Solutions for Exercise session I

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1 Solutions for Exercise session I 1. The maximally polarisation-entangled photon state can be written as Ψ = 1 ( H 1 V V 1 H ). Show that the state is invariant (i.e. still maximally entangled) after a rotation of 45. Hint: Express Ψ in terms of diagonally polarised photon states. The definitions of polarised light states in +45 and -45 are P = 1 ( H + V ) and M = 1 ( H V ), respectively. Entangled state in H/V basis is: Ψ = 1 ( H 1 V V 1 H ) Using the definitions of plus and minus 45 degree polarised light we have: H = 1 ( P + M ) and V = 1 ( P M ). Replacing H and V in the expression of Ψ we get: Ψ = 1 ( 1 1 (( P 1 + M 1 ) ( P M ) ( P 1 M 1 ) ) ( P + M )) Multiplying out the tensor product: Ψ = 1 ( 1 ( P 1 P + M 1 P P 1 M M 1 M ) ( P 1 P M 1 P + P 1 M M 1 M )) Finally some terms will cancel and we arrive at Ψ in the ±45 basis: Ψ = 1 ( M 1 P P 1 M ) The state is still entangled in this basis and shows the same correlations.. Exercise 4.13 form course book Quantum Optics. A helium-neon laser consists of a laser tube of length 0.3 m with mirrors bonded to the end of the tube. The output coupler has a reflectivity of 99%. The laser operates on the 63.8 nm transition of neon (relative atomic mass 1

2 0.18), which has an Einstein A coefficient of s 1. The tube runs at 00 C and the laser transition is Doppler- broadened. On the assumption that the only loss in the cavity is through the output coupler, that the average refractive index is equal to unity, and that the laser operates at the line center, calculate: (a) the gain coefficient in the laser tube; (b) the population inversion density. (a) R = 0.99; L = 0.3 m; λ = 3.8 nm; A = s 1 ; T = 00 C= 473 K; m = kg. Laser oscillation equation: ( ) 1 R R 1 R ε e γl = 1 γ = ln, L where R 1 = 1 and ε = 1. The gain is hence γ = m 1. (b) Equation for laser gain is: γ(ω) = λ N g(ω) 4 n τ setting n = 1 and τ = 1 we obtain for the population inversion A density: N = 4 γ. λ A g(ω) Line shape is given by Doppler broadening. At the line center g(ω) becomes: g(ω 0 ) = c λ m = 9.13 π c π k b T (rad/s) 1. Finally we obtain for the population inversion density: N = m Exercise 5.6 form course book Quantum Optics. (a) the mean photon number per pulse: The mean photon number ( n) per pulse is: n = E pulse, where E h ω pulse is the energy per pulse. E pulse = P N pulse, where P is the optical power and N pulse is the number of pulses per second. P P λ So we have, n = h ω N pulse = h π c N pulse = (b) the standard deviation per pulse: The photon statistics of a laser is Poissonian, so n = n =

3 4. Exercise 5.7 form course book Quantum Optics. The laser described in the previous question is attenuated by a factor For the attenuated beam calculate: (a) the mean photon number per pulse: n = n old 10 9 = 0.04 (b) the the fraction of pulses containing one photon: From Poission distribution we have: P (1) = ne n = 0.04 e 0.04 = (c) the the fraction of pulses containing more than one photon: P (n > 1) = 1 P (0) P (1) = 1 e n ne n = One pulse in every 184 pulses has more than 1 photon. 5. Exercise 5.8 form course book Quantum Optics. Average photon number impinging on detector is: n = Φ T = 10 4, where Φ is the photon flux and T is the time interval. Average detected count rate N = η n = (a) the light has Poissonian statistics: Mean photon number: N = Standard deviation: From equation 5.56 we have: ( N) = η n = N. N = N = (b) the light has super-poissonian statistics with n = n P oissonian : Mean photon number: N = Standard deviation: ( N) = η 4 n+η(1 η) n = 0.3 n = 300. N = (c) the light is a photon number state: Mean photon number: N = Standard deviation:( N) = η(1 η) n = 0.16 n = N = Exercise 5.13 form course book Quantum Optics. An LED emitting light at 800 nm is driven b a 9 V battery through a resistor with R = 1000Ω. The LED has a quantum efficiency of 40% and 80% of the photons emitted are focussed onto a photodiode detector with a quantum efficiency of 90%. 3

