Prove Heisenberg uncertainty principle from Compton Effect
The Heisenberg Uncertainty Principle states that it is impossible to simultaneously measure the exact position ($x$) and momentum ($p$) of a particle. This can be demonstrated by attempting to "see" an electron using a high-power microscope.
1. The Resolving Power (Uncertainty in Position)
To locate an electron, we must illuminate it with light (photons). According to optics, the resolution of a microscope—and thus the uncertainty in the electron's position ($\Delta x$)—is limited by the wavelength ($\lambda$) of the light and the aperture angle ($\theta$) of the microscope lens:
$$\Delta x \approx \frac{\lambda}{2 \sin \theta}$$
2. The Compton Effect (Uncertainty in Momentum)
For the electron to be seen, at least one photon must strike the electron and scatter into the microscope's lens. This is where the Compton Effect comes in. When the photon collides with the electron, it transfers some of its momentum to the electron.
The initial momentum of the photon is $p = \frac{h}{\lambda}$. After scattering, the photon can enter the microscope anywhere within the cone of angle $2\theta$. Therefore, the horizontal component of the photon's momentum can vary between $+\frac{h}{\lambda} \sin \theta$ and $-\frac{h}{\lambda} \sin \theta$.
This introduces an uncertainty in the electron's momentum ($\Delta p_x$) due to the recoil:
$$\Delta p_x \approx \frac{2h}{\lambda} \sin \theta$$
3. Combining the Uncertainties
To find the product of the uncertainties, we multiply the two expressions:
$$\Delta x \cdot \Delta p_x \approx \left( \frac{\lambda}{2 \sin \theta} \right) \cdot \left( \frac{2h \sin \theta}{\lambda} \right)$$
Notice that the wavelength ($\lambda$) and the sine of the angle ($\sin \theta$) cancel out, leaving us with:
$$\Delta x \cdot \Delta p_x \approx h$$
Conclusion
This result shows that the product of the uncertainties is on the order of Planck’s constant. More rigorous derivations show that the minimum product is actually:
$$\Delta x \cdot \Delta p_x \geq \frac{\hbar}{2}$$
Thus, by using the physics of Compton scattering, we prove that the more accurately we try to measure position (by using a shorter wavelength), the more we disturb the momentum, and vice versa.
(END SEMESTER EXAMINATION)
CLASS: IMSc SEMESTER: I
BRANCH: MATHEMATICS SESSION: MO/2024
SUBJECT: CH111 CHEMISTRY-I