Griffiths - Problem 1.5

Adar Kahiri
May 2020

Problem: Consider the wave function \[\Psi (x,t)=Ae^{-\lambda |x|}e^{-i\omega t},\] where \(A\), \(\lambda\), and \(\omega\) are positive real constants. Normalize \(\Psi\), determine the expectation values of \(x\) and \(x^2\), and find the standard deviation (\(\sigma\)) of \(x\). What is the probability the particle would be found outside the range \(-\sigma \leq x \leq \sigma\).

Solution: First, to normalize \(\Psi\) means to find \(A\) such that \[\int_{-\infty}^{\infty} |\Psi (x, t)|^2 dx = 1,\] so let's do that. \[\int_{-\infty}^{\infty} |\Psi (x, t)|^2 dx = 1,\] \[\int_{-\infty}^{\infty} \Psi^* \Psi dx = 1,\] remembering that \(\Psi^*\) is the complex conjugate of the wave function, \[\int_{-\infty}^{\infty} Ae^{-\lambda |x|}e^{-i\omega t} Ae^{-\lambda |x|}e^{i\omega t} dx = 1\] \[A^2 \int_{-\infty}^{\infty} e^{-2 \lambda |x|} dx = 1\] \[A^2 \int_{-\infty}^0 e^{2 \lambda x}dx + A^2 \int_0^{\infty} e^{-2 \lambda x}dx = 1\] \[\left. \frac{A^2}{2\lambda} e^{2 \lambda x}\right|_{-\infty}^{0} - \left. \frac{A^2}{2\lambda} e^{-2 \lambda x} \right|_0^{\infty} = 1\] \[\frac{A^2}{2\lambda} - \lim\limits_{x\rightarrow -\infty} \frac{A^2}{2\lambda} e^{2 \lambda x} - \lim\limits_{x\rightarrow \infty} \frac{A^2}{2\lambda} e^{-2 \lambda x} + \frac{A^2}{2\lambda} = 1\] Evaluating this at the limits, its clear that \(e^{-\infty}\) approaches zero, and so we're left with: \[\frac{A^2}{2\lambda} + \frac{A^2}{2\lambda} = 1\] \[\frac{A^2}{\lambda} = 1\] \[A = \sqrt{\lambda}\] Now, we must find the expecation value of \(x\). If \(|\Psi (x, t)|^2\) gives us the probability of finding a particle at a point \(x\) at time \(t\), then the expectation value of \(x\) is defined as the average point \(x\) you'd find that particle at given the probability distribution (that's perhaps not the best way to think about it, but explaining the significance of the expectation value is outside the scope of this solution). And so, the expectation value is defined as \[\langle x \rangle = \int_{-\infty}^{\infty} x |\Psi (x, t)|^2 dx\] Now, solving for it, \[\langle x \rangle = A^2 \int_{-\infty}^{\infty} x e^{-2 \lambda |x|} dx = \lambda \int_{-\infty}^0 x e^{2 \lambda x} dx + \lambda \int_0^{\infty} x e^{-2 \lambda x} dx\] \[\langle x \rangle = \lambda \left[ \frac{x}{2\lambda} e^{2\lambda x} - \int \frac{1}{2\lambda} e^{2\lambda x}dx \right]_{- \infty}^0 + \lambda \left[ -\frac{x}{2\lambda} e^{- 2\lambda x} - \int \frac{1}{-2\lambda} e^{-2\lambda x}dx \right]_0^{\infty} \] \[\langle x \rangle = \lambda \left[ \frac{x}{2\lambda} e^{2\lambda x} - \frac{e^{2\lambda x}}{4\lambda^2} \right]_{- \infty}^0 + \lambda \left[ -\frac{x}{2\lambda} e^{- 2\lambda x} - \frac{e^{-2\lambda x}}{4\lambda^2} \right]_0^{\infty}\] \[\langle x \rangle = - \frac{1}{4\lambda} + \lim\limits_{x\rightarrow -\infty} \lambda \left(\frac{x}{2\lambda} e^{2\lambda x} - \frac{e^{2\lambda x}}{4\lambda^2}\right) + \lim\limits_{x\rightarrow \infty} \lambda \left(-\frac{x}{2\lambda} e^{- 2\lambda x} - \frac{e^{-2\lambda x}}{4\lambda^2}\right) + \frac{1}{4\lambda}\] \[\langle x \rangle = 0\] If you think about it, this makes sense, since funtions of the form \(a^{b|x|}\) are generally perfectly symmetrical about the vertical axis (I encourage you to check it out on Desmos for yourself).

