1

Quantum box with infinite well potential Masatsugu Sei Suzuki

Department of Physics, State University of New York at Binghamton (Date: December 02, 2013)

1. 1D one-dimensional well potential

x

Vx

E

V0=¶

0 a

I II III

m

pH

2

ˆˆ2

)(2

)()(2

)(22

2

22

xm

kxEx

dx

d

mxH

The solution of this equation is

)cos()sin()( kxBkxAx where

m

kE

2

22

Using the boundary condition:

0)()0( axx

2

we have

B = 0 and A≠0.

0)sin( ka

nka (n = 1, 2, …) Note that n = 0 is not included in our solution because the corresponding wave function becomes zero. The wave function is given by

)sin(2

)sin()(a

xn

aa

xnAxx nnn

with

22

2

a

n

mEn

((Normalization))

2

0

22

2)(sin1 n

a

n Aa

dxa

xnA

2. Mathematica

)(sin2

)]sin(2

[)( 222

a

xn

aa

xn

axn

3

n=1n=2n=3

0.2 0.4 0.6 0.8 1.0x

0.5

1.0

1.5

2.0

y 2

Fig. Plot of

2)(xn with a = 1, as a function of x. n = 1 (red), 2 (yellow), 3 (green), 4

(blue), and 5 (dark blue). There are n peaks for the state n .

The expectation values and uncertainty

a

nma

nm

nm dx

a

xnx

adxxxxx

00

* )(sin2

)()(

a

n

m

nm dxx

xixp

0

* )()(

Since

0x , )3

2(6 22

22

n

ax

0p , 2

2222

a

np

we have

)3

2(6

122

22

naxxx

4

a

nppp

22

Then

)3

2(6

122

n

nxp

When n = 1,

xp = 1.67029 ħ>0.5 ħ ((Mathematica))

5

Clear"Global`"; n_, x_ :2

aSinn x

a;

xav1 0

ax n, x2 x Simplify, n Integers &

a

2

xav2 0

ax2 n, x2 x Simplify, n Integers &

1

6a2 2

3

n2 2

pav1 —

0

an, x Dn, x, x x

Simplify, n Integers &

0

pav2 —

2

0

an, x Dn, x, x, 2 x

Simplify, n Integers &

n2 2 —2

a2

xp xav2 pav2 Simplify, — 0, a 0 &

n2 3n2 2 2 —

6

xp

—. n 1 N

1.67029 3. Exercise: Townsend 6-16 problem

A particle of mass m is in lowest energy (ground) state of the infinite potential energy well

0)( xV for 0<x<L and ∞ elsewhere.

6

At time t = 0, the wall located at x = L is suddenly pulled back to a position at x = 2 L. This change occurs so rapidly that instantaneously the wave function does not change. (a) Calculate the probability that a measurement of the energy will yield the ground-

state energy of the new well. What is the probability that a measurement of the energy will yield the first excited energy of the new well?

(b) Describe the procedure you would use to determine the time development of the system. Is the system in a stationary state?

((Solution)) The old wave function of the ground state is given by

)sin(2

)(1 a

x

ax

only for 0<x<a (0 otherwise).

The new wave function is given by

)2

sin(1

)2

sin(2

2)()(

a

xn

aa

xn

axn

new

with the energy of

22)(

22

a

n

mE n

ewn

(a)

n

nnewn xcx )()( )(

1

)2

sin()4(

24

sin)2

sin(2

)()(

2

0

2

0

1

*)(

n

n

dxa

x

a

xn

a

dxxxc

a

an

newn

Note that

2

12 c

7

360253.09

322

2

1

c , 5.02

12

2 c

(b) The system is not stationary since )0( t is not an eigenstate of the new Hamiltonian

newH , but is a superposition of the eigenstates )(nnew with various kinds of n.

n

nnewnct )(

1)0(

n

nnew

nnewn

n

nnewnewn

new

tEi

c

tHi

c

ttHi

t

)()(

)(

)exp(

)ˆexp(

)0()ˆexp()(

or

n

nnew

nnewn xtE

ictxtx )()exp()(),( )()(

where cn are determined as in (a)

)2

sin()4(

242

n

ncn

and

22)(

22

a

n

mE n

ewn

,

)2

sin(1

)2

sin(2

2)()(

a

xn

aa

xn

axn

new

Then we get

8

mn

nnew

nnew

nnew

mnewnm

n

nnew

nnewn

m

mnew

mnewm

tEEi

xxcc

xtEi

cxtEi

ctx

,

)()()(*)(*

)()(*)()(*2

])(exp[)()(

)()exp()()exp(),(

((Mathematica))

We use m = ħ = 1. a = 1. Red (At t = 0). The Plot of 2

),( tx as a function of x (0<x<2a),

where t is changed as parameter; t = 0 - 3 with t = 0.1. The summation over n (n = 1 – 10). (a) t = 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9

(b) t = 1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9

9

(c) t = 2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9

10

4. 2D well potential

Next we consider a particle in a 2D well potential The potential:

V(x,y) = 0 for 0≤x≤a and 0≤y≤a. V(x,y) = ∞ otherwise.

