Quantum Theory of Light and Matter

Transcription

1 Quantum Theory of Light and Matter Field quantization Paul Eastham February 23, 2012

2 Quantization in an electromagnetic cavity Quantum theory of an electromagnetic cavity e.g. planar conducting cavity, E n = 0, B.n = 0 E x (r, t) = 2ω 2 m sin(kz)q n (t). V ɛ n 0 B y (r, t) = ɛµ 2ω 2 m cos(kz) q n (t) k V ɛ n 0

3 Quantization in an electromagnetic cavity Electromagnetic cavity: mode decomposition E x (r, t) = 2ω 2 m sin(kz)q n (t). V ɛ n 0 B y (r, t) = ɛµ 2ω 2 m cos(kz) q n (t) k V ɛ n 0 Normalization of q arbitrary in classical theory Introduce some mass m so q is a length Any EM wave consistent with boundary conditions can be decomposed in this way Functions in decomposition are mode functions

4 Quantization in an electromagnetic cavity Electromagnetic cavity: energy in terms of modes H = 1 2 d 3 r ( ɛ 0 Ex ) By 2 µ 0 = n = n mω 2 nq 2 n 2 mω 2 nq 2 n 2 + m q2 n 2 + p2 n 2m. Energy of the electromagnetic field = energy of set of harmonic oscillators ( radiation oscillators ) Similarly Maxwell s equations become ṗ n = mω 2 nq n, q n = p m.

5 Quantization in an electromagnetic cavity Electromagnetic cavity: quantization H = n mω 2 nq 2 n 2 + p2 n 2m. Hypothesis: quantum theory by replacing q n, p n ˆq n, ˆp n, with [ˆq n, ˆp n ] = i δ n,n Introduce ladder operators as before, (ˆp n, ˆq n ) (â n, â n) H = n ( ω n â nân + 1 ). 2

6 Quantization in an electromagnetic cavity Eigenstates of the field H = n ( ω n â nân + 1 ). 2 set of quantum Harmonic oscillators one for each mode of the field Eigenstates first Harmonic oscillator in eigenstate n 1, second in n 2... denote n 1 n 2 n 3... n 1, n 2, n 3... Energy is ω i (n i + 1/2) i

7 Quantization in an electromagnetic cavity Field operators E x (r, t) = n E x (r) = n = n 2ωnm 2 sin(k n z)q n (t). V ɛ 0 ω n (a n + a V ɛ n) sin(k n z) 0 E n (a n + a n) sin(k n z). also B y (r, t) = n ie n c (a n a n) cos(k n z)

8 Quantization in an electromagnetic cavity Field operators E x (r) = n = n ω n V ɛ 0 (a n + a n) sin(k n z) E n (a n + a n) sin(k n z). Mass disappears Normalization E n natural quantum scale of electric field E n made from, ω electric field of a photon E x and B y the position and momentum of a quantum harmonic oscillator

9 Other geometries: quantization in general Schematic: generalization to other geometries Can generalize this procedure to other geometries. We expect to get and field operators like e.g. H EM = Ê(r) = modes,i modes,i ω i â i âi, E i (r)â + h.c. (Normalization of mode function E i (r) EM energy is ω i when one photon in mode)

10 Other geometries: quantization in general Example: free-space quantization Introduce periodic boundary conditions in box of side L. mode functions are plane waves e k,s e ik.r. labelled by wavevector k = (lπ, mπ, nπ)/l + polarization e k,s (two for every k) Giving Ê(r) = k,s ω k (â k,s e ik.r + â 2ɛ 0 V k,s e ik.r)

11 Physical consequences Electric field distribution in a Fock state Electric field in Fock states What would we find if we measure the electric field of a single mode of the cavity? Depends on state suppose we had an energy eigenstate n Result E x with probability E x n 2 basically the position-space wavefunction of the harmonic oscillator E x n e E 2 x /(2E2 sin 2 kz) H n ( H n are Hermite polynomials : H 0 = 1, H 1 (x) = 2x, H 2 (x) = 4x E x E sin(kz) ).

12 Physical consequences Electric field distribution in a Fock state Electric field in Fock states

13 Physical consequences Electric field distribution in a Fock state Distribution in Fock states Single-mode cavity field in a Fock state Ê x (r) = E(â + â ) sin(kz) E = ω V ɛ 0 n Êx n n (â + â ) n = 0 and n Ê2 x n = E 2 sin 2 (kz) n (â + â ) 2 n = E 2 sin 2 (kz) n ââ + â â n = E 2 sin 2 (kz)(2n + 1).

14 Physical consequences Electric field distribution in a Fock state Distribution in Fock states So n Êx n = 0 n Ê2 x n = E 2 sin 2 (kz)(2n + 1) σ 2 E Expected field is always zero Field fluctuates, variance σe 2 follows mode profile More photons stronger fluctuations σe 2 n Spatial average of σe 2 associated with one photon is E2 Range of electric fields E even when n = 0 vacuum fluctuations

15 Physical consequences Electric field distribution in a Fock state Distribution in Fock states: origins Fluctuations because [E x, B y ] 0 (like position and momentum of harmonic oscillator) H involves E x and B y [H, E x ] 0 So an eigenstate of energy is not an eigenstate of E x

16 Summary Summary Quantum theory of light is constructed by Identifying the normal modes of the EM field Treating these normal modes as quantum harmonic oscillators This reproduces the Planck ansatz and gives us expressions we can work with for the electric and magnetic field operators The eigenstates of the field involve Fock states n in which the electric field is zero on average but will not always be zero in a measurement Even in the lowest energy state 0 we will measure E x 0 often vacuum fluctuations