quantization of the electromagnetic field · dutch theoretical physicist hendrik casimir...
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Quantization of the electromagnetic field
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The classical electromagnetic field
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Maxwell Equations
Gauss’s law r · E = ⇢ " 0
Gauss’s law for magnetism r · B = 0
Maxwell-Faraday equation r⇥ E = _ 1 c @B @t
(Faraday’s law of induction)
Ampere’s circuital law r⇥ B = µ0J + µ0 " 0 @E @t
(with Maxwell’s correction)
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Maxwell Equations p• In empty space (c = 1/ µ0 " 0)
Gauss’s law r · E = 0
Gauss’s law for magnetism r · B = 0
Maxwell-Faraday equation r⇥ E = _ 1 c @@B t
Ampere’s circuital law r⇥ B = 1 @E c @t
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Wave Equations
1 @2E r2E � = 0 c2 @t2
1 @2B r2B � = 0 c2 @t2
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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
1 @ ✓ 1 @E
◆
r(r · E) r2E = c @t c @t
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Derivation of wave equations
• Use Gauss Law
1 @ ✓ 1 @E
◆
r(⇠⇠⇠r · E) r2E = c @t c @t
• Obtain wave equation
1 @2E �r2E = � c2 @t
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�
Wave equation 1 @2E r2E = 0 c2 @t2
• Eigenvalue equation from separation of
t) =X
,x~ ~~E( (t)um(x) m
d2fm2 2k2 r um = �k2 um + c fm(t) = 0 m mdt2
variables: fm
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Normal modes
• {um} are eigenfunctions of the wave equation
• Boundary conditions (from Maxwell eqs.)
r · um = 0, ~ ⇥n um = 0
• Orthonormality condition Z
~um(x)~un(x)d3 x = 6n,m
• They form a basis.
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B-field
• Electric field in {um} basis:
)~x(t)um( m
~
• Magnetic field in {um} basis
B(x, t) = X
hm(t) (r⇥ um(x)) m
t) =X
, E(x~ fm
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B-field solution
• What are the coefficients hn?
• We still need to satisfy Maxwell equations:
1 r⇥ E = � @tB ! c
1X fn(t)r⇥ un = �
X @thn(t)r⇥ un
c n n
• Solution: d hn = -cfnd t
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Eigenvalues of hn
• Find equation for hn only:Ampere’s law
1 @E r⇥ B = c @t
1 d fnX hn(t)r⇥ (r⇥ un) =
X un
c d t n n
2 1 d fn! � X
hnr un = X
un c d t
n n
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Eigen-equations
• Eigenvalue equation for hn
d2 2k2hn(t) + c hn(t) = 0 ndt2
• Eigenvalue equation for fn
d2fn + c (t) = 0dt2
2 2k fnn
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E.M. field Hamiltonian
• Total energy:
1 Z
H / (E2 + B2)d3 x
2
• Substituting, integrating and using orthonormality conditions:
1 ✓ Z Z ◆
H = X
fnfm un(x)um(x)d3 x + hnhm (r⇥ un) · (r⇥ um)d3
x 8⇡
n,m
1 H = X
(f2 + k2 h2 )n n n8⇡ n 14
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E.M. field as H.O.
• Hamiltonian looks very similar to a sum of harmonic oscillators:
1 X(p2n+ !2nq 2n) , He.m. =
1 X 1 2 4⇡
(f2n+ k Hh.o. = h )
2 nn
• hn is derivative of fn ⇒ identify with momentum
2n
2n
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Quantized electromagnetic field
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Operators • We associate quantum operators to the
coefficients fn, fn ! f n
• We write this operator in terms of annihilation and creation operators
†f n =
p2⇡!n ~(a + an)n
that create or destroy one mode of the e.m. field
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Operator fields
• Electric field
†E(x, t) = Xp
2~⇡!n[a (t) + an(t)]un(x)nn
• Magnetic field r
2⇡~ †B(x, t) = X
icn [a � an]r⇥ un(x)n!n n
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Hamiltonian
• The Hamiltonian is then simply expressed in terms of the an operators
✓ † 1
◆
H = X
!n a an +n 2 n
• The frequencies are
!n(k) = c|~kn|
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Gauges
Lorentz (scalar potential ' = 0 ) ~Coulomb (vector potential r · A=0)
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Zero-Point Energy
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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,xrx) sin(kn,yry) sin(kn,zrz )
n↵⇡• with kn,↵ = , n↵ 2 N L↵
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Polarization
• Because of the boundary condition,
r · ~un = 0
• the coefficients A must satisfy: A
x
kn,x + A
y
kn,y + A
z
kn,z = 0
• For each set {nx
, ny
, nz } there are 2 solutions
Two polarizations per each mode
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Electric field in cavity
• The electric field has a simple form
E(x, t) = X
(E↵ + E↵† ) ↵=1,2
E† X
† i(~kn ·r~�!t)• with ↵ = e↵ Enane n
r
• and En = ~!n the field of one photon2✏0V
of frequency !n
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Energy density kc ⌧
1 �
†E = hHi = 2 X
~!k akak + 2
k=1
• The Zero-point energy density is then
kc2 1 E0 =
X ~!k
V 2 k=1
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Energy density
• If cavity is large, wavevector is almost continuous
X 1 8
Z d3k⇢(k)
k>0
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Zero-point energy
• Integrating over the positive octant Z kc2 2V 4⇡ 1
E0 = dk ~k3 c V ⇡3 8 k=0 2
• setting a cutoff kc,we have
Z kcc~ ~ck4 cE0 = dkk3 =
2⇡2 k=0 8⇡2
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Zero-point energy • It’s huge!
Cutoff at visible frequency
Ac = 2⇡/k = 0.4 ⇥ 10-6 m
332.7 ⇥ 10 8J/m @ 1m 23 J/m
• But is it ever seen?
Image by MIT OpenCourseWare.
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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
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Casimir Effect
• Cavity bounded by L conductive walls
WL
• Add a conductive plate @ distance R
• Change in energy is:
�W = (WR + WL�R) � WL
R
WL-RWR
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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
⇡2 L2
�W = �~c 720 R3
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Casimir Force
• The difference in energy corresponds to an attractive force
@�W ⇡2 L2
F = � = �~c @R 240 R4
• or a pressure ⇡2 ~c
P = � 240 R4
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Casimir in MEMS
Quantum Mechanical Actuation of Microelectromechanical Systems by the Casimir ForceH. 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
© Source unknown. All rights reserved. This content is excluded from our CreativeCommons license. For more information, see http://ocw.mit.edu/fairuse.
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MIT OpenCourseWarehttp://ocw.mit.edu 22.51 Quantum Theory of Radiation InteractionsFall 2012 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.