Quantum mechanics course microsoftpowerpoint-harmonicoscillator-operatorapproach

8/14/2019 Quantum mechanics course microsoftpowerpoint-harmonicoscillator-operatorapproach

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The OperatorApproach


2/28

Previously we solved the Schrdinger equation to obtain the wavefunctions andenergy levels of a quantum-mechanical HARMONIC OSCILLATOR

* The energy levels are QUANTIZED in terms of the oscillator frequency w while the wavefunctions correspond to HERMITE POLYNOMIALS with aFINITE range

The Operator Approach

)27.13(,2,1,0,21 nn E n

ENERGY LEVELS , WAVEFUNCTIONS (LEFT) AND PROBABILITY DENSITY (RIGHT) FOR THE QUANTUM HARMONIC OSCILLATOR THE PICTURE BOOK OF QUANTUM MECHANICS, S. BRANDT and H-D. DAHMEN, SPRINGER-VERLAG, NEW YORK (1995)


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Motivated by Equation 14.2 we define the new OPERATORS

* It is easy to show that MULTIPLICATION of these two operators yieldsthe following result

* If we REVERSEthe order of multiplication however we obtain the followingresult


)3.14(2

1xim

xima

)4.14(21

)(21 2

2

xm xim

aa

)5.14(21

)(21 2

2

xm xim

aa

: KEEP THE ORDER OF THE OPERATORS AND NOTE THAT d (fx )dx = x (df /dx ) + f


5/28

Comparison of Equations 14.4 & 14.5 reveals that the result of multiplying theoperators depends on the ORDERin which they are multiplied

* With these definitions our original Schrdinger equation (Equation 14.1) cannow be REWRITTEN as

* We now notice something interesting if we perform the following operation


)6.14( aaaa

)7.14()()(21

x E xaa

)8.14()(21

xaaa


6/28

We can EXPAND Equation 14.8 so that it yields

* Equation 14.9 shows that if (x ) satisfies the Schrdinger equation withenergy E then a + (x ) ALSO satisfies the Schrdinger equation but with energy E + h

* If we know one initial solution to the Schrdinger equation then we can usethe operator a + to determine ALL OTHERsolutions!


)(21

)(21

)(21

xaaa xaaaa xaaa

)()()()(21 x x E a x xaaa

)9.14()()( xa E


7/28

ALTERNATIVELY we could use the operator a - to demonstrate that

* Equation 14.10 shows that if (x ) satisfies the Schrdinger equation withenergy E then a - (x ) ALSO satisfies the Schrdinger equation but with energy E -h

* If we know one initial solution to the Schrdinger equation then we canequally use the operator a - to determine ALL OTHERsolutions


)(21

)(21

)(21

xaaa xaaaa xaaa

)()()()(21 x x E a x xaaa

)10.14()()( xa E


8/28

The operators a are referred to as LADDER OPERATORSsince they allow us todetermine ALLallowed energy levels of the harmonic oscillator from a knowledgeof just ONE level

* a + is known as the CREATION or RAISING operator since it allows us todetermine HIGHER levels from an initial energy solution

* a - is known as the ANNIHILATION or LOWERING operator since it allowsus to determine LOWERlevels from an initial energy solution


E

E +

E + 2

E + 3

E - 3

E - 2

E -

a +

a -

THE BASIC CONCEPT OF THE LADDER OPERATORS IS ILLUSTRATED SCHEMATICALLY IN THIS FIGURE

THE IDEA IS THAT WE ASSUME THAT WE CAN START FROM SOME INITIAL KNOWN SOLUTION WITH ENERGY E

SUCCESSIVE APPLICATION OF THE CREATION OPERATOR a + THEN ALLOWS US TO DETERMINE HIGHER ENERGY LEVELS WHILE APPLICATION OF THE ANNIHILATION OPERATOR ALLOWS US TO DETERMINE LOWER ENERGY LEVELS


9/28

Now the preceding discussion suggests that if we repeatedly apply theannihilation operator we should eventually reach a set of states with NEGATIVEenergy!

