next up previous contents
Next: Potential scattering Up: Scattering states in one Previous: Scattering states in one   Contents

Plane waves

The simplest case is the one where the potential $ V(x)$ is zero throughout: the two independent solutions of the Schrödinger equation then are plane waves
$\displaystyle -\frac{\hbar^2\partial_x^2}{2m}\psi(x)= E\psi(x)\leadsto \psi_+(x)=e^{i kx},
\quad \psi_-(x)= e^{-i kx},
\quad k= \sqrt{(2m/\hbar^2)E}$     (114)

with positive energy $ E>0$. We denote both solutions as $ \psi_k(x)=e^{ikx}$ with $ k$ either positive or negative. They are plane waves with fixed wave vector $ k$ and therefore fixed momentum $ p=\hbar k$. We can have no physically meaningful solutions with negative energy $ E<0$ because in this case the wave function would become infinite either for $ x\to \infty$ or $ x\to -\infty$.

A problem arises, however, because $ \psi_k(x)$ can not be normalized over the whole $ x$-axis according to

$\displaystyle \int_{-\infty}^{\infty}dx \vert\psi_k(x)\vert^2 =1,$     (115)

because this integral is infinite: the probability density, i.e. the square $ \vert\psi_k(x)\vert^2$ is constant, i.e. $ 1$ everywhere.

In particular, we have for the mean square deviation of the momentum and the position

$\displaystyle \langle \Delta p ^2 \rangle$ $\displaystyle =$ $\displaystyle \lim_{L \to \infty} \left(
\frac{[\int_{-L/2}^{L/2}\vert a\vert^2...
...hbar\partial_x)e^{ikx}} {\int_{-L/2}^{L/2}dx \vert a\vert^2}
\right]^2 \right )$  
  $\displaystyle =$ $\displaystyle \hbar^2k^2-\hbar^2k^2=0$  
$\displaystyle \langle \Delta x ^2 \rangle$ $\displaystyle =$ $\displaystyle \lim_{L \to \infty} \left(
\frac{[\int_{-L/2}^{L/2}\vert a\vert^2...
...vert^2e^{-ikx}xe^{ikx}} {\int_{-L/2}^{L/2}dx \vert a\vert^2}
\right]^2 \right )$  
  $\displaystyle =$ $\displaystyle \infty.$ (116)

There is no uncertainty in the momentum of the particle, but there is maximum uncertainty in its position: the wave function $ \psi_k(x)$ describes a particle with fixed momentum $ p=\hbar k$ which is completely delocalized (spread) over the $ x$-axis.

A practical solution is to consider a large, but finite interval $ [-L/2,L/2]$ instead of the total $ x$-axis, and to normalize the wave functions according to

$\displaystyle \psi_k=\frac{1}{\sqrt{L}}e^{ikx},\quad \int_{-L/2}^{L/2}dx \vert\psi_k(x)\vert^2 =1.$     (117)

Then, the boundary conditions at $ x=\pm L/2$ have to be specified. Again, a convenient (but not necessary the only) choice are periodic boundary conditions: we bend the interval $ [L/2,L/2]$ into a ring such that the points $ \pm L/2$ coincide. Demanding continuity of $ \psi_k$, i.e. $ \psi_k(L/2)= \psi_k(-L/2)$, we obtain a quantization condition for the possible $ k$-values,
$\displaystyle 1=e^{ikL} \leadsto k_n = \frac{2\pi n}{L}, \quad n=0, \pm 1, \pm 2, ...,$     (118)

i.e. a discrete set of possible $ k$ values and therewith discrete energies $ E(k)= \hbar^2k^2/2m$. To each energy $ E=E_n>0$ there are two linearly independent plane waves with wave vectors $ 2\pi n/L$ and $ -2\pi n/L$. One says the energy value $ E_n$ is two-fold degenerate. If $ L$ is very large, the possible values for $ k$ are still discrete but subsequent $ k$-values get very close to each other.


next up previous contents
Next: Potential scattering Up: Scattering states in one Previous: Scattering states in one   Contents
Tobias Brandes 2004-02-04