Q20P Quantum dots. Consider a particl... [FREE SOLUTION]

Chapter 7: Q20P (page 313)

Quantum dots. Consider a particle constrained to move in two dimensions in the cross-shaped region.The 鈥渁rms鈥� 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
$E_{t h r e s h o l d} = \frac{蟺^{2} h^{2}}{8 m a^{2}}$
any solution with energy less than that has to be a bound state. Hint: Go way out one arm (say $x 鈮� a$ ), and solve the Schr枚dinger 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):
$蠄 (x, y) = A {\begin{matrix} [\cos (蟺虫 / 2 a) + \cos (蟺测 / 2 a)] e^{- 伪} & |x| 鈮� a and |y| 鈮� a \\ \cos (蟺 x / 2 a) e^{- 伪} |y| / a & |x| 鈮� a and |y| > a \\ \cos (蟺 y / 2 a) e^{- 伪} |y| / a & |x| . a and |y| 鈮� a \\ 0 & elsewhere \end{matrix} \begin{array}{l} \end{array}$
Normalize it to determine A, and calculate the expectation value of H.
Answer:
$< H > = \frac{h^{2}}{m a^{2}} [\frac{蟺^{2}}{8} - (\frac{1 - (伪 / 4)}{1 + (8 / 蟺^{2}) + (1 / 2 伪)})]$
Now minimize with respect to 伪, 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鈥� 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鈥攖here are 鈥渞oof-lines鈥� at the joins, and you will need to exploit the technique of Example 8.3.

Short Answer

Expert verified

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

Step by step solution

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

The lowest Energy that can propagate off to infinity

$- \frac{h^{2}}{2 m} (\frac{{鈭�}^{2} 蠄}{鈭� x^{2}} + \frac{{鈭�}^{2} 蠄}{鈭� y^{2}}) = E 蠄$

Let

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

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" $Y (卤 a) = 0 yield k_{y} = \frac{n 蟺}{2 a}$ with minimum $\frac{蟺}{2 a}$ .

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

So $E 鈮� \frac{h^{2}}{2 m} (k_{x}^{2} + \frac{蟺^{2}}{4 a^{2}})$ .

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

(a) The ground state has energy less than Ethreshold

Here鈥檚 a graph of $蠄 (x, y)$ :

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

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 ${|蠄 (x, y)|}^{2}$ 2 over regions I and II (in the figure), and multiply by 8.

$l_{i i} = A^{2} {鈭�}_{x - a}^{鈭�} {鈭�}_{y - 0}^{a} \cos^{2} (\frac{蟺测}{2 a}) e^{- 2 伪 x / a} d x d y . l_{i} = \frac{1}{2} A^{2} {鈭�}_{x - a}^{鈭�} {鈭�}_{y - 0}^{a} {[\cos (\frac{蟺虫}{2 a}) + \cos (\frac{蟺 y}{2 a})]}^{2} e^{- 2 伪} d x d y . = \frac{1}{2} A^{2} e^{- 2 伪} \{2 {鈭�}_{0}^{a} \cos^{2} (\frac{蟺虫}{2 a}) d x {鈭�}_{0}^{a} d y + 2 {鈭�}_{0}^{a} \cos (\frac{蟺虫}{2 a}) d x {鈭�}_{0}^{a} \cos (\frac{蟺虫}{2 a}) d y\} . = A^{2} e^{- 2 伪} \{(\frac{a}{2}) a + {[\frac{2 a}{蟺} \sin {(\frac{蟺虫}{2 a})}_{0}^{a}]}^{2}\} = A^{2} \frac{a^{2}}{2} e^{- 2 伪} (1 + \frac{8}{蟺^{2}}) .$

Normalizing:

