LECTURE 17

THE PARTICLE IN A BOX

PHYSICS 420SPRING 2006Dennis Papadopoulos

CLASSICAL STANDING WAVES

2 4 6 8 10

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

1

2

1 2

( , ) ( )( , ) ( )

( , ) ( , ) ( , ) ( ) cos( )2

x t Bsin kx tx t Bsin kx tx t x t x t Asin kx t

A B

Nodes

Nodes at n

String clamped between two points separated by a distance L. What possible standing waves can fit it?

Only the ones that have nodes at the end 2L/n.

Quantum mechanical version- the particle is confined by an infinite potential on either side. The boundary condition- the probability of finding the particle outside of the box is ZERO!

LxattLxatt

0),(00),0(

=2L

=L

=3L/2

For the guitar string:For the quantum mechanical case:

Assume a general case:

( ) ( )1 2

( 2 / / ) ( 2 / / )1 2

( / ) ( / )1 2

( 2 / / ) ( 2 / / )1 2

/ 2 / 2 /

1( , ) 2( , )

0

0

0

0

2 sin( 2 / ) 0

2 /

i kx t i kx t

i mEx Et i mEx Et

i Et i Et

i mEL Et i mEL Et

iEt i mEL i mEL

x t x t

Ae A e

Ae A eat x

Ae A eat x L

Ae A e

Ae e e

iA mEL

mEL

h h h h

h h

h h h h

h h h

h

2 2 2

2 1,2,3...2n

n

nE nmL

hh

2 2

2

/ 2 ( ) / 2

2 /

k p

E p m k m

k mE

hh

h

Fig. 6-7, p. 202

TIME INDEPENDENT SE

2 2

2

( , ) ( ) ( )

( )

( ) exp( / )

( ) ( ) ( )2

x t x tdi E tdtt iEt

d U x x E xm dx

hh

h

Free particle U(x)=0

2 2

( ) exp( )

/ 2

x ikx

E k m

hConstant E is the total particle energy

Notice that

2 2

( , ) ( ) exp( / )

( , ) ( )

x t x iEt

x t x

h

FOR STATIONARY STATES ALL PROBABILITIES ARE STATIC AND CAN BE CALCULATED BY THE TIME INDEPENDENT SE

Quantum mechanical version- the particle is confined by an infinite potential on either side. The boundary condition- the probability of finding the particle outside of the box is ZERO!

LxattLxatt

0),(00),0(

For the quantum mechanical case:

here, we assumed the walls were

infinite—there was no

possibility the particle could

escape.

22

2

22

( )

2

d k xdx

mEk

h

2 2 2 2 2

2

( ) exp( ) exp( )(0) 0

( ) 0[exp( ) exp( )] 2 ( ) 0

2 2n

n

x A ikx B ikx

A BL

A ikL ikL AiSin kLkL n

k nEm mL

h h

ODD AND EVEN

ALTERNATIVE SOLUTION

2 2 2 2 2

2

( ) sin( ) cos( )(0) 0

0( ) 0

( ) 0

2 2n

n

x A kx B kx

BL

ASin kLkL n

k nEm mL

h h

2 2

0

( ) sin( )

01, 2,3,..

1 sin ( )

2 /

n

L

n xx AL

x Ln

n xA dxL

A L

What is the probability that a particle will be found between L/4 and 3L/4 in the ground state?

Note odd and even solutions

Fig. 6-15, p. 209

2 22 ( ) /m U E h2 22 ( ) /m U E h

I II IIIU

0 L

E2

2 2

( ) 2 ( ) ( )d x m E U xdx

h

Since E<U:

2

2 2

( ) 2 ( ) ( )d x m U E xdx

h

Generally, solutions are then: 2 ( ) / 2 ( ) /1 2 1 2( ) m U E x m U E xx xx C e C e C e C e h h

Setting

22

2 ( )m U E h )()( 22

2

xdxxd

This begins to look like the familiar undamped SHO equation:

But remember the conditions imposed on wave-functions so that they make physical sense:

I II IIIU

-L/2 L/2

E

2 ( ) / 2 ( ) /1 2 1 2( ) m U E x m U E xx xx C e C e C e C e h h

C1 must be 0 in Region III and C2 must be zero in Region I, otherwise, the probabilities

would be infinite in those regions.

Now we must use the condition of continuity (the wavefunction must be continuous at the boundary, and so must its first derivative).

(x) d/dx

L/2 L/2x x

Here, the acceleration

would be infinite. Uh-oh!

Suppose we had a

discontinuous function…

Note that the wavefunction is not necessarily 0 in Regions I and II. (It is 0 in the limit of an infinite well.) How is this possible

when U>E?? The uncertainty principle.