Next, the questions asks us to find the expectation of \(x^2\) (the average of the squares of \(x\)), which, similar to the the previous one, is defined as \[\langle x^2 \rangle = \int_{-\infty}^{\infty} x^2 |\Psi (x, t)|^2 dx\] \[\langle x^2 \rangle = A^2 \int_{-\infty}^{\infty} x^2 e^{-2 \lambda |x|} dx = \lambda \int_{-\infty}^0 x^2 e^{2 \lambda x} dx + \lambda \int_0^{\infty} x^2 e^{-2 \lambda x} dx\] I won't include the entire integration process as it isn't too difficult (read: I don't want to type it all out!) \[\langle x^2 \rangle = \left. \frac{(4\lambda^2 x^2 - 4\lambda x + 2) e^{2\lambda x}}{8\lambda^2} \right|_{-\infty}^0 - \left.\frac{(4\lambda^2 x^2 + 4\lambda x + 2) e^{-2\lambda x}}{8\lambda^2} \right|_0^{\infty}\] \[\langle x^2 \rangle = \frac{1}{4\lambda^2} - \lim\limits_{x\rightarrow -\infty} \frac{(4\lambda^2 x^2 - 4\lambda x + 2) e^{2\lambda x}}{8\lambda^2} - \lim\limits_{x\rightarrow \infty} \frac{(4\lambda^2 x^2 + 4\lambda x + 2) e^{-2\lambda x}}{8\lambda^2} + \frac{1}{4\lambda^2}\] \[\langle x^2 \rangle = \frac{1}{2\lambda^2}\] Next, the question asks us to find the standard deviation of the wave function. Given a probability distribution, the standard deviation is defined as \[\sigma = \sqrt{ \langle x^2 \rangle - \langle x \rangle^2}\] Fortunately, we have these two values for our wave function and so we can easily calculate this. \[\sigma = \sqrt{\frac{1}{2\lambda^2}} = \frac{1}{\sqrt{2}\lambda}\] Now, we must determine the probability that we'll find the particle outside the range (which we'll call \(r\)) \(-\sigma \leq x \leq \sigma\). To do this, we can calculate the probability of the particle being found within this range, and subtract it from 1 (which is the total probability of the particle being anywhere). \[r = 1 - A^2 \int_{-1/\sqrt{2}\lambda}^{1/\sqrt{2}\lambda} e^{-2\lambda|x|} dx\] \[r = 1 - \lambda \int_{-1/\sqrt{2}\lambda}^0 e^{2\lambda x} dx - \lambda \int_0^{1/\sqrt{2}\lambda} e^{-2\lambda x} dx\] \[r = 1 - \lambda \int_{-1/\sqrt{2}\lambda}^0 e^{2\lambda x} dx - \lambda \int_0^{1/\sqrt{2}\lambda} e^{-2\lambda x} dx\] \[r = 1 - \lambda + \frac{e^{-\sqrt{2}}}{2} - \frac{e^{\sqrt{2}}}{2} + \lambda\] \[r=1 - \lambda \left.\frac{e^{2\lambda x}}{2\lambda} \right|_{-1/\sqrt{2}\lambda}^0 + \lambda \left.\frac{e^{-2\lambda x}}{2\lambda} \right|_0^{1/\sqrt{2}\lambda}\] \[r = 1 - \frac{1}{2} + \frac{e^{-\sqrt{2}}}{2} + \frac{e^{-\sqrt{2}}}{2} - \frac{1}{2}\] \[r = e^{-\sqrt{2}} \approx 0.2431\] So, the probability of the particle being found outside of the specified range is about 24.31%.