),(2

),(),()(2

),(22

2

2

2

22

yxm

kyxEyx

dy

d

dx

d

myxH

)(2

222

yx kkm

E

),()(),()( 22

2

2

2

2

yxkkyxdy

d

dx

dyx

11

We use the method of the separation variables. Suppose that

)()(),( yYxXyx

)()(

)("

)(

)(" 22yx kk

yY

yY

xX

xX

We assume that

)()(" 2 xXkxX x

)()(" 2 yYkyyY

Using the boundary condition

0)()0( axXxX and

0)()0( ayYyY

Then we have

)sin()sin(2

),(

2

, a

yn

a

xn

ayx yx

nynx

5. Mathematica

12

A particle in a two dimensional box

Clear"Global`";

2

a

2

bSinn x

a Sin m y

b;

prb 2 . a 1, b 1;

p13D1 Plot3Dprb . n 4, m 4, x, 0, 1, y, 0, 1,

PlotPoints 100

cont1 ContourPlotprb . n 4, m 4, x, 0, 1,

y, 0, 1, PlotPoints 100

13

6. Standing wave solutions with a fixed boundary condition

We consider a free particle inside a box with length Lx, Ly, Lz along the x, y, and z axes, respectively. The Schrödinger equation of the system is given by

),,(),,(2

),,( 22

zyxEzyxm

zyxH

under the boundary condition;

0)0,,(),,(

0),0,(),,(

0),,0(),,(

zyxLzyx

zyxzLyx

zyxzyLx

zx

z

x

We use the method of separation variables. We assume that

)()()(),,( zZyYxXzyx with

0)()0( xLXX , 0)()0( yLYY , 0)()0( zLZZ

The substitution of the solution into the Schrödinger equation yields

2

2

)(

)(''

)(

)("

)(

)(''

mE

zZ

zZ

yY

yY

xX

xX

We assume that

2

)(

)(''xk

xX

xX , 2

)(

)(''yk

yY

yY , 2

)(

)(''zk

zZ

zZ

The solution of these differential equations can be obtained as a standing wave solution,

)sin()( xkxX x , )sin()( ykyY y , )sin()( zkzZ z

under the boundary conditions, where kx, ky, and kz are constants. The resulting wave function is

)sin()sin()sin(),,( zkykxkAzyx zyx

The condition that 0 at x = Lx requires that

14

x

xx L

nk

.

The values for the kx, ky, and kz are

x

xx L

nk

,

y

yy L

nk

,

z

zz L

nk

where nx, ny, and nz are positive integers. ((Mathematica)) ContourPlot3D

Fig. ContourPlot3D of constzkykxk zyx )(sin)(sin)(sin 222 in the 3D real space.

xL

nxk x

x

. y

L

nyk y

y

. z

L

nzk z

z

. L = 1 for simplicity. nx = 1, ny = 1, nz = 1

15

Fig. ContourPlot3D of constzkykxk zyx )(sin)(sin)(sin 222 in the 3D real space.

xL

nxk x

x

. y

L

nyk y

y

. z

L

nzk z

z

. L = 1 for simplicity. nx = 2, ny = 1, nz = 1

((Density of states))

)(2

)(22

),,(

2

2

2

2

2

222

2222

22

z

z

y

y

x

x

zyxzyx

L

n

L

n

L

n

m

kkkm

km

kkkE

.

There is one state per volume of the k-space;

zyx LLL

.

16

In the region of k - k + dk, the number of states is

dmV

dkkV

LLL

dkkdD

zyx

2/3

22

23

3

2

2

2

4)2(

2

4

8

12)(

where the factor 2 comes from the two allowed state and for the spin quantum

number (S = 1/2); fermions such as electron. The density of state )(D is obtained as

17

The total particle number N and total energy E can be described by

2/32/3

220

2/3

220

2

23

22

2)( F

mVd

mVdDN

FF

and

F F

F

mVd

mVdDE

0

2/52/3

220

2/32/3

22

2

25

22

2)(

.

