Question

The question is 1.3 but the Psi(x,0) and phi(k) are given in the previous two parts

1.1 Consider a free particle in one-dimension with a wavefunction at t-0 Show that Ψ is normalized. 1.2 Show that the momentum wavefunction is Notice that lin (k-ak-k) 1.3 we can find Ψ(r) at any time t by: (ie. Ψ(x,t) is given by the Schrodinger' egn and the initial condition Ψ(x Show that iht 2mA イAr /2 2mA

Answer 1

1.1

The wave function is

$\Psi(x)=(2\pi \Delta^2)^{-1/4}e^{-\frac{x^2}{4\Delta^2}}e^{i k_o x}$

The conjugate of the wave function is

$\Psi^*(x)=(2\pi \Delta^2)^{-1/4}e^{-\frac{x^2}{4\Delta^2}}e^{-i k_o x}$

Therefore the probability density is

$P(x)=\Psi(x)\Psi^*(x)=(2\pi \Delta^2)^{-1/2}e^{-\frac{x^2}{2\Delta^2}}$

The norm of the wave function is

$\int_{-\infty}^\infty \Psi(x)\Psi^*(x)\,dx=(2\pi \Delta^2)^{-1/2}\int_{-\infty}^\infty e^{-\frac{x^2}{2\Delta^2}}\,dx$

Now use the Gaussian integral result

$\int_{-\infty}^\infty e^{-a x^2}\,dx=\frac{\sqrt{\pi }}{\sqrt{a}}$

to obtain

$\small \int_{-\infty}^\infty \Psi(x)\Psi^*(x)\,dx=(2\pi \Delta^2)^{-1/2}\int_{-\infty}^\infty e^{-\frac{x^2}{2\Delta^2}}\,dx=(2\pi \Delta^2)^{-1/2}\times (2\pi \Delta^2)^{1/2}=1$

Therefore the wave function is normalised