Statement#

Derive the Heisenberg Uncertainty Principle for position and momentum: $\Delta x \cdot \Delta p \geq \frac{\hbar}{2}$

where $\Delta x$ and $\Delta p$ are the standard deviations of position and momentum measurements.

Required Topics#

• Quantum operators
• Commutation relations
• Expectation values and standard deviations
• Cauchy-Schwarz inequality
• Operator algebra

Key Definitions#

Standard deviation: $\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2}$

Commutator of position and momentum: $[\hat{x}, \hat{p}] = \hat{x}\hat{p} - \hat{p}\hat{x} = i\hbar$

Cauchy-Schwarz inequality: For any states $|\phi\rangle$ and $|\psi\rangle$: $\langle\phi|\phi\rangle \langle\psi|\psi\rangle \geq |\langle\phi|\psi\rangle|^2$

What to Prove#

1. Express $(\Delta x)^2$ and $(\Delta p)^2$ in terms of expectation values
2. Define $\Delta\hat{x} = \hat{x} - \langle\hat{x}\rangle$ and $\Delta\hat{p} = \hat{p} - \langle\hat{p}\rangle$
3. Show that $[\Delta\hat{x}, \Delta\hat{p}] = i\hbar$
4. Apply Cauchy-Schwarz to states $\Delta\hat{x}|\psi\rangle$ and $\Delta\hat{p}|\psi\rangle$
5. Derive the uncertainty relation

Solution#

Solution coming soon.