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Question

The wave function for a Harmonic oscillator described by $Nxexp(-ax^2/2)$ has

The correct answer is
one maximum, one minimum only

Wave Function Analysis

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.

Finding Critical Points

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 $x = -\frac{1}{\sqrt{a}}$.

Determining Maxima and Minima

We examine the behavior of $\psi'(x)$ around these critical points to classify them.

  • When $x < -\frac{1}{\sqrt{a}}$: $x^2 > \frac{1}{a}$, so $1 - ax^2 < 0$. If $N > 0$, $\psi'(x)$ is negative, meaning $\psi(x)$ is decreasing.
  • When $-\frac{1}{\sqrt{a}} < x < \frac{1}{\sqrt{a}}$: $x^2 < \frac{1}{a}$, so $1 - ax^2 > 0$. $\psi'(x)$ is positive, meaning $\psi(x)$ is increasing.
  • When $x > \frac{1}{\sqrt{a}}$: $x^2 > \frac{1}{a}$, so $1 - ax^2 < 0$. $\psi'(x)$ is negative, meaning $\psi(x)$ is decreasing.

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.

Conclusion

Based on the derivative analysis, the wave function has exactly one local maximum and one local minimum.

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Important Questions from Harmonic Oscillator

  1. 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 _________________

  2. The wavefunction of a 1-D harmonic oscillator between $x =+\infty$ and $x = -\infty$ is given by $\psi(x) = N(2x^2 -1)e^{-x^2/2}$. The value of $N$ that normalizes the function $\psi(x)$ is
    (Given: $\int_{-\infty}^{+\infty} x^{2n} e^{-x^2} dx = \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2^n} \sqrt{\pi}$)
  3. The correct option for the average value of kinetic energy and the average value of potential energy of a one-dimensional harmonic oscillator with frequency $\nu$ in its ground state is
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