4. Spontaneous Emission october 2016

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1 4. Spontaneous Emission october 016 References: Grynberg, Aspect, Fabre, "Introduction aux lasers et l'optique quantique". Lectures by S. Haroche dans "Fundamental systems in quantum optics", cours des Houches 1990, Dalibard, Raimond, Zinn-Justin Eds. Hulet, Hilfer, Kleppner, "Inhibited spontaneous emission by a Rydberg atom", Phys. Rev. Lett. 55, 137 (1985). A Introduction In the last lecture, we saw how the inclusion of a single mode quantized electromagnetic field in the atom-field Hamiltonian leads naturally to the concepts of spontaneous and stimulated emission. In this lecture we will show how to treat the more realistic situation of a multimode field. This treatment will allow us to calculate the spontaneous decay rate of an excited atomic state. It will be necessary to treat a multimode field and include a sum over all possible decay modes. The mode counting arguments we discussed in the 1st lecture will reappear. We will also give an example of spontaneous decay in a confined space. A sum over modes can be tricky when the space in which the decay takes place is something other than an infinitely large box. B Review: transitions induced by a single mode field Recall the inputs to the single mode calculation we did in the last lecture. We begin with a quantized electric field: E = i ħω ϵ 0 V a ϵ e i(k z(ω t) ( a ϵ + e ( i(k z(ω t) We write the field as a vector (vectors will be in boldface), characterized by the polarization vector ϵ. The polarization can elliptical and thus complex. The field interacts with a two level atom with levels f and e. The interaction Hamiltonian, in the Hilbert space α, n, α = f or e, n = 0, 1,,... is: V(t) = (D E (t) = / E D ϵ a e i ω t ( a e ( i ω t, with E = ħ ω ϵ 0 V Using 1st order, time dependent perturbation theory, we can find the transition amplitude between two levels in the atom: c f, m (t) = ( 1 ħ E 0 t d t f, m D a e, n e i ω(ω 0 t ( f, m D a e, n e (i ω+ω 0 t Transition probabilities in the rotating wave approximation are given by:

2 17 OQ 4 SpontEm.nb P f, n6e, n(1 = d E ħ ω(ω 0 sin (ω ( ω 0 ) t n P e, n6f, n+1 = d E ħ ω(ω 0 sin (ω ( ω 0 ) t (n + 1) where d = f D e is known as the electric dipole matrix element. We see that stimulated (n) and spontaneous emission (1) probabilities emerge naturally. The time dependence for short times is quadratic. An important point in what follows is that after summing over large number of modes, a continuum, the transition probability becomes linear for short times. C Transitions with multimode, quantum field A real atom is seldom coupled to only one mode of the radiation field. To treat the problem of the decay of isolated atom we must treat a multimode field. The field operator is a simple sum over modes: E = / Σ k, p ħω k ϵ 0 V a k,p ϵ p e i k r(ω k t ( a k,p ϵ p + e ( i k r(ω k t where p always has possible values corresponding to, orthogonal, possibly complex polarizations. For each field mode we can compute, using the formulae above, a transition amplitude. To include the contributions of all the modes, we sum probabilities rather than amplitudes associated with the individual modes. We do this because the modes are all orthogonal. Orthogonality guarantees that different final states of the field are distinguishable and so there can be no interference between different quantum mechanical paths. The probability sum is: ω sin ω(ω 0 t P e,06f,{1} = Σ k, p ħϵ 0 V ω(ω 0 f D ϵ p e The notation {1} means exactly one photon in one of the modes in the sum. We convert the sum into an integral over ω = c k by using the mode counting argument from the 1st lecture: P e, 06f, {1} = L3 ( π) 3 Σ p AΩ Aω ω c 3 ω ħ ϵ 0 V sin ω(ω 0 t ω(ω 0 f D ϵ p e I cannot carry out the integral over angles (AΩ = Aϕ A(cos θ)) unless I know the angular dependence of D ϵ p. We will do this later. First we will look at the integral over ω. Here is a plot of sin ω(ω 0 t ω(ω 0. Notice that the height of the peak is t 4 π, its width is, and its area is π t : t

