Book edition 2nd Edition

Author(s) Randy Harris

Pages 633 pages

ISBN 9780805303087

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Short Answer

Question: The 2D Infinite Well: In two dimensions the Schrödinger equation is
$(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}}) ψ (x, y) = \frac{- 2 m (E - U)}{h^{2}} ψ (x, y)$
(a) Given that U is a constant, separate variables by trying a solution of the form $ψ (x, y) = f (x) g (y)$ , then dividing by $f (x) g (y)$ . Call the separation constants C_X and C_Y .
(b) For an infinite well
role="math" localid="1659942086972" $U = {0 0 < x < L, 0 < y < L \infty o t h e r w i s e$
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 C_X and C_Y positive, negative or zero? Imposing appropriate conditions find the allowed values of C_X and C_Y .
(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?

Answer:

(a) $C_{x} + C_{y} = \frac{2 m (E - U)}{ℏ^{2}}$

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

$f (x) = A \sin (\sqrt{C_{x}} x) + B \cos (\sqrt{C_{x}} x) g (y) = C \sin (\sqrt{C_{y}} y) + D \cos (\sqrt{C_{y}} y)$

Allowed values of the constants are

$C_{x} = {(\frac{n_{x} π}{L})}^{2} n_{x} = 1, 2, 3, . . . C_{y} = {(\frac{n_{y} π}{L})}^{2} n_{y} = 1, 2, 3, . . .$

(d) The allowed energy values are $E = \frac{h^{2} π^{2}}{2 m L^{2}} (n_{x}^{2} + n_{y}^{2})$

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The Schrodinger equation in two dimensions is

$(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}}) ψ (x, y) = \frac{- 2 m (E - U)}{h^{2}} ψ (x, y) \cdot \cdot \cdot \cdot \cdot \cdot (1)$ $(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}}) ψ (x, y) = \frac{- 2 m (E - U)}{h^{2}} ψ (x, y) \cdot \cdot \cdot \cdot \cdot \cdot (1)$

The potential of a 2D infinite well is

$U = \{0 0 < x < L, 0 < y < L \infty otherwise$

Step 2: Solution to differential equation

The general solution to the differential equation

$\frac{d^{2} X}{d x^{2}} = - k^{2} X \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)$

is of the form

$X = A \sin (k x) + B \cos (k x) \cdot \cdot \cdot \cdot \cdot \cdot (3)$ $X = A \sin (k x) + B \cos (k x) \cdot \cdot \cdot \cdot \cdot \cdot (3)$

where A and B are constants

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

Let

$ψ (x, y) = f (x) g (y)$

Substitute this in equation (1) to get

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

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

$\frac{1}{f (x)} \frac{\partial^{2} f (x)}{\partial x^{2}} + \frac{1}{g (y)} \frac{\partial^{2} g (y)}{\partial y^{2}} = \frac{- 2 m (E - U)}{h^{2}}$ $\frac{1}{f (x)} \frac{\partial^{2} f (x)}{\partial x^{2}} = - C_{x} \cdot \cdot \cdot \cdot \cdot \cdot (4)$

Call

$\frac{1}{f (x)} \frac{\partial^{2} f (x)}{\partial x^{2}} = - C_{x} \cdot \cdot \cdot \cdot \cdot \cdot (4)$

and

$\frac{1}{g (y)} \frac{\partial^{2} g (y)}{\partial y^{2}} = - C_{y} \cdot \cdot \cdot \cdot \cdot \cdot (5)$

Finally, $C_{x} + C_{y} = \frac{2 m (E - U)}{ℏ^{2}} \cdot \cdot \cdot \cdot \cdot \cdot (6)$

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

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

$f (x) = A \sin (\sqrt{C_{x}} x) + B \cos (\sqrt{C_{x}} x) g (y) = C \sin (\sqrt{C_{y}} y) + D \cos (\sqrt{C_{y}} y)$

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

The complete wave function becomes

$ψ (x, y) = \{A \sin (\sqrt{C_{x}} x) + B \cos (\sqrt{C_{x}} x)\} \{C \sin (\sqrt{C_{y}} y) + D \cos (\sqrt{C_{y}} y)\}$

The first set of boundary conditions are

$ψ (x, 0) = 0 ψ (0, y) = 0$

Substitute these to get

$ψ (x, 0) = \{A \sin (\sqrt{C_{x}} x) + B \cos (\sqrt{C_{x}} x)\} D = 0 D = 0$

and

$ψ (0, y) = B \{C \sin (\sqrt{C_{y}} y) + D \cos (\sqrt{C_{y}} y)\} = 0 B = 0$

Combine C to A and get

$ψ (x, y) = A \sin (\sqrt{C_{x}} x) \sin (\sqrt{C_{y}} y)$

The second set of boundary conditions are

$ψ (x, L) = 0 ψ (L, y) = 0$

The first one gives

$A \sin (\sqrt{C_{x}} x) \sin (\sqrt{C_{y}} L) = 0 \sin (\sqrt{C_{y}} L) = 0 \sqrt{C_{y}} L = n_{y} π n_{y} = 1, 2 . . . . \sqrt{C_{y}} = \frac{n_{y} π}{L} C_{y} = {(\frac{n_{y} π}{L})}^{2}$

The second one gives

$A \sin (\sqrt{C_{x}} L) \sin (\sqrt{C_{y}} y) = 0 \sin (\sqrt{C_{x}} L) = 0 \sqrt{C_{x}} L = n_{x} π n_{x} = 1, 2 . . . . \sqrt{C_{x}} = \frac{n_{x} π}{L} C_{x} = {(\frac{n_{x} π}{L})}^{2}$

The final wave function becomes

$ψ (x, y) = A \sin (\frac{n_{x} π x}{L}) \sin (\frac{n_{y} π y}{L})$

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

$f (x) = A \sin (\sqrt{C_{x}} x) + B \cos (\sqrt{C_{x}} x) g (y) = C \sin (\sqrt{C_{y}} y) + D \cos (\sqrt{C_{y}} y)$

Allowed values of the constants are

$C_{x} = {(\frac{n_{x} π}{L})}^{2} n_{x} = 1, 2, 3, . . C_{y} = {(\frac{n_{y} π}{L})}^{2} n_{y} = 1, 2, 3, . .$

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

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

$C_{x} + C_{y} = \frac{2 m E}{h^{2}} \frac{n_{x}^{2} π^{2}}{L^{2}} + \frac{n_{y}^{2} π^{2}}{L^{2}} = \frac{2 m E}{h^{2}} E = \frac{h^{2} π^{2}}{2 m L^{2}} (n_{x}^{2} + n_{y}^{2})$

Thus the allowed values of energy are $\frac{h^{2} π^{2}}{2 m L^{2}} (n_{x}^{2} + n_{y}^{2})$ with two independent quantum numbers n_x and n_y .

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

The energies for which n_x and n_y 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 $ψ (x, y) = A \sin (\frac{π x}{L}) \sin (\frac{2 π y}{L})$ and $ψ (x, y) = A \sin (\frac{2 π x}{L}) \sin (\frac{π y}{L})$ have quantum numbers (1 , 2) and (2 ,1) but they both have energy $\frac{ℏ^{2} π^{2}}{2 m L^{2}} (1^{2} + 2^{2}) = \frac{5 ℏ^{2} π^{2}}{2 m L^{2}}$ .

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Modern Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Solution to differential equation

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

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

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

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

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Modern Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Solution to differential equation

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

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

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

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

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