130_notes.dvi

which automatically satisfies the BC at 0. To satisfy the BC atx=awe need the argument of sine
to benπthere.
un=Csin

(nπx

a

)

Plugging this back into the Schr ̈odinger equation, we get

− ̄h^2

2 m

(−

n^2 π^2

a^2

)Csin(kx) =ECsin(kx).

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

En=

n^2 π^2 ̄h^2

2 ma^2

We have solutions to the Schr ̈odinger equation that satisfy the boundary conditions. Now we need
to set the constantCtonormalizethem to 1.

〈un|un〉=|C|^2

∫a

0

sin^2

(nπx

a

)

dx=|C|^2

a

2

Remember that the average value of sin^2 is one half (over half periods). So we setCgiving us the
eigenfunctions

un=

√

2

a

sin

(nπx

a

)

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

Particle in a Box Eigenfunctions

012

x

u(x)

n=1

n=2

n=3

n=4