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Benedict Irwin edited Quantum.tex
over 9 years ago
Commit id: 3cc6d7840424ef59d866b8e7e235264b19fc8470
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If we let $d/\tau=v$ and let the iterative system go it actually converges on $2mv^2$ and not $\frac{1}{2}mv^2$... This can be fixed by setting $d/\tau=v/2$, then it converges to the right answer for many values of $m$ and $v$ checked. Curious.
Now take an example with a potential \begin{equation}
\epsilon=\frac{1}{2}m\omega^2x^2
\end{equation}
Solving for $x(t)$ gives $exp(i\omega t)$, which has real part $cos(\omega t)$. Therefore substituting and rearranging and integrating gives \begin{equation}
E_i=\frac{1}{\alpha\tau}\int_0^{E_{i-1}}\frac{1}{\omega}cos^{-1}\bigg(\sqrt{\frac{2\epsilon}{m\omega^2}}\;\bigg) \; \mathrm{d\epsilon}\\
E_i=\frac{1}{\alpha\tau}\Bigg[ \frac{\sqrt{\frac{2\epsilon}{m\omega^2}}\epsilon(\frac{2\epsilon}{m\omega^2}\epsilon-1)+\sqrt{\epsilon}\sqrt{1-\frac{2\epsilon}{m\omega^2}\epsilon}sin^{-1}\bigg(\sqrt{\frac{2\epsilon}{m\omega^2}}\;\bigg)}{2\sqrt{\frac{2\epsilon}{m\omega^2}}\sqrt{-\frac{2\epsilon}{m\omega^2}\epsilon(\frac{2\epsilon}{m\omega^2}\epsilon-1)}} +\epsilon cos^{-1}\bigg(\sqrt{\frac{2\epsilon}{m\omega^2}}\;\bigg)\Bigg]_0^{E_{i-1}}
\end{equation}
Now take a quantum system with a Hamiltonian $\hat{H}$, and wavefunctions $\Psi_i$ we know that \begin{equation}