Chapter 2 The Harmonic Oscillatorsite.iugaza.edu.ps/bsaqqa/files/2010/10/Ch.21.pdf · 2 2 E classical 1 kA That is the classical energy is continuous: it allows all values between

Chapter 2 The Harmonic Oscillator

.

It is well-known that any particle executes S.H.O if it is acted upon be a

restoring force of the form

constantforcetheis)1( kkxxF

)2(,221 VFwherekxxV

Now applying Newton’s 2nd law for such a problem we get

The potential energy of such a problem is given by

The Classical H.O:

)3(0 xm

kxxmF

The solution of the above differential equation is

constantsarbitaryare&and,with BAmk

If, for instance, at t=0, x=A and v=0 then

)4(sincos tBtAx

amplitudetheforstandswithcos AtAx

The total energy of the Oscillator is

2212

21 xmkxKVE

tAx sinthatKnowing

221222

21222

2122

21 sincossincos kAttkAtAmtkAE

(5)constant2212

212

21 kAxmkxEor

For the velocity we have from Eq.(5)

(6)22 xAxor

It is clear now, from Eq.(6), that the particle can’t exceeds the point x=A,

otherwise we will have an imaginary speed.

From Eq.(5) we conclude that E has continuous values.

If the probability of finding the particle in a region of length x to be P(x) x, and

let t be the time required for the particle to cross x.

Since each particle crosses x twice during each complete oscillation, so we

have.

periodtheiswith,2

TT

txxP

Using Eq.(6) and the fact that T=2/ we get for the classical probability density

It is clear now, from Eq.(7), that the

probability density is minimum at x=0 and

increases with increases the

displacement. It goes asymptotically to

infinity when x=A.

xT

t

xT

xP

22

)7(1

2

2

2222 xAxAxP

This means that the probability density is inversely proportional to the speed.

Clearly, you are more likely to find the particle in regions where it is moving

slowly, and vice versa.

The potential energy of the linear Harmonic Oscillator, from Eq.(2), is given by

2221 xxV

The Schrodinger equation now becomes

)8(2

22212

2

xExx

x

dxd

d

d

d

d

dx

d

dx

d

2

2

2

2

d

d

dx

d

Eq.(8) now becomes

)9(0

2 2

2

2

E

d

d

Define the dimensionless parameter as

The Quantum H.O:

Which is not a 2-terms recurrence relation, i.e., Eq.(9) doesn't lead to a two-

terms recurrence relation.

To solve it we first note that as Eq.(9) approximated to

02

2

2

d

d

The direct series solution of Eq.(9) is given by

0k

kka

Substituting back in Eq.(9) we get

02

10

2

00

2

k

kk

k

kk

k

kk aa

Eakk

Equating the coefficient of to zero k

02

2 22 kkk aaE

kak

22 22 ee

But the second term violates the boundary condition

0

So we propose a solution for Eq.(9) of the form

)10(22

ve

d

dveve

d

dNow 22 22

2

222222

2

2 2222

2

d

vde

d

dveveve

d

d

Substitute back in Eq.(9) we get

)11(01

22

2

2

v

E

d

dv

d

vd

Equation is called Hermit's differential equation where its solution is

represented by infinite series of the form

0k

kkav

0

1

k

kkka

d

dv

0

22

2

1k

kkakk

d

vdand

Substituting back into equation (11) we get

012

21000

2

k

kk

k

kk

k

kk a

Ekaakk

Equating the coefficient of to zero k

01

2212 2 kkk a

Ekaakk

)12(

12

212

2 kk akk

kE

a

If we set a1 =0 and choosing ao arbitrary we generate the even solution.

In the other hand if we set a0 =0 and choosing a1 arbitrary we generate the

odd solution.

)13(22

oddeven vve

It is clear from Eq.(12) that ka

a

kk

k 22

Now let us examine the series

00

2

!2

!

2

k

k

j

j

kje

kkk

k

a

a

kk

k 2

12

1

!12

!22

The function v and the function behave alike as 2e

both the even and the odd series violate the boundary condition

0

To solve this dilemma we have to terminate the series after a finite number of

terms, say n,

02na from Eq.(12) we get

)14(212

nE

If n is even and all higher terms vanish and since the odd series get

unacceptable solution we set a1 =0 2na

If n is odd we retain only the odd series by letting a0 =0 . the solution exits

for integer n .

from Eq.(14) we get

)15(21 nEn

It is clear from Eq.(15) that 21

21

1 1 nnEE nn

The energy levels of the harmonic oscillator are evenly spaced, regardless of n.

and the Eigen functions are

vNe 22

)16(

2

2

xHNexor n

x

Where Hn are the Hermit's polynomials with the following properties:

