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Question:Let V=0in the Schrodinger equation (3.22) and separate variables in 2-dimensional rectangular coordinates. Solve the problem of a particle in a 2-dimensional square box, 0<x<l,0<y<lThis means to find solutions of the Schrodinger equation which are 0 for x=0,x=I,y=0,y=I, that is, on the boundary of the box, and to find the corresponding energy eigenvalues. Comments: If we extend the idea of a 鈥減article in a box鈥 (see Section 3, Example 3) to two or three dimensions, the box in 2D might be a square (as in this problem) or a circle (Problem 8); in 3D it might be a cube (Problem 7.17) or a sphere (Problem 7.19). In all cases, the mathematical problem is to find solutions of the Schrodinger equation with V=0inside the box and =0on the boundary of the box, and to find the corresponding energy eigenvalues. In quantum mechanics, describes a particle trapped inside the box and the energy eigenvalues are the possible values of the energy of the particle.

Short Answer

Expert verified

The solution is (x,y,t)=n=1m=1sin苍蟺Ixsin尘蟺Iye-颈丑蟺2(n2+m2)2ml2t .

Step by step solution

01

Given Information.

An expression has been given as 0<x<1,0<y<1.

02

Definition of Schrodinger equation:

A linear partial differential equation that determines the wave function of a quantum-mechanical system is known as the Schr枚dinger equation.

The Schrodinger equation is -h22m2+痴蠄=iht.

03

Use Schrodinger equation:

Use the Schrodinger equation.

-h22m2+痴蠄=iht

Assume a solution of the form mentioned below.

x,y,t=(x,y)T(t)

The equation can be separated into a time equation with the solution given below.

Tt=e-iEth

It can be separated into a space equation (with V = 0 )

-h22m2=贰蠄

Put a solution of the form given below.

=X(x)Y(y)

Then divide byX(x)Y(y) .

1xd2Xdx2+1Yd2Ydy2=-2mEh2=-kx2-ky21xd2Xdx2+kx2+1Yd2dy2+ky2=0


The solutions are a trigonometric solutions.
X(x)=cos(kxx)sin(kxx)Y(y)=cos(kyy)sin(kyy)

04

Use boundary condition:

Y(0)=0

On the boundary=0 .
(0,y,t)=0(l,y,t)=0(x,0,t)=0(x,I,t)=0

x(0)=00=Csin(kx0)+Dcos(kx0)
D=0
X(I)=00=sin(kxI)
kxI=苍蟺kx=苍蟺I

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0=Esin(ky0)+Fcos(ky0)

F=0

Y(I)=0
0=sin(kyl)
kyl=尘蟺
ky=尘蟺l

05

Calculate Energy:

Calculate the energy.

nx2+ny2=2mEh2=苍蟺I2+尘蟺l2=2l2(n2+m2)

E=h22(n2+m2)2ml2

It can be seen that the energies are a degenerate. n and m can vary to get the same value.

Write the solution.
N=sin苍蟺lxsin尘蟺lye-iENthEN=h22(n2+m2)2ml2
(x,y,t)=n=1m=1sin苍蟺Ixsin尘蟺lye-ih2(n2+m2)2ml2t
Hence, the final solution is(x,y,t)=n=1m=1sin苍蟺Ixsin尘蟺lye-ih2(n2+m2)2ml2t

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