40.3.3 Solve I: Find the Wave Functions

At all points inside the box the potential energy is $U (x) = 0$ . Thus the Schrödinger equation inside the box is

\frac{d^{2} ψ}{d x^{2}} = - \frac{2 m E}{ℏ^{2}} ψ (x)

(40.12)

There are two aspects to solving this equation:

For what values of $E$ does Equation 40.12 have physically meaningful solutions?
What are the solutions $ψ (x)$ for those values of E?

To begin, let’s simplify the notation by defining $β^{2} = 2 m E / ℏ^{2}$ . Equation 40.12 is then

\frac{d^{2} ψ}{d x^{2}} = - β^{2} ψ (x)

(40.13)

We’re going to solve this differential equation by guessing! Can you think of any functions whose second derivative is a negative constant times the function itself? Two such functions are

ψ_{1} (x) = sin β x and ψ_{2} (x) = cos β x

(40.14)

Both are solutions to Equation 40.13 because

\begin{array}{l} \frac{d^{2} ψ_{1}}{d x^{2}} = \frac{d^{2}}{d x^{2}} (sin β x) = - β^{2} sin β x = - β^{2} ψ_{1} (x) \\ \frac{d^{2} ψ_{2}}{d x^{2}} = \frac{d^{2}}{d x^{2}} (cos β x) = - β^{2} cos β x = - β^{2} ψ_{2} (x) \end{array}

Furthermore, these are independent solutions because $ψ_{2} (x)$ is not a multiple or a rearrangement of $ψ_{1} (x)$ . Consequently, according to Equation 40.8, the general solution to the Schrödinger equation for the particle in a rigid box is

ψ (x) = A sin β x + B cos β x

(40.15)

where

β = \frac{\sqrt{2 m E}}{ℏ}

(40.16)

The constants $A$ and $B$ must be determined by using the boundary conditions of Equation 40.11. First, the wave function must go to zero at $x = 0$ . That is,

ψ (at x = 0) = A \cdot 0 + B \cdot 1 = 0

(40.17)

This boundary condition can be satisfied only if $B = 0$ . The $cos β x$ term may satisfy the differential equation in a mathematical sense, but it is not a physically meaningful solution for this problem because it does not satisfy the boundary conditions. Thus the physically meaningful solution is

ψ (x) = A sin β x

The wave function must also go to zero at $x = L$ . That is,

ψ (at x = L) = A sin β L = 0

(40.18)

This condition could be satisfied by $A = 0$ , but then we wouldn’t have a wave function at all! Fortunately, that isn’t necessary. This boundary condition is also satisfied if $sin β L = 0$ , which requires

β L = n π or β_{n} = \frac{n π}{L} n = 1, 2, 3, \dots

(40.19)

Notice that $n$ starts with 1, not 0. The value $n = 0$ would give $β = 0$ and make $ψ = 0$ at all points, a physically meaningless solution.

Thus the solutions to the Schrödinger equation for a particle in a rigid box are

ψ_{n} (x) = A sin β_{n} x = A sin (\frac{n π x}{L}) n = 1, 2, 3, \dots

(40.20)

We’ve found a whole family of solutions, each corresponding to a different value of the integer $n$ . These wave functions represent the stationary states of the particle in the box. The constant $A$ remains to be determined.