# The differential equation for the ground state of the quantum harmonic oscillator

For a quantum harmonic oscillator, write down a differential equation for the wave function of the ground state, by directly deriving it from the defining property $a|0\rangle = 0$. Solve such a differential equation (up to normalisation).

Attempt: $$a = \sqrt{\frac{m\omega}{2\hbar}} \left( x + \frac{i}{m\omega}p \right) \\ p = -i\hbar \frac{d}{dx} \\ \left( x + \frac{\hbar}{ m \omega} \frac{d}{dx} \right) | 0 \rangle = 0$$

How do I proceed from here?

Ultimately you want to separate $|0>$ from the differential equation. For simplicity, let's just write the ground state as $\Psi_0$. Then our equation is
$$\frac{h}{m\omega} \Psi_0'(x) \;\; =\;\; - x \Psi_0$$
$$\frac{d\Psi_0}{\Psi_0} \;\; =\;\; -\frac{m\omega}{h} x dx.$$
Do you see where to go from here? Once you solve for an explicit function $\Psi(x)$ (with integration constant) you'll need to normalize it by noting that $\int_{-\infty}^\infty |\Psi_0(x)|^2dx = 1$. This should be standard since you'll get that $\Psi(x)$ is a well-known function. One advantage to this problem is that if you have the other ladder operator (the raising operator) $a_+$, you can find all of the eigenstates of the oscillator by successive iteration of this: $\Psi_n(x) = \left (a_+\right )^n \Psi_0(x)$.
• Is this correct ? $ln \psi_{0} = \frac{-m\omega}{2h}x^{2}$ $\psi_{0} = ae-{\frac{m \omega}{2h}x^{2}}$ – italy Jan 12 '17 at 22:06
• I would add a constant of integration, but yes your answer is correct otherwise. Once you solve for $\Psi_0$ you get a Gaussian with unknown scalar coefficient. Use the normalization requirement to find the coefficient. – Mnifldz Jan 12 '17 at 22:08