3 17 OQ 4 SpontEm.nb 3 Plot Sin (ω $ ω 0 ) t (ω $ ω 0 ) (. {ω 0 * 1, t * 6.8}, {ω, 0.7, 1.3}, PlotRange * All For a long interaction times, this function is sharply peaked compared to everything else. Thus we can replace it by π t δ(ω ( ω 0). The probability P e6f after summation is linear rather than quadratic in t. If we assume that the probability to remain in the excited state decays is exponentially e (Γ t = 1 ( Γ t +... we can identify the decay rate in our perturbation calculation as Γ = P e f. It is not necessary to assume exponential decay, in the exercise you will use a slight t elaboration on the above calculation to demonstrate exponential decay (the Weisskopf-Wigner approach). To get a quantitative answer we still need to evaluate the matrix element f D ϵ p e. We will consider a transition in a spinless 1 electron atom (hydrogen): n =, l = 1, m = 0 6 n = 1, l = 0, m = 0. Recall that the dipole operator is q r, were r = x ϵ x + y ϵ y + z ϵ z is the position operator of the electron. The electronic wave functions of the excited and ground states are: with: ψ e (r) = R,1 (r) Y 1,0 (θ, ϕ), ψ f (r) = R 1,0 (r) Y 0,0 (θ, ϕ) R 10 (r) = a 3I e(ri a, Y 00 = 1 4 π R 1 (r) = 1 6 a 3I e(ri ( a) r a, Y 10 = 3 4 π cos θ If you rewrite x, y, z in polar coordinates, (x = r sin θ cos ϕ etc.) you can see that for these two states:

4 4 17 OQ 4 SpontEm.nb f x e = 0 = f y e and q f z e = d = q a where a is the Bohr radius. Therefore for these two states, the dipole operator is proportional to ϵ z, i.e. the dipole is polarized along the z axis. Now we need Σ s ϵ z ϵ p for an arbitrary plane wave. The polarizations must be orthogonal to the k vector. Consider a k vector k = ω c (cos θ ϵ z + sin θ(ϵ x cos ϕ + ϵ y sin ϕ)) I must choose orthogonal polarizations for the ϵ p. It is convenient to chose ϵ 1 = sin θ ϵ z ( cos θ (ϵ x cos ϕ + ϵ y sin ϕ) ϵ = ϵ x sin ϕ ( ϵ y cos ϕ These vectors are shown in the diagram below. I have chosen ϵ to be in the x ( y plane. You can check that these are orthogonal to each other and to k. With this choice of polarizations we have ϵ z ϵ 1 = sin θ, and ϵ z ϵ = 0. So that p f D ϵ p e = d sin θ. Below is an experimental illustration of a dipole radiation pattern. Atoms were made to spontaneously emit with a dipole matrix element aligned along the x axis. Instead of recording the direction of the emitted photons, the experiment recorded velocity of the recoiling atoms. The quantity v rec = ħ k I m is the recoil velocity. The circular pattern has a width because of the residual velocity width of the atom cloud.

5 17 OQ 4 SpontEm.nb 5 Exercise : show that p f D ϵ p e = d cos θ for a transition 1 l = 1, m = 1 6 l = 0, m = 0 (dipole (ϵ x + i ϵ y )). We can now put all this together to find: Γ = 1 8 π d ħ ϵ 0 ω 0 3 c 3 Aϕ A(cos θ) sin θ = 1 3 π d ω 3 0 ħ ϵ 0 c 3. This calculation gives a lifetime Γ (1 of 1.6 ns for the P-1S transition in hydrogen (λ=11.5 nm), in good agreement with observations. The decay rate of the m = ±1 states is the same and you can verify by doing the same calculation as above for an initial state n =, l = 1, m = 1. You will need to know that for the m = ±1 state, the angular wave function is given by: Y 11 (θ, ϕ) = 3 8 π sin θ e/ ϕ. You must also show that p f D ϵ p e = d cos θ for a transition 1 l = 1, m = 1 6 l = 0, m = 0 (dipole (ϵ x + i ϵ y )). Via the "golden rule" ("Fermi's nd golden rule") With a quantum field, spontaneous emission amounts to the coupling of a discrete state e ; 0, to a continuum f, {1} (1 photon, but in any mode). One can write the result in the following form:

6 6 17 OQ 4 SpontEm.nb Γ = 4 ħ π δ(ω ( ω 0) f D E i = π ħ AΩ f H 1 (θ, ϕ) i d ρ(e) d Ω (1) H 1 = ħω k ϵ 0 V Σ s D ϵ s a k,s d ρ(e) d ρ(ħ ω) V ω AE A Ω = d(ħ ω) d Ω = d Ω d Ω ( π) 3 ħ c 3 d(ħω) d Ω = V k ( π) 3 d(ħ c k) d Ω ħ c The quantity ρ(e, Ω) is called the density of states, and Eq. (1) is often referred to as Fermi's (nd) golden rule. One must have Λ d ρ(e) d Ω space. Exercise: verify that the above formula indeed gives Γ = 1 d E d Ω = number of states in the volume Λ of phase 3 π d ω 3 ħϵ 0 c 3. D. When the walls are not far away, the Purcell effect Sometimes it is not legitimate to take the limit of an infinitely large quantization volume. In this case, properties such as the spontaneous emission rate can change. As an example, we will consider spontaneous emission between metal plates separated along the z axis by a distance a λ V = L a, L big. See the diagram below. With the appropriate orientation of the atomic dipole, and a sufficiently small separation between the plates, it is possible to stop spontaneous emission. In other situations it is possible to increase the decay rate. We shall study both possibilities in this section. Experiment and qualitative analysis The experiment by Hulet, Hilfer et Kleppner, PRL 1985, gives a clear demonstration that nearby walls can dramatically alter the spontaneous emission rate of an atom. In the

