Question #b61e8

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Question #b61e8

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Answer 1 · 2017-09-01T07:22:58

Consider the one dimensional potential well,

$V = 0$ $V = 0$ for $- \frac{a}{2} < x < \frac{a}{2}$ $-a/2 < x < a/2$
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,

$ˆ H ψ = E ψ$ $hat Hpsi = Epsi$

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

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

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

$\frac{{ˆ p}^{2}}{2 m} ψ = E ψ$ $hat p^2/(2m)psi = Epsi$

Writing down the momentum operator in 1D,

$- \frac{{¯ h}^{2}}{2 m} \frac{d^{2} ψ}{{d x}^{2}} = E ψ$ $-barh^2/(2m)(d^2psi)/dx^2 = Epsi$

Putting, $k^{2} = \frac{2 m E}{{¯ h}^{2}}$ $k^2 = (2mE)/bar h^2$ , the equation reads,

$\frac{d^{2} ψ}{{d x}^{2}} + k^{2} ψ = 0$ $(d^2psi)/dx^2 + k^2psi = 0$

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

$ψ (x) = A sin (k x) + B cos (k x)$ $psi(x) = ASin (kx) + BCos (kx)$

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

Thus, $ψ (\frac{a}{2}) = A sin (\frac{k a}{2}) + B cos (\frac{k a}{2}) = 0$ $psi(a/2) = ASin ((ka)/2) + BCos ((ka)/2) = 0$

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 -

$sin (\frac{k a}{2}) = 0$ $Sin ((ka)/2) = 0$ then $B = 0$ $B = 0$ and we get an anti-symmetric wavefunction of the form,

$ψ (x) = A sin (k x)$ $psi(x) = ASin(kx)$

But, since, $sin (\frac{k a}{2}) = 0 \Rightarrow \frac{k a}{2} = \frac{n π}{2}$ $Sin ((ka)/2) = 0 implies (ka)/2 = (npi)/2$ 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.

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Question #b61e8

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