The correct expression for the ground state eigenfunction $\psi(x)$ of a linear harmonic oscillator is:

$\psi(x) = \left(\frac{a}{\pi}\right)^{1/4} e^{-a x^2 / 2}$

This is option (b).

Explanation:

1. The ground state eigenfunction for a one-dimensional quantum harmonic oscillator is generally written as:

   $\psi_0(x) = \left( \frac{m \omega}{\pi \hbar} \right)^{1/4} e^{- \frac{m \omega}{2 \hbar} x^2}$

   Here, $m$ is the mass, $\omega$ is the angular frequency, and $\hbar$ is the reduced Planck constant.

2. In the problem, the parameter $a$ is defined as $a = \sqrt{\frac{mk}{\hbar^2}}$, where $k$ is the force constant. This gives the simplified form of the ground state eigenfunction:

   $\psi(x) = \left( \frac{a}{\pi} \right)^{1/4} e^{-a x^2 / 2}$

This matches option (b).

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Related Questions:

1. How is the ground state energy of the harmonic oscillator calculated?
2. What is the general form of the eigenfunctions for excited states of a harmonic oscillator?
3. How does the force constant $k$ affect the frequency and energy levels of the oscillator?
4. What is the role of the parameter $a$ in the wavefunction's shape and normalization?
5. How is the expectation value of position calculated for the ground state?

Tip:

In quantum mechanics, the ground state wavefunction is always real and normalized, and it describes the lowest energy state of the system with no nodes.