lecture 13: the classical limit - michigan state university · wavepacket evolution •for a...
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Lecture 13: The classical limit
Phy851/fall 2009€
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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( )
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=12M
XP 2 − PXP + PXP − P 2X( )
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=12M
X,P[ ]P + P X,P[ ]( )
M
Pih=
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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 +=
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X,H[ ] =12M
X,P 2[ ] + X,V (X)[ ]
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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
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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