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.

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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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