The wave function for a harmonic oscillator is given in the form:
$ \psi(x) = Nx \exp(-ax^2/2) $
Here, '$N$' is a normalization constant and '$a$' is a positive parameter related to the oscillator's properties.
To identify potential maxima and minima (extrema), we calculate the first derivative of the wave function, $\psi'(x)$, and find where it equals zero.
Using the product rule for differentiation:
$ \psi'(x) = \frac{d}{dx} [Nx \exp(-ax^2/2)] $
$ \psi'(x) = N \exp(-ax^2/2) + Nx \left( \exp(-ax^2/2) \cdot (-ax) \right) $
Factor out the common terms:
$ \psi'(x) = N \exp(-ax^2/2) [1 - ax^2] $
Set the derivative to zero to find critical points:
$ N \exp(-ax^2/2) [1 - ax^2] = 0 $
Since $N$ is non-zero and $\exp(-ax^2/2)$ is always positive, the equation holds true only if:
$ 1 - ax^2 = 0 $
Solving for $x$:
$ ax^2 = 1 \implies x^2 = \frac{1}{a} \implies x = \pm \frac{1}{\sqrt{a}} $
The critical points are $x = \frac{1}{\sqrt{a}}$ and $x = -\frac{1}{\sqrt{a}}$.
We examine the behavior of $\psi'(x)$ around these critical points to classify them.
At $x = -\frac{1}{\sqrt{a}}$, the function transitions from decreasing to increasing, signifying a local minimum.
At $x = \frac{1}{\sqrt{a}}$, the function transitions from increasing to decreasing, signifying a local maximum.
Note that $\psi(0)=0$. The function tends to zero as $x \to \pm \infty$. The identified extrema are the only local maximum and minimum.
Based on the derivative analysis, the wave function has exactly one local maximum and one local minimum.
The lowest energy of a quantum mechanical one-dimensional simple harmonic oscillator is $300$ cm$^{-1}$. The energy (in cm$^{-1}$) of the next higher level is _________________