Download Lecture 13: The classical limit

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Lecture 13: The classical limit
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Phy851/fall 2009
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Wavepacket Evolution
• For a wavepacket in free space, we have
already seen that
x = x0 +
p0
t
M
p = p0
– So that the center of the wavepacket obeys
Newton’s Second Law (with no force):
p
d
x =
dt
m
d
p =0
dt
• Assuming that:
– The wavepacket is very narrow
– spreading is negligible on the relevant timescale
• Would Classical Mechanics provide a
quantitatively accurate description of the
wavepacket 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?
Equation of motion for expectation
value
• How do we find equations of motion for
expectation values of observables?
– Consider a system described by an arbitrary
Hamiltonian, H
– Let A be an observable for the system
– Question: what is:
d
A
dt
?
• Answer:
d
d
A =
ψ (t ) A ψ (t )
dt
dt
d
=  ψ (t )
 dt
=

 ∂A 
 A ψ (t ) + ψ (t )   ψ (t ) + ψ (t )

 ∂t 
i
i
∂A
ψ (t ) HA ψ (t ) − ψ (t ) AH ψ (t ) +
h
h
∂t
d
i
∂A
A = − [ A,H ] +
dt
h
∂t
€
d

A ψ (t ) 
 dt

€
Example 1: A free particle
• Assuming that
∂A
=0
∂t
• The basic equation of motion is:
d
i
A = − [ A,H ]
dt
h
P2
• For a free particle, we have: H =
2M
€
1
X,H
=
X,P 2 ]
[ ]
[
[H,P ] = 0
2M
1
2
2
=
XP
−
P
X)
(
2M
Very common
1
2
2
=
XP − PXP€+ PXP − P X ) trick in QM to
(
2M
become familiar
1
with
=
[ X,P ] P + P [ X,P ])
(
€
2M
P
=
ih
€
M
€
P
d
i
P
X = − ih
=
dt
h
m
m
d
P =0
dt
Adding the Potential
• Let
P2
H=
+V (X )
2M
1
2
X,P
+ [ X,V (X)]
[ X,H ] =
[
]
2M
=
0
1
=
X,P 2 ]
[
2M
P
= ih
M
€
€
[H,P ] =
1
P 2 ,P ] + [V (X),P ]
[
2M
=
0
= −[ P,V (X)]
€
• How do we handle this commutator?
€
One possible approach
[P,V ( X )]
• We want an expression for:
• We can instead evaluate:
x [ P,V (X)] ψ
Theorem: If 〈x|A|ψ〉=〈x|B|ψ〉 is true for any x and
€
|ψ〉, then it follows
that A=B.
• Will need to make use of
x P ψ = −ih
d
xψ
dx
You should think of
this as the defining
equation for how to
handle P in x-basis
x [ P,V (X)] ψ = x PV (X) ψ − x V (X)P ψ
= −ih x V ′( X ) ψ
€
– Where
d
V ′( x) = V ( x)
dx
• Thus we have:
[P,V ( X )]= −ihV ′( X )
You will derive this
in the HW
Equations of motion for 〈X〉 and 〈P〉
• As long as:
P2
H=
+V (X )
2M
P
d
• The we will have:
X =
dt
m
– Not just true for a wavepacket
• For the momentum we have:
€
d
i
P = − [ P,V (X)]
dt
h
i
= − − ihV ′( X )
h
= − V ′( X )
= F(X )
F(X) is the Force
operator
F ( x) = −
d
V ( x)
dx
• The QM form of the Second Law is Thus:
d2
M 2 X = F(X )
dt
Difference between classical and QM forms
of the 2nd Law of Motion
• Classical:
d2
M 2 x = F ( x)
dt
• Quantum:
d2
M 2 X = F(X )
dt
• Classical Mechanics would be an accurate
description of the motion of the center of a
wavepacket, defined as x (t ) = X , if:
F(X) ≈ F ( X
• So that
€
)
d2
M 2 X = F( X )
dt
• This condition is satisfied in the limit as the
width of the wavepacket goes to zero
• ALWAYS TRUE for a constant (e.g. gravity) or
linear force (harmonic oscillator potential),
regardless of the shape of the wavefunction
Narrow wavepacket
• The expectation value of the Force is:
F ( X ) = ∫ dx F ( x)ψ ( x)
2
• Let us assume that the force F(x) does not
change much over the length scale σ
F (x)
ψ (x)
2
〈X 〉 = x0
x0
• In this case we can safely pull F(x) out of
integral:
F ( X ) ≈ F ( x0 ) ∫ dx | ψ ( x) |2
≈ F ( x0 )
≈ F( X
)
• So CM is a valid description if the wavepacket
is narrow enough
x
A More Precise Formulation
F ( X ) = ∫ dx F ( x)ψ ( x)
2
• Expand F(x) around x = 〈X〉:
F(X) = F ( X ) + F ′( X
)( X − X ) + F ′′( X
)
(X −
X
2
)
2
+K
• Then take the expectation value:
F(X) = F ( X
)I
+ F ′( X
) X−
X + F ′′( X
)
(X − X
2
I =1
X − X = X − X =0
€
(X − X
€
)
2
= X 2 − 2X X + X
= X2 − X
:= (ΔX )
€
2
2
2
This is known as the Quantum
Mechanical Variance. We will
study it more formally later.
)
2
+K
• We have:
F(X) = F ( X ) + F ′′( X
)
(ΔX )
2
2
+K
• For CM to be a good approximation, it is
therefore necessary that:
€
F( X
)
ΔX )
(
>> F ′′( X )
2
2
– The width of the wavepacket squared times the
curvature of the force should be small
€ compared to the force itself
• CAUTION: Even if the width of the
wavepacket is small enough at one instant to
satisfy this inequality, we also need to
consider the rate of spreading
Example 1: Baseball
• Lets consider a baseball accelerating under
the force of gravity:
– m = 1 kg
– let ΔX = 10-10 m
• Will then be stable for 30 million years
– The force is:
F(X ) = −
GMm
X2
2GMm
X3
6GMm
F ′′( X ) = −
X4
F ′( X ) =
• Requirement for CM validity is:
F( X
) >> F ′′( X ) (ΔX )2
GMm
6GMm
2
>>
ΔX
(
)
2
4
X
X
€
X
2
>> 6(ΔX )
2
15
−14
• Since X ≈ Re = 6_107 m this gives 10 >> 10
€
• So CM€should work pretty well for a baseball
Example 2: Hydrogen atom
• Consider the very similar problem of an
electron orbiting a proton:
– Use ground state parameters
e2
F( x ) = −
4 πε0 x
F ′′( X ) =
6F ( X
X
€
2
)
4
• Thus applicability of classical mechanics
requires:
€
X
2
>> 6(ΔX )
2
• In the ground state we have:
Identical result as the
baseball, due to 1/r2
law for gravity and
coulomb forces
X →0
ΔX ~ a0 = 10 −10 m
€
• Which gives
0 >> 10 −20
– So CM will not be valid for an electron in the
hydrogen ground state
• Highly excited `wavepacket’ states can be
described classically  Rydberg States
– Because 〈X〉 can get extremely large
Rydberg States in Hydrogen
• The coulomb potential is
High quantum numbers:
motion very classical.
Kepplerian orbits.
RYDBERG STATES
Low quantum
numbers:
motion very
quantum
mechanical