Question

3. (10 pts) Heisenberg Uncertainty Principle

The uncertainty principle places a limit on specifying the location and momentum of a particle simultaneously. Δ?Δ? ≥ ℏ/2 This is a consequence of the wave nature of particles, which we can see by examining the uncertainty in the single-slit diffraction of light.

(a) In single-slit diffraction, the width of the slit ? represents the uncertainty in x position of the beam, ∆?:

∆? = ?

We can imagine that we can try to limit the uncertainty ∆? by choosing a narrower slit so that the beam lands in smaller spot. The momentum of the beam would be in the direction of the screen, but we know that the beam fans out and forms a central bright max. The uncertainty in the x momentum ∆?? is related to the angle subtended by the half maximum:

∆??/?? = ????

Using the single-slit equation for interference minima, the small angle approximation, the de Broglie wave equation, and the equations above, show that ∆?∆?? ≈ ℎ, which is consistent with the Heisenberg uncertainty equation.

(b) The uncertainty principle also puts a limit on the precision of energy measurements.

∆?∆? ≥ ℏ/2

Due to energy-mass equivalence, it also puts a limit on mass measurements. Many elementary particles are unstable. For example, the delta baryon (∆) has a lifetime of 3 × 10^−24?. The average mass of the ∆ is 1232 ???/?^2 . What is the uncertainty in the mass of the ∆ in units of eV/c^2?

Answer 1

a) The single slit diffraction condition yields us the condition

a is the width of the slit

hence

which is the uncertainity in the position in the x direction

it is also given that

for small angle approximation we can write

hence

we can rewrite combining the equation as

from debroglies equation we can write

substituting in the equation results

hence proved

b) We know

∆?∆? ≥ ℏ/2

hence uncertainity in energy,

hence