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Photons Where We Consider a Radiation field Whose Vector Potential a Ω = |K|C

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Photons Where We Consider a Radiation field Whose Vector Potential a Ω = |K|C Photons where We consider a radiation field whose vector potential A ω = |k|c. satisfies the transversality condition Now ∇ · A = 0. A(x, t) 1 X X Because the electric and magnetic fields √ (α) ik·x ∗ (α) −ik·x = (ck,α(t)ˆ e + c (t)ˆ e ) V k,α 1 ∂ k α E = − A 1 X X c ∂t √ (α) ik·x ∗ (α) −ik·x = (ck,α(0)ˆ e + c (0)ˆ e ), B = ∇ × A V k,α k α satisfy the free space Maxwell equations where we have employed the four-vector notation ∇ · E = 0 k · x = k · x − ωt = k · x − |k|ct. ∂B ∇ × E = − ∂t The Hamiltonian function of the classical radiation field is ∇ · B = 0 Z 1 ∂E 1 2 2 3 ∇ × B = , H = (|B| + |E| ) d x c ∂t 2 1 Z = |∇ × A|2 + |(1/c)(∂A/∂t)|2 d3x. the vector potential satisfies the wave equation 2 1 ∂2A ∇2A − = 0. A straightforward calculation shows that c2 ∂t2 X X H = 2(ω/c)2c∗ c . We write the vector potential at the moment t = 0 as a k,α k,α α superposition of the periodically normalized plane waves k in an L-sided cube, Because the coefficients u (x) = ˆ(α)eik·x, −iωt k,α ck,α(t) = ck,α(0)e like: satisfy the equation of motion 1 X X 2 √ c¨ = −ω c , A(x, t)|t=0 = (ck,α(0)uk,α(x) k,α k,α V k α=1,2 it would look like the classical radiation field were +c∗ (0)u∗ (x)). k,α k,α composed of independent harmonic oscillators. We define the variables Here V = L3 and ˆ(α), α = 1, 2 are real polarization 1 vectors. Q = (c + c∗ ) Due to the transversality condition we have k,α c k,α k,α iω P = − (c − c∗ ). ˆ(α) · k = 0, k,α c k,α k,α so the polarization can chosen so that the vectors With the help of these the Hamiltonian function can be (ˆ(1), ˆ(2), k/|k|) form a righthanded rectangular written as coordinate system. The Fourier components uk,α satisfy X X 1 H = (P 2 + ω2Q2 ). the orthogonality conditions 2 k,α k,α k α Z 1 3 ∗ d x u · u 0 = δ 0 δ 0 . V k,α k ,α0 kk αα Since ∂H Due to the periodicity conditions the wave vectors can ˙ = −Pk,α take the values ∂Qk,α ∂H = +Q˙ , kx, ky, kz = 2nπ/L, n = ±1, ±2,.... k,α ∂Pk,α At the moment t 6= 0 the vector potential is obtained the variables Pk,α and Qk,α are canonically conjugated simply by setting and the Hamiltonian function the sum of the total energies of the corresponding harmonic oscillators. c (t) = c (0)e−iωt k,α k,α Thus the classical radiation field can be thought to be a c∗ (t) = c∗ (0)eiωt, k,α k,α collection of independent harmonic oscillators. There • every oscillator is characterized by the wave vector k Thus we can write and the polarization ˆ(α), † a |nk,α i = c+|nk,α + 1i • the dynamic variables of every oscillator are k,α combinations of Fourier coefficients. ak,α|nk,α i = c−|nk,α − 1i. We quantize these oscillators by postulating that Pk,α Because the states |nk,α i are normalized we can calculate and Qk,α are not any more pure numbers but operators the coefficients as which satisfy the canonical commutation rules 2 2 |c+| = |c+| hnk,α + 1|nk,α + 1i [Q ,P 0 ] = i¯hδ 0 δ 0 k,α k ,α0 kk αα = hn |a a† |n i k,α k,α k,α k,α [Q ,Q 0 ] = 0 k,α k ,α0 = hn |N + [a , a† ]|n i [P ,P 0 ] = 0. k,α k,α k,α k,α k,α k,α k ,α0 = nk,α + 1, † We define dimensionless combinations a and a of 2 † k,α ,α |c−| = hn |a a |n i = n . k k,α k,α k,α k,α k,α the operators Pk,α and Qk,α as We choose the phase of the coefficients so that at the 1 √ moment t = 0 we have ak,α = (ωQk,α + iPk,α) 2¯hω q 1 a† |n i = n + 1|n + 1i † √ k,α k,α k,α k,α a = (ωQk,α − iPk,α). k,α 2¯hω p ak,α|nk,α i = nk,α |nk,α − 1i. It is easy to see that they satisfy the