The Particle in a 1D Box

As a simple example, we will solve the 1D Particle in a Box problem. That is a particle confined to a region \bgroup\color{black}$0<x<a$\egroup. We can do this with the (unphysical) potential which is zero with in those limits and \bgroup\color{black}$+\infty$\egroup outside the limits.

\begin{displaymath}\bgroup\color{black} V(x)=\left\{ \matrix{0 & 0<x<a \cr \infty & \mathrm{elsewhere} }\right. \egroup\end{displaymath}

Because of the infinite potential, this problem has very unusual boundary conditions. (Normally we will require continuity of the wave function and its first derivative.) The wave function must be zero at \bgroup\color{black}$x=0$\egroup and \bgroup\color{black}$x=a$\egroup since it must be continuous and it is zero in the region of infinite potential. The first derivative does not need to be continuous at the boundary (unlike other problems), because of the infinite discontinuity in the potential.

The time independent Schrödinger equation (also called the energy eigenvalue equation) is

\begin{displaymath}\bgroup\color{black}Hu_j=E_ju_j \egroup\end{displaymath}

with the Hamiltonian (inside the box)

\begin{displaymath}\bgroup\color{black}H=-{\hbar^2\over 2m}{d^2\over dx^2} \egroup\end{displaymath}

Our solutions will have

\begin{displaymath}\bgroup\color{black}u_j=0 \egroup\end{displaymath}

outside the box.

The solution inside the box could be written as

\begin{displaymath}\bgroup\color{black}u_j=e^{ikx} \egroup\end{displaymath}

where \bgroup\color{black}$k$\egroup can be positive or negative. We do need to choose linear combinations that satisfy the boundary condition that \bgroup\color{black}$u_j(x=0)=u_j(x=a)=0$\egroup.

We can do this easily by choosing

\begin{displaymath}\bgroup\color{black}u_j=C\sin(kx) \egroup\end{displaymath}

which automatically satisfies the BC at 0. To satisfy the BC at \bgroup\color{black}$x=a$\egroup we need the argument of sine to be \bgroup\color{black}$n\pi$\egroup there.

\begin{displaymath}\bgroup\color{black}u_n=C\sin\left({n\pi x\over a}\right) \egroup\end{displaymath}

Plugging this back into the Schrödinger equation, we get

\begin{displaymath}\bgroup\color{black}{-\hbar^2\over 2m}(-{n^2\pi^2\over a^2})C\sin(kx)= EC\sin(kx) .\egroup\end{displaymath}

There will only be a solution which satisfies the BC for a quantized set of energies.

\begin{displaymath}\bgroup\color{black}E_n={n^2\pi^2\hbar^2\over 2ma^2} \egroup\end{displaymath}

We have solutions to the Schrödinger equation that satisfy the boundary conditions. Now we need to set the constant \bgroup\color{black}$C$\egroup to normalize them to 1.

\begin{displaymath}\bgroup\color{black}\langle u_n\vert u_n\rangle=\vert C\vert^...
...\left({n\pi x\over a}\right)dx=\vert C\vert^2{a\over 2} \egroup\end{displaymath}

Remember that the average value of \bgroup\color{black}$\sin^2$\egroup is one half (over half periods). So we set \bgroup\color{black}$C$\egroup giving us the eigenfunctions

\begin{displaymath}\bgroup\color{black}u_n=\sqrt{2\over a}\sin\left({n\pi x\over a}\right) \egroup\end{displaymath}

The first four eigenfunctions are graphed below. The ground state has the least curvature and the fewest zeros of the wavefunction.


Note that these states would have a definite parity if \bgroup\color{black}$x=0$\egroup were at the center of the box.

The expansion of an arbitrary wave function in these eigenfunctions is essentially our original Fourier Series. This is a good example of the energy eigenfunctions being orthogonal and covering the space.

Jim Branson 2013-04-22