91Ó°ÊÓ

A differential equation describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer relates to a particle's probability density in space and time.

The equation in the book is:

$z_0=\frac{a}{\sout{h}}\sqrt{2m{V}_0}$

We want $\mathrm{Î\pm }$Area of the potential held constant as$\left(\mathrm{Î\pm }=2a{V}_0\right)$.

The potential will be:

$V_0=\frac{\mathrm{Î\pm }}{2a}$

So $z_0$will be :

$$\begin{gathered} z_0=\frac{a}{\sout{h}}\sqrt{2m\frac{\mathrm{Î\pm }}{2a}} \\ =\frac{1}{\sout{h}}\sqrt{m\mathrm{Î\pm }a}→0 \end{gathered}$$

So $z_0$is small, and the intersection in Fig 2.18 occures at very small z.

Solve Eqfor very smallas:

$$\begin{gathered} \tan zâ–¡z \\ =\sqrt{\frac{z_0^2}{z^2}-1} \\ =\frac{1}{z}\sqrt{{z_0}^2-z^2} \end{gathered}$$

From Eqns2.146, 2.148, and 2.155 in the book:

${z_0}^2-z^2=k^2{a}^2$

So .$z^2=ka$

But $z_0^2-z^2=z^4≪1$. That will gives us :

$zâ‰\dots z_0$

Thus, now$z_0^2=ka$

But, we found that:

role="math" localid="1658122728379" $z_0=\frac{1}{\sout{h}}\sqrt{m\mathrm{Î\pm }a}$

or

$$\begin{gathered} ka=\frac{m\mathrm{Î\pm }a}{{\sout{h}}^2} \\ k=\frac{m\mathrm{Î\pm }}{{\sout{h}}^2} \end{gathered}$$

For limit $a→0$:

$$\begin{gathered} \frac{\sqrt{-2mE}}{\sout{h}}=\frac{m\mathrm{Î\pm }}{{\sout{h}}^2} \\ E=\frac{m{\mathrm{Î\pm }}^2}{{\sout{h}}^2} \end{gathered}$$

That is in agreement with Eq 1.129.

From Eq 2.169

$V_0≫E→T^{-1}≃\frac{1+V_0^2}{4E{V}_0}{\sin }^2\left(\frac{2a}{\sout{h}}\sqrt{2m{V}_0}\right)$

But $V_0=\frac{\mathrm{Î\pm }}{2a}$so, the argument of the sin is small then the expression will be:

$$\begin{gathered} T^{-1}=1+\frac{V_0}{4E}\left(\frac{2a}{\sout{h}}\right)2m{V}_0 \\ =1+{\left(2m{V}_0\right)}^2\frac{m}{2\sout{h}2E} \end{gathered}$$

Substitutting $\left(\mathrm{Î\pm }=2a{V}_0\right)$in the above expressiiopn, and we get,

$T^{-1}=1+\frac{m{\mathrm{Î\pm }}^2}{2{\sout{h}}^{2}E}$

That is in agreement with 2.141.

Thus, the delta function well (Equation 2.114) is weak potential (even though it is infinitely deep), in the sense that $z_0→0$and bound state energy for the delta function potential is $E=\frac{-m{\mathrm{Î\pm }}^2}{2{\sout{h}}^2}$.