Question

Question #b61e8

Answer 1

Consider the one dimensional potential well,

`V = 0` for `-a/2 < x < 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 Hpsi = Epsi`

where, `hat H` is the Hamiltonian operator and `E` is corresponding eigen-energy.

Now, `hat H = hat p^2/(2m) + V`

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

`hat p^2/(2m)psi = Epsi`

Writing down the momentum operator in 1D,

`-barh^2/(2m)(d^2psi)/dx^2 = Epsi`

Putting, `k^2 = (2mE)/bar h^2`, the equation reads,

`(d^2psi)/dx^2 + k^2psi = 0`

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

`psi(x) = ASin (kx) + BCos (kx)`

The boundaries have infinite potential implies that the wavefunction vanishes at the boundaries which gives us,
`psi(-a/2) = psi(a/2) = 0`

Thus, `psi(a/2) = ASin ((ka)/2) + BCos ((ka)/2) = 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 -

`Sin ((ka)/2) = 0` then `B = 0` and we get an anti-symmetric wavefunction of the form,

`psi(x) = ASin(kx)`

But, since, `Sin ((ka)/2) = 0 implies (ka)/2 = (npi)/2` where `n = 2,4,6,8,....`

Thus, `k=(npi)/a`

Then Corresponding Eigen-energy,
`E_n = (bar h^2pi^2n^2)/(2ma^2)`

And from Normalization condition, one gets,

`int_(-a/2)^(a/2) |psi(x)|^2dx = 1`

Evaluating the integral,

`A = sqrt (2/a)`

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

`psi_n(x) = sqrt(2/a)Sin ((npix)/a)` where `n=2,4,6,8,.....`

Case II -

`Cos ((ka)/2) = 0` `implies A=0`

In this case the wavefunction is symmetric.

Thus, `(ka)/2 = (npi)/2` where `n=1,3,5,7,......`

Then corresponding Eigen-energy as before is,

`E_n = (bar h^2pi^2n^2)/(2ma^2)`

Yet again, Normalization condition gives,

`B=sqrt(2/a)`

Thus, normalized symmetric wavefunctions in this case are,

`psi_n(x) = sqrt(2/a)Cos ((npix)/a)` where `n=1,3,5,7,......`

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