* Since such states are physically MEANINGLESS there must instead be aMINIMUM state with positive energy that represents the lowest rung of ourenergy ladder

* That is we assume the existence of a wavefunction solution y 0(x ) such that

* Expanding Equation 14.11 we arrive at the following condition on y 0(x )


)11.14(0)(2

10)( 00 x xim xim

or xa

)12.14()()( 00 x xm x

x


10/28

Equation 14.12 can be easily solved to yield the GROUND-STATE wavefunction

* We now need to determine the ENERGYassociated with this wavefunctionwhich we do by substitution into the Schrdinger equation

Since by definition a - 0 = 0 Equation 14.14 reduces to the SIMPLEresult


20

0

0

2)(ln

)()(

xm

x xdxm

x xd

)13.14(2

exp)( 200 xm A x

)14.14()()(2

1000 x E xaa

)14.14(21

)()(21

0000 E x E x


11/28

Having determined the ground-state energy of the harmonic oscillator thecreation operator can be used to generate the quantized SET of energy levels

* This result follows by remembering that each time we apply the creationoperator we INCREASE the energy by h w

* The wavefunctions of the excited energy levels can also be reconstructed inthe same way

These solutions are in fact IDENTICAL to those obtained previously(Equations 13.27 & 13.32) by analytic solution of the Schrdinger equation


)15.14(,2,1,0,21

nn E n

)16.14(2

exp)()()()( 20 xm

a A xa x nn

n

n


12/28

Determine the normalization constant 0(0) in the ground-state wavefunction

Solution: For normalization we require an expression for 0(0) such that

By exploiting the fact that

We arrive at the following condition for the normalization constant

Examples

)13.13(2

exp)( 200 xm

A x

1exp 20*0

dx x

m A A

adxax / exp 2

4 / 1

0

2 / 1

0*0 1

m A

m A A


13/28

Use the following operator relation to determine the GENERALnormalizationcoefficient An in Equation 14.16

Solution: We begin by making the change of variables n n 1

We can use this last relation to generate FURTHER wavefunctionsolutions

)17.14()()1()( 1 xni xa nn

)()()()( 11 xani

x xni xannnn

)(2

)( 12 xai

x

)(1

)( 01 xai

x

)()2(

)( 32 xani

xnn

)()1(

)( 21 xani

xnn

Examples


14/28

By COMBINING the series of wavefunctions generated above we obtain anexpression for n (x ) in terms of 0(x )

We have seen already however that the n th wavefunction can be written as

From a comparison of Equations 14.16 & 14.18 we see that thenormalization constant

)18.14(2

exp)()(!

)()()(

)(!

)()( 2

4 / 1

0 xm

am

n

i xa

n

i x n

n

nn

n

n

n

)16.14(2

exp)()( 2xm

a A x nnn

)19.14()(!

)(4 / 1

m

n

i A

n

n

n

Examples (cont d)


15/28

Thus far we have focused on the solutions of the TIME-INDEPENDENT Schrdingerequation for a particle that moves in a HARMONIC potential

* The wavefunctions and probability density obtained for this problem take the form

Superposition States

THE PROBABILITY DENSITY ASSOCIATED WITH THE FIRST FIVE ENERGY LEVELS OF A HARMONIC OSCILLATOR

THESE PROBABILITY DISTRIBUTIONS ARE INDEPENDENT OF TIME SINCE THEY DERIVE FROM WAVEFUNCTIONS THAT ARE STATIONARY STATES

THE PICTURE BOOK OF QUANTUM MECHANICS S. BRANDT and H-D. DAHMEN, SPRINGER-VERLAG,NEW YORK (1995)

)32.13(

!2

1)( 2 /

4 / 12 xm

nnne x

m H

n

m x

)1.15(!2

1)()( / 2

2 / 1* 2 xm

nnnne x

m H

n

m x x


16/28

We now wish to discuss the DYNAMICS of a quantum-mechanical particle thatmoves in the presence of a harmonic potential

* We immediately run into a PROBLEMhowever since the wavefunctions ofEquation 13.32 are STATIONARY STATES whose probability density is timeINDEPENDENT

Recall that the stationary states have time-dependent wavefunctions of

the form

The probability density associated with this wavefunction is easily

calculated


)2.15(exp)0,(),( t E

i xt x nnn

t E

i xt E

i xt xt x nnn

nnn exp)0,(exp)0,(),(),(**

)3.15()0,()0,(* x xn

THE PROBABILITY DENSITY IS INDEPENDENT OF TIME!