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

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

$J_{i i} = A^{2} {鈭�}_{x - a}^{鈭�} {鈭�}_{y - 0}^{a} [\cos (\frac{蟺测}{2 a}) e^{- 伪 x / a}] (\frac{{鈭�}^{2}}{鈭� x^{2}} + \frac{{鈭�}^{2}}{鈭� y^{2}}) [\cos (\frac{蟺测}{2 a}) e^{- 伪 x / a}] d x d y = [{(\frac{伪}{a})}^{2} - {(\frac{蟺}{2 a})}^{2}] l_{l l} = A^{2} \frac{1}{4 伪} (伪^{2} - \frac{蟺^{2}}{4}) e^{- 2 伪} . J_{l} = \frac{1}{2} A^{2} {鈭�}_{x - a}^{鈭�} {鈭�}_{y - 0}^{a} [\cos (\frac{蟺 x}{2 a}) + \cos (\frac{蟺测}{2 a})] e^{- 伪} (\frac{{鈭�}^{2}}{鈭� x^{2}} + \frac{{鈭�}^{2}}{鈭� y^{2}}) [\cos (\frac{蟺 x}{2 a}) + \cos (\frac{蟺测}{2 a})] e^{- 伪} d x d y = - \frac{蟺^{2}}{4 a^{2}} l_{l} = - A^{2} (1 + \frac{蟺^{2}}{8}) e^{- 2 伪}$

So far them,

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

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

$J_{l l l} = {鈭�}_{x = a - 鈭�}^{a + \overset{藱}{o}} {鈭�}_{y = 0}^{a} 蠄 (x, y) (\frac{{鈭�}^{2}}{鈭� x^{2}}) 蠄 (x, y) d x d y . = A {鈭�}_{y = 0}^{a} 蠄 (a, y) \{\frac{鈭�}{鈭� x} {[\cos (\frac{蟺测}{2 a}) e^{- 伪 x / a}]}_{x = a}^{} - \frac{鈭�}{鈭� x} {[\cos (\frac{蟺测}{2 a}) e^{- 伪}]}_{x = a}\} d y . = A^{2} {鈭�}_{y = 0}^{a} [\cos (\frac{蟺测}{2 a}) e^{- 伪}] \{- \frac{伪}{a} \cos (\frac{蟺测}{2 a}) e^{- 伪} + \frac{蟺}{2 a} \sin (\frac{蟺}{2}) e^{- 伪}\} d y . = A^{2} \frac{1}{a} e^{- 2 伪} {鈭�}_{0}^{a} [\frac{蟺}{2} \cos (\frac{蟺测}{2 a}) - 伪 \cos^{2} (\frac{蟺测}{2 a})] d y = A^{2} e^{- 2 伪} (1 - \frac{伪}{2}) . {<H>}_{b} = - 8 \frac{h^{2}}{2 m} A^{2} e^{- 2 伪} (1 - \frac{伪}{2}) S 0, = \frac{h^{2}}{2 m} A^{2} e^{- 2 伪} 4 (伪 - 2)$

Putting it all together,

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

Minimizing:

$\frac{d <H>}{d 伪} = 0 鈬� \frac{d}{d 伪} (\frac{4 - 伪}{{(1 + (8 / 蟺^{2}) + (1 / 2 伪))}^{2}}) = 0 \frac{- 1}{1 + (8 / 蟺^{2}) + (1 / 2 伪)} - \frac{4 - 伪}{(1 + (8 / 蟺^{2}) + {(1 / 2 伪))}^{2}} (\frac{- 1}{2 伪^{2}}); 1 + (8 / 蟺^{2}) + (1 / 2 伪) - \frac{4 - 伪}{2 伪^{2}} = 0; 伪^{2} (1 + \frac{8}{蟺^{2}}) + 伪 - 2; 伪 = \frac{1 卤 \sqrt{1 + 8 [1 + (8 / 蟺^{2})]}}{2 [1 + (8 / 蟺^{2})]} = \frac{- 1 + \sqrt{9 + {(8 / 蟺)}^{2}}}{2 [1 + (8 / 蟺^{2})]} .$

(Note that 伪 must be positive)

(b) The ground state

Putting all together.

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

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

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

Therefore

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

That means that particle won't escape to infinity.

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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

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