quantumcryptography with microwaves

Quantumcryptography withMicrowaves

byPia Doring

Bachelor Thesis in PhysicsPresented to the Faculty for Mathematics, ComputerSciences and Natural Sciences at the RWTH Aachen

September 2017

presented toInstitute for Quantum Information

under supervision ofProf. Dr. David P. DiVincenzo

Acknowledgements

I hereby thank my supervisor Prof. Dr. David P. DiVincenzo for his guidance,help and ideas throughout the thesis.I am thanking Prof. Dr. Fabian Hassler for being the second assessor of mythesis.I am also grateful to the proof readers of my thesis for avoiding embarrassingmistakes.

Statutory Declaration in Lieu of an Oath/ Ei-desstattliche Versicherung

I hereby declare in lieu of an oath that I have completed the present Bachelorthesis independently and without illegitimate assistance from third parties. Ihave used no other than the specifed sources and aids. The thesis has not beensubmitted to any examination body in this, or similar, form.

Ich versichere hiermit an Eides Statt, dass ich die vorliegende Bachelorarbeitselbststandig und ohne unzulassige fremde Hilfe erbracht habe. Ich habe keineanderen als die angegebenen Quellen und Hilfsmittel benutzt. Die Arbeit hatin gleicher oder ahnlicher Form noch keiner Prufungsbehorde vorgelegen.

Aachen, 11th September 2017

Contents

1 Introduction 1

2 Theoretical Principles 22.1 Quantumcryptography . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Black Waveguide (Nyquist Theorem) . . . . . . . . . . . . . . . . 2

2.2.1 Method of Nyquist . . . . . . . . . . . . . . . . . . . . . . 32.2.2 Method of Reif . . . . . . . . . . . . . . . . . . . . . . . . 4

2.3 Transmission Line Theory . . . . . . . . . . . . . . . . . . . . . . 52.4 Thevenin and Norton Equivalent . . . . . . . . . . . . . . . . . . 62.5 Detection of Photons . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.5.1 Decoy State Method . . . . . . . . . . . . . . . . . . . . . 72.5.2 Quantum Nondemolition Detection . . . . . . . . . . . . . 82.5.3 Catch and Release . . . . . . . . . . . . . . . . . . . . . . 82.5.4 Harmonic Oscillators as Intermediate Stage . . . . . . . . 8

3 Calculations and Results 103.1 Solutions for the Tranmission Line Theory . . . . . . . . . . . . . 103.2 Noise by Thevenin or Norton Equivalent . . . . . . . . . . . . . . 113.3 Distortionless Case . . . . . . . . . . . . . . . . . . . . . . . . . . 113.4 Transferring Voltage into Number of Photons . . . . . . . . . . . 123.5 Di↵erence in Temperature . . . . . . . . . . . . . . . . . . . . . . 143.6 Power and Number of Photons for Finite Length . . . . . . . . . 153.7 Detector Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.8 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4 Conclusion 27

1 INTRODUCTION

1 Introduction

Since several years physicists research the field of quantum information. As ap-plications of quanta like quantum computing become more and more reachable,an interesting field to explore is the field of quantum information transfer. Itseems to be a good alternative to today’s technologies due to the properties ofphotons such as the indivisibility of quanta and the quantum no-cloning theo-rem.Particularly nowadays secure communication becomes more important as sev-eral spy a↵airs became public in the last years. The quantumcryptography is away of encoding information that ensures that only the parties which participatein the exchange of information know the content. Therefore it is advantageousto find a way of using quantumcryptography for information transfer.In 2007, an experiment with a transfer of optical photons was realized. Thedetection of optical photons over a distance of 144 km through air was success-ful [1]. A recent experiment in China shows that quantum communication iseven possible over distances of 1200 km. The experiment called QUESS man-aged to securely transfer photons of the ultraviolet range from a satellite toground stations which has been reported in June 2017 from several websitessuch as http://www.sciencemag.org and http://www.nature.com.In a paper from 2017, it is stated that the transfer of microwave photons througha fiber can only be managed if the fiber was cooled. Otherwise the disturbancedue to thermal photons would become too big to ensure a secure transfer [2].But the need to cool to milli-Kelvin which was stated in earlier works is abol-ished and replaced by a need to cool to about 4 K.Two important aspects which have to be improved and are being discussed inthis thesis are the distance over which the transfer is possible and the temper-ature of the setup. We will find out if and over which lengths the transfer ofphotons is possible at room temperature.The frequency range in which we calculate is the microwave range. The numberof thermal photons is supposed to decrease for higher frequencies which makesmicrowave photons very advantageous.