17/28

To overcome this problem we note that a more GENERALsolution to theSchrdinger equation is obtained by taking a LINEAR SUPERPOSITION ofstationary-state wavefunctions

* The TIME EVOLUTION of this wavefunction is now given by

* As we will discuss in further detail an important property of the stationarystate wavefunctions is that they define a so-called ORTHONORMAL SET

The condition of orthonormality can be expressed as


)3.15()()(n

nn xc x

)4.15(exp)(),(n

nnn t

E i xct x

)5.15(

,0

,1

,)()(*

k n

k n

dx x x nk nk k n

d nk IS THE KRONECKER DELTA

FUNCTION


18/28

By exploiting the orthonormality of the stationary-state solutions the expansioncoefficients in Equation 15.3 can be determined

* Another important result follows from the NORMALIZATION condition forthe wavefunction


dx xc xdx x xn

nnnn )()()()(**

)6.15()()(* nnnn cdx x xc

1)()(1)()(***

dx xc xcdx x x n nnn nn

)7.15(11)()( *** n

nnn

nnnn ccdx x xcc

NORMALIZATION RELATION FOR THE EXPANSION COEFFICIENTS


19/28

An important feature of the linear superposition of stationary states is that itresults in a WAVE PACKET whose properties now EVOLVEwith time

* Consider for example the expectation value of the POSITION of the wavepacket

By introducing Equation 15.3 for the wavefunction this expectation maybe written as

Harmonic Particle Motion

dxt x xt xt x xt xt x ),(),()),(),,(()( *

dxe x xce xct xk

t iE k k

n

t iE nn

k n / / ** )()()(

dx x x xecc k nn k t E E i

k nk n

)()(* / )(*

)8.15( / )(* nk n k

t E E i

k n xecc k n

NOTE HOW WE DEFINE THE MATRIX ELEMENT x nk


20/28

To compute the expectation value of the position we now have to evaluate theMATRIX ELEMENT x nk that appears in Equation 15.8

* By exploiting the properties of Hermite polynomials (see Appendix) it can beshown that the matrix element reduces to

* Substituting this expression into Equation 15.8 we arrive at the followingresult for the expectation value of the position

We have exploited here the fact that E n E k = (n k )h so that E n E n - 1 =h


)9.15(2

12 1,1, nk nk nk

nnm

x

)10.15()(2)(*

11*2 / 1

n

t inn

t inn ecceccnmt x


21/28

Since the expansion coefficients c n may be COMPLEXquantities we can writethem in POLARform

* With this definition we can then rewrite Equation 15.10 as

* If the PHASE ANGLE n n 1 = where is a CONSTANT independent ofn and if the expansion coefficients c n are of EQUAL magnitude then Equation15.12 becomes


)11.15(ni

nnecc

)12.15()cos(2

)( 1*

1*

12n

nnnnnn t ccccnmt x

)13.15()cos()cos(2

)( *2 t xt cc E

mt x o

nnnn

THIS TERM IS THE EXPECTATION VALUE

OF E


22/28

Equation 15.13 shows that the expectation value of the wavepacket in theparabolic potential OSCILLATES as a function of time

* The oscillations occur at the SAME frequency ( ) of the classical oscillator* From the figures below we see that the WIDTH of the wavepacket also

oscillates but at TWICE the frequency of the classical oscillatorThe width of the packet is narrowest at the classical TURNING

POINTS where the velocity of the classical particle is zero


MOTION OF A WAVE PACKET IN A HARMONIC POTENTIAL CIRCLES SHOW THE MOTION OF A CLASSICAL PARTICLE THE SAME MOTION IS SHOWN IN MORE DETAIL OVER HALF AN OSCILLATION CYCLE

THE PICTURE BOOK OF QUANTUM MECHANICS

S. BRANDT and H-D. DAHMEN, SPRINGER-VERLAG, NEW YORK (1995)


23/28

Instead of a wavepacket what is the corresponding variation of the positionexpectation in one of the STATIONARY states?

