Heisenberg uncertainty principle

What is Going On?\n1. Wave-Particle Duality\n Tiny particles act like waves. A wave that is tightly packed in space (well-known position) will have many frequencies mixed in (uncertain momentum).\n\n2. Fourier Idea\n Mathematically, a short wave packet needs a wide range of wavelengths. Momentum $p$ is linked to wavelength $\lambda$ by de Broglie, $p = \frac{h}{\lambda}$. Hence, spread in $\lambda$ (and therefore in $p$) is unavoidable.\n\n3. Heisenberg’s Statement\n $\Delta x \; \Delta p \; \ge \; \frac{\hbar}{2}$\n Another useful form is \n $\Delta E \; \Delta t \; \ge \; \frac{\hbar}{2}$\n\n4. Why the Factor 1/2?\n A more careful Fourier analysis of a Gaussian wave packet gives the exact factor $\frac{1}{2}$.\n\n5. Practical Meaning\n - Not a defect of instruments, but a basic rule of nature.\n - Limits how small micro-chips can be, affects electron microscopes, and explains why electrons don’t fall into the nucleus.\n\n6. In JEE Problems\n You generally get questions to (a) calculate minimum uncertainty in momentum or position, (b) show why an electron can’t stay inside a nucleus, (c) relate lifetime of an excited state to energy width, etc.

Easy Story Version\nImagine you are trying to click a fast moving cricket ball with your phone camera at night.\n- If you keep the flash on for a very short time, you catch the position of the ball sharply, but the picture becomes blurred about its speed.\n- If you use a long exposure, you get a clear idea of its speed (a streak), but the exact spot of the ball in the photo is fuzzy.\n\nNature behaves the same way for very tiny particles like electrons.\n- The more exactly we know where (position) a particle is, the less exactly we can know how fast (momentum) it is moving, and vice-versa.\n\nMathematically, the rule is:\n$\Delta x \; \Delta p \; \ge \; \frac{\hbar}{2}$\nwhere $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum, and $\hbar$ (read "h-bar") is a very small constant.\n\nSo, the universe itself tells us: “You can not measure both perfectly at the same time.”

Step-by-Step Derivation (Gaussian Wave Packet)\n1. Start with a Gaussian wave packet in position space:\n $\psi(x) = A \exp\!\left( -\frac{x^{2}}{2\sigma^{2}} \right)$\n Here, $\sigma$ is the spread (standard deviation) in position. Hence\n $\Delta x = \sigma$\n\n2. Fourier Transform to Momentum Space\n $\phi(p) = \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} \psi(x)\, e^{-ipx\/\hbar}\,dx$\n For a Gaussian, the transform is also Gaussian with width \n $\Delta p = \frac{\hbar}{2\sigma}$\n\n3. Multiply the Uncertainties\n $\Delta x \; \Delta p = \sigma \; \frac{\hbar}{2\sigma} = \frac{\hbar}{2}$\n\n4. General Result\n For any wave function, the mathematics of Fourier transforms shows\n $\boxed{\;\Delta x \; \Delta p \; \ge \; \frac{\hbar}{2}\;}$\n Equality holds only for ideal Gaussian packets.\n\n5. Energy–Time Form\n Using $E = \frac{p^{2}}{2m}$ (or direct operator method), a similar derivation leads to\n $\Delta E \; \Delta t \; \ge \; \frac{\hbar}{2}$