Heisenberg's Uncertainty Principle — Question 10

Question 10

Exponential lifetime produces a Lorentzian linewidth rather than a finite rms energy spread

In an ideal decay model the amplitude is $A(t)=\Theta(t)e^{-t/(2\tau)}e^{-i\omega_0t}$, so survival probability decays as $e^{-t/\tau}$. The spectral amplitude is proportional to $\int_0^\infty A(t)e^{i\omega t}dt$. Take $\tau=5.00\,\mathrm{ns}$ and $\hbar=6.582\times10^{-16}\,\mathrm{eV\,s}$.

1. Derive the line shape and its full width at half maximum in energy.

2. Calculate the linewidth and determine whether it can be identified with a finite energy standard deviation for this ideal Lorentzian.

Question 10 - Solution

Exponential lifetime produces a Lorentzian linewidth rather than a finite rms energy spread

(a) Method and calculation

$$\widetilde A(\omega)\propto\frac1{1/(2\tau)-i(\omega-\omega_0)},\quad S(\omega)\propto\frac1{(\omega-\omega_0)^2+[1/(2\tau)]^2}.$$$$S=S_{\max}/2\text{ at }|\omega-\omega_0|=1/(2\tau)\quad\Rightarrow\quad\boxed{\Gamma_E=\hbar/\tau}.$$

(b) Interpretation and further calculation

$$\boxed{\Gamma_E=1.3164\times10^{-7}\,\mathrm{eV}}.$$$$S(\omega)\sim(\omega-\omega_0)^{-2}\quad\Rightarrow\quad\boxed{\int(\omega-\omega_0)^2S(\omega)d\omega=\infty}.$$

Physical check

The ideal Lorentzian has a finite FWHM but no finite variance. This model does not establish a universal rule equating measurement duration with a position-momentum-style uncertainty variable.