1

2 THEORETICAL PRINCIPLES

2 Theoretical Principles

In this section I am going to give an overview on the important principles thatare needed to understand and describe the di�culties and opportunities of quan-tum information transfer.

2.1 Quantumcryptography

In this section we will have a look at the encoding of quantum information. Itensures that the transfer of information is secure.If two parties, in the following Alice and Bob, communicate with one another nothird party should be allowed to get the information that is sent. To secure this,the photons have to be cryptographied. To achieve a highly secure exchangeof information, there are several protocols that allow to get keys for Alice andBob.The most common one is the BB84 protocol [3, chapter 12.6.3] where Alicegenerates (4 + �)n with � 2 Z random data bits, string a, to get a key of thelength n. She encodes string a with another random (4 + �)n-bit string b as

{|0i, |1i} if b = 0 and {|+i, |�i} if b = 1

and sends the encoded bits to Bob. He himself generates (4 + �)n random databits b0 and decodes the received bits with b0 in the same manner. Alice an-nounces b and Bob deletes every bit where b 6= b0. Of the remaining bits hekeeps 2n. The protocol will be aborted if there remain less than 2n bits becausein that case there would not be enough bits to compare and get a n-bits longkey in the end.Alice and Bob compare the values of n of the 2n bits. If there are more dis-agreements than a respectable number, the protocol will be aborted. A highnumber of disagreements is a sign for too much noise in the transfer channel. Ifnot, Alice and Bob will use the uncompared n bits as a key.Another way to get a common key is called the EPR protocol [3, chapter 12.6.3]where Alice and Bob both receive a set of photons from n entangled pairs. Theyselect a random subset of the photons and test their fidelity e.g. with Bell’s in-equation. If they pass the test, Alice and Bob will get a lower bound on fidelityand secret key bits by measuring in jointly set random bases.

2.2 Black Waveguide (Nyquist Theorem)

While Planck’s law for black body radiation is used to describe the ideal formof a black body absorbing and emitting photons of every frequency, we need aformula for the one-dimensional case to calculate the radiation on a transmissionline on which the photons are supposed to be transferred. The formula helps usto determine the number of thermal photons which disturb the detection of thesignal photons. It was deduced by Nyquist and connects the energy of a photonh⌫ with the voltage V on the transmission line. The equation can especially bederived in two di↵erent ways.

2


2.2.1 Method of Nyquist

One method given in a paper written by Nyquist himself [4] is using a construc-tion of a circuit consisting of two conductors with resistances R (see fig. 1).The current in the circuit with two resistances in series can then be calculatedas I = V

2R with the voltage due to the thermal agitation in the conductors V .Because of the first law of thermodynamics, it is given that the power trans-ported from I to II equals the power transported from II to I. This especiallyapplies to every chosen frequency range. The reason for this is that if we blockall frequencies except from frequencies between ⌫ and ⌫ + d⌫, the first law ofthermodynamics will apply, too.

R R R R

l

Figure 1: Circuits with two resistances R

At any time t after the equilibrium the transmission line of length l canbe isolated from the conductors, whereby the energy on the waveguide will beconserved and the wave is completely reflected at the ends. The line can thenoscillate with frequencies ⌫ = n · v

2l , where v is the velocity of propagation.With the number of modes between ⌫ and ⌫ + d⌫

Nd⌫ =2l

vd⌫, (1)

we get the total energy within d⌫ as

Ed⌫ = Nd⌫ · kB

T =2l

vkB

Td⌫ (2)

with the Boltzmann constant kB

and the temperature of the transmission line T .Because the power can be calculated with P = E

t

= E

l/v

, the power transferredto the line by each of the two conductors is

Pd⌫ =E

2l/vd⌫ = k

B

Td⌫. (3)

This results with P = I2R =�

V

2R

�2R = V

2

4R in the Nyquist theorem

V 2d⌫ = 4RkB

Td⌫ (4)

respectivelyV 2 = 4Rk

B

T�⌫, (5)

3


where �⌫ is the bandwidth over which the voltage is measured. This is thenthe formula which describes the voltage in one conductor with resistance R andat temperature T .

