Lecture 13: The classical limit - Michigan State University · Wavepacket Evolution •For a wavepacket in free space, we have already seen that –So that the center of the wavepacket

Lecture 13: The classical limit

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Wavepacket Evolution

• For a wavepacket in free space, we havealready seen that

– So that the center of the wavepacket obeysNewton’s Second Law (with no force):

• Assuming that:– The wavepacket is very narrow– spreading is negligible on the relevant time-

scale• Would Classical Mechanics provide a

quantitatively accurate description of thewavepacket evolution?– How narrow is narrow enough?

• What happens when we add a potential, V(x)?– Will we find that the wavepacket obeys

Newton’s Second Law of Motion?

0

00

pp

tM

pxx

=

+=

0=

=

pdt

dm

px

dt

d

Equation of motion for expectationvalue

• How do we find equations of motion forexpectation values of observables?– Consider a system described by an arbitrary

Hamiltonian, H– Let A be an observable for the system– Question: what is:

• Answer:

?Adt

d

)()( tAtdt

dA

dt

dψψ=

+

∂

∂+

= )()()()()()( tdt

dAtt

t

AttAt

dt

dψψψψψψ

€

ddt

A = −ih

A,H[ ] +∂A∂t

t

AtAHt

itHAt

i

∂

∂+−= )()()()( ψψψψ

hh

Example 1: A free particle

• Assuming that

• The basic equation of motion is:

• For a free particle, we have:M

PH

2

2

=

€

X,H[ ] =12M

X,P 2[ ]

€

ddt

A = −ih

A,H[ ]

€

=12M

XP 2 − P 2X( )

€

=12M

XP 2 − PXP + PXP − P 2X( )

€

=12M

X,P[ ]P + P X,P[ ]( )

M

Pih=

€

H,P[ ] = 0

m

P

m

Pi

iX

dt

d=−= h

h0=P

dt

d

0=∂

∂

t

A

Very commontrick in QM to

become familiarwith

Adding the Potential

• Let

• How do we handle this commutator?

)(2

2

XVM

PH +=

€

X,H[ ] =12M

X,P 2[ ] + X,V (X)[ ]

€

H,P[ ] =12M

P 2,P[ ] + V (X),P[ ]

= 0

€

=12M

X,P 2[ ]

M

Pih=

€

= − P,V (X)[ ]

= 0

One possible approach

• We want an expression for:

• We can instead evaluate:

• Will need to make use of

– Where

• Thus we have:

€

x P,V (X)[ ]ψ

€

x P,V (X)[ ]ψ = x PV (X)ψ − x V (X)Pψ

ψ)(XVxi ′−= h

[ ] )()(, XViXVP ′−= h

ψψ xdx

diPx h−=

You should think ofthis as the definingequation for how tohandle P in x-basis

)()( xVdx

dxV =′

You will derive thisin the HW

[ ])(, XVP

Theorem: If 〈x|A|ψ〉=〈x|B|ψ〉 is true for any x and|ψ〉, then it follows that A=B.

Equations of motion for 〈X〉 and 〈P〉

• As long as:

• The we will have:

– Not just true for a wavepacket

• For the momentum we have:

• The QM form of the Second Law is Thus:

m

PX

dt

d=

€

ddt

P = −ih

P,V (X)[ ]

)(XVii

′−−= hh

)()( xVdx

dxF −=

)(XF=

)(2

2

XFXdt

dM =

)(2

2

XVM

PH +=

)(XV ′−=

F(X) is the Forceoperator

Difference between classical and QM formsof the 2nd Law of Motion

• Classical:

• Quantum:

• Classical Mechanics would be an accuratedescription of the motion of the center of awavepacket, defined as , if:

• So that

• This condition is satisfied in the limit as thewidth of the wavepacket goes to zero

• ALWAYS TRUE for a constant (e.g. gravity) orlinear force (harmonic oscillator potential),regardless of the shape of the wavefunction

)(2

2

xFxdt

dM =

)(2

2

XFXdt

dM =

Xtx =)(

)(2

2

XFXdt

dM =

€

F(X) ≈ F X( )

• The expectation value of the Force is:

• Let us assume that the force F(x) does notchange much over the length scale σ

• In this case we can safely pull F(x) out ofintegral:

• So CM is a valid description if the wavepacketis narrow enough

Narrow wavepacket

2)(xψ

)(xF

2)()()( xxFdxXF ψ∫=

)( 0xF≈

20 |)(|)()( xdxxFXF ψ∫≈

( )XF≈

x0

x〈X 〉 = x0

A More Precise Formulation

• Expand F(x) around x = 〈X〉:

• Then take the expectation value:

2)()()( xxFdxXF ψ∫=

€

F(X) = F X( ) + ′ F X( ) X − X( ) + ′ ′ F X( )X − X( )2

2+K

€

F(X) = F X( ) I + ′ F X( ) X − X + ′ ′ F X( )X − X( )2

2+K

€

I =1

€

X − X = X − X = 0

€

X − X( )2 = X 2 − 2X X + X 2

= X 2 − X 2

:= ΔX( )2

This is known as the QuantumMechanical Variance. We willstudy it more formally later.

• We have:

• For CM to be a good approximation, it istherefore necessary that:

– The width of the wavepacket squared times thecurvature of the force should be smallcompared to the force itself

• CAUTION: Even if the width of thewavepacket is small enough at one instant tosatisfy this inequality, we also need toconsider the rate of spreading

€

F(X) = F X( ) + ′ ′ F X( ) ΔX( )2

2+K

€

F X( ) >> ′ ′ F X( ) ΔX( )2

2

Example 1: Baseball

• Lets consider a baseball accelerating underthe force of gravity:

– m = 1 kg

– let ΔX = 10-10 m• Will then be stable for 30 million years

– The force is:

• Requirement for CM validity is:

• Since X ≈ Re = 6_107 m this gives

• So CM should work pretty well for a baseball

2)(

X

GMmXF −=

3

2)(

X

GMmXF =′

4

6)(

X

GMmXF −=′′

€

F X( ) >> ′ ′ F X( ) ΔX( )2

€

GMmX 2 >>

6GMmX 4 ΔX( )2

€

X 2>> 6 ΔX( )2

1415 1010 −>>

Example 2: Hydrogen atom

• Consider the very similar problem of anelectron orbiting a proton:– Use ground state parameters

• Thus applicability of classical mechanicsrequires:

• In the ground state we have:

• Which gives

– So CM will not be valid for an electron in thehydrogen ground state

• Highly excited `wavepacket’ states can bedescribed classically Rydberg States– Because 〈X〉 can get extremely large

€

F x( ) = −e2

4πε0 x2

€

′ ′ F X( ) =6F X( )

X 4

€

X 2>> 6 ΔX( )2

20100 −>>

m10~

010

0−=Δ

→

aX

X

Identical result as thebaseball, due to 1/r2

law for gravity andcoulomb forces

Low quantumnumbers:

motion veryquantum

mechanical

Rydberg States in Hydrogen

• The coulomb potential is

High quantum numbers:motion very classical.

Kepplerian orbits.RYDBERG STATES