Quantum Harmonic Oscillator via Separation of Variables and/or Normalization

Question

I tried to solve the problem of the Quantum Harmonic Oscillator in one dimension. $$\frac{-\hbar ^2}{2m} \Psi _{xx} + \frac{1}{2}m(\omega x)^2\Psi = i\hbar \Psi _t$$ I set up the following condition: let $$\Psi = e^{-\beta t}u$$ Thus, $$\Psi_{xx} = e^{-\beta t}u_{xx}$$ and $$\Psi_t = e^{-\beta t}u_t-\beta e^{-\beta t}u$$ I plug this into the Schrodinger PDE formula to get, $$\frac{-\hbar ^2}{2m} e^{-\beta t}u_{xx} + \frac{1}{2}m(\omega x)^2e^{-\beta t}u = i\hbar e^{-\beta t}u_t-\beta e^{-\beta t}u$$ I add one more condition(I am not entirely sure if this condition is valid but I assume it is), $$\frac{-\hbar ^2}{2m} u _{xx} - i\hbar u _t = 0$$ I set up this condition mainly because this equation is very easy to solve. (then again maybe my error is here and requires the boundary conditions at $\pm \infty$ to approach 0. Factoring out the common term on the substituted equation results in, $$\frac{-\hbar ^2}{2m} u_{xx}-i\hbar u_t + \frac{1}{2}m(\omega x)^2u = -i\hbar\beta u$$ Using the second condition listed above, the result is, $$\frac{1}{2}m(\omega x)^2u = -i\hbar\beta u$$ And thus $$\beta = \frac{i}{2\hbar }m(\omega x)^2$$ Thus my solution is, $$\Psi = e^{-\frac{i}{2\hbar }m(\omega x)^2t}u$$ Is this a practical approach for solving QHO problem? How should I normalize from here so that I end up with the Hermite Polynomials? By the way I expect to solve $u$ with Separation of Variables.

Answer 1

The idea is to separate

$$ \Psi(t, x) = \color{blue}{e^{-\beta t}}\color{red}{u(x)} \tag{1} $$

where $\beta$ is a constant. With this in mind, Schrodinger's equation becomes

$$ \color{blue}{e^{-\beta t}}\left[-\frac{\hbar^2}{2m}\frac{{\rm d}^2 \color{red}{u}}{{\rm d}x^2} + \frac{1}{2}m\omega^2 x^2 \color{red}{u}\right] = \color{red}{u}\left[i\hbar \frac{{\rm d}\color{blue}{e^{-\beta t}}}{{\rm d}t} \right] $$

Or rearranging a bit

$$ \frac{1}{\color{red}{u}}\left[-\frac{\hbar^2}{2m}\frac{{\rm d}^2 \color{red}{u}}{{\rm d}x^2} + \frac{1}{2}m\omega^2 x^2 \color{red}{u}\right] = -i\hbar \beta \tag{2} $$

which is just an ordinary equation, and can be solved via Hermite polynomials. It is customary to write

$$ \beta = -i \frac{E}{\hbar} \tag{3} $$

where $E$ is the energy of the system.