Solutions for Exercise set III

1. Exercise 5.6 form course book ”Quantum Optics”.

(a) the mean photon number per pulse:

The mean photon number (n) per pulse is: n =Epulse

h ω, where Epulse

is the energy per pulse.Epulse = P

Npulse, where P is the optical power and Npulse is the

number of pulses per second.So we have, n = P

h ω Npulse= P λ

h 2π cNpulse= 4.0× 107

(b) the standard deviation per pulse:The photon statistics of a laser is Poissonian, so ∆n =

√n =

6.4× 103.

2. Exercise 5.7 form course book ”Quantum Optics”. The laser describedin the previous question is attenuated by a factor 109. For the attenu-ated beam calculate:

(a) the mean photon number per pulse:n = nold

109= 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 =0.038

(c) the the fraction of pulses containing more than one photon:P (n > 1) = 1− P (0)− P (1) = 1− e−n − ne−n = 7.8× 10−4. Onepulse in every 1284 pulses has more than 1 photon.

3. Exercise 5.8 form course book ”Quantum Optics”.Average photon number impinging on detector is: n = ΦT = 104,where Φ is the photon flux and T is the time interval. Average detectedcount rate N = η n = 2× 103.

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(a) the light has Poissonian statistics:Mean photon number: N = 2× 103.Standard deviation: From equation 5.56 we have: (∆N)2 = η n =

N .⇒ ∆N =√N = 44.7.

(b) the light has super-Poissonian statistics with ∆n = 2×∆nPoissonian:Mean photon number: N = 2× 103.Standard deviation: (∆N)2 = η2 4n+η(1−η)n = 0.32n = 3200.⇒∆N = 56.6.

(c) the light is a photon number state:Mean photon number: N = 2× 103.Standard deviation:(∆N)2 = η(1− η)n = 0.16n = 1600.⇒ ∆N =40.0.

4. Exercise 5.13 form course book ”Quantum Optics”. An LED emittinglight at 800 nm is driven b a 9 V battery through a resistor withR = 1000Ω. The LED has a quantum efficiency of 40% and 80% ofthe photons emitted are focussed onto a photodiode detector with aquantum efficiency of 90%.

(a) Calculate the average drive current, given that the voltage dropacross the LED is approximately equivalent to the photon energyin eV in normal operating conditions:Photon energy is h 2π c

λ= 1.55 eV. Voltage drop at LED is 1.55V.Voltage

across resistor is 9− 1.55 = 7.45 V. Current is UR

= 7.45 mA.

(b) Calculate the Fano factor of the drive current for T = 293 K:

Fdrive =(∆i)2Johnson

(∆i)2Shot−noise

= 4 kb T ∆fR 2 e<i>∆f

= 7.0× 10−3.

(c) Calculate the average photocurrent in the detection circuit:Average photocurrent is:< iphoto >= ηDET eΦ = ηDET e

<idrive>e

ηLED ηoptics = 2.15 mA.

(d) Calculate the Fano factor of the photocurrent:Fphoto = ηTOT Fdrive + (1− ηTOT ) = 0.714.

(e) Compare the photocurrent noise power in a 50Ω load resistor withthe shot noise level for a bandwidth of 10 kHz: Shot-noise power:Pshot−noise = 2 eRL ∆f < i >= 0.34 fW. Fphoto = Pmeasured

Pphoto=

0.714.⇒ Pmeasured = 0.25 fW.

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