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Expert-verified Found in: Page 1 ### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087 # Question: The 2D Infinite Well: In two dimensions the Schrödinger equation is $\mathbf{\left(}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)}{\mathbit{\psi }}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}{\mathbf{=}}\frac{\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{\left(}\mathbf{E}\mathbf{-}\mathbf{U}\mathbf{\right)}}{{\mathbf{h}}^{\mathbf{2}}}{\mathbit{\psi }}\mathbf{\left(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\right)}$(a) Given that U is a constant, separate variables by trying a solution of the form ${\mathbit{\psi }}\left(x,y\right){\mathbf{=}}{\mathbit{f}}\left(x\right){\mathbit{g}}\left(y\right)$, then dividing by${\mathbit{f}}\left(x\right){\mathbit{g}}\left(y\right)$ . Call the separation constants CX and CY .(b) For an infinite well role="math" localid="1659942086972" ${\mathbit{U}}{\mathbf{=}}\mathbf{\left\{}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{L}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{<}\mathbf{y}\mathbf{<}\mathbf{L}\phantom{\rule{0ex}{0ex}}\mathbf{\infty }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{o}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{r}\mathbf{w}\mathbf{i}\mathbf{s}\mathbf{e}\phantom{\rule{0ex}{0ex}}$What should f(x) and g(y) be outside the well? What functions should be acceptable standing wave solutions f(x) for g(y) and inside the well? Are CX and CY positive, negative or zero? Imposing appropriate conditions find the allowed values of CX and CY .(c) How many independent quantum numbers are there?(d) Find the allowed energies E .(e)Are there energies for which there is not a unique corresponding wave function?

(a) ${C}_{x}+{C}_{y}=\frac{2m\left(E-U\right)}{{\hslash }^{2}}$

(b)The functions and are zero outside the wall. Inside the wall the general solutions are

$f\left(x\right)=A\mathrm{sin}\left(\sqrt{{C}_{x}}x\right)+B\mathrm{cos}\left(\sqrt{{C}_{x}}x\right)\phantom{\rule{0ex}{0ex}}g\left(y\right)=C\mathrm{sin}\left(\sqrt{{C}_{y}}y\right)+D\mathrm{cos}\left(\sqrt{{C}_{y}}y\right)\phantom{\rule{0ex}{0ex}}$

Allowed values of the constants are

${C}_{x}={\left(\frac{{n}_{x}\pi }{L}\right)}^{2}{n}_{x}=1,2,3,...\phantom{\rule{0ex}{0ex}}{C}_{y}={\left(\frac{{n}_{y}\pi }{L}\right)}^{2}{n}_{y}=1,2,3,...\phantom{\rule{0ex}{0ex}}$

(c) There are two independent quantum numbers nx and ny .

(d) The allowed energy values are $E=\frac{{h}^{2}{\pi }^{2}}{2m{L}^{2}}\left({n}_{x}^{2}+{n}_{y}^{2}\right)$

(e) The energies for which nx and ny are not equal have no unique corresponding wave functions.

See the step by step solution

## Step 1: Given data

The Schrodinger equation in two dimensions is

$\left(\frac{{\partial }^{2}}{\partial {x}^{2}}+\frac{{\partial }^{2}}{\partial {y}^{2}}\right)\psi \left(x,y\right)=\frac{-2m\left(E-U\right)}{{h}^{2}}\psi \left(x,y\right)······\left(1\right)$$\left(\frac{{\partial }^{2}}{\partial {x}^{2}}+\frac{{\partial }^{2}}{\partial {y}^{2}}\right)\psi \left(x,y\right)=\frac{-2m\left(E-U\right)}{{h}^{2}}\psi \left(x,y\right)······\left(1\right)$

The potential of a 2D infinite well is

$U=\left\{00

## Step 2: Solution to differential equation

The general solution to the differential equation

$\frac{{d}^{2}X}{d{x}^{2}}=-{k}^{2}X·······\left(2\right)$

is of the form

$X=A\mathrm{sin}\left(kx\right)+B\mathrm{cos}\left(kx\right)······\left(3\right)$ $X=A\mathrm{sin}\left(kx\right)+B\mathrm{cos}\left(kx\right)······\left(3\right)$