* For such a state ALLcoefficients but one are equal to zero so that Equation15.10 reduces to

The expectation value is ZERO for ALLtimes as we expect for thestationary states which have SYMMETRICprobability densities


)14.15(0)(2

)( * 11*2 / 1

n

t inn

t inn ecceccnm

t x

THE PROBABILITY DENSITY ASSOCIATED WITH THE FIRST FIVE ENERGY LEVELS OF A HARMONIC OSCILLATOR

NOTE THAT THESE PROBABILITY DENSITIES ARE ALL SYMMETRIC ABOUT THE ORIGIN OF MOTION AND SO GIVE RISE TO AN EXPECTATION VALUE FOR THE POSITION THAT IS EQUAL TO ZERO AT ALL TIMES

THE PICTURE BOOK OF QUANTUM MECHANICS S. BRANDT and H-D. DAHMEN, SPRINGER-VERLAG,NEW YORK (1995)


24/28

The other quantity in which we are interested is the expectation value of theMOMENTUM

* The SIMPLEST way to determine this quantity is to note that we expect thefollowing relation to hold

* By differentiating Equation 15.10 with respect with time we obtain thefollowing result

The time dependence of this equation is IDENTICAL to that ofEquation 15.10We therefore expect that the expectation value of the momentum

should also OSCILLATE as a function of time


)15.15()()()( t xdt d

mt vmt p

)10.15()(2

)( * 11*2 / 1

n

t inn

t inn ecceccn

mit p


25/28

Under conditions where the assumptions that lead to Equation 15.13 hold theexpectation value of the momentum can be APPROXIMATEDas

* Note that the oscillations of the position and momentum are OUT of phaseThis is just what we EXPECT for the harmonic oscillator which is atREST when its displacement is MAXIMALand moves FASTEST when itsdisplacement is ZERO

)13.15()sin()sin(2)( * t pt cc E mt p on

nnn


A

A B

THE TWO POINTS LABELED A CORRESPOND TO THE MAXIMUM DISPLACEMENT OF THE HARMONIC OSCILLATOR AND AT THESE POINTS THE PARTICLE IS INSTANTANEOUSLY STATIONARY

THE POINT LABELED B CORRESPONDS TO THE CENTER OF THE HARMONIC MOTION WHERE THE DISPLACEMENT OF THE PARTICLE IS EQUAL TO ZERO

THE PARTICLE IS MOVING WITH ITS HIGHEST VELOCITY HERE AND SO THE EXPECTATION VALUE OF THE MOMENTUM IS CONSEQUENTLY MAXIMAL


26/28

In this section we consider how to evaluate the MATRIX ELEMENTS

* As we have seen the wavefunctions y n (x ) correspond to HERMITE

POLYNOMIALS

* So that Equation A15.1 can be rewritten as

)1.15()()(* Adx x x x x k nnk

Appendix

)32.13(,)(!2

1)( 2 /

4 / 12

xm

e H n

m x nnn

)2.15()()(!!2

1 2 Ad e H H

mk n x k nk nnk


27/28

To evaluate the integral in Equation A15.2 shall need our DEFINITIONS of thegenerating function

* Rather than compute Equation A15.2 directly we first construct the moreGENERALintegral

* Equation A15.3 may be rewritten using Equations A13.2 & A13.9 as

Appendix

)3.15(),(),(22 Ad et F sF I

)9.13(),(222 )(2

AeeesF sss

)2.13(!

)(),(0

Ans

H sF n

n

n

)4.15()()(!!

222222 22)()( Ad e H H k nt s

d eee k nn k

k nt s


28/28

The LHS of Equation A15.4 can be evaluated explicitly to yield

* Comparing the coefficients of equal powers of s n

t k

l p

we obtain the value of auseful integral

* For example the integral in Equation A15.2 is obtained by setting p = 1 which yields

Using Equation A15.7 the matrix element of Equation A15.2 can becomputed and hence the expectation value of the position can be determined(Equation 15.10)

Appendix

)5.15()(2)()(22222

Aed ee t sst t st sst

)6.15()()( 2 Ad e H H pk n

)7.15())1(2(!2)()( 1,1,12

Annd e H H nk nk

n

k n

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Quantum mechanics course microsoftpowerpoint-harmonicoscillator-operatorapproach