Chapter 7: Q20P (page 313)

Quantum dots. Consider a particle constrained to move in two dimensions in the cross-shaped region.The “arms” of the cross continue out to infinity. The potential is zero within the cross, and infinite in the shaded areas outside. Surprisingly, this configuration admits a positive-energy bound state
(a) Show that the lowest energy that can propagate off to infinity is
$E_{t h r e s h o l d} = \frac{π^{2} h^{2}}{8 m a^{2}}$
any solution with energy less than that has to be a bound state. Hint: Go way out one arm (say $x ≫ a$ ), and solve the Schrödinger equation by separation of variables; if the wave function propagates out to infinity, the dependence on x must take the form $\exp (i k_{x} x) with k_{x} > 0$
(b) Now use the variation principle to show that the ground state has energy less than $E_{t h r e s h o l d}$ . Use the following trial wave function (suggested by Jim Mc Tavish):
$ψ (x, y) = A {\begin{matrix} [\cos (πx / 2 a) + \cos (πy / 2 a)] e^{- α} & |x| \leq a and |y| \leq a \\ \cos (π x / 2 a) e^{- α} |y| / a & |x| \leq a and |y| > a \\ \cos (π y / 2 a) e^{- α} |y| / a & |x| . a and |y| \leq a \\ 0 & elsewhere \end{matrix} \begin{array}{l} \end{array}$
Normalize it to determine A, and calculate the expectation value of H.
Answer:
$< H > = \frac{h^{2}}{m a^{2}} [\frac{π^{2}}{8} - (\frac{1 - (α / 4)}{1 + (8 / π^{2}) + (1 / 2 α)})]$
Now minimize with respect to α, and show that the result is less than $E_{t h r e s h o l d}$ . Hint: Take full advantage of the symmetry of the problem— you only need to integrate over 1/8 of the open region, since the other seven integrals will be the same. Note however that whereas the trial wave function is continuous, its derivatives are not—there are “roof-lines” at the joins, and you will need to exploit the technique of Example 8.3.

Short Answer

Expert verified

$(a) E_{m i n} = \frac{h^{2} π^{2}}{8 m a^{2}} = 1.234 \frac{h^{2}}{m a^{2}} (b) {<H>}_{m i n} = 1.058 \frac{h^{2}}{m a^{2}}$

Step by step solution

(a) Showing the lowest energy that can propagate off to infinity

The lowest Energy that can propagate off to infinity

$- \frac{h^{2}}{2 m} (\frac{\partial^{2} ψ}{\partial x^{2}} + \frac{\partial^{2} ψ}{\partial y^{2}}) = E ψ$

Let

localid="1658386907422" $ψ (x, y) = X (x) Y (y) . Y \frac{d^{2} X}{d x^{2}} + X \frac{d^{2} Y}{d y^{2}} = - \frac{2 m E}{h^{2}} X Y; \frac{1}{X} \frac{d^{2} X}{d x^{2}} + \frac{1}{Y} \frac{d^{2} Y}{d y^{2}} = - \frac{2 m E}{h^{2}} \frac{d^{2} X}{d x^{2}} = - K_{x}^{2} X; \frac{d^{2} Y}{d y^{2}} = - k_{y}^{2} Y k_{x}^{2} + k_{y}^{2} = \frac{2 m E}{h^{2}}$

The general solution to the y equation is $Y (y) = A \cos k_{y} y + B \sin k_{y} y$ ; the boundary conditions localid="1658387119190" $Y (\pm a) = 0 yield k_{y} = \frac{n π}{2 a}$ with minimum $\frac{π}{2 a}$ .

[Note that $k_{y}^{2}$ has to be positive, or you cannot meet the boundary conditions at all.]

So $E \geq \frac{h^{2}}{2 m} (k_{x}^{2} + \frac{π^{2}}{4 a^{2}})$ .

For a traveling wave $k_{x}^{2}$ has to be positive. Conclusion: Any solution with $E < \frac{π^{2} h^{2}}{8 m a^{2}}$ will be a bound state.

(a) The ground state has energy less than Ethreshold

Here’s a graph of $ψ (x, y)$ :

$a = 0.9 0.9 g [x_{-}, y_{-}] = P i e c e w i s e (C o s [πx / 2] + C o s [πy / 2]) e^{^} (- a), A b s [x] < 1 A b s [y] < 1, (\cos [πx / 2] e^{^} (- a A b s [y]), A b s [x] < 1 A b s [y] > 1, P l o t 3 D [g [x, y], \{x, - 4, 4\}, \{y, - 4, 4\}, P l o t R a n g e \to \{θ, 085\}] .$

It has roof-lines along the edges of the central square, as you can see by plotting g[x; 0.5] across the kink:

$j [x_{-}] : = g [x, . 5] : = g [x, . 5] . Plot [j [x]], \{x, . 8, 1, 2\}] .$

To normalize, we integrate ${|ψ (x, y)|}^{2}$ 2 over regions I and II (in the figure), and multiply by 8.

