Question

As Fig. 39-8 suggests, the probability density for the region

0 < x < L for the finite potential well of Fig. 39-7 is sinusoidal, being given by

${\mathbf{\psi }}^{\mathbf{2}}\left(x\right)\mathbf{=}\mathbf{B}\mathbf{}{\mathbf{sin}}^{\mathbf{2}}\mathbf{kx}$ , in which B is a constant. (a) Show that the wave function $\mathit{\psi }\left(x\right)$

may be found from this equation is a solution of Schrodinger’s equation in its one-dimensional form. (b) Express an equation for that makes this true.

Short answer

(a) The proposed function satisfies Schrodinger’s equation.

(b)$k=\pm \frac{2\mathrm{\pi }}{h}\sqrt{2mE}$

Step-by-step solution

Identification of the given data

The given data is listed below as-

The probability density for the region is, 0 < x < L

The sinusoidal function

Thesinusoidal functionis given by-

$\mathbf{\psi }\left(x\right)\mathbf{=}\mathbf{B}\mathbf{}{\mathbf{sin}}^{\mathbf{2}\mathbf{}}\mathbf{kx}$

Here, B is the constant.

To show that the wave function ψ(x) may be found from the equation which is a solution of Schrodinger’s equation in its one-dimensional form  (a)

The region is 0 < x <L .

The Schrodinger’s equation for this region is $\frac{d^2\psi }{d{x}^2}+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}E\psi =0$

Here, E > 0 .

Now, if ${\psi }^2\left(x\right)=B{\sin }^{2}kx$

Then, $\psi \left(x\right)=B\text{'}\sin kx$

Here, B’ is another constant.

Satisfying $B{\text{'}}^2=B$

Therefore,

$$\begin{gathered} \frac{d^2\psi }{d{x}^2}=-k^{2}B\text{'}\sin kx \\ =-k^2\psi \left(x\right) \end{gathered}$$

And, $\frac{d^2\psi }{d{x}^2}+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}E\psi =-k^2\psi \left(x\right)+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}E\psi$

The above equation tends to zero provided that $k^2=\frac{8{\mathrm{\pi }}^2\mathrm{mE}}{h^2}$

Thus, the proposed function satisfies the Schrodinger’s equation as the right side of the above equation is positive and so k is real.

Step 4: Express an equation for k . (b)

The value of k can be positive or negative.

$k^2=\frac{8{\mathrm{\pi }}^2\mathrm{mE}}{h^2}$

Solving the above will give final value of k.

$k=\pm \frac{2\mathrm{\pi }}{h}\sqrt{2mE}$

Thus, value of $k=\pm \frac{2\mathrm{\pi }}{h}\sqrt{2mE}$.