Then we have

F

F

F

N

E

5

3

3

25

2

2/3

2/3

((Note)) Fermi-Dirac distribution function

The Fermi-Dirac distribution gives the probability that an orbital at energy will be occupied in an ideal gas in thermal equilibrium

1

1)(

)(

ef , (12)

where is the chemical potential and = 1/(kBT). (i) F

T

0lim .

(ii) f() = 1/2 at = . (iii) For - »kBT, f() is approximated by )()( ef . This limit is called the

Boltzman or Maxwell distribution. (iv) For kBT«F, the derivative -df()/d corresponds to a Dirac delta function having a

sharp positive peak at = . ________________________________________________________________________ 7. Plane wave solution with a periodic boundary condition A. Energy level in 1D system

We consider a free electron gas in 1D system. The Schrödinger equation is given by

)()(

2)(

2)(

2

222

xdx

xd

mx

m

pxH kk

kkk

, (1)

where

18

dx

d

ip

,

and k is the energy of the electron in the orbital.

The orbital is defined as a solution of the wave equation for a system of only one electron:one-electron problem.

Using a periodic boundary condition: )()( xLx kk , we have the plane-wave

solution

ikxk ex ~)( , (2)

with

222

2 2

22

n

Lmk

mk

,

1ikLe or nL

k2

,

where n = 0, ±1, ±2,…, and L is the size of the system. B. Energy level in 3D system

We consider the Schrödinger equation of an electron confined to a cube of edge L.

kkkkk

p 222

22 mmH

. (3)

It is convenient to introduce wavefunctions that satisfy periodic boundary conditions.

Boundary condition (Born-von Karman boundary conditions).

),,(),,( zyxzyLx kk ,

),,(),,( zyxzLyx kk ,

),,(),,( zyxLzyx kk . The wavefunctions are of the form of a traveling plane wave.

rkk r ie)( , (4)

with

kx = (2/L) nx, (nx = 0, ±1, ±2, ±3,…..), ky = (2/L) ny, (ny = 0, ±1, ±2, ±3,…..), kz = (2/L) nz, (nz = 0, ±1, ±2, ±3,…..).

19

The components of the wavevector k are the quantum numbers, along with the quantum number ms of the spin direction. The energy eigenvalue is

22

2222

2)(

2)( kk

mkkk

m zyx

. (5)

Here

)()()( rkrrp k kkk i

. (6)

So that the plane wave function )(rk is an eigenfunction of p with the eigenvalue k . The ground state of a system of N electrons, the occupied orbitals are represented as a point inside a sphere in k-space.

Because we assume that the electrons are noninteracting, we can build up the N-electron ground state by placing electrons into the allowed one-electron levels we have just found. ((The Pauli’s exclusion principle))

The one-electron levels are specified by the wavevectors k and by the projection of the electron’s spin along an arbitrary axis, which can take either of the two values ±ħ/2. Therefore associated with each allowed wave vector k are two levels:

,k , ,k .

In building up the N-electron ground state, we begin by placing two electrons in the one-electron level k = 0, which has the lowest possible one-electron energy = 0. We have

3

2

3

3

3

33

4

)2(2 FF k

Vk

LN

where the sphere of radius kF containing the occupied one-electron levels is called the Fermi sphere, and the factor 2 is from spin degeneracy.

The electron density n is defined by

3

23

1Fk

V

Nn

The Fermi wavenumber kF is given by

3/123 nkF . (9) The Fermi energy is given by

20

3/222

32

nmF

. (10)

The Fermi velocity is

3/123 nmm

kv F

F . (11)

((Note)) The Fermi energy F can be estimated using the number of electrons per unit volume as

F = 3.64645x10-15 n2/3 [eV] = 1.69253 n02/3 [eV],

where n and n0 is in the units of (cm-3) and n = n0×1022. The Fermi wave number kF is calculated as

kF = 6.66511×107 n01/3 [cm-1].

The Fermi velocity vF is calculated as

vF = 7.71603×107 n01/3 [cm/s].

________________________________________________________________________ REFERENCES John S. Townsend, A Modern Approach to Quantum Mechanics, second edition

(University Science Books, 2012). Bruce C. Reed, Quantum Mechanics (Jones and Bartlett Publications, Boston, 2008). David. Bohm, Quantum Theory (Dover Publication, Inc, New York, 1979).