2.2.2 Method of Reif

Reif derives the formula in his book in another way [5, chapter 15.16]. Heuses the e↵ective fluctuations of the voltage at a resistance due to randomthermal movements of the electrons V (t) and its spectral density J+(!) whichis a measurement for the power of V (t). To calculate this, we picture a resistorconnected with a linear amplifier which blocks out frequencies out of the intervalbetween ! and ! +�!.The mean of the square of the voltage is given by

hV (t)2i =Z

!+�!

!

J+(!)d!. (6)

The relation between the resistance and the voltage is constituted by

R =1

2kB

T

Z 1

�1K(s)ds =

1

2kB

T

Z 1

�1hV (0)V (s)i0ds =

⇡

2kB

TJ+(0) (7)

where K(s) = hV (0)V (s)i0 is the correlation function of V (t).Because K(s) = 0 for |s| � correlation time ⌧⇤ and K(s) has a sharp maximumat s = 0, the term e�i!s in J+(!) =

1⇡

R1�1 K(s)e�i!sds has to equal 1 in the

area where K(s) 6= 0 if !⌧⇤ 1.

J+(!)

!

J+(0)

1⌧

⇤

Figure 2: Spectral density in dependency on the frequency

This leads to the estimate that J+(!) = J+(0) for |!| ⌧ 1⌧

⇤ (see fig. 2)which results in

J+(!) =2

⇡Rk

B

T for |!| ⌧ 1

⌧⇤. (8)

4


By integrating we get the similar formula to equation (5)

hV (t)2i =Z

!+�!

!

2

⇡Rk

B

Td! =2

⇡Rk

B

T�! = 4kB

TR�⌫. (9)

The di↵erence of the two derivations is that in the second derivation in contrastto the first, it is clear that we calculate the mean of the squared voltage andnot the squared voltage. The voltage can be an arbitrary function of time.

2.3 Transmission Line Theory

The transmission line theory [6, chapter 2.1] allows us to calculate the volt-age along the line. With the voltage we can later on derive the number ofthermal photons. The theory describes a transmission line as a two-wire line.Infinitesimal pieces �z of the line are modeled as circuits with a resistance R,an inductance L, a conductance R and a capacitance C. These properties areall given per unit length. Each of them describes a di↵erent characteristic of thetransmission line. Given in the same order as the properties, the characteristicsare self-inductance, close proximity, finite conductivity and dielectric losses be-tween the conductors. The whole transmission line is then modeled by addinginfinitesimal circuits in a row.

R�z L�z

G�z C�z

+

�

+

�

i(z, t) i(z +�z, t)

v(z, t) v(z +�z, t)

�z

Figure 3: Transmission line scheme

With the Kirchho↵ laws for current and voltage we get

v(z, t)�R�z i(z, t)� L�z@i(z, t)

@t� v(z +�z, t) = 0 (10)

i(z, t)�G�z v(z +�z, t)� C�z@v(z +�z, t)

@t� i(z +�z, t) = 0 . (11)

For this circuit we can now assume a time dependency v(z, t) = V (z) · ei!t and

5


i(z, t) = I(z) · ei!t which simplifies the equations to

V (z +�z)� V (z)

�z�z!0=

@V (z)

@z= �(R+ i!L)I(z) (12)

I(z +�z)� I(z)

�z�z!0=

@I(z)

@z= �(G+ i!C)V (z) . (13)

By di↵erentiating equations (12) and (13) a second time with respect to z andinserting the original equations we get

@2V (z)

@z2= ��2V (z) (14)

@2I(z)

@z2= ��2I(z) (15)

with � =p(R+ i!L)(G+ i!C).