7 17 OQ 4 SpontEm.nb 7 experiment, a beam of atoms was prepared in a highly excited state (n =, l = 1, m = 1). The only allowed electric dipole transition for this state is to the (n = 1, l = 0, m = 0). This transition corresponds to a wavelength of λ = 460 μm, and has a decay rate in vacuum of Γ = s (1. When the separation between the plates was less than λ, it was possible render the decay rate immeasurably small. We will begin our analysis by identifying the modes that exist between the two plates. The possible k vectors are: k = π L (n x ϵ x + n y ϵ y ) + π a n z ϵ z, k = ω c We are free to impose periodic boundary conditions in the x and y directions, but not along z. The quantities n x and n y are large integers, but n z is small. Indeed if a < λ we must have n z = 0 and the only allowed modes propagate in the x ( y plane. What about the polarizations? If n z = 0, the only possible polarization is ϵ z, because any electric field parallel to the plates will be cancelled by currents in the plates. (Indeed, if the plates are shorted together, even the ϵ z polarization is forbidden, but that would require a true short at THz frequencies.) The next question is what is the polarization of the transition dipole, e D f between the states used in the experiment? If you have studied dipole selection rules a little, you may recognize that a transition matrix element between magnetic quantum numbers m and m ( 1 is always (circularly) polarized in the plane orthogonal to the quantization axis, z. If you do not recognize this, you can convince yourself by considering the z component of the angular momentum operator l z = x p y ( y p x. This operator commutes with z: [l z, z] = 0. Therefore we can write: l f, m f [l z, z] l i, m i = 0 which implies : (m f ( m i ) l f, m f z l i, m i = 0 Therefore, if m f ( m i 0 then l f, m f z l i, m i = 0 and the dipole matrix element is orthogonal to ϵ z. Thus, Hulet et al., constructed a situation in which the dipole matrix element governing the spontaneous decay and the only allowed field polarization are orthogonal, and this should turn off spontaneous emission. Decay rate for any plate separation The paper also mentions that when the plate separation is slightly larger than the half wavelength, the decay rate is three times larger than in vacuum. We will now analyze this situation. The first task is to write the operator for the field between two parallel plates. As described above, the modes in the z direction must be standing waves. Quantization of a standing wave One can transform between running and standing waves by making a change of basis: a c = 1 (a k + a (k), a s = 1 (a k ( a (k) You can easily check that with these definitions, the new operators obey the commutation relations: a c, a c = 1, a s, a s = 1, [a c, a s ] = 0 etc. Thus we can conclude that the new mode operators also correspond to independent quantum harmonic oscillators. It is therefore straightforward to rewrite a field operator corresponding to

8 8 17 OQ 4 SpontEm.nb two oppositely running waves in terms of two standing waves: E = i ħω k ϵ 0 V a c, p ϵ p e (i ω k t ( a c, p ϵ p + e i ω k t cos k z + i a s, p ϵ p e (i ω k t + a s, p ϵ p + e i ω k t sin k z In three dimensions, a multimode field between two conducting plates (z = 0, z = a) is written: E = / K,kz>0, p E ω ϵ p a K,kz, s,p e (i ω t e i K R + ϵ p + a K,kz,s, p e i ω t e ( i K R sin k z z + / K,kz=0 ϵ z E ω a K, kz=0 e (i ω t e i K R ( a K, kz=0 e i ω t e (i K R with, K R = k x x + k y y, capital letters designate a vector in the x-y plane. The frequency of the mode is given by ω c = K + k z, and the polarisation, p is TE (electric field in the x ( y plane) or TM (magnetic field in the x ( y plane). We have eliminated the cos k z z term because it does not satisfy the boundary conditions, except for the case k z = 0, which is accounted for separately. The a K, kz, s operators create and anihilate fields corresponding to standing (sine) waves along the z axis and traveling waves in the x-y plane. If the transition dipole is in the x ( y plane, the k z = 0 part of the field never contributes. The quantity E ω is given by E ω = ħω ϵ 0 L a. To compute the decay rate we must again sum over all possible modes. Because the box is large in the x ( y plane, part of the sum can be converted to integral Σ K = L π K AK Aϕ. But along z the sum must remain discrete: Σ K, k z, p = Σ n z, p L π K AK Aϕ A(cos θ) δ cos θ ( n z π k a with k = ω c = π a n z + K. The expression in {} brackets simply expresses the fact that the choice of k and n z imposes an angle on the emission direction, cos θ = k z. We now express the k integral in terms of the frequency: ω c = n z π a + K and K AK = ω Aω c and we again make the approximation that sin ω t ω π t δ(ω): Γ = Σ n z, p 1 ħ L π ω c Aω AΩ π δ(ω ( ω 0) δ cos θ ( n z Transition dipole ħω ϵ 0 L a sin π z a n z e D ϵ p f π k a Now we need the explicit form of the transition dipole. We already know it is in the x ( y plane, and you may be able to guess that it corresponds to a circularly polarized dipole. To

