Question

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

${\mathit{E}}_{\mathbf{t}\mathbf{h}\mathbf{r}\mathbf{e}\mathbf{s}\mathbf{h}\mathbf{o}\mathbf{l}\mathbf{d}}\mathbf{=}\frac{{\mathbf{\pi }}^{\mathbf{2}}{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{8}\mathbf{m}{\mathbf{a}}^{\mathbf{2}}}$

any solution with energy less than that has to be a bound state. Hint: Go way out one arm (say $\mathit{x}\mathbf{\gg }\mathit{a}$), 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$\mathbf{exp}\mathbf{\left(}\mathit{i}{\mathit{k}}_{\mathbf{x}}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{with}\mathbf{}{\mathit{k}}_{\mathbf{x}}\mathbf{>}\mathbf{0}$

(b) Now use the variation principle to show that the ground state has energy less than ${\mathit{E}}_{\mathbf{t}\mathbf{h}\mathbf{r}\mathbf{e}\mathbf{s}\mathbf{h}\mathbf{o}\mathbf{l}\mathbf{d}}$. Use the following trial wave function (suggested by Jim Mc Tavish):

$\mathit{\psi }\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{=}\mathit{A}\{\begin{array}{cc}\left[\cos (\mathrm{\pi x}/2a)+\cos (\mathrm{\pi y}/2a)\right]e^{-\alpha }& \left|x\right|\le a\mathrm{and}\left|y\right|\le a\\ \cos (\pi x/2a)e^{-\alpha }\left|y\right|/a& \left|x\right|\le a\mathrm{and}\left|y\right|>a\\ \cos (\pi y/2a)e^{-\alpha }\left|y\right|/a& \left|x\right|.a\mathrm{and}\left|y\right|\le a\\ 0& \mathrm{elsewhere}\end{array}\begin{array}{}\end{array}$

Normalize it to determine A, and calculate the expectation value of H.
Answer:

$<H>\mathbf{=}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{m}{\mathbf{a}}^{\mathbf{2}}}\left[\frac{{\pi }^2}{8}-\left(\frac{1-(\alpha /4)}{1+(8/{\pi }^2)+(1/2\alpha )}\right)\right]$

Now minimize with respect to α, and show that the result is less than${\mathit{E}}_{\mathbf{t}\mathbf{h}\mathbf{r}\mathbf{e}\mathbf{s}\mathbf{h}\mathbf{o}\mathbf{l}\mathbf{d}}$. 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

Step-by-step solution

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

The lowest Energy that can propagate off to infinity

$-\frac{{\sout{h}}^2}{2m}\left(\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}\right)=E\psi$

Let

localid="1658386907422" $$\begin{gathered} \psi (x,y)=X\left(x\right)Y\left(y\right).Y\frac{d^{2}X}{d{x}^2}+X\frac{d^{2}Y}{d{y}^2} \\ =-\frac{2mE}{{\sout{h}}^2}XY;\frac{1}{X}\frac{d^{2}X}{d{x}^2}+\frac{1}{Y}\frac{d^{2}Y}{d{y}^2} \\ =-\frac{2mE}{{\sout{h}}^2} \\ \frac{d^{2}X}{d{x}^2}=-K_x^{2}X;\frac{d^{2}Y}{d{y}^2} \\ =-k_y^{2}Y \\ {k}_x^2+k_y^2=\frac{2mE}{{\sout{h}}^2} \end{gathered}$$

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

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

So $E\ge \frac{{\sout{h}}^2}{2m}\left(k_x^2+\frac{{\mathrm{\pi }}^2}{4a^2}\right)$.

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

(a) The ground state has energy less than Ethreshold

Here’s a graph of $\psi (x,y)$:

$$\begin{gathered} a=0.9 \\ 0.9 \\ g\left[x_{-},y_{-}\right]=Piecewise(Cos\left[\mathrm{\pi x}/2\right]+Cos\left[\mathrm{\pi y}/2\right])e^{^}(-a),Abs\left[x\right]<1Abs\left[y\right]<1, \\ \left(\cos \left[\mathrm{\pi x}/2\right]e^{^}\right(-aAbs\left[y\right]),Abs\left[x\right]<1Abs\left[y\right]>1, \\ Plot3D[g\left[x,y\right],\left\{x,-4,4\right\},\left\{y,-4,4\right\}, \\ PlotRange\to \left\{\theta ,085\right\}]. \end{gathered}$$