2.4 Thevenin and Norton Equivalent

Due to the finite temperature of the transmission line the resistance and theconductance cannot be considered to be noiseless. To calculate the noise we canreplace the resistance and the conductance by Thevenin or Norton equivalents.

R

V

R0

�+

U(t)

V

R00

I(t)

i

Figure 4: Left: to be replaced resistance, Middle: Thevenin equivalent, Right: Nor-ton equivalent

For the Thevenin equivalent a resistance R is replaced by a noiseless resis-tance R0 and a noisy voltage source U(t) in series connection so that the tappedvoltage does not change [7]:

V = RI = R0I + U(t). (16)

The Norton equivalent circuit replaces a resistance R by a noiseless resistanceR00 and a noisy current source I(t) in parallel connection so that the tappedcurrent does not change [8]:

i =V

R=

V

R00 + I(t) (17)

The same replacements can be made for the conductance G:

V =I

G=

I

G0 + U(t) (18)

i = GV = G00V + I(t) (19)

With those equations we can then replace the resistance and conductance infigure 3 by both either Thevenin or Norton equivalents.

6


2.5 Detection of Photons

An important part of the quantum transfer is the detection of the sent photons.There are several di↵erent possibilities to achieve a proper detection discussedin latest papers. The papers cover di↵erent problems of photon detection. Forexample the problems of thermal photons, photon losses during the transfer anddark counts have to be taken into account in calculations. Dark counts herebymean that the detector declares a detection despite no signal photon arriving.Every detector has properties that describe how good the problems are com-pensated. In this section we discuss some of these properties.One main characteristic of a detector is its e�ciency ⌘. The e�ciency describesthe probability of actually detecting a received photon. Hence a detector with alow e�ciency does not detect the photons reaching it and is therefore not useful.The time interval in which a detector can not detect two received photons iscalled the recovery time. It describes the time a detector needs to restore afterreceiving a photon before it can register another one.In addition, the detector has to be built in such a way that it can reduce lossesand minimze dark counts, so that the probability of registering an actucallyarriving photon is still high enough. A possibility is to minimize the probabilityof a dark count in contrast to the probability of a real count [9, App. D].The fidelity is another important value for the characterization of a detector. Itdescribes how good the state of the sending qubit is transferred to the receivingqubit. If the fidelity is high meaning above 2

3 , the transfer of the state can beseen as quantum.Also crucial for a detector is the linewidth of the photon which is limited byits generation time. The linewidths of Alice who generates the photon and Bobwho detects the photon has to match so that the photon detection is possible.The linewidths also imply the range of thermal photons that has to be includedas a disturbing factor in the detection.

Di↵erent simulations of detectors in the range of microwaves can be found inseveral current papers which we will look at in the following subsections. Thisoverall view is just to get to know di↵erent realizations and ideas of detectors.Later on we are going to use values from a di↵erent paper [9] to describe theprobabilities of the detector. These values are typical for modern detectors.

2.5.1 Decoy State Method

The decoy state method was used in an actual experiment [1] and is not just atheoretical consideration of how the detection of photons would work. It tookplace on the Canary Islands Teneri↵e and LaPalma. In the experiment, opticalphotons were encoded with the BB84 method and sent and received over alength of 144 km.The physicists did not only send signal photons but they also used decoy photonswhich they sent between the beams of signal photons. With the help of thesedecoy photons the detector got to know an upper boundary for the fraction ofphotons that is tagged by an eavesdropper. This avoids that the laser pulseswith the signal photons have to be too heavily attenuated to be attractive fora measurement or that they are too little attenuated so that eavesdropping isstill possible.

7


2.5.2 Quantum Nondemolition Detection

Another group developed a simluation for the detection of photons which doesnot destroy them during the process [10]. For this simulation they used Ntransmons which are basically noninteracting artificial atoms. They can bedescribed as a three-level system.Two fields are used in the simulation, a control field which is the one supposedto be measured and a probe field. They are both close to resonance with thetransmons. The control field can excite the first transmon so that it then excitesthe next which continues through the N transmons. Thus the probe field isdisturbed at the end. In the absence of a control field the probe field can passthe system undisturbed. The disturbance can be measured by comparing thefrequencies of the probe field before passing the system and afterwards.

