I am reviewing some quantum mechanics and and have come across a solution to a differential equation that I do not understand in the derivation of the quantum harmonic oscillator. The Schrodinger equation for a 1D harmonic oscillator is:


with the introduction of:

$$K=\frac{2E}{\hbar\omega}, \xi=\sqrt{\frac{m\omega}{\hbar}}x$$

for $\xi>>K$, we get:

$$\frac{d^2 \psi }{d\xi ^2} \approx \xi^2\psi$$

With solution:

$$\psi = A\exp\left(-\dfrac{\xi^2}{2}\right) + B\exp\left(\dfrac{\xi^2}{2}\right )$$

Where I was expecting the solution to be:

$$\psi = A\exp(-{\xi}) + B\exp({\xi}).$$

Can someone point out what I am missing? Thanks in advance

  • 3
    $\begingroup$ $\xi$ is your variable hence you cannot use the "normal" method of solving $\lambda^2 = \xi^2$. $\endgroup$ – user88595 Jun 5 '14 at 21:51
  • $\begingroup$ Could you explain how the above solution is obtained? $\endgroup$ – Stripers247 Jun 5 '14 at 22:05
  • 1
    $\begingroup$ Despite it's simplicity it's not that easy and don't think the answer is correct. I suggest you have a look at WA. Where did you get your solution from? $\endgroup$ – user88595 Jun 5 '14 at 22:25
  • $\begingroup$ The eqn is in all the standard text books on QM. $\endgroup$ – Stripers247 Jun 5 '14 at 22:32
  • 4
    $\begingroup$ In the limit $|\xi| \gg 1$ when you can approximate the Schrodinger equation by $\frac{d^2\psi}{d\xi^2} = \xi^2\psi$. You can also approximate it by $\left(\frac{d}{d\xi} + \xi\right)\left(\frac{d}{d\xi} - \xi\right)\psi = 0$ and/or $\left(\frac{d}{d\xi} - \xi\right)\left(\frac{d}{d\xi} + \xi\right)\psi = 0$. The first one has solution $e^{\xi^2/2}$ while the second one has solution $e^{-\xi^2/2}$. In fact, the second one give you the ground state for the quantum harmonic oscillators. $\endgroup$ – achille hui Jun 5 '14 at 22:46

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