Momentum Eigenfunctions

We can also look at the eigenfunctions of the momentum operator.

\begin{displaymath}\bgroup\color{black}p_{op}u_p(x)=pu_p(x) \egroup\end{displaymath}

\begin{displaymath}\bgroup\color{black}{\hbar\over i}{d\over dx}u_p(x)=pu_p(x) \egroup\end{displaymath}

The eigenstates are

\begin{displaymath}\bgroup\color{black}u_p(x)=C e^{ipx/\hbar} \egroup\end{displaymath}

with \bgroup\color{black}$p$\egroup allowed to be positive or negative.

These solutions do not go to zero at infinity so they are not normalizable to one particle.

\begin{displaymath}\bgroup\color{black}\langle p\vert p\rangle=\langle u_p\vert u_p\rangle=\infty \egroup\end{displaymath}

This is a common problem for this type of state.

We will use a different type of normalization for the momentum eigenstates (and the position eigenstates).

\begin{displaymath}\bgroup\color{black}\langle p'\vert p\rangle=\vert C\vert^2\i...
...i(p-p')x/\hbar} dx=2\pi\hbar \vert C\vert^2\delta(p-p') \egroup\end{displaymath}

Instead of the Kronecker delta, we use the Dirac delta function. The momentum eigenstates have a continuous range of eigenvalues so that they cannot be indexed like the energy eigenstates of a bound system. This means the Kronecker delta could not work anyway.

These are the momentum eigenstates

\bgroup\color{black}$\displaystyle u_p(x)={1\over \sqrt{2\pi\hbar}} e^{ipx/\hbar} $\egroup
satisfying the normalization condition
\bgroup\color{black}$\displaystyle \langle p'\vert p\rangle=\delta(p-p') $\egroup

For a free particle Hamiltonian, both momentum and parity commute with \bgroup\color{black}$H$\egroup. So we can make simultaneous eigenfunctions.

\begin{eqnarray*}[H,p]=0 \\
{[H,P]=0} \\

We cannot make eigenfunctions of all three operators since

\begin{displaymath}\bgroup\color{black} [P,p]\neq 0 .\egroup\end{displaymath}

So we have the choice of the \bgroup\color{black}$e^{ikx}$\egroup states which are eigenfunctions of \bgroup\color{black}$H$\egroup and of \bgroup\color{black}$p$\egroup, but contain positive and negative parity components. or we have the \bgroup\color{black}$\sin(kx)$\egroup and \bgroup\color{black}$\cos(kx)$\egroup states which contain two momenta but are eigenstates of \bgroup\color{black}$H$\egroup and Parity. These are just different linear combinations of the same solutions.

Jim Branson 2013-04-22