Question

Find the energy levels for the following 1D potentials, using the Bohr Sommerfeld quantization condition

V(x) =kx²/2

Answer 1

The energy levels for the potential
`\( V(x) = (kx^2)/(2) \)` using the Bohr Sommerfeld quantization condition are
`\( E_n = (\hbar \omega(n + (1)/(2)))/(2) \)` where
`\( \omega = \sqrt{(k)/(m)} \).`

Step-by-step explanation:

For a 1D potential given by
`\( V(x) = (kx^2)/(2) \)`, the Bohr Sommerfeld quantization condition helps determine the energy levels. This potential represents a harmonic oscillator. According to this condition, the energy levels
`\( E_n \)` are quantized and given by
`\( E_n = (\hbar \omega(n + (1)/(2)))/(2) \),` where
`\( \omega \)` is the angular frequency and is calculated as
`\( \omega = \sqrt{(k)/(m)} \)` with k as the spring constant and m as the mass of the particle.

This quantization condition arises from the classical harmonic oscillator's quantization process applied to quantum mechanics. The potential
`\( V(x) = (kx^2)/(2) \)` represents the potential energy of a particle undergoing simple harmonic motion in one dimension. By applying the Bohr Sommerfeld condition, the energy levels of this quantum mechanical system are derived, revealing a quantized energy spectrum that corresponds to discrete energy states.