A particle is subject to a potential energy that has an essentially infinitely high wall at the origin, like the infinite well, but for positive values of x is of the form U(x)= -b/ x, where b is a constant

(a) Sketch this potential energy.

(b) How much energy could a classical particle have and still be bound by such a potential energy?

(c) Add to your sketch a plot of E for a bound particle and indicate the outer classical tuning point (the inner being the origin).

(d) Assuming that a quantum-mechanical description is in order, sketch a plausible ground-state wave function, making sure that your function's second derivative is of the proper sign when U(x)is less than E and when it is greater.

The given function is,

$U\left(x\right)=-b{x}^{-1}$.

Features of quantum mechanical model:

1. The energy of an electron is quantized i.e. an electron can only have certain specific values of energy.

2. The quantized energy of an electron is the allowed solution of the Schrödinger wave equation and it is the result of wave like properties of electron.

3. As per Heisenberg's Uncertainty principle, the exact position and momentum of an electron cannot be determined.

(a)

In order to sketch the ground state wave function, the function's second derivative should be of proper sign when U(x) is less than E and when it is greater.

The plot obtained is as follows,

(b)

In the case that the particle is in the box, the total energy is given by:

$\mathrm{E}=\frac{1}{2}{\mathrm{mx}}^2+\mathrm{U}\left(\mathrm{x}\right)=\frac{1}{2}{\mathrm{mx}}^2-\frac{\mathrm{b}}{\mathrm{x}}$

Hence, the energy is equal to E = 0 when the particle is free and $\mathrm{E}\le 0$ to be still bounded by the given potential energy.

(c)

The sketch of the energy for a bound particle with the outer classical turning point can be shown as ${\mathrm{x}}_1$and ${\mathrm{x}}_2$are two bounds and turning points.

(d)

Using the WKB method, we have:

$$\begin{gathered} k\left(x\right)=\sqrt{\frac{2m\left(r_2-U\left(x\right)\right)}{{\sout{h}}^2}} \\ \Psi \left(x\right)=A{e}^{\pm ikx} \\ \Psi \left(x\right)=A{e}^{\pm ikx}=\sqrt{\frac{2m\left(r_2-U\left(x\right)\right)}{{\sout{h}}^2}} \\ \Psi \left(x\right)=A{e}^{\pm ikx}=\sqrt{\frac{2m\left(r_2-{\displaystyle \frac{b}{x}}\right)}{{\sout{h}}^2}} \end{gathered}$$

Hence, the graph of the wave function can be shown as: