Quantization of the electromagnetic field
1
The classical electromagnetic field
2
Maxwell Equations
r·E=
Gauss’s law
⇢
"0
r·B=0
Gauss’s law for magnetism
Maxwell-Faraday equation
(Faraday’s law of induction)
Ampere’s circuital law
(with Maxwell’s correction)
3
r ⇥ E = _ 1c @B
@t
r ⇥ B = µ0 J + µ0 "0 @E
@t
Maxwell Equations
• In empty space
p
(c = 1/ µ0 "0 )
Gauss’s law
Gauss’s law for magnetism
Maxwell-Faraday equation
r·E=0
r·B=0
r ⇥ E = _ 1c @@Bt
r⇥B=
Ampere’s circuital law
4
1 @E
c @t
Wave Equations
2
r E
1 @2E
=
0
c2 @t2
r2 B
1 @2B
=
0
c2 @t2
5
Derivation of wave equations
• Curl of Maxwell Faraday equation
1 @r ⇥ B
r ⇥ (r ⇥ E) = _
c
@t
• Use Ampere’s Law
and vector identity r ⇥ (r ⇥ ~
v
) = r(r ·
~v
) - r2
~v
r(r · E)
2
r E=
6
1 @
c @t
✓
1 @E
c @t
◆
Derivation of wave equations
• Use Gauss Law
⇠
⇠
r(⇠
r · E)
1 @
c @t
2
r E=
• Obtain wave equation
2
r E=
2
1 @ E
c2 @t
7
✓
1 @E
c @t
◆
Wave equation
2
r E
2
1 @ E
=0
2
2
c @t
from separation of • Eigenvalue equationX
x
, t) =
variables:
E(~
fm
(t)u
~
m (x)
~
m
2
r um =
d2 fm
2 2
+
c
k
f
(t)
=
0
m
m
dt2
2
km um
8
Normal modes
• {u } are eigenfunctions of the wave equation
• Boundary conditions (from Maxwell eqs.)
m
r · um = 0,
~n ⇥ um = 0
condition
• Orthonormality
Z
3
~um (x)~un (x)d x = 6n,m
• They form a basis.
9
B-field
• Electric field in {u
E(x
~
, t) =
m}
X
basis:
~
m (~x
)
fm
(t)u
m
•
Magnetic field in {um } basis
B(x, t) =
X
m
hm (t) (r ⇥ um (x))
10
B-field solution
• What are the coefficients h ?
n
•
We still need to satisfy Maxwell equations:
r⇥E =
X
n
•
fn (t)r ⇥ un =
Solution:
1
@t B
c
!
1X
@t hn (t)r ⇥ un
c n
d hn
= -cfn
dt
11
Eigenvalues of hn
• Find equation for h
n
only:Ampere’s law
1 @E
r⇥B =
c @t
X
n
!
1 X d fn
hn (t)r ⇥ (r ⇥ un ) =
un
c n dt
X
n
X d fn
1
2
h n r un =
un
c n dt
12
Eigen-equations
• Eigenvalue equation for h
n
2
d
2 2
hn (t) + c kn hn (t) = 0
2
dt
• Eigenvalue equation for f
n
d2 fn
2 2
+
c
k
f
(t)
=
0
n
n
dt2
13
E.M. field Hamiltonian
• Total energy:
1
H/
2
Z
2
2
3
(E + B )d x
• Substituting, integrating and using
orthonormality conditions:
1
H=
8⇡
✓
X
n,m
fn fm
Z
un (x)um (x)d3 x + hn hm
X 1
H=
(fn2 + kn2 h2n
)
8⇡
n
14
Z
(r ⇥ un ) · (r ⇥ um )d3 x
◆
E.M. field as H.O.
• Hamiltonian looks very similar to a sum of
harmonic oscillators:
X
1
1
1
X
2
2 2
2
2 2
(p
n
+ !
n
q
n
) ,
He.m.
=
(f
n
+ kn
h
n
)
Hh.o. =
2 n
4⇡
2
n
•h
n
is derivative of fn
⇒ identify with momentum
15
Quantized electromagnetic field
16
Operators
to the
• We associate quantum operators
ˆ
coefficients fn , fn ! fn
• We write this operator in terms of
annihilation and creation operators
p
†
ˆ
fn = 2⇡!n ~(an + an )
that create or destroy one mode of the e.m. field
17
Operator fields
• Electric field
Xp
†
E(x, t) =
2~⇡!n [an (t) + an (t)]un (x)
n
• Magnetic field
B(x, t) =
X
n
icn
r
2⇡~
†
[an
!n
18
an ]r ⇥ un (x)
Hamiltonian
• The Hamiltonian is then simply expressed
in terms of the an operators
◆
✓
X
1
†
H=
!n an an +
2
n
• The frequencies are
!n (k) = c|~kn |
19
Gauges
Lorentz (scalar potential ' = 0 )
~ )
Coulomb (vector potential r · A=0
20
Zero-Point Energy
21
Field in cavity
• Field in a cavity of volume
V = Lx Ly Lz
• Given the boundary
conditions,
the normal modes are:
un,↵ = A↵ cos(kn,x rx ) sin(kn,y ry ) sin(kn,z rz )
• with
kn,↵
n↵ ⇡
=
, n↵ 2 N
L↵
22
Polarization
• Because of the boundary condition,
r · ~un = 0
• the coefficients A must satisfy:
Ax kn,x + Ay kn,y + Az kn,z = 0
• For each set {n , n , n } there are 2 solutions
x
y
z
Two polarizations per each mode
23
Electric field in cavity
has a simple form
• The electric field X
E(x, t) =
(E↵ +
↵=1,2
• with
•
E↵†
and En =
= ê↵
X
n
r
~!n
2✏0 V
†
E↵ )
† i(~
En an e kn ·r~ !t)
the field of one photon
of frequency !n
24
Energy density
E = hHi = 2
kc
X
k=1
⌧
1
†
~!k ak ak +
2
• The Zero-point energy density is then
kc
X
2
1
E0 =
~!k
V
2
k=1
25
Energy density
• If cavity is large, wavevector is almost
continuous
X
1
8
k>0
26
Z
3
d k⇢(k)
Zero-point energy
• Integrating over the positive octant
2 2V 4⇡
E0 =
V ⇡3 8
Z
kc
1 3
dk ~k c
2
k=0
• setting a cutoff k ,we have
c
c~
E0 =
2⇡ 2
Z
kc
3
dkk =
k=0
27
4
~ckc
8⇡ 2
Zero-point energy
• It’s huge!
Cutoff at visible frequency
Ac = 2⇡/k = 0.4 ⇥ 10-6 m
Image by MIT OpenCourseWare.
2.7 ⇥ 10
8
3
3
23 J/m
J/m @ 1m
• But is it ever seen?
28
Casimir Effect
• Dutch theoretical physicist Hendrik Casimir
(1909–2000) first predicted in 1948 that when
two mirrors face each other in vacuum,
fluctuations in the vacuum exert “radiation
pressure” on them
29
Casimir Effect
• Cavity bounded by
L
conductive walls
WL
• Add a conductive plate
@ distance R
WR WL-R
R
• Change in energy is:
W = (WR + WL
30
R)
WL
Casimir effect
• Each term is calculated from zero-point
energy
• Continuous approximation is not valid if R
is small
• Thus the difference ∆W is not zero
W =
⇡ 2 L2
~c
720 R3
31
Casimir Force
• The difference in energy corresponds to an
attractive force
F =
@ W
=
@R
• or a pressure
P =
2
⇡ L
~c
240 R4
⇡ 2 ~c
240 R4
32
2
Casimir in MEMS
© Source unknown. All rights reserved. This content is excluded from our Creative
Commons license. For more information, see http://ocw.mit.edu/fairuse.
Quantum Mechanical Actuation of Microelectromechanical Systems by the Casimir Force
H. B. Chan, V. A. Aksyuk, R. N. Kleiman, D. J. Bishop and Federico Capasso
Science 9 March 2001: Vol. 291 no. 5510 pp. 1941-1944
33
MIT OpenCourseWare
http://ocw.mit.edu
22.51 Quantum Theory of Radiation Interactions
Fall 2012
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