Griffiths - Problem 1.17

Adar Kahiri
June 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.

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, \(\tau = \hbar/2\Gamma\).