# Momentum Wavefunction

You are given a particle that is free, having its state at $t=0$ equal that of the n-th energy level in the inifinite well.

$\varphi=\sqrt{2/L}\sin (n\pi x/L)$ when $0<x<L$

a) Please find the momentum space wavefunction of this particle ($t=0$).

b) If the particle is in an energy eigenstate in an infinite square well, is it also in a momentum eigenstate?

c) Is possible that the particle be in a momentum eigenstate in an infinite square well? Explain.

Given position wavefunction

$ψ(x,0)= \sqrt{2/L}*\sin(nπ/L*x)$ for $0<x<L$

The corresponding momentum wave function is ($k=p/\hbar$)

$ϕ(k)=\sqrt{2/L} ∫_0^L \sin(nπ/L*x)*\exp(-ikx)*dx$ for $0<k<2π/L$

For $n=1$, $n=2$ and $n=10$ the value of the expression above is

$ϕ_1 (k)=-\sqrt{\frac{2}{L}} \frac{πL}{k^2*L^2-π^2} [1+exp(-ikL) ]$

$ϕ_2 (k)=\sqrt{\frac{2}{L}} \frac{2πL}{k^2 L^2-4π^2} [exp(-ikL)-1]$

$ϕ_{10} (k)=\sqrt{\frac{2}{L}} \frac{10πL}{k^2 L^2-100π^2} [exp(-ikL)-1]$

b) The answer is NO. If the particle is in an energy eigenstate, it cannot be in a momentum eigenstate.

First way to show: In classical mechanics the relation between energy and momentum is

$E=p^2/2m$

So that a fixed energy (an energy eigenstate) implies just the existence of a “square of momentum” eigenstate.

Second way to show: The energy eigenstates are the eigenvalues of Hamiltonian (inside infinite well, $V=0$)

$(-\hbar^2/2m)*(d^2/dx^2 )ψ=E*ψ$

While the momentum eigenstates need to satisfy:

$(\hbar/i)*(d/dx)*ψ=p*ψ$

By comparing the above two equations one can say again that an energy eigenstate is just also an eigenstate of $p^2$ , so that the particle will be also in an eigenstate corresponding to $p^2$ (not p).

d) From the Heisenberg incertitude principle one has

$Δx*Δp≥\hbar$

So that since the particle is inside the well (somewhere) between $0<x<L$ then $Δx=L$,that is $Δp≥\hbar/L$.

To have a momentum eigenstate it means a precise $p$, fact excluded by the equation above inside a finite length (and infinite potential) well.

Reference