 Solution

ID: past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2014/paper-1/1/a/solution

∣ A ∣^{2} = mω / (π ℏ)

E_{0} = ℏ ω /2

x = \pm ℏ/ (mω)

. With

y = x mω /ℏ

P = \frac{2}{π} \int_{1}^{\infty} e^{- y^{2}} d y = 1 - erf (1) ≃ 0.16.

The stated rounded error function gives

0.16

; the more precise Gaussian integral is

0.15730

 New to topics? Read the docs here!