where A and B are constants

## Step 3(a): Determining the general solution using separation of variables

Let

$\psi \left(x,y\right)=f\left(x\right)g\left(y\right)$

Substitute this in equation (1) to get

$\left(\frac{{\partial }^{2}}{\partial {x}^{2}}+\frac{{\partial }^{2}}{\partial {y}^{2}}\right)f\left(x\right)g\left(y\right)=\frac{-2m\left(E-U\right)}{{h}^{2}}f\left(x\right)g\left(y\right)\phantom{\rule{0ex}{0ex}}g\left(y\right)\frac{{\partial }^{2}f\left(x\right)}{\partial {x}^{2}}+f\left(x\right)\frac{{\partial }^{2}g\left(y\right)}{\partial {y}^{2}}=\frac{-2m\left(E-U\right)}{{h}^{2}}f\left(x\right)g\left(y\right)\phantom{\rule{0ex}{0ex}}$$\left(\frac{{\partial }^{2}}{\partial {x}^{2}}+\frac{{\partial }^{2}}{\partial {y}^{2}}\right)f\left(x\right)g\left(y\right)=\frac{-2m\left(E-U\right)}{{h}^{2}}f\left(x\right)g\left(y\right)\phantom{\rule{0ex}{0ex}}g\left(y\right)\frac{{\partial }^{2}f\left(x\right)}{\partial {x}^{2}}+f\left(x\right)\frac{{\partial }^{2}g\left(y\right)}{\partial {y}^{2}}=\frac{-2m\left(E-U\right)}{{h}^{2}}f\left(x\right)g\left(y\right)\phantom{\rule{0ex}{0ex}}$

Divide throughout by f(x) g(y) to get

$\frac{1}{f\left(x\right)}\frac{{\partial }^{2}f\left(x\right)}{\partial {x}^{2}}+\frac{1}{g\left(y\right)}\frac{{\partial }^{2}g\left(y\right)}{\partial {y}^{2}}=\frac{-2m\left(E-U\right)}{{h}^{2}}$$\frac{1}{f\left(x\right)}\frac{{\partial }^{2}f\left(x\right)}{\partial {x}^{2}}=-{C}_{x}······\left(4\right)$

Call

$\frac{1}{f\left(x\right)}\frac{{\partial }^{2}f\left(x\right)}{\partial {x}^{2}}=-{C}_{x}······\left(4\right)$

and

$\frac{1}{g\left(y\right)}\frac{{\partial }^{2}g\left(y\right)}{\partial {y}^{2}}=-{C}_{y}······\left(5\right)$

Finally, ${C}_{x}+{C}_{y}=\frac{2m\left(E-U\right)}{{\hslash }^{2}}······\left(6\right)$

Hence, the solution is ${C}_{x}+{C}_{y}=\frac{2m\left(E-U\right)}{{h}^{2}}$

## Step 4(b): Determining the solution for an infinite well

Since the well is infinite there should be no probability of any particle of staying outside it. Hence f(x) and g(y) should be 0 outside the wall.

The equations (IV) and (V) are of the form (II). Their general solutions inside the well are of the form of equation (III) as follows

$f\left(x\right)=A\mathrm{sin}\left(\sqrt{{C}_{x}}x\right)+B\mathrm{cos}\left(\sqrt{{C}_{x}}x\right)\phantom{\rule{0ex}{0ex}}g\left(y\right)=C\mathrm{sin}\left(\sqrt{{C}_{y}}y\right)+D\mathrm{cos}\left(\sqrt{{C}_{y}}y\right)\phantom{\rule{0ex}{0ex}}$

Here A and B are functions of x and C and D are functions of y. In general Cx and CY can be positive, negative or zero.

The complete wave function becomes

$\psi \left(x,y\right)=\left\{A\mathrm{sin}\left(\sqrt{{C}_{x}}x\right)+B\mathrm{cos}\left(\sqrt{{C}_{x}}x\right)\right\}\left\{C\mathrm{sin}\left(\sqrt{{C}_{y}}y\right)+D\mathrm{cos}\left(\sqrt{{C}_{y}}y\right)\right\}$

The first set of boundary conditions are

$\psi \left(x,0\right)=0\phantom{\rule{0ex}{0ex}}\psi \left(0,y\right)=0$

Substitute these to get

$\psi \left(x,0\right)=\left\{A\mathrm{sin}\left(\sqrt{{C}_{x}}x\right)+B\mathrm{cos}\left(\sqrt{{C}_{x}}x\right)\right\}D\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}D=0\phantom{\rule{0ex}{0ex}}$

and

$\psi \left(0,y\right)=B\left\{C\mathrm{sin}\left(\sqrt{{C}_{y}}y\right)+D\mathrm{cos}\left(\sqrt{{C}_{y}}y\right)\right\}\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}B=0$

Combine C to A and get

$\psi \left(x,y\right)=A\mathrm{sin}\left(\sqrt{{C}_{x}}x\right)\mathrm{sin}\left(\sqrt{{C}_{y}}y\right)$

