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