4 (a) Calculate the average drive current, given that the voltage drop across the LED is approximately equivalent to the photon energy in ev in normal operating conditions: h π c Photon energy is = 1.55 ev. Voltage drop at LED is 1.55V.Voltage λ across resistor is = 7.45 V. Current is U = 7.45 ma. R (b) Calculate the Fano factor of the drive current for T = 93 K: F drive = ( i) Johnson ( i) Shot noise = 4 k b T f R e<i> f = (c) Calculate the average photocurrent in the detection circuit: Average photocurrent is: < i photo >= η DET e Φ = η DET e <i drive> η e LED η optics =.15 ma. (d) Calculate the Fano factor of the photocurrent: F photo = η T OT F drive + (1 η T OT ) = (e) Compare the photocurrent noise power in a 50Ω load resistor with the shot noise level for a bandwidth of 10 khz: Shot-noise power: P shot noise = e R L f < i >= 0.34 fw. F photo = P measured P photo = P measured = 0.5 fw. 7. Exercise 6.9 form course book Quantum Optics. (a) A single atom is irradiated with a powerful beam from a continous wave laser which can promoto the atom to an excited state with a radiative lifetime of τ R. On the assumption that the excitation time is negligibly small, calculate the probability that the atom emits two photons in a time T. We assume that the atom has just emitted a photon at time t = 0 and is re-excited immediately. We have to find now the probability (P ( T )) that the atom emits a second photon within time T. This probability is given by the ratio of the number of atoms decayed within T to the total number of atoms: P ( T ) = N decayed ( T ) N all. Number of decayed atoms after T is given by spontaneous emission. N decayed ( T ) = N all N all e T τr, where τ r is the natural lifetime of the atom. The probability to emit two photons is hence given by P ( T ) = 1 e T τr. (b) In a Hanbury Brown-Twiss (HBT) experiment, single-photon counting detectors with a response time of τ D are connected to the start 4

5 and stop inputs of a timer. The finite response time of the detectors implies that two events separated in time by τ d will be registered as simultaneously. Use this fact, together with the result of part (a), to estimate the value of g () (0) that would be expected in such a HBT experiment. To evaluate the g () function events are categorised according to the time difference between the start (detector A) and the stop (detector B) trigger. The time difference is T. If the detector response time was zero then we would have a continuous distribution of T. However with a finite detector response time τ D, this is the smallest time interval we can distinguish. The g () function is hence discreet and can be represented as a histogram, where the minimum size of the bin-width is equal to τ D, see fig 1. All events registered in time by less than τ D will be counted towards the same bin. For example all events with τ D < T τ D will be in bin 1. Bin 0 is twice as large ( τ D to τ D ) as we cannot make the difference which detector fired first, hence this bin contains positive and negative T values. The number of events in bin 0 (N 0 ) is equal to the total number of photons (N tot ) times 50% (only events where the photons went to different detectors are registered) times the probability that the atom emitted two photons during the time 1 ( T = τ D ), so: N 0 = N tot (1 eτ D/τ r ). If we just take look at positive T values then right side of bin 0 has N 0 / counts. We now need to normalise this by dividing the counts in each bin by the counts of a bin with large T. Remember tat for large times the g () function is 1. The counts in a such a bin is equal to: 1 1 N 0 = N tot. The second factor of 1 comes from the fact that we can now distinguish between events where detector A fired first from when detector B fired first. After normalisation the number of events in bin 0 is given by: N0 = (1 e τ D/τ r ). This results implies that the g () will not be zero even when a single atom is investigated due to the time uncertainty of the detectors. 5

6 N Bin 0 Bin 1 Bin τ D ΔT Figure 1: Histogram of the g () function. 6

Correlation functions in optics; classical and quantum 2. TUW, Vienna, Austria, April 2018 Luis A. Orozco

Correlation functions in optics; classical and quantum 2. TUW, Vienna, Austria, April 2018 Luis A. Orozco www.jqi.umd.edu Correlations in optics Reference that includes pulsed sources: Zheyu Jeff Ou Quantum