The Dirac delta function can be bought off as the limiting case of a rectangle area 1, as the height goes to infinity and the width goes to Zero. Show that the delta function well (Equation 2.114) is weak potential (even though it is infinitely deep), in the sense that ${\mathit{Z}}_{\mathbf{0}}\mathbf{\to }\mathbf{0}$. Determine the bound state energy for the delta function potential, by treating it as the limit of a finite square well. Check that your answer is consistent with equation 2.129. Also, show that equation 2.169 reduces to Equation 2.141 in the appropriate limit.

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 $\alpha$Area of the potential held constant as$\left(\alpha =2a{V}_0\right)$.

The potential will be:

$V_0=\frac{\alpha }{2a}$

So $z_0$will be :

$$\begin{gathered} z_0=\frac{a}{\sout{h}}\sqrt{2m\frac{\alpha }{2a}} \\ =\frac{1}{\sout{h}}\sqrt{m\alpha a}\to 0 \end{gathered}$$

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

Solve Eqfor very smallas:

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

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 $z_0^2-z^2=z^4\ll 1$. That will gives us :

$z\cong z_0$

Thus, now$z_0^2=ka$

But, we found that:

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

or

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

For limit $a\to 0$:

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

That is in agreement with Eq 1.129.

From Eq 2.169

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

But $V_0=\frac{\alpha }{2a}$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 $\left(\alpha =2a{V}_0\right)$in the above expressiiopn, and we get,

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

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 $z_0\to 0$and bound state energy for the delta function potential is $E=\frac{-m{\alpha }^2}{2{\sout{h}}^2}$.