# Solution to second order differential equation Quantum harmonic oscillator

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:

$$-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\psi=E\psi$$

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

• $\xi$ is your variable hence you cannot use the "normal" method of solving $\lambda^2 = \xi^2$. – user88595 Jun 5 '14 at 21:51
• Could you explain how the above solution is obtained? – Stripers247 Jun 5 '14 at 22:05
• 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? – user88595 Jun 5 '14 at 22:25
• The eqn is in all the standard text books on QM. – Stripers247 Jun 5 '14 at 22:32
• 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. – achille hui Jun 5 '14 at 22:46