## Problem 1.17

May 2020

Problem: Suppose you wanted to describe an unstable particle, that spontaneously disintegrates with a 'lifetime' $$\tau$$. In that case the total probability of finding the particle somewhere should not be constant, but should decrease at an exponential rate:

$P(t) = \int_{-\infty}^{\infty} |\Psi(x, t)|^2 dx = e^{-t/\tau}.$

If

$V = V_0 - i\Gamma,$

where $$V_0$$ is the true potential energy and $$\Gamma$$ is a positive real constant.

• (a) Show that, in place of $$dP/dt = 0$$, as with a "stable particle", we now get $$\frac{dP}{dt} = - \frac{2\Gamma}{\hbar}P$$
• (b) Solve for $$P(t)$$, and find the lifetime of the particle in terms of $$\Gamma$$

Solution: Starting with (a), we must evaluate the derivative of the probability function.

$\frac{dP}{dt} = \frac{d}{dt} \int_{-\infty}^{\infty} |\Psi(x, t)|^2 dx = \int_{-\infty}^{\infty} \frac{\partial}{\partial t} (\Psi^* \Psi) dx$

By the product rule,

$= \int_{-\infty}^{\infty} \frac{\partial \Psi^*}{\partial t}\Psi + \frac{\partial \Psi}{\partial t}\Psi^* dx$

Substituting in the Schrödinger equation for the derivatives with respect to time,

$= \int_{-\infty}^{\infty} - \frac{i\hbar}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}\Psi + \frac{i}{\hbar}V\Psi^*\Psi + \frac{i\hbar}{2m}\frac{\partial ^2 \Psi}{\partial x^2}\Psi^* - \frac{i}{\hbar} V\Psi^* \Psi dx$

Remembering that $$V = V_0 - i\Gamma$$ (and that $$V = V_0 + i\Gamma$$ for the complex conjugate),

$= \int_{-\infty}^{\infty} - \frac{i\hbar}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}\Psi + \left(\frac{i}{\hbar}V_0 - \frac{\Gamma}{\hbar}\right)\Psi^*\Psi + \frac{i\hbar}{2m}\frac{\partial ^2 \Psi}{\partial x^2}\Psi^* - \left(\frac{i}{\hbar}V_0 + \frac{\Gamma}{\hbar}\right)\Psi^* \Psi dx$

$= \int_{-\infty}^{\infty} - \frac{i\hbar}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}\Psi + \frac{i\hbar}{2m}\frac{\partial ^2 \Psi}{\partial x^2}\Psi^* - \frac{2\Gamma}{\hbar} \Psi^* \Psi dx$

$= - \int_{-\infty}^{\infty} \frac{\partial}{\partial x} \left(\frac{i\hbar}{2m}\frac{\partial \Psi^*}{\partial x}\Psi - \frac{i\hbar}{2m}\frac{\partial \Psi}{\partial x}\Psi^*\right)dx - \frac{2\Gamma}{\hbar} \int_{-\infty}^{\infty} \Psi^* \Psi dx$

$= -\left[\frac{i\hbar}{2m}\frac{\partial \Psi^*}{\partial x}\Psi - \frac{i\hbar}{2m}\frac{\partial \Psi}{\partial x}\Psi^*\right]_{-\infty}^{\infty} - \frac{2\Gamma}{\hbar} P$

If $$\Psi$$ is to be normalizable, it must go to zero as $$x$$ goes to $$\pm \infty$$, and so we are left with the ordinary differential equation,

$\frac{dP}{dt} = -\frac{2\Gamma}{\hbar}P.$

Part (b) asks us asks us to solve this equation which we can easily do via separation of variables.

$\frac{dP}{dt} = -\frac{2\Gamma}{\hbar}P$

$\frac{dP}{P} = -\frac{2\Gamma}{\hbar}dt$

$\int \frac{dP}{P} = -\frac{2\Gamma}{\hbar} \int dt$

$\ln(P) = -\frac{2\Gamma}{\hbar} t + c$

$P = e^{-\frac{2\Gamma}{\hbar} t + c} = e^{-\frac{2\Gamma}{\hbar} t}e^c = C e^{-\frac{2\Gamma}{\hbar} t}$

where $$C = e^c$$. Relating this back to the given exponential function, $$\tau = \hbar/2\Gamma$$.