0

2

!,

2

n

n

ntxt

n

txHetxg generating function

22

1 xn

nxn

n edx

dexH

Rodrigues formula

xHxH nn

n 1 Parity

022 xnHxHxxH nnn Differential equation

nmn

mnx ndxxHxHe !2

2

Orthogonality

022

02

022

1

11

xnHxHxxH

xnHxH

xnHxxHxH

nnn

nn

nnn Recurrence relations

Now using the Rodrigues formula one can prove that

10 xH xxH 21

24 22 xxH xxxH 128 3

3

Now since the wave function given by Eq.(16) must be normalized we have

1222

dxxHxHeNdxxx nnx

with

Using the orthogonally property we get

1!22

nN n !2

1

!2

41

nnN

nn

xHe

nx n

x

nn

2

41

2

!2

1

The normalized wafefunctions of the Harmonic oscillator is given by

2

41

20

xex

21oE

24

1

23

33

14 x

xex

23

1 E

25

2 E

24

1

222 1

44

xexx

Let us draw the wave function and the probability density for n=0,1,2 and make

correspondence with the classical harmonic oscillator.

221 kAEclassical

That is the classical energy is continuous: it allows all values between –A & A.

But from Eq.(15) it is clear that the energy of the quantum mechanically H.O is

quantized: it has discrete values.

Recall that the classical harmonic oscillator has an energy given by

Also in classical case the probability is inversely proportional to the speed, i.e.,

it is minimum about x = 0 and high at the turning points.

In quantum case the situation is the opposite for lower states and approaches

the classical limit as n→∞.

The ground state for the classical oscillator has zero energy (and zero motion),

whereas the quantum oscillator in the ground state has an energy of 21

0 E

This quantum fact is a consequence of the Heisenberg uncertainty principle.

The probability density for the

quantum oscillator “leaks out”

beyond the x = ±A classical

limits (tunneling).

the quantum probability density will have n+1 maxima and n minima. These

minima correspond to zero probability! This means that for a particular quantum state n there will be exactly n forbidden locations where the wave function goes to zero (nodes).

This is very different from the classical case, where the mass can be found at

any location within the limits −A < x < A.

The region of non-zero

probability outside the

classical limits drops very

quickly for high energies, so

that approaches the classical

limit as n→∞.

In classical case the particle

can never exceeds these

limits, since if it did it would

have more potential energy

than the total energy.

Exercise: Prove that

1,1, 12

)1( nknkknnnx

1,1,12

)2( nknkknnnip

212)3( nx

212)4( np

21)5( npx

(6) Estimate the ground state energy of a H.O. by minimizing

22

21

2

2x

pE

Subject to the uncertainty restriction 21 px

The Dirac Notation Method

Let be the eigenket of the system n

)1(nEnH n

)2(2

2221

2

xp

Hwith

Let us introduce the dimensionless coordinate and momenta operators

xx2

ppand

2

1

The Hamiltonian of Eq.(2) becomes

22 xpH

2

,2

1,

ipxpxbut

pixpixH pxipixpixH ,

2122 pixpixxpH

)3(apixLetting

)4(†apixand

)5(21† aaH

Substituting back in Eq. (1)

nEnaa n21†

)6(21† nEnnaa n

)7(† nnaaLetting n

Eq.(6) becomes

nEn nn 21

)8(21 nnEor

To find the values of n we operate on Eq.(7) by a we get

)9(† nanaaa n

From the definition of Eqs.(3&4) it is easy to prove that

)10(1, † aa

aaaa †† 1

Substituting back in eq.(9) we get

nanaaa n†1

Comparing eqs (7) and Eq.(11) we conclude that

)11(1† nanaaa n

n naand are eigenkets for the operator aa†

The first with eigenvales of n 1nWhile the second with eigenvales of

)12(1 nCna n

By operating on Eq.(7) by one can prove that †a

)13(1† nCna n

Now multiplying Eq.( 12) by its bra form

)7(† nnaaLetting n

)14(112† nnCnaan n

Recalling Eq.(7)

)15(2 nn C

Similarly, multiplying Eq.( 13) by its bra form

112† nnCnaan n

)7(† nnaa n

Now, from Eq.(7) and Eq.( 14)

aaaa †† 1But from Eq.(10) we have

The last equation becomes

1112† nnCnaan n

Using Eq.(7) we get

)16(12 nn C

From Eq.(15) we conclude that 0n

there must be a nonnegative minimum value such that , from Eq.(12)

)12(1 nCna n

1minminmin nCna 0minC

From Eq.(15) we conclude that

0min

By applying the operators repeatedly on Eq.(7) we can generate from

any given eigenket new eigenkets with different eigen values that is

integrally spaced.