commutation relations Because † † n = hn |N |n i = hn |a a |n i [a , a 0 ] = δ 0 δαα0 k,α k,α k,α k,α k,α k,α k,α k,α k,α k ,α0 kk † † [a , a 0 ] = [a , a 0 ] = 0. and because the norm of a vectors is always non-negative k,α k ,α0 k,α k ,α0 we must have Note In these relations the operators must be evaluated n ≥ 0. † k,α at the same moment, i.e. [a , a 0 ] stands in fact for k,α k ,α0 From this we can deduce that the only possible † the commutator [a (t), a 0 (t)]. eigenvalues are k,α k ,α0 We further define the Hermitean operator nk,α = 0, 1, 2,.... N = a† a . We interprete k,α k,α k,α • the state |nk,α i to contain exactly nk,α photons, It is easy to see that each of which is characterized by a wave vector k and a polarization ˆ(α). [a ,N 0 ] = δ 0 δ 0 a k,α k ,α0 kk αα k,α † † † • the operator a to create a photon with the wave [a ,N 0 0 ] = −δ 0 δαα0 a . k,α k,α k ,α kk k,α vector k and the polarization ˆ(α). Due to the Hermiticity the eigenvalues nk,α of the • the operator a to destroy a photon with the wave operator Nk,α are real and the eigenvectors k,α vector k and the polarizartion ˆ(α). Nk,α|nk,α i = nk,α |nk,α i • the operator Nk,α to count the number of photons form an orthonormal complete basis. with the wave vector k and the polarization ˆ(α) in With the help of the commutation rule the state † † [a ,N 0 0 ] = −δ 0 δαα0 a k,α k ,α kk k,α The state composed of various kind of photons is a direct product of individual vectors |nk ,α i: we see that i i |n , n , . , n ,...i N a† |n i = (a† N + a† )|n i k1,α1 k2,α2 ki,αi k,α k,α k,α k,α k,α k,α k,α = |n i ⊗ |n i ⊗ · · · ⊗ |n i ⊗ · · · . k1,α1 k2,α2 ki,αi = (n + 1)a† |n i. k,α k,α k,α The vector |0i stands for the state that has no kind of Similarly we can show that photons, i.e. N a |n i = (n − 1)a |n i. |0i = |0 i ⊗ |0 i ⊗ · · · . k,α k,α k,α k,α k,α k,α k1,α1 k2,α2 Application of any operator ak,α onto this results always where zero. We say that |0i represents the vacuum. ω = |k|c. It is easy to see that a general normalized photon state can be constructed applying operations a† When we choose the energy scale so that k,α consecutively: H|0i = 0, † n (a ) ki,αi Y k ,α the Hamiltonian takes the form |n , n ,...i = i i |0i. k1,α1 k2,α2 q n ! X X k ,α ki,αi i i H = ¯hωNk,α. α † † k Note Since the operators a and a 0 commute the k,α k ,α0 order of operators does not matter. The many photon When it acts on a many photon state the result is states are symmetric with respect to the exchange of H|n , n ,...i photons. We say that the photons obey the Bose-Einstein k1,α1 k2,α2 statistics or that they are bosons. X = n ¯hωi|n , n ,...i. ki,αi k1,α1 k2,α2 Since the numbers nk,α tell us the number of photons of i type (k, α) in the volume under consideration we call them the occupation numbers of the state. The quantum mechanical momentum operator is exactly Correspondingly the space spanned by the state vectors is of the same form as the classical function (the Poynting called the occupation number space. vector): In the quantum theory the Fourier coefficients of a 1 Z classical radiation field must be replaced by the P = (E × B) d3x corresponding non-commuting creation and annihilation c X X 1 operators. Substituting = ¯hk(a† a + a a† ) 2 k,α k,α k,α k,α α c 7→ cp¯h/2ω a (t) k k,α k,α X X 1 p † = ¯hk(N + ). c∗ 7→ c ¯h/2ω a (t) k,α 2 k,α k,α k α we get Since here the summation goes over all wave vectors the r term associated with the factor 1/2 will not appear in the 1 X ¯h h √ (α) ik·x final result the terms ¯hk and −¯hk cancelling each other. A(x, t) = c ak,α(t)ˆ e V 2ω k,α For the momentum operator we get thus i +a† (t)ˆ(α)e−ik·x .
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