$l_{i i} = A^{2} \int_{x - a}^{\infty} \int_{y - 0}^{a} \cos^{2} (\frac{πy}{2 a}) e^{- 2 α x / a} d x d y . l_{i} = \frac{1}{2} A^{2} \int_{x - a}^{\infty} \int_{y - 0}^{a} {[\cos (\frac{πx}{2 a}) + \cos (\frac{π y}{2 a})]}^{2} e^{- 2 α} d x d y . = \frac{1}{2} A^{2} e^{- 2 α} \{2 \int_{0}^{a} \cos^{2} (\frac{πx}{2 a}) d x \int_{0}^{a} d y + 2 \int_{0}^{a} \cos (\frac{πx}{2 a}) d x \int_{0}^{a} \cos (\frac{πx}{2 a}) d y\} . = A^{2} e^{- 2 α} \{(\frac{a}{2}) a + {[\frac{2 a}{π} \sin {(\frac{πx}{2 a})}_{0}^{a}]}^{2}\} = A^{2} \frac{a^{2}}{2} e^{- 2 α} (1 + \frac{8}{π^{2}}) .$

Normalizing:

$1 = 8 (l_{l} + l_{n}) = 8 A^{2} a^{2} e^{- 2 α} [\frac{1}{4 α} + \frac{1}{2} (1 + \frac{8}{π^{2}})] = 4 A^{2} a^{2} e^{- 2 α} (1 + \frac{8}{π^{2}} + \frac{1}{2 α}) So, A^{2} = \frac{e^{2 α}}{4 a^{2} (1 + \frac{8}{π^{2}} + \frac{1}{2 α})} .$

Next, ignoring the roof-lines for the moment, we calculate ${<H>}_{a} = - 8 \frac{h^{2}}{2 m} (J_{l} + J_{l l})$ , where

$J_{i i} = A^{2} \int_{x - a}^{\infty} \int_{y - 0}^{a} [\cos (\frac{πy}{2 a}) e^{- α x / a}] (\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}}) [\cos (\frac{πy}{2 a}) e^{- α x / a}] d x d y = [{(\frac{α}{a})}^{2} - {(\frac{π}{2 a})}^{2}] l_{l l} = A^{2} \frac{1}{4 α} (α^{2} - \frac{π^{2}}{4}) e^{- 2 α} . J_{l} = \frac{1}{2} A^{2} \int_{x - a}^{\infty} \int_{y - 0}^{a} [\cos (\frac{π x}{2 a}) + \cos (\frac{πy}{2 a})] e^{- α} (\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}}) [\cos (\frac{π x}{2 a}) + \cos (\frac{πy}{2 a})] e^{- α} d x d y = - \frac{π^{2}}{4 a^{2}} l_{l} = - A^{2} (1 + \frac{π^{2}}{8}) e^{- 2 α}$

So far them,

${<H>}_{a} = - 8 \frac{h^{2}}{2 m} [- A^{2} (1 + \frac{π^{2}}{8}) e^{- 2 α} + A^{2} \frac{1}{4 α} (α^{2} + \frac{π^{2}}{4}) e^{- 2 α}] = \frac{h^{2}}{2 m} A^{2} e^{- 2 α} (8 + π^{2} - 2 α + \frac{π^{2}}{2 α})$

Now the roof-lines, along the four sides of the central square: ${<H>}_{b} = - 8 \frac{h^{2}}{2 m} J_{l l l}$ , where