†& aa

n

composed of a set of positive integers and zero. n

contradiction

Setting and using Eq.(8) we obtain nn

)17(21 nEn

Now from Eqs.(15 & 16) we have

nCn 1 nCn

Eqs.(12 & 13) now read

)18(1 nnna

)19(11† nnna

Equations (18) and (19) specify completely the operators as the

lowering (annihilation) and raising (creation) operators, respectively.

†& aa

Recalling Eqs.(3&4) we have

apix †apix

apix

2

1

2†

2

1

2& apix

Adding and subtracting the above two equations we get, respectively

†2aax

†2aapi

)20(2

†aax

)21(2

† aaip

Matrix Representation of Some Operators

Let is now evaluate the matrix elements of some operators

naamnxmXmn†

2

namnamXmn†

2

111

2 nmnnmnXmn

1,1, 12

nmnmmn nnX

0300

3020

0201

0010

2x

For the lowering operator we have

1 nmnnamamn 1, nmmn na

0000

3000

0200

0010

a

For the raising operator we have a†

11†† nmnnama mn 1,† 1 nmmn na

0300

0020

0001

0000

†a

Now for the Hamiltonian operator we have

nmnaamnaamnHmHmn 21†

21†

mnmn nnamnH 21† 1

7000

0500

0030

0001

2

H

Let us generate some eigen functions:

Using Eq.(17) we have 00 a

Expressing a in terms of x and p

002

1

2

pix

022

0

dx

dx

01

0

dx

dx

002

0 xdx

d xdx

d 2

0

0

20

22

41 x

e

To generate the first excited state we have

10† a

12

1

2

opix

12

2

2

1

2

o

dx

dx

But 20

22

41 x

e

and 0

20 x

dx

d

1

22

22

ox

2

1

22

41

22

x

o xex

Exercise: Prove that for the H.O 21 npx

22 AAAwhere

The Motion of the Wave Packets:

If the initial state is known, that is, if is known we can find the state at

any time by Eq.(21) provided that you can find Cn. To find Cn we have from

Eq.(21)

)21(,

0nn

tEi

n xeCtxn

0

0,n

nn xCx

Multiply both sides by and integrate we get

)22(,0

2

n

ntin

n

ti

xeCetx

Let us now show how the oscillator evolves with time. It is known that

0,x

m

mn

mnnn

nmnm CCdxxCdxx

00

0,

Substituting for En from Eq.(16) into Eq.(21) we get for the state at any time

Now if is normalized we have 0,x

)23(10

2

nnC

Let us now find the expectation value of the Hamiltonian, we have

dxtxHtxH ,,

dxtxxtx ,2

, 22212

2

Probability of finding the oscillator in the nth state which is time-independent.

00

0,0,n

nmnmm

dxxxCCdxxx

00

1n

mnnmm

CC

Substituting for tx, From Eq.(22) into the above equation we get

0

22212

22

2nnnn dxxxxCH

From the Schrödinger equation we have

xExx nnn

22

212

2

2

)24(0

2

0

2

nnn

nnnnn ECdxxxECH

From Eqs. (23) and (24) we conclude the two facts :

(1) The only possible results of measuring the energy of the system are the

energy values En.

(2) If the system is in the state tx,

the probability of obtaining the result En is equal to 2

nC

As stated in postulate no. (IV).

Let us now find the expectation value of x with respect to the state tx,

dxtxxtxx kn ,,

0 0n kkn

tknikn dxxxxeCC

1,1, 12

nknkkn nndxxxx

But we proved that

011

2 n

tinn

tinn eCCeCCnx

ninn eCC

Setting

011

11

2 n

tinn

tinn

nnnn eCCeCCn

x

)25(cos

2

011

nnnnn tCC

nx

0

11 12 n

tinn

tinn eCCneCCnx

Letting n+1=n in the 2nd term we get

nn 1

Let us test the last result at the limit n → ∞. At this limit we can assume

nn CCand 1

Again for n → ∞ we can write for the energy nEn

Then Eq.(25) now reads

tECxn

nn cos2

0

2

2

EECthatKnowingn

nn0

2 )26(cos2

2

t

Ex

In classical mechanics we have

tAx cos 22

21 AEand

)27(cos2

2

t

Ex

Which match Eq.(26).

t

AHA

iA

dt

d,

1

In the first chapter we proved that

Hxi

xdt

d,

1

22

21

2

2,, x

pxHx

px

pppx ,

2,

2

1

pi

)29(2 xpdt

d

Hpi

pdt

dsimilarly ,

1

22

21

2

2,, x

ppHp

xpxxxp ,, 2

212

21 2

2i

)28(

px

dt

d

Differentiate Eq.(28) with respect to t we get

dt

pdx

dt

d

12

2

Using the result of Eq.(29) we get

xxdt

d 2

2

2

)30(cos0

txx

Which is similar to Eq.(26).