# Question #b61e8

##### 1 Answer

Consider the one dimensional potential well,

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,

where,

Now,

Therefore, for the region inside the box where

Writing down the momentum operator in 1D,

Putting,

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

The boundaries have infinite potential implies that the wavefunction vanishes at the boundaries which gives us,

Thus,

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** -

But, since,

Thus,

Then Corresponding Eigen-energy,

And from Normalization condition, one gets,

Evaluating the integral,

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

**Case II** -

In this case the wavefunction is symmetric.

Thus,

Then corresponding Eigen-energy as before is,

Yet again, Normalization condition gives,

Thus, normalized symmetric wavefunctions in this case are,

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