Quantization of the elect r omagnetic field

The classical elect r omagnetic field

Max w ell Equations

Max w ell Equations

• In empty space ( c = 1 / p µ 0 ε 0 )

G au ss’ s l a w r · E = 0 G au ss’ s l a w f or magn et i sm r · B = 0 M ax w el l - F ar ad a y eq u at i on r ⇥ E = _ 1 @ B

Amp er e’ s ci r cu i t al l a w r ⇥ B = 1 @ E

W a v e Equations

∇ 2 E —

∇ 2 B —

1 @ 2 E

c 2 @ t 2 = 0

1 @ 2 B

c 2 @ t 2 = 0

Derivation of w a v e equations

• Curl of Maxw ell Farad a y equation

r ⇥ ( r ⇥ E ) = _ 1 @ r ⇥ B

c @ t

• Use Amper e ’ s L a w

and v ector identity r ⇥ ( r ⇥ ~ v ) = r ( r · ~ v ) - r 2 ~ v

∇ ( ∇ · E ) — ∇ 2 E = — 1 @ 1 @ E

c @ t c @ t

Derivation of w a v e equations

• Use Gauss L a w

∇ ( ⇠ ∇ ⇠ · ⇠ E ) — ∇ 2 E = — 1 @ 1 @ E

c @ t c @ t

• Obtain w a v e equation

— ∇ 2

1 @ 2 E

E = — c 2 @ t

W a v e equation

∇ 2 E

1 @ 2 E

c 2 @ t 2 = 0

• Eigen value equation f r om separation of

variables:

E ( ~ x ,

t ) = X

m

f m ( t ) ~ u m ( ~ x )

2 2 d 2 f m 2 2

∇ u m = — k m u m

dt 2

+ c k m f m ( t ) = 0

Normal modes

• { u m } ar e eigenfunctions of the w a v e equation

• Boundar y conditions (f r om Maxw ell eqs.)

r · u m = 0 , ~ n ⇥ u m = 0

• Or thonormality condition

Z ~ u m ( x ) ~ u n ( x ) d 3 x = 6 n,m

• The y f orm a basis.

B- field

• Electric field in { u m } basis:

E ( ~ x ,

t ) = X

m

f m ( t ) ~ u m ( ~ x )

• Magnetic field in { u m } basis

B ( x, t ) = X h m ( t ) ( r ⇥ u m ( x ) )

m

B- field solution

• What ar e the coefficients h n ?

• W e still need to satisfy Maxw ell equations:

1

∇ × E = — @ t B →

c

X f n

( t ) ∇ × u n

= — 1

c

X @ t

h n ( t ) ∇ × u n

n n

• Solution: d h n = - cf

d t n

Eige n values of h n

• Find equation f or h n onl y: Amper e ’ s l a w

r ⇥ B =

1 @ E

c @ t

X h ( t ) r ⇥ ( r ⇥ u

) = 1 X

d f n u

n n c d t n

n n

→ — X h

∇ 2 u

= 1 X

d f n u

n n c d t n

n n

Eigen-equations

• Eigen value equation f or h n

d 2 2 2

dt 2

h n ( t ) + c

k n h n ( t ) = 0

• Eigen value equation f or f n

d 2 f n

dt 2

+ c 2 k 2 f n ( t ) = 0

E.M. field Hamiltonian

• T otal energ y:

H / 1

2

( E 2 + B 2 ) d 3 x

• Substituting, integrating and using or thonormality conditions:

1

H = 8 ⇡ f n f m

n,m

u n ( x ) u m

( x ) d 3 x + h n h m Z

( r ⇥ u n

) · ( r ⇥ u m

) d 3 x ◆

H = X

1 ( f 2 + k 2 h 2 ) 8 ⇡

E.M. field as H. O .

• Hamiltonian looks v er y similar to a sum of harmonic oscillators:

H = 1 X (p 2

+ ω 2 q 2 ) e H

= 1 X 1

(f 2

+ k 2 h 2 )

h. o. 2 n n n

n

e . m .

