Let's solve the integral step by step: ### **Given:** \[ \langle p'|\alpha\rangle = \left(\frac{1}{\sqrt{2\pi\hbar}}\right) \left(\frac{1}{\pi^{1/4}\sqrt{d}}\right) \int_{-\infty}^{\infty} dx' \exp\left(-\frac{ip'x'}{\hbar} + ikx' - \frac{x'^2}{2d^2}\right) \] ### **Step 1: Combine Exponents** Let's combine the terms in the exponent: \[ -\frac{ip'x'}{\hbar} + ikx' - \frac{x'^2}{2d^2} = \left(-\frac{ip'}{\hbar} + ik\right)x' - \frac{x'^2}{2d^2} \] Let \( A = -\frac{ip'}{\hbar} + ik \): \[ A x' - \frac{x'^2}{2d^2} \] ### **Step 2: Complete the Square** Write the exponent in the form \(-\frac{1}{2d^2}(x'^2 - 2A d^2 x')\): \[ -\frac{x'^2}{2d^2} + A x' = -\frac{1}{2d^2}\left(x'^2 - 2A d^2 x'\right) \] Complete the square inside the parenthesis: \[ x'^2 - 2A d^2 x' = (x' - A d^2)^2 - (A d^2)^2 \] So, \[ -\frac{1}{2d^2} \left[(x' - A d^2)^2 - (A d^2)^2\right] = -\frac{(x' - A d^2)^2}{2d^2} + \frac{(A d^2)^2}{2d^2} = -\frac{(x' - A d^2)^2}{2d^2} + \frac{A^2 d^2}{2} \] ### **Step 3: Evaluate the Gaussian Integral** Now, the integral becomes: \[ \int_{-\infty}^{\infty} dx' \exp\left(-\frac{(x' - A d^2)^2}{2d^2}\right) \] This is a standard Gaussian integral with mean \(A d^2\) and width \(d\): \[ \int_{-\infty}^{\infty} dx' \exp\left(-\frac{(x' - \mu)^2}{2\sigma^2}\right) = \sqrt{2\pi}\sigma \] Here, \(\sigma = d\), \(\mu = A d^2\): \[ \int_{-\infty}^{\infty} dx' \exp\left(-\frac{(x' - A d^2)^2}{2d^2}\right) = \sqrt{2\pi}d \] ### **Step 4: Collect All Terms** Include the exponential factor from completing the square: \[ \exp\left(\frac{A^2 d^2}{2}\right) \] Substitute back \(A = -\frac{ip'}{\hbar} + ik\): \[ A = -\frac{ip'}{\hbar} + ik \] \[ A^2 = \left(-\frac{ip'}{\hbar} + ik\right)^2 = -\frac{p'^2}{\hbar^2} - \frac{2k p'}{\hbar} i - k^2 \] So, \[ \frac{A^2 d^2}{2} = -\frac{(p' - \hbar k)^2 d^2}{2\hbar^2} \] ### **Step 5: Substitute Everything Back** Now, putting all together: \[ \langle p'|\alpha\rangle = \left(\frac{1}{\sqrt{2\pi\hbar}}\right) \left(\frac{1}{\pi^{1/4}\sqrt{d}}\right) \sqrt{2\pi}d \exp\left(-\frac{(p' - \hbar k)^2 d^2}{2\hbar^2}\right) \] Simplify coefficients: \[ \left(\frac{1}{\sqrt{2\pi\hbar}}\right) \left(\frac{1}{\pi^{1/4}\sqrt{d}}\right) \sqrt{2\pi}d = \frac{d}{\hbar^{1/2} \pi^{1/4} d^{1/2}} = \sqrt{\frac{d}{\hbar \sqrt{\pi}}} \] So the final answer is: --- \[ \boxed{ \langle p'|\alpha\rangle = \sqrt{\frac{d}{\hbar \sqrt{\pi}}} \exp\left[-\frac{(p' - \hbar k)^2 d^2}{2\hbar^2}\right] } \] --- Which matches the boxed result in your image.

\(\boxed{\langle p'|\alpha\rangle = \sqrt{\frac{d}{\hbar \sqrt{\pi}}}\ \exp\left[-\frac{(p' - \hbar k)^2 d^2}{2 \hbar^2}\right]}\)