The second set of boundary conditions are

$\psi \left(x,L\right)=0\phantom{\rule{0ex}{0ex}}\psi \left(L,y\right)=0$

The first one gives

$A\mathrm{sin}\left(\sqrt{{C}_{x}}x\right)\mathrm{sin}\left(\sqrt{{C}_{y}}L\right)=0\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(\sqrt{{C}_{y}}L\right)=0\phantom{\rule{0ex}{0ex}}\sqrt{{C}_{y}}L={n}_{y}\pi {n}_{y}=1,2....\phantom{\rule{0ex}{0ex}}\sqrt{{C}_{y}}=\frac{{n}_{y}\pi }{L}\phantom{\rule{0ex}{0ex}}{C}_{y}={\left(\frac{{n}_{y}\pi }{L}\right)}^{2}\phantom{\rule{0ex}{0ex}}$

The second one gives

$A\mathrm{sin}\left(\sqrt{{C}_{x}}L\right)\mathrm{sin}\left(\sqrt{{C}_{y}}y\right)=0\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(\sqrt{{C}_{x}}L\right)=0\phantom{\rule{0ex}{0ex}}\sqrt{{C}_{x}}L={n}_{x}\pi {n}_{x}=1,2....\phantom{\rule{0ex}{0ex}}\sqrt{{C}_{x}}=\frac{{n}_{x}\pi }{L}\phantom{\rule{0ex}{0ex}}{C}_{x}={\left(\frac{{n}_{x}\pi }{L}\right)}^{2}\phantom{\rule{0ex}{0ex}}$

The final wave function becomes

$\psi \left(x,y\right)=A\mathrm{sin}\left(\frac{{n}_{x}\pi x}{L}\right)\mathrm{sin}\left(\frac{{n}_{y}\pi y}{L}\right)$

Hence, The functions and are zero outside the wall. Inside the wall the general solutions are

$f\left(x\right)=A\mathrm{sin}\left(\sqrt{{C}_{x}}x\right)+B\mathrm{cos}\left(\sqrt{{C}_{x}}x\right)\phantom{\rule{0ex}{0ex}}g\left(y\right)=C\mathrm{sin}\left(\sqrt{{C}_{y}}y\right)+D\mathrm{cos}\left(\sqrt{{C}_{y}}y\right)$

Allowed values of the constants are

${C}_{x}={\left(\frac{{n}_{x}\pi }{L}\right)}^{2}{n}_{x}=1,2,3,..\phantom{\rule{0ex}{0ex}}{C}_{y}={\left(\frac{{n}_{y}\pi }{L}\right)}^{2}{n}_{y}=1,2,3,..$

## Step 5(c) and (d): Determining the quantum numbers and allowed energies

From equation (VI) and the value of potential inside the well

${C}_{x}+{C}_{y}=\frac{2mE}{{h}^{2}}\phantom{\rule{0ex}{0ex}}\frac{{n}_{x}^{2}{\pi }^{2}}{{L}^{2}}+\frac{{n}_{y}^{2}{\pi }^{2}}{{L}^{2}}=\frac{2mE}{{h}^{2}}\phantom{\rule{0ex}{0ex}}E=\frac{{h}^{2}{\pi }^{2}}{2m{L}^{2}}\left({n}_{x}^{2}+{n}_{y}^{2}\right)$

Thus the allowed values of energy are $\frac{{h}^{2}{\pi }^{2}}{2m{L}^{2}}\left({n}_{x}^{2}+{n}_{y}^{2}\right)$ with two independent quantum numbers nx and ny .

## Step 6(e): Determine there are energies for which there is not a unique corresponding wave function

The energies for which nx and ny are not equal have no unique corresponding wave functions.

Separate set of quantum numbers that have the same ${n}_{x}^{2}+{n}_{y}^{2}$ have the same energy. For example the wave functions $\psi \left(x,y\right)=A\mathrm{sin}\left(\frac{\pi x}{L}\right)\mathrm{sin}\left(\frac{2\pi y}{L}\right)$ and $\psi \left(x,y\right)=A\mathrm{sin}\left(\frac{2\pi x}{L}\right)\mathrm{sin}\left(\frac{\pi y}{L}\right)$ have quantum numbers (1 , 2) and (2 ,1) but they both have energy $\frac{{\hslash }^{2}{\pi }^{2}}{2m{L}^{2}}\left({1}^{2}+{2}^{2}\right)=\frac{5{\hslash }^{2}{\pi }^{2}}{2m{L}^{2}}$ . ### Want to see more solutions like these? 