Question

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

• A. \(\dfrac{\hbar}{4}\)
• B. \(\dfrac{\hbar}{2}\)
• C. \(\hbar\)
• D. \(\dfrac{\hbar\omega}{2}\)

Hint:

Phase-space area equals \(\Delta x\,\Delta p\); its floor is the uncertainty bound \(\hbar/2\).

Answer 1

The Correct Option is B

Step 1: The Heisenberg uncertainty relation for conjugate variables position and momentum is
\[\Delta x\,\Delta p \ge \frac{\hbar}{2}.\]
Step 2: A phase-space cell (area in the \(x\)-\(p\) plane) has size \(\Delta x\,\Delta p\), so its minimum value is set by the equality.

Step 3: The minimum-uncertainty state (a Gaussian, exactly the ground state of the harmonic oscillator) saturates this bound.

Step 4: Hence the smallest phase-space volume the oscillator can occupy is
\[\boxed{\dfrac{\hbar}{2}}\]