Momentum Eigenfunctions

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

$$\bgroup\color{black}p_{op}u_p(x)=pu_p(x) \egroup$$

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

The eigenstates are

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

with $p$ allowed to be positive or negative.

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

$$\bgroup\color{black}\langle p\vert p\rangle=\langle u_p\vert u_p\rangle=\infty \egroup$$

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).

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

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

$$u_p(x)={1\over \sqrt{2\pi\hbar}} e^{ipx/\hbar}$$

satisfying the normalization condition

$$\langle p'\vert p\rangle=\delta(p-p')$$

For a free particle Hamiltonian, both momentum and parity commute with $H$. So we can make simultaneous eigenfunctions.

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

We cannot make eigenfunctions of all three operators since

$$\bgroup\color{black} [P,p]\neq 0 .\egroup$$

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