Question

Find the uncertainty of \(x\) for the wave function \(\Psi(x, t)=A e^{-\lambda x^{2}} e^{-i \omega t}\).

Short answer

Based on the given wave function, \(\Psi(x,t)\), the uncertainty of \(x\) is calculated through a step-by-step process. This involves finding the expectation value of \(x\) and \(x^2\), and then using those values to compute the uncertainty. After the calculations, the uncertainty of \(x\) is found to be \(\Delta x = \frac{1}{2\sqrt{2\lambda^2}}\).

Step-by-step solution

Compute the expectation value of \(x\)

To compute the expectation value of \(x\), we need to integrate the probability density function times \(x\). The probability density function is the wave function squared, \(|\Psi|^2 = \Psi^* \Psi\). Therefore, the expectation value of \(x\) is given by: \(\langle x \rangle = \int_{-\infty}^{\infty} x |\Psi|^2 dx = \int_{-\infty}^{\infty} x A^2 e^{-2\lambda x^2} dx\) Now we can compute the integral using the substitution method. Let \(u=2\lambda x^2\) so that \(du = 4\lambda x dx\). Hence, \(\langle x \rangle = \int_{-\infty}^{\infty} \frac{1}{4\lambda} A^2 e^{-u} du = \frac{A^2}{4\lambda} \left(\int_{-\infty}^{\infty} e^{-u} du\right)\) Now integrating, we get \(\langle x \rangle = \frac{A^2}{4\lambda} \cdot \left[ -e^{-u} \right]_{-\infty}^{\infty} = \frac{A^2}{4\lambda}(-1 + 1) = 0\).

Compute the expectation value of \(x^2\)

To compute the expectation value of \(x^2\), we need to integrate the probability density function times \(x^2\). So, \(\langle x^2 \rangle = \int_{-\infty}^{\infty} x^2 |\Psi|^2 dx = \int_{-\infty}^{\infty} x^2 A^2 e^{-2\lambda x^2} dx\) Using the same substitution method (\(u = 2\lambda x^2\) and \(du =4\lambda x dx\)), we have: \(\langle x^2 \rangle = \int_{-\infty}^{\infty} \frac{1}{4\lambda} A^2 x e^{-u} du\) Now, we integrate by parts, with \(dv = x e^{-u} du\) and \(v = -\frac{1}{2\lambda} e^{-u}\). Thus, \(\langle x^2 \rangle = \frac{A^2}{4\lambda} \left[\left(-\frac{1}{2\lambda} e^{-u} x\right)_{-\infty}^{\infty} -\int_{-\infty}^{\infty} -\frac{1}{2\lambda} e^{-u} du\right]\) Therefore, \(\langle x^2 \rangle = \frac{A^2}{8\lambda^2} = \frac{1}{8\lambda^2}\) because \(A^2 = 1\).

Find the uncertainty of \(x\)

The uncertainty of \(x\) is given by the square root of the variance, \(\Delta x = \sqrt{\langle x^2\rangle - \langle x \rangle^2}\). Using the expectation values we calculated for \(\langle x \rangle\) and \(\langle x^2 \rangle\), we can find the uncertainty of \(x\): \(\Delta x = \sqrt{\frac{1}{8\lambda^2}- 0} = \frac{1}{2\sqrt{2\lambda^2}}\) Thus, the uncertainty of \(x\) for the given wave function is \(\Delta x = \frac{1}{2\sqrt{2\lambda^2}}\).