For a quantum harmonic oscillator, the energy levels are given by:

$E_n = \left(n + \frac{1}{2}\right)\hbar\omega$

where $n = 0, 1, 2, \dots$, $\hbar$ is the reduced Planck's constant, and $\omega$ is the angular frequency of the oscillator.

The uncertainty in energy $\Delta E$ for a particular state can be interpreted as the standard deviation of the energy distribution for that state. However, in a single energy eigenstate of a quantum harmonic oscillator (such as the ground state $n = 0$), the uncertainty in energy $\Delta E$ is zero because the energy is precisely defined. Therefore, in an ideal scenario where only a single energy state is occupied, $\Delta E = 0$.

Given the options provided:

a) $\hbar\omega$ b) $3\hbar\omega$ c) $\left(\frac{\hbar}{2}\right) \sqrt{\frac{2m\omega^2}{h}}$ d) None of the above

Since the uncertainty in energy for the ground state is theoretically zero and none of the provided options reflect this, the correct answer is:

d) None of the above.

Would you like further clarification or have any other questions? Here are some related topics:

1. What is a quantum harmonic oscillator?
2. How is the energy of a quantum harmonic oscillator quantized?
3. What are energy eigenstates and eigenvalues?
4. How is the uncertainty principle applied in quantum mechanics?
5. What is the significance of $\hbar$ (reduced Planck's constant)?
6. How do quantum harmonic oscillators relate to real-world physical systems?
7. What is the ground state of a quantum system?
8. How does the concept of zero-point energy arise from quantum mechanics?

Tip: The ground state energy in a quantum harmonic oscillator is non-zero due to quantum fluctuations, a phenomenon known as zero-point energy.