2 4 π n n n n

• h n is derivativ e of f n

* identify with momentum

Quantized elect r omagnetic field

Operators

• W e associate quantum operators to the coefficients f n , f n ! f ˆ n

• W e write this operator in terms of annihilation and cr eation operators

f ˆ n

= p 2 πω

~ ( a †

+ a n )

that cr eate or dest r o y one mode of the e .m. field

Operator fields

• Electric field

E ( x, t ) = X p 2 ~ π ω n [ a †

( t ) + a n ( t )] u n ( x )

n

• Magnetic field

B ( x, t ) = X ic

r 2 π ~ [ a † — a

] ∇ × u

( x )

Hamiltonian

• The Hamiltonian is then simpl y expr essed in terms of the a n operators

H = X ω n ✓ a † a n +

n

• The fr equencies ar e

ω n ( k ) = c | → k n |

Gauges

Lor entz (scalar potential ϕ = 0 ) Coulomb (v ector potential r · A ~ =0 )

Ze r o- P oint Energ y

Field in c a vity

• Field in a c a vity of v olume V = L x L y L z

• Giv en the boundar y conditions,

the normal modes ar e:

u n, ↵ = A ↵ cos( k n,x r x ) s i n ( k n,y r y ) s i n ( k n,z r z )

• with k n, ↵ =

n ↵ ⇡ L ↵

, n ↵ 2 N

P olarization

• Because of the boundar y condition,

r · ~ u n = 0

• the coefficients A m ust satisfy:

A x k n,x + A y k n,y + A z k n,z = 0

• F or each set { n x , n y , n z } ther e ar e 2 solutions T w o polarizations per each mode

Electric field in c a vity

• The electric field has a simple f orm

E ( x, t ) = X ( E ↵ + E † )

↵ =1 , 2

• with

† e ˆ α

X E n

n

† i ( → k n · → r — u t ) n

• and E n =

~ ω n the field of one photon

2 ϵ 0 V

of fr equency ω n

Energ y density

X k c ⌧ 1

• The Ze r o-point energ y density is then

2 k c 1

E 0 = X ~ ω k

k =1

Energ y density

• If c a vity is large , w a v e v ector is almost contin uous

Ze r o-point energ y

• Integrating o v er the positiv e octant

2 2 V 4 ⇡

Z k c 1

E 0 = dk ~ k 3 c

V ⇡ 3 8 k =0 2

• setting a cutoff k c ,w e h a v e

c ~

Z k c

~ ck 4

2 ⇡ 2 k =0 8 ⇡ 2

Ze r o-point energ y

• It ’ s huge!

Cutoff at visible fr equency

A c = 2 ⇡ /k = 0 . 4 ⇥ 10 - 6 m

Image by MIT OpenCourseWare.

2 . 7 ⇥ 10 — 8 J / m 3

@ 1m 23 J / m 3

• But is it e v er seen?

Casimir Eff ect

• Dutch theor etical p h ysicist Hendrik Casimir (1909–2000) first pr edicted in 1948 that when

tw o mir r ors face each other in vacuum, fluctuations in the vacuum e x er t “radiation pr essur e” on them

Casimir Eff ect

• C a vity bounded b y L conductiv e walls

W L

• Ad d a conductiv e plate

@ distance R R

• Change in energ y is:

∆ W = ( W R + W L — R ) — W L

Casimir eff ect

• Each term is calculated f r om ze r o-point energ y

• Contin uous a pp r o ximation is not valid if R is small

• Thus the diff er ence ∆ W is not ze r o

⇡ 2 L 2

∆ W = — ~ c 720 R 3

Casimir F o r ce

• The diff er ence in energ y cor r esponds to an attractiv e f o r ce

@ ∆ W ⇡ 2 L 2

F = —

@ R

• or a pr essur e

= — ~ c 240 R 4

⇡ 2 ~ c

P = — 240 R 4

Casimir in MEMS

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Fall 2012

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