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

$$\begin{gathered} j\left[x_{-}\right]:=g\left[x,.5\right]: \\ =g\left[x,.5\right].\mathrm{Plot}\left[\mathrm{j}\left[\mathrm{x}\right]\right],\left\{\mathrm{x},.8,1,2\right\}]. \end{gathered}$$

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

$$\begin{gathered} l_{ii}=A^2{\int }_{x-a}^{\infty }{\int }_{y-0}^a{\cos }^2\left(\frac{\mathrm{\pi y}}{2a}\right)e^{-2\alpha x/a}dxdy. \\ {l}_i=\frac{1}{2}{A}^2{\int }_{x-a}^{\infty }{\int }_{y-0}^a{\left[\cos \left(\frac{\mathrm{\pi x}}{2a}\right)+\cos \left(\frac{\pi y}{2a}\right)\right]}^2{e}^{-2\alpha }dxdy. \\ =\frac{1}{2}{A}^2{e}^{-2\alpha }\left\{2{\int }_0^a{\cos }^2\left(\frac{\mathrm{\pi x}}{2a}\right)dx{\int }_0^{a}dy+2{\int }_0^a\cos \left(\frac{\mathrm{\pi x}}{2a}\right)dx{\int }_0^a\cos \left(\frac{\mathrm{\pi x}}{2a}\right)dy\right\}. \\ =A^2{e}^{-2\alpha }\left\{\left(\frac{a}{2}\right)a+{\left[\frac{2a}{\mathrm{\pi }}\sin {\overline{)\left(\frac{\mathrm{\pi x}}{2a}\right)}}_0^a\right]}^2\right\}=A^2\frac{a^2}{2}{e}^{-2\alpha }\left(1+\frac{8}{{\mathrm{\pi }}^2}\right). \end{gathered}$$

Normalizing:

$$\begin{gathered} 1=8\left(l_l+l_n\right)=8A^2{a}^2{e}^{-2\alpha }\left[\frac{1}{4\alpha }+\frac{1}{2}\left(1+\frac{8}{{\mathrm{\pi }}^2}\right)\right]=4A^2{a}^2{e}^{-2\alpha }\left(1+\frac{8}{{\mathrm{\pi }}^2}+\frac{1}{2\alpha }\right) \\ \mathrm{So},A^2=\frac{e^{2\alpha }}{4a^2\left(1+\frac{8}{{\mathrm{\pi }}^2}+\frac{1}{2\alpha }\right)}. \end{gathered}$$

Next, ignoring the roof-lines for the moment, we calculate , where

$$\begin{gathered} J_{ii}=A^2{\int }_{x-a}^{\infty }{\int }_{y-0}^a\left[\cos \left(\frac{\mathrm{\pi y}}{2a}\right)e^{-\alpha x/a}\right]\left(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}\right)\left[\cos \left(\frac{\mathrm{\pi y}}{2a}\right)e^{-\alpha x/a}\right]dxdy \\ =\left[{\left(\frac{\alpha }{a}\right)}^2-{\left(\frac{\mathrm{\pi }}{2a}\right)}^2\right]l_{ll}=A^2\frac{1}{4\alpha }\left({\alpha }^2-\frac{{\mathrm{\pi }}^2}{4}\right)e^{-2\alpha }. \\ {J}_l=\frac{1}{2}{A}^2{\int }_{x-a}^{\infty }{\int }_{y-0}^a\left[\cos \left(\frac{\pi x}{2a}\right)+\cos \left(\frac{\mathrm{\pi y}}{2a}\right)\right]e^{-\alpha }\left(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}\right)\left[\cos \left(\frac{\pi x}{2a}\right)+\cos \left(\frac{\mathrm{\pi y}}{2a}\right)\right]e^{-\alpha }dxdy \\ =-\frac{{\mathrm{\pi }}^2}{4a^2}{l}_l=-A^2\left(1+\frac{{\mathrm{\pi }}^2}{8}\right)e^{-2\alpha } \end{gathered}$$

So far them,

Now the roof-lines, along the four sides of the central square: , where

Putting it all together,

Minimizing:

(Note that α must be positive)

(b) The ground state

Putting all together.

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

Therefore

That means that particle won't escape to infinity.