# Harmonic Oscillator Trial Wavefunction

I was learning today about trial wave functions for a harmonic oscillator.

We learnt that the solution to Schrödinger equation for a harmonic oscillator is a Gaussian curve, i.e. $$f(x) = e^{-x^2} .$$

Testing a trial function such as:

$$\psi = N_{0}e^{-ax^2}$$

where $$x$$ is position gave $$\frac{d^2\psi}{dx^2} = N_{0} \ (4a^2x^2 - 2a)\cdot e^{-ax^{2}} .$$

Applying this to Schrödinger's equation using reduced mass $$\mu$$ $$-N_{0} \cdot \frac{\hbar^2}{2\mu}(4a^2x^2 - 2a)e^{-ax^2}+ \frac{1}{2}kx^2\ \cdot N_{0}e^{-ax^2} = E\ \cdot \ N_{0}e^{-ax^2}$$ simplified to $$- \frac{\hbar^2}{2\mu}(4a^2x^2 - 2a)+ \frac{1}{2}kx^2 = E.$$

The lecturer mentioned that as the total energy $$E$$ was constant, $$E$$ cannot be dependent on position $$x$$ which made sense from studies on Harmonic Motion.

Then he continued to state:

We therefore have a solution to the Schrödinger equation if the terms in $$x$$ are equal and opposite and cancel.

Suddenly the equation becomes: $$\frac{\hbar^2}{2\mu} \cdot 4a^2x^2 = \frac{1}{2}kx^2$$ and solving for $$a$$: $$a = \frac{\sqrt{k\mu}}{2\hbar}.$$ My question is how did the equation $$- \frac{\hbar^2}{2\mu}(4a^2x^2 - 2a)+ \frac{1}{2}kx^2\ = E$$ suddenly transform into $$\frac{\hbar^2}{2\mu} \cdot 4a^2x^2 = \frac{1}{2}kx^2$$ in just one line?

• The answer by LonelyProf is right, but just to help... The equation didn't "transform". The operation is not the same thing as changing, for example, $x+2=2x-1$ to $x=3$. This is actually a new equation based on the reasoning you give. In other words, it's not like we changed the original equation by setting $a=E=0$. It's a new equation. Nov 4, 2018 at 10:34
• "We learnt that the solution to Schrödinger equation for a harmonic oscillator is a Gaussian curve." Keep in mind that this is a solution, not the solution. Nov 4, 2018 at 10:39

## 1 Answer

The basic point is that the equation involving $$E$$ is an identity, which must hold for all values of $$x$$, not just particular values of $$x$$. So, all the $$x$$-dependent parts must cancel identically. Sometimes, identities are distinguished from simple equations by using the symbol $$\equiv$$ rather than $$=$$.

Slightly more generally, if you bring all the terms onto the left and rearrange your equation into the form of a polynomial identity $$C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \equiv 0$$ which must hold for all values of $$x$$, then it follows that all the coefficients $$C_i$$ must vanish. You can show this by setting $$x=0$$ (hence $$C_0=0$$); then by differentiating with respect to $$x$$ and setting $$x=0$$ (hence $$C_1=0$$); and so on.