$J_{l l l} = \int_{x = a - \in}^{a + \dot{o}} \int_{y = 0}^{a} ψ (x, y) (\frac{\partial^{2}}{\partial x^{2}}) ψ (x, y) d x d y . = A \int_{y = 0}^{a} ψ (a, y) \{\frac{\partial}{\partial x} {[\cos (\frac{πy}{2 a}) e^{- α x / a}]}_{x = a}^{} - \frac{\partial}{\partial x} {[\cos (\frac{πy}{2 a}) e^{- α}]}_{x = a}\} d y . = A^{2} \int_{y = 0}^{a} [\cos (\frac{πy}{2 a}) e^{- α}] \{- \frac{α}{a} \cos (\frac{πy}{2 a}) e^{- α} + \frac{π}{2 a} \sin (\frac{π}{2}) e^{- α}\} d y . = A^{2} \frac{1}{a} e^{- 2 α} \int_{0}^{a} [\frac{π}{2} \cos (\frac{πy}{2 a}) - α \cos^{2} (\frac{πy}{2 a})] d y = A^{2} e^{- 2 α} (1 - \frac{α}{2}) . {<H>}_{b} = - 8 \frac{h^{2}}{2 m} A^{2} e^{- 2 α} (1 - \frac{α}{2}) S 0, = \frac{h^{2}}{2 m} A^{2} e^{- 2 α} 4 (α - 2)$

Putting it all together,

$<H> = {<H>}_{a} + {<H>}_{b} = \frac{h^{2}}{2 m} A^{2} e^{- 2 α} (8 + π^{2} - 2 α + \frac{π^{2}}{2 α}) + \frac{h^{2}}{2 m} A^{2} e^{- 2 α} (4 α - 8) = \frac{h^{2}}{m a^{2}} [\frac{π^{2}}{8} - (\frac{1 - (α / 4)}{1 + (8 / π^{2}) + (1 / 2 α)})]$

Minimizing:

$\frac{d <H>}{d α} = 0 \Rightarrow \frac{d}{d α} (\frac{4 - α}{{(1 + (8 / π^{2}) + (1 / 2 α))}^{2}}) = 0 \frac{- 1}{1 + (8 / π^{2}) + (1 / 2 α)} - \frac{4 - α}{(1 + (8 / π^{2}) + {(1 / 2 α))}^{2}} (\frac{- 1}{2 α^{2}}); 1 + (8 / π^{2}) + (1 / 2 α) - \frac{4 - α}{2 α^{2}} = 0; α^{2} (1 + \frac{8}{π^{2}}) + α - 2; α = \frac{1 \pm \sqrt{1 + 8 [1 + (8 / π^{2})]}}{2 [1 + (8 / π^{2})]} = \frac{- 1 + \sqrt{9 + {(8 / π)}^{2}}}{2 [1 + (8 / π^{2})]} .$

(Note that α must be positive)

(b) The ground state

Putting all together.

$<H> = - \frac{2 h^{2} α e^{- 2 α}}{3 m} A^{2} + \frac{2 h^{2} e^{- 2 α}}{m α} A^{2} + \frac{2 h^{2} e^{- 2 α}}{m} A^{2} + \frac{2 h^{2} (2 α - 1) e^{- 2 α}}{3 m} A^{2} n = \frac{2 h^{2} e^{- 2 α}}{3 m α} (α^{2} + 2 α + 3) A^{2} = \frac{3 h^{2}}{m a^{2}} \frac{α^{2} + 2 α + 3}{6 + 11 α} \frac{d <H>}{d α} = 0 \frac{3 h^{2}}{m a^{2}} \frac{(2 α + 2) . (6 + 11 α) - 11 . (α^{2} + 2 α + 3)}{{(6 + 11 α)}^{2}} = 0 (2 α + 2) . (6 + 11 α) - 11 . (α^{2} + 2 α + 3) = 0 12 α + 22 α^{2} + 12 + 22 α - 11 α^{2} - 22 α - 33 = 0 11 α^{2} + 12 α - 21 = 0 \Rightarrow α_{1, 2} = \frac{- 12 \pm \sqrt{144 + 924}}{22}$

Because $α$ has to be positive, we have Now, minimum of expected value of Hamiltonian is:

${<H>}_{m i n} = \frac{h^{2}}{m a^{2}} - \frac{3 (α^{2} + 2 α + 3)}{6 + 11 α} = 1.058 \frac{h^{2}}{m a^{2}}$

Therefore

$a) E_{t h r e s h o l d} = \frac{π^{2}}{8} \frac{h^{2}}{m a^{2}} = 1.2337 \frac{h^{2}}{m a^{2}} b) {<H>}_{m i n} = 1.058 \frac{h^{2}}{m a^{2}}, which is less than E_{m i n}$

That means that particle won't escape to infinity.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Short Answer

Step by step solution

(a) Showing the lowest energy that can propagate off to infinity

(a) The ground state has energy less than Ethreshold

(b) The ground state

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Medical Physics

Engineering Physics

Physics of Motion

Measurements

Fields in Physics

Study anywhere. Anytime. Across all devices.