# Position and momentum-space wave function

## Position-space and momentum-space wavefunction

Given the position-space wavefunction

$\psi(x,0)=A*\begin{Bmatrix}

0 & \text{for }x<0\\

x(L-x) & \text{for }0\geq x\geq L\\

0 & \text{for }x\geq L

\end{Bmatrix} \text{with }A =\sqrt {30/L^5}$

a) Expand this position-space wavefunction in terms of free particle momentum eigenstates to find the momentum-space wave function$\Phi(p)$

b) Write down, but do not evaluate the integral expression for$\psi(x,t)$

The free particle momentum eigenstates are the functions

$u(x)=C*\exp(-px/\hbar)=C*\exp(-ikx)$

Because they need to satisfy the equation

$\hbar/i*[du(x)/dx] =p*u(x)$

To find the momentum-space wavefunction $\Phi(p)$ one needs just to find the Fourier transform of $\psi(x)$

Working in the k domain since $k=2\pi/x$ one has

$\phi(p)=A*\begin{Bmatrix}

0 & \text{for } k<0\\

\int_0^L x(L-x)\exp(-ikx)*dx & \text{for } 0<k<2\pi/L\\

0 & \text {for } k>2\pi/L

\end{Bmatrix}$

The value of the integral computed is:

$\int_0^L x(L-x)\exp(-ikx)*dx=- (1/k^3)[(kL-2i)\exp(-ikL)+(kL+2i)]$

To find $\psi(x,t)$ being given $\psi(x,0)$ one needs to use the time dependent Schroedinger equation.

$i\hbar*(d\psi/dt)= H\psi$

So that

$\psi(x,t)= (-i/\hbar) \int_0^t \hat H \psi(x,0)*dt$

If one knows the Hamiltonian H than this equation can be solved. For a free particle ($V(x) =0$) one has

$\psi (x,t)=\psi (x,0)*exp(-i\omega t)$