Fig. 6-16, p. 210

2 2 2

2

12 ( )

2 ( 2 )n

m U E

nEm L

h

h

Even solutions Odd solutions

A “real world example” of a finite box would be a neutron in a nucleus.

Note that the particle has negative kinetic energy outside of the well: K=E-U

The wave-function decreases rapidly- 1/e in a space of 1/

1/ is a penetration depth. To observe the particle in this region, you must measure the the position with an accuracy of less than 1/

2

1/2 ( )

xm U E

h

From the uncertainty principle:

2

22

2 ( )/

8 ( )2

m U Ep h x h

pK m U E

m

h

The uncertainty in the measurement is larger than the negative kinetic energy.

Note that the allowed energies are inversely proportional to the length of the box.

The energy levels of a finite box are lower than for an infinite well because the box is effectively larger. There is less confinement energy.

This is analogous to a classical system, such as a spring, where potential energy is being exchanged for kinetic energy.

At x=0, all of a particle’s energy is potential energy, as it approaches the boundary, its kinetic energy becomes less and less until it all of the particles energy is potential energy- it stops and is reflected back. An example would be a vibrating diatomic molecule.

In contrast to the square well, where the particle moves with constant kinetic energy until it hits a wall and is reflected back, in the parabolic potential well of the harmonic oscillator, the kinetic energy decreases (wavelength increases) as the boundary is approached.

Kinetic energy: )(22

2

UEmpUEmpK

)(2 UEmhp

The wavelength is position dependent:

Making an approximation-assuming small penetration depth and high frequency, the condition for an infinite number of half wavelengths as in an infinite well must be recast as an integral to account for a variable wavelength:

...23,1,

21)]([2

b

adxxUEm

Note that the width of the well is greater for higher energies. As the energy increases, the “confinement energy” decreases. The levels are evenly spaced. We have Planck’s quantization condition! xat0

Boundary conditions:

1( ) 1,2,3,...2

E n n h

22 2

2 2

2 1( ) ( )2

/ 2

d m m x E xdxm

h

h

In classical physics, the “block on a spring” has the greatest probability of being observed near the endpoints of its motion where it has the least kinetic energy. (It is moving slowly here.)

This is in sharp contrast to the quantum case for small n.

In the limit of large n, the probabilities start to resemble each other more closely.

Fig. 6-18, p. 214

Fig. 6-19, p. 215

Fig. 6-20, p. 216

•We can find allowed wave-functions.

•We can find allowed energy levels by plugging those wavefunctions into the Schrodinger equation and solving for the energy.

•We know that the particle’s position cannot be determined precisely, but that the probability of a particle being found at a particular point can be calculated from the wave-function.

•Okay, we can’t calculate the position (or other position dependent variables) precisely but given a large number of events, can we predict what the average value will be? (If you roll a dice once, you can only guess that the number rolled will be between 1 and 6, but if you roll a dice many times, you can say with certainty that the fraction of times you rolled a three will converge on 1 in 6…)

If you roll a dice 600 times, you can average the results as follows:

600...43556421

...)3(600104)2(

60097)1(

60099

Alternatively, you can count the number of times you rolled a particular number and weight each number by the the number of times it was rolled, divided by the total number of rolls of the dice:

After a large number of rolls, these ratios converge on the probability for rolling a particular value, and the average value can then be written:

xxPx

This works any time you have discreet values.

What do you do if you have a continuous variable, such as the probability density for you particle?

It becomes an integral….

2

( , )xx xP dx x x t dx

Table 6-1, p. 217

2

2.5 3.7 1.4 .... 5.3 5.4618

1.4(1/18) 2.5(1/18) ... 5.4(3 /18) 6.2(2 /18) ... 8.8(1/18) 5.46

,

( , )

x

x

x xP

Expectation value

x x x t dx

The expectation value can be interpreted as the average value of x that we would expect to obtain from a large number of measurements. Alternatively it could be viewed as the average value of position for a large number of particles which are described by the same wave-function.

We have calculated the expectation value for the position x, but this can be extended to any function of positions, f(x).

For example, if the potential is a function of x, then:

dxtxxUU 2),()(

the potentialexpression for kinetic energy

kinetic plus potential energy gives the total energy

2

;2pKE p km

h

position x xmomentum p

potential energy U U(x)

kinetic energy K

total energy E

i x

h

2 2

22m x

h

it

h

observable

operator

dxQQ

*

dxxUdxxUdxUU2

** )()(

2 2

* *22

K K dx dxm x

h

In general to calculate the expectation value of some observable quantity:

We’ve learned how to calculate the observable of a value that is simply a function of x:

But in general, the operator “operates on” the wave-function and the exact order of the expression becomes important:

Table 6-2, p. 222