Disclaimer - this is just an attempt - it never finishes. Blogging this so that I don't forget where I were:
Recall the Quantum Harmonic Oscillator problem as follow:
$ \begin{eqnarray*} \hat{H}\psi &=& E\psi \\ (\frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2x^2)\psi &=& E\psi \\ (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2)\psi &=& E\psi \\ \end{eqnarray*} $
The so called series method is basically assume $ \psi $ as an analytic function, as such, it can be (locally) represented as a power series.
$ \begin{eqnarray*} \psi(x) &=& \sum\limits_{k=0}^{\infty}{a_kx^k} \\ \end{eqnarray*} $
Substituting it back to the problem, we have
$ \begin{eqnarray*} E\psi &=& (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2)\psi \\ E(\sum\limits_{k=0}^{\infty}{a_kx^k}) &=& (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2)(\sum\limits_{k=0}^{\infty}{a_kx^k}) \\ \end{eqnarray*} $
So we simply match the coefficients.
$ \begin{eqnarray*} Ea_0 &=& \frac{-\hbar^2}{2m}(2)a_2 \\ Ea_1 &=& \frac{-\hbar^2}{2m}(3)(2)a_3 \\ Ea_k &=& \frac{-\hbar^2}{2m}(k+1)(k+2)a_{k+2} + \frac{1}{2}m\omega^2x^2(a_{k-2}) \\ \end{eqnarray*} $
This apparently lead to the recurrence that we can use to solve for $ a_k $, but I am stuck on the next step that I can foresee, how do I know I have found the Gaussian?
Recall the Quantum Harmonic Oscillator problem as follow:
$ \begin{eqnarray*} \hat{H}\psi &=& E\psi \\ (\frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2x^2)\psi &=& E\psi \\ (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2)\psi &=& E\psi \\ \end{eqnarray*} $
The so called series method is basically assume $ \psi $ as an analytic function, as such, it can be (locally) represented as a power series.
$ \begin{eqnarray*} \psi(x) &=& \sum\limits_{k=0}^{\infty}{a_kx^k} \\ \end{eqnarray*} $
Substituting it back to the problem, we have
$ \begin{eqnarray*} E\psi &=& (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2)\psi \\ E(\sum\limits_{k=0}^{\infty}{a_kx^k}) &=& (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2)(\sum\limits_{k=0}^{\infty}{a_kx^k}) \\ \end{eqnarray*} $
So we simply match the coefficients.
$ \begin{eqnarray*} Ea_0 &=& \frac{-\hbar^2}{2m}(2)a_2 \\ Ea_1 &=& \frac{-\hbar^2}{2m}(3)(2)a_3 \\ Ea_k &=& \frac{-\hbar^2}{2m}(k+1)(k+2)a_{k+2} + \frac{1}{2}m\omega^2x^2(a_{k-2}) \\ \end{eqnarray*} $
This apparently lead to the recurrence that we can use to solve for $ a_k $, but I am stuck on the next step that I can foresee, how do I know I have found the Gaussian?
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