9 17 OQ 4 SpontEm.nb 9 see this, consider the operator x + = 1 (x + / y). It is straightforward to show that, [l z, x + ] = ħ x +. Therefore we have: This leads to l f, m f [l z, x + ] ( ħ x + l i, m i = 0 (m f ( (m i + 1)) l f, m f x + l i, m i = 0. Therefore for the transition m i 6 m f = m i ( 1, the transition dipole x + vanishes. The only dipole remaining is x ( = 1 (x ( / y). You can adapt the above argument to show that indeed the m i 6 m f = m i ( 1 is allowed. We conclude that the only important component of the dipole operator is f D e = q (x ϵ x ( / y ϵ y ). The cylindrical symmetry of the dipole requires that q f x e = q f y e d. Therefore f D e = d 1 (ϵ x ( / ϵ y ) = d ϵ (. To do the polarization sum we can use the same basis vectors as in section C: 1 1 (ϵ x ( / ϵ y ) ϵ 1 = (cos θ 1 (ϵ x ( / ϵ y ) ϵ = 1 (sin ϕ ( / cos ϕ) (cos ϕ ( / sin ϕ) The result is Σ p ϵ ( ϵ p = cos θ. This is the radiation pattern of a circularly polarized dipole. We can put everything together to find the decay rate for an atom between the two plates. To keep the calculation simple we will restrict ourselves to the situation λ < a < λ, which means that only n z = 1 is possible, higher n z will not obey energy conservation (recall that the n z = 0 mode has been shown not to contribute). Γ = d ω 0 1 ħϵ 0 c a π (k a) sin π a z. If we suppose that a is slightly larger than λ, then π Γ = 1 π d ω 0 3 ħϵ 0 c 3 sin π a z. (k a) 1 and we find: For an atom in the midplane between the two plates z λ 4, We have Γ = 3 Γ vacuum, whereas the decay rate vanishes for an atom close to the walls. We also see that the radiation is entirely in the direction θ 0. In the experiment of Hulet et al., the beam of atoms fills the entire space between the plates. The average decay rate is found by averaging sin (k z z) 6 1. Therefore the observed spontaneous emission rate was Γ = 3 Γ vacuum. In other geometries the rate can be increased by a large factor. This factor is called the Purcell factor. Another point of view : density of states in a box We can recast the problem of nearby walls in terms of Fermi s golden rule. The transition matrix element is no different in a small box; all the difference lies in the density of

10 10 17 OQ 4 SpontEm.nb states. Consider a wave vector, k = π (n x ϵx + n y ϵy) + π n z ϵz. For k < π, we necessarily have L a a n z = 0. For larger k a small number of n z can be involved. We can count modes for different values of n z. For n z = 0, TE polarization is forbidden: N n z=0 TE = 0 because of the boundary conditions N n z=0 TM = L π k δ(cos θ) Ak Aϕ Acos θ 6 d ρ nz=0 = L dω π k δ(cos θ) Not surprisingly, this formulation introduces an angular dependence in the mode density. For n z =1, The two polarizations are possible: N n z=1 TE = L π k δ cos θ ( π Ak Aϕ A cos θ a k N n z=1 TM = L π k δ cos θ ( π Ak Aϕ A cos θ a k And so on: N tot =N n z=0+n n z=1+... up until n z < k a If we integrate over dϕ d cos θ: ρ TE = L π k Int k a π ρ TM = L π (TE) k a k 1 + Int (TM) π π n z = Int k a π In the limit a 6, we have Int k a 6 k a, and therefore π π ρ TE + ρ TM 6 L a π k = V π k This is the same result as in free space if we count both polarizations. Below is a comparison of ρ between two plates and in free space. TM

11 17 OQ 4 SpontEm.nb 11 Plot k (1 + IntegerPart[k]), k, {k, 0, 3.5}

12 Spontaneous emission between plates ESO 3/M optique quantique

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14 Expérience de Hulet, Hilfer et Kleppner PRL, 55, 17 (1985) L = 6 cm 1 cm séparation 0.3 mm (C ~ 3pF) channeltrons Cs atomes dans l état n=, l=1, m=1, desexcitation uniquement vers n=1, l=0, m=0, D ε x +iε y λ = 460 µm, Γ = s -1, T = 6.5 K

15 Atoms detected in the excited state data from de H, H et K

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17 Enhancing emission in a quantum dot