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, $$\tau = \hbar/2\Gamma$$.