The One-Dimensional Infinite Square Well• defined over from x = 0 to x = a, so ‘width’ of well is a• potential V(x) is even about center of well x = a/2• evenness/oddness of wavefunctions about x = a/2• theme: if potential is even, lowest energy wavefunction (ground state) will be even too.. then O, E…• this follows because the second derivative of a function has the same evenness/oddness as the function itself

• inside the well V = 0: free particle wavefunctions work• outside the well V = ∞: the wavefunction must be zero• the monstrous (INFINITE) jump in V at the well edges [x = 0, x = a] implies that the wavefunction’s slope is not continuous at those places• the wavefunction must always be continuous, no matter what V(x)!!

axx

axxVd

or 0

00)( is potentialSW 1

Wavefunction inside the 1d ∞ Square Well

• rightward-moving and leftward-moving free wavicles have the same energy E, so a linear combination of them also solves the TISE with that same E• choose a linear combination that satisfies the boundary conditions (0) = (a) = 0

mE

kikxAikxAx2

: expexp)( RL

odd/even torise gives thisand 1exp

:onquantizati so 22

0)2(sin and 1 )2cos( so

0 havet can' wesince 12exp

expexp 00)(

00)0(

L

L

LRRL

ikaa

nknka

kaka

Aika

ikaikaAL

AAAA

n

Even or Odd about a/2?

• these are even solutions about well middle (includes ground state)• they are labeled, however, by odd integers n = 1,3,5…

• These are odd solutions about well middle (first excited state…) • (labeled by even integers)

• lower sign (–)• let AL = A/2i

• upper sign (+)• Let AL = A/2

• Both:28

22

22

22

2

22

2

so ma

nh

ma

n

mnk

na

nn Ek

6,4,21)cos(

)cos( expexp2

)(

nnakka

kxAikxikxA

x

n

5,3,11)sin(

)sin( expexp2

)(

nnakka

kxAikxikxi

Ax

n

Probability densities for 1d∞SW

E3 = 9 E1

E2 = 4 E1

• the wavefunctions are standing waves, so regardable as the sum of two oppositely-traveling wavicles with equal amplitudes• wavefunctions are sinusoidal, so probability densities are sine-squared functions with integral number of cycles: area underneath is half the maximum times the width. NORMALIZE:

• typical value for a = 1 nm: E1 ≈ (50 x 10-68)/(10)(10-30)(10-18) ≈ 5 x 10-20 J ≈ 1/3 eV [since 1 eV = 1.602 x 10-19 J]

Hydrogen atom is about 20 times smaller (.53 Å) E = – 13.6 eV

a

aAa

AdxAdxxxa

a

xn 22

||)(sin||)()(1 2

0

22*

E1 = h2/8ma2

Things to note regarding this example• walls are ‘infinitely high’ and correspond to impenetrable• infinite jump in V causes a ‘kink’ in – unusual!• energy spectrum climbing like n2 is the MOST rapid climb it can possibly be for ANY well since well ‘sides’ are ∞-ly steep!• crucial helpful fact: eigenfunctions are ‘orthonormal’:

a

xn

a

xm

a

xn

a

xm

a

xnm sinsincoscos

)(cos since

mn

a

a

a

a

a

nm

nma

adx

a

xm

a

nma

xnm

nma

xnm

nm

dxa

xnm

a

xnm

a

dxa

xn

a

xm

adxxx

for 12

2sin

2

for 0)(

sin)(

1)(sin

)(

1

)(cos

)(cos

1

sinsin2

)()(

0

2

0

0

*

Expectation values and uncertainties I• Expectation value for position should be middle of the well

• this tells us very little about ‘where’ the particle is• <x2> requires two integrations by parts; result is

• this is evidently a measure of the typical ‘distance’from the center, and grows for larger n; in the limit, <x2> a2/3

well theof middle 28

1

8

2cos

4

2

8

2cos

4

2sin

4

2sin

22

:once partsby integrate and 2

2cos1sinuse

so let ;sin

22

22

0

2

220

2

2

0

22

ann

n

a

yyyy

n

adyyy

n

a

ax

dyn

adx

a

xn y dxxx

na

a

a

xna

22

2

0

222

2

1

3

1sin

2

nadx

a

xnx

ax

a

Expectation values and uncertainties II• uncertainty in position is x = (<x2> – <x>2)1/2

• turning to the momentum expectations and uncertainty

• ground state has lowest product, which still exceeds minimum uncertainty, and the product grows with n• begs the question of how to achieve minimum uncertainty

22

2

222 2

3

1

242

1

3

1

n

aa

nax

232

2

3

1

2

22

2222

n

n

n

a

npp pxp

idea trigaverage'' theused we where2

2

sin2

:ˆ

0sincos2

:ˆ

22

222

3

222

0

222

22

222

0

a

na

a

n

dxa

xn

a

n

ap

xp

dxa

xn

a

xn

a

n

a

ip

xip

a

a

Using these orthonormal wavefunctions as building blocks: Fourier Series

• wavefunctions {n}, are orthonormal, as we have seen• they are complete: any odd [f(–x) = – f(x); f(0) = 0] periodic [f(x + 2a) = f(x), so period is 2a] (this is not the most general situation – stay tuned) function can be expressed as a linear combination

11

sin2

)()(n

nn

nn a

xnc

axcxf

• Fourier’s Trick: to get the {cm}, multiply expansion of f(x) by m*(x) and integrate term by term from 0 to a:

)()(2

2

sinsin2

sin2

sin2

0

*

11

1 010

a

mmmn

nmnn

nmn

nn

a

nn

a

dxxfxcccca

a

dxa

xnc

a

xm

adx

a

xnc

aa

xm

a

More general remarks about the Fourier series

• for any [odd or even or neither] periodic [f(x + 2a) = f(x), so period is 2a] function can be expressed as a linear combination• conventionally, the limits are now taken to be from –a to a so the new normalization is 1/√a• both cosines and sines must be included

• Fourier’s Trick is now very similar—the cosines and the constant are both even contributions, so

a

d

a

xnd

a

xnc

a

dxdxcxf

nnn

nnnnn

2cossin

1

2)()()( 0

11

0

0,1,2,3... for )(cos1

1,2,3... for )(sin1

mdxxfa

xm

ad

mdxxfa

xm

ac

a

a

m

a

a

m