Q47E Consider the delta well potentia... [FREE SOLUTION]

Chapter 5: Q47E (page 189)

Consider the delta well potential energy:
$U (x) = {\begin{cases} 0 & x 鈮� 0 \\ - 鈭� & x = 0 \end{cases}$
Although not completely realistic, this potential energy is often a convenient approximation to a verystrong, verynarrow attractive potential energy well. It has only one allowed bound-state wave function, and because the top of the well is defined as U = 0, the corresponding bound-state energy is negative. Call its value -E₀.
(a) Applying the usual arguments and required continuity conditions (need it be smooth?), show that the wave function is given by
$蠄 (x) = {(\frac{2 m E_{0}}{h^{2}})}^{1 / 4} e^{- (\sqrt{2 m E_{0}} / 鈩�) | x |}$
(b) Sketch $蠄 (x)$ and U(x) on the same diagram. Does this wave function exhibit the expected behavior in the classically forbidden region?

Short Answer

Expert verified

(a) It is verified that the wave function for the delta well potential is given by $蠄 (x) = {(\frac{2 {mE}_{0}}{{鈩�}^{2}})}^{\frac{1}{4}} e^{鈭� (\sqrt{2 {mE}_{0}} / 鈩�) | x |}$

(b) The plot of the wave function is given below. It exhibits expected behavior in the classically forbidden region.

Step by step solution

Step 1: Given data

There is a delta well potential of the form

$U (x) = \{\begin{cases} \begin{array}{l} 0 & x 鈮� 0 \end{array} \\ \begin{array}{l} 鈭� 鈭� & x = 0 \end{array} \end{cases}$ ....................................(I)

The top of the well is defined as U = 0 and the corresponding bound-state energy is negative (-E₀).

Wave function outside a finite potential well

The wave function of a particle of mass m and energy E outside a finite potential well of height U₀ is

$蠄 (x) = \{\begin{cases} A e^{\frac{\sqrt{2 m (U_{0} - E)}}{鈩�} x} & x < 0 \\ B e^{- \frac{\sqrt{2 m (U_{0} - E)}}{鈩�} x} & x > 0 \end{cases}$ .....(II)

Here is the reduced Planck's constant.

Step 3: Determining the wave function for the delta well potential

For the delta potential, $U_{0}$ = 0 and E = $- E$ and the wave function in equation (II) reduces to

localid="1660047013782" $蠄 (x) = \{\begin{cases} A e^{\frac{\sqrt{2 m (U_{0} - E)}}{鈩�} x} & x < 0 \\ B e^{- \frac{\sqrt{2 m (U_{0} - E)}}{鈩�} x} & x > 0 \end{cases}$

The function has to be continuous at x = 0 and thus

A = B

The function thus becomes

localid="1660047017169" $蠄 (x) = A e^{- \frac{\sqrt{2 m E_{0}}}{鈩�} |x|}$

Normalize this to get

localid="1660047021009" $\begin{array}{rcl} 1 & = & {鈭�}_{- 鈭�}^{鈭�} {(A e^{- \frac{\sqrt{2 m E_{0}}}{鈩�} |x|})}^{2} d x \\ = & 2 A^{2} {鈭�}_{0}^{鈭�} e^{- \frac{2 \sqrt{2 m E_{0}}}{鈩�} x} d x \end{array}$

Let

localid="1660047024841" $\begin{array}{rcl} \frac{2 \sqrt{2 m E_{0}}}{鈩�} x & = & z \\ d x & = & \frac{鈩�}{2 \sqrt{2 m E_{0}}} d z \end{array}$

Thus

localid="1660047032303" $\begin{array}{rcl} 1 & = & 2 A^{2} \frac{鈩�}{2 \sqrt{2 m E_{0}}} {鈭�}_{0}^{鈭�} e^{- z} d z \\ = & \frac{A^{2} 鈩�}{\sqrt{2 m E_{0}}} \\ A & = & {(\frac{\sqrt{2 m E_{0}}}{鈩�})}^{1 / 2} \end{array}$