Question:easy

According to the uncertainty principle, what is the minimum possible phase space volume that a quantum harmonic oscillator can occupy?

Show Hint

Phase-space area equals \(\Delta x\,\Delta p\); its floor is the uncertainty bound \(\hbar/2\).
Updated On: Jul 2, 2026
  • \(\dfrac{\hbar}{4}\)
  • \(\dfrac{\hbar}{2}\)
  • \(\hbar\)
  • \(\dfrac{\hbar\omega}{2}\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Recall that the uncertainty principle limits how finely a state can be localized in the position-momentum plane: the product $\sigma_x\sigma_p$ cannot be smaller than $\hbar/2$.

Step 2: The equality $\sigma_x\sigma_p=\hbar/2$ is achieved only by a Gaussian wave packet.

Step 3: For a harmonic oscillator the ground-state wave function is precisely such a Gaussian, so it is the minimum-uncertainty state and occupies the smallest possible cell.

Step 4: Note the distractor $\hbar\omega/2$ is the ground-state energy, not a phase-space area, so it is dimensionally wrong here.

Step 5: Therefore the minimum phase-space volume is
\[\boxed{\dfrac{\hbar}{2}}\]
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