# Question #b61e8

Sep 1, 2017

Consider the one dimensional potential well,

$V = 0$ for $- \frac{a}{2} < x < \frac{a}{2}$
and infinite otherwise.

That is how the potential looks like for a one dimensional box with boundaries that you've supplied.

The time independent Schrodinger equation is,

$\hat{H} \psi = E \psi$

where, $\hat{H}$ is the Hamiltonian operator and $E$ is corresponding eigen-energy.

Now, $\hat{H} = {\hat{p}}^{2} / \left(2 m\right) + V$

Therefore, for the region inside the box where $V = 0$, the TISE reads,

${\hat{p}}^{2} / \left(2 m\right) \psi = E \psi$

Writing down the momentum operator in 1D,

$- {\overline{h}}^{2} / \left(2 m\right) \frac{{d}^{2} \psi}{\mathrm{dx}} ^ 2 = E \psi$

Putting, ${k}^{2} = \frac{2 m E}{\overline{h}} ^ 2$, the equation reads,

$\frac{{d}^{2} \psi}{\mathrm{dx}} ^ 2 + {k}^{2} \psi = 0$

which is known to have plane wave solutions on the form,

$\psi \left(x\right) = A S \in \left(k x\right) + B C o s \left(k x\right)$

The boundaries have infinite potential implies that the wavefunction vanishes at the boundaries which gives us,
$\psi \left(- \frac{a}{2}\right) = \psi \left(\frac{a}{2}\right) = 0$

Thus, $\psi \left(\frac{a}{2}\right) = A S \in \left(\frac{k a}{2}\right) + B C o s \left(\frac{k a}{2}\right) = 0$

Now since sine and cosine functions cannot be simultaneously zero, the we must consider two different cases with either of them zero where the other is not.

Case I -

$S \in \left(\frac{k a}{2}\right) = 0$ then $B = 0$ and we get an anti-symmetric wavefunction of the form,

$\psi \left(x\right) = A S \in \left(k x\right)$

But, since, $S \in \left(\frac{k a}{2}\right) = 0 \implies \frac{k a}{2} = \frac{n \pi}{2}$ where $n = 2 , 4 , 6 , 8 , \ldots .$

Thus, $k = \frac{n \pi}{a}$

Then Corresponding Eigen-energy,
${E}_{n} = \frac{{\overline{h}}^{2} {\pi}^{2} {n}^{2}}{2 m {a}^{2}}$

And from Normalization condition, one gets,

${\int}_{- \frac{a}{2}}^{\frac{a}{2}} | \psi \left(x\right) {|}^{2} \mathrm{dx} = 1$

Evaluating the integral,

$A = \sqrt{\frac{2}{a}}$

Thus, the normalized anti-symmetric wavefunctions in this case are,

${\psi}_{n} \left(x\right) = \sqrt{\frac{2}{a}} S \in \left(\frac{n \pi x}{a}\right)$ where $n = 2 , 4 , 6 , 8 , \ldots . .$

Case II -

$C o s \left(\frac{k a}{2}\right) = 0$ $\implies A = 0$

In this case the wavefunction is symmetric.

Thus, $\frac{k a}{2} = \frac{n \pi}{2}$ where $n = 1 , 3 , 5 , 7 , \ldots \ldots$

Then corresponding Eigen-energy as before is,

${E}_{n} = \frac{{\overline{h}}^{2} {\pi}^{2} {n}^{2}}{2 m {a}^{2}}$

Yet again, Normalization condition gives,

$B = \sqrt{\frac{2}{a}}$

Thus, normalized symmetric wavefunctions in this case are,

${\psi}_{n} \left(x\right) = \sqrt{\frac{2}{a}} C o s \left(\frac{n \pi x}{a}\right)$ where $n = 1 , 3 , 5 , 7 , \ldots \ldots$

Thus wavefunctions in this potential box are alternatively anti-symmetric and symmetric.