Quantum Optics An Introduction (Oxford Master Series in Physics No. 15) di: Mark Fox Editore: Oxford University Press

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Quantization of the e.m. field

Classical harmonic oscillators

�

EHO = px2

2m+ 12mω 2x 2 = mω 2x0

2

2cos2ωt + sin2ωt( )

Mass attached to spring

�

F = m˙ x = −kx

�

⇒ ˙ x = −ω 2x, ω = km

�

x t( ) = x0 sinωt; px t( ) = mωx0 cosωt

Light waves as classical harmonic oscillators I

EM standing wave in cavity

By z, t( ) = ε0µ0ε0ωk

coskzcosωt = ε0

ccoskzcosωt = B0 coskzcosωt

∇⋅ε = ρε0

∇⋅B = 0

∇×ε = − ∂B∂t

∇× B = µ0 j + ε0∂ε∂t

⎛⎝⎜

⎞⎠⎟

Maxwell, mon amour

ε x z, t( ) =ε0 sinkzsinωt

−∂By

∂z= ε0µ0

∂ε x

∂t= ε0µ0ε0ω sinkzcosωt c = ω

k= 1

ε0µ0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Light waves as classical harmonic oscillators II

Energy of EM standing wave in cavity

Emag =14µ0

VB02 cos2ωt

Local energy density – to be integrated in space across cavity volume U = 1

2ε0ε 2 + B

2

µ0

⎛⎝⎜

⎞⎠⎟

Eel =12ε0A ε0

2 sin2 kzsin2ωt dz0

L

∫ = 14ε0Vε0

2 sin2ωt

Eem = V4

ε0ε02 sin2ωt + B0

2

µ0cos2ωt

⎛⎝⎜

⎞⎠⎟= V4ε0ε0

2 sin2ωt + cos2ωt( )

Light waves as classical harmonic oscillators III

New coordinates

q t( ) = ε0V2ω 2ε0 sinωt

p t( ) = V2µ0

B0 cosωt =V2µ0

ε0

ccosωt = ε0V

2ε0 cosωt

q t( ) = mx t( )

p t( ) = 1mpx t( )

p = qq = p = −ω 2q

Eem = 12p2 +ω 2q2( )

Light as a quantum harmonic oscillator

En = n + 12

⎛⎝⎜

⎞⎠⎟ ω

ΔxΔpx ≥2

⇒ ΔqΔp ≥ 2

q t( ) = mx t( )

p t( ) = 1mpx t( )

Vacuum field

U = 12

ε0ε 2 + B2

µ0

⎛⎝⎜

⎞⎠⎟

E0 =12ω = 2 × 1

2ε0εvac

2 dV∫

εvac =ω2ε0V

Casimir force

FCasimir =π 2c240L4

A

Numerical examples: For A = 1 m2 and L = 10-3 m, F = 1.3 10-15 N. For A = 10-4 m2 and L = 10-6 m, F = 1.3 10-7 N.

Between parallel perfect mirrors

Statistics of light

Operator solution of the harmonic oscillator I

a = 12mω

mω x + ipx( )

a† = 12mω

mω x − ipx( )

H = px2

2m+ 12mω 2 x2

Hψ x( ) = Eψ x( )

x = 2mω

a + a†( )

px = −i mω2

a − a†( )

Operator solution of the harmonic oscillator II

Operator solution of the harmonic oscillator II

a†a = 1ω

px2

2m+ 12mω 2 x2 − 1

2ω

⎛⎝⎜

⎞⎠⎟

aa† = 1ω

px2

2m+ 12mω 2 x2 + 1

2ω

⎛⎝⎜

⎞⎠⎟

H = ω a†a + 12

⎛⎝⎜

⎞⎠⎟�

ˆ a , ˆ a †[ ] = ˆ a a † − ˆ a † ˆ a = 1

Operator solution of the harmonic oscillator III

= ω a† aa† − a†a( ) = ω a† a†, a⎡⎣ ⎤⎦ = ω a†

H, a†⎡⎣ ⎤⎦ = ω a†a + 12

⎛⎝⎜

⎞⎠⎟ , a

†⎡⎣⎢

⎤⎦⎥= ω a†aa† − a†a†a( ) =

Hψ n = Enψ n

Ha†ψ n = H, a†⎡⎣ ⎤⎦ + a†H( )ψ n = ω a

† + a†H( )ψ n = ω + En( ) a†ψ n

Haψ n = −ω + En( ) aψ n

H, a⎡⎣ ⎤⎦ = −ω a

are still eigenfunctions of H nn aa ψψ ˆ;ˆ†

Ground state

aψ 0 = 0

Hψ 0 = E0ψ 0

aψ 0 =1

2mωmω x + ipx( )ψ 0 =

12mω

mω xψ 0 + i −i( ) ∂∂ x

ψ 0 = 0

dψ 0

dx= −mω

xψ 0

ψ 0 = Ce−mω x

2

2 e−ax2

dx = πa−∞

∞

∫

Hψ 0 = ω a†a + 12

⎛⎝⎜

⎞⎠⎟ψ 0 =

12ωψ 0

�

E0 = 12ω

Energy ladder and number operator

En = E0 + nω = n + 12

⎛⎝⎜

⎞⎠⎟ ω

ψ n x( ) = Cn a†( )nψ 0 x( )

Hψ n = Enψ n

ω a†a + 12

⎛⎝⎜

⎞⎠⎟ψ n = n + 1

2⎛⎝⎜

⎞⎠⎟ ωψ n

a†aψ n = nψ n

a†a = n

Photon number states

H n = En n = n + 12

⎛⎝⎜

⎞⎠⎟ ω n

n ′n = δn ′n

a† n = n +1 n +1

a n = n n −1

n = 1n!

a†( )n 0

n n = a†a n = na† n −1 = n n −1+1 n = n n

Electric field of a photon

x = 2mω

a + a†( )

ε x z, t( ) =ε0 sinkzsin ωt +φ( )

ε x z, t( ) = 2ω 2

ε0Vq t( )sinkx

q t( ) = ε0V2ω 2ε0 sinωt; x t( ) = 1

mq t( )

ε x =2ω 2

ε0Vm

2mωa + a†( )sinkx = ω

ε0Va + a†( )sinkx

n ε x n = 0