> \right). So in the new basis, the state vector becomes
\frac{1}{\sqrt{2}}\left( <br /> \begin{array}{cc} <br /> \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & 1\\ <br /> \mathrm{e}^{-\mathrm{i}(\omega - \omega_{21})t} & -1<br /> \end{array} <br /> \right)\left(<br /> \begin{array}{cc} <br /> c_1(t)\\ <br /> c_2(t)<br /> \end{array} <br /> \right):=\left(<br /> \begin{array}{cc} <br /> d_1(t)\\ <br /> d_2(t)<br /> \end{array} <br /> \right).
Therefore, we have
\mathrm{i}\hbar\left(<br /> \begin{array}{cc} <br /> \dot{d}_1(t)\\ <br /> \dot{d}_2(t)<br /> \end{array} <br /> \right)=\left(<br /> \begin{array}{cc} <br /> \gamma & 0\\ <br /> 0 & -\gamma<br /> \end{array} <br /> \right)\left(<br /> \begin{array}{cc} <br /> d_1(t)\\ <br /> d_2(t)<br /> \end{array} <br /> \right)
which can be solved easily. Once the d's are found, the c's are easily computed.
Problem is, this doesn't give the right answer. I suspect it has something to do with using a time-dependent change of basis matrix, but I'm not sure.