2.5.3 Catch and Release

The authors of another paper [11], a group from the University of California, arediscussing several coupled elements which are used to catch and shortly storephotons before they are then released. For this theoretical simulation a qubitis used to initiate photons which are resonant to a cavity. The frequency ofthe qubit f

q

can be tuned in between 6 and 7 GHz. This then determines thecoupling of the qubit and the resonator which has the frequency f

r

= 6.57 GHz.The coupling can be completely turned of by setting f

q

= fr

� 400 MHz' 6.57GHz �400 MHz.In addition, there is a coupling

C

between the resonator and the transmissionline. It can be varied between zero and maximum coupling

max

= 15ns . The

variation is controlled with a coupler bias whose rising time limits the coupling.If the coupling is o↵ or weak, the photon would decay in the resonators due tolosses. If the coupling is strong enough, the photon would be emitted.The coupling between the resonator and the transmission line determines thedecay time in the resonator. This can be shown by exciting the resonator at adistinct coupling strength

C

, setting the coupling to zero for a storage time ⌧s

and then back to C

which leads to a coherent release of photons. The signaldecays after the sharp onset with a decay time T

d

⇡ 2

C

.

2.5.4 Harmonic Oscillators as Intermediate Stage

In this part, we will have a look at a paper which discusses the general im-provement of the transfer of photons and in particular the improvement of thefidelity [2]. The paper treats the problem that the transfer channel is warmerthan the refridgerators in which the qubits are situated and which are at a fewmilli-Kelvin. Although they only talk about T ⇡ 4 K which is not the temper-atures we want to have a look in, it is actually an advantage to the assumptionthat the channel has to be cooled to mK to minimize the number of thermalphotons which was made before.They consider two qubits of which each has a ground state |0i and an excitedstate |1i with a di↵erence ~!0. The qubits are connected via an unidirectionalchannel. The task would then be to transfer a state | i which is prepared atthe first node to the second node.The fidelity decreases for temperatures T

ch

> ~!0k

B

because the thermal photonsof the channel would then disturb the transfer. Instead of a direct connection

8


between the qubits and the channel, harmonic oscillators are used as an interme-diate stage. This is advantageous because their swap operation is independentof the number of states in the channel.They mention more operations to improve the transfer. But these are not usefulfor the subject of this thesis. To get to know the other improvements I wouldhighly recommend to read the paper [2].

9

3 CALCULATIONS AND RESULTS

3 Calculations and Results

3.1 Solutions for the Tranmission Line Theory

The transmission line problem we look at is a transmission line terminated atboth ends [6, chapter 2.3]. The transmission line itself is at a temperature T

high

,while the terminating resistances are at a temperature T

low

. Over the length Lof the transmission line we string together infinitesimal pieces �z of the typeshown in figure 3.

z = 0 z = L

I(0) at Thigh

I(L)

ZT

at Tlow

ZT

at Tlow

V (0) V (L)

Figure 5: Terminated transmission line

The solutions of the di↵erential equations for the voltage (14) and the current(15) in the transmission line theory can be solved with an ansatz

V (z) = V +0 e��z + V �

0 e�z (20)

I(z) = I+0 e��z + I�0 e�z, (21)

where V ±0 and I±0 are the amplitudes of the waves travelling in ⌥z-direction

and � = ↵ + i� =p(R+ i!L)(G+ i!C). The factor � consists of a real

part ↵ which describes the attenuation of the wave and the imaginary part �which characterizes the oscillation of the voltage respectively the current. Theamplitudes correlate which can be derived from equation (12) or equation (13):

@V (z)

@z= �� V +

0 e��z + � V �0 e�z = �(R+ i!L)(I+0 e��z + I�0 e�z) (22)

, I(z) =�

R+ i!L| {z }=Z

�10

(V +0 e��z � V �

0 e�z) (23)

, I(z) =V +0

Z0e��z � V �

0

Z0e�z (24)

with the characteristic impedance Z0 =R+ i!L

�=

rR+ i!L

G+ i!C.

By looking at a line terminated on both ends with impedances ZT

, we can derivea relation between the amplitudes of the waves in +z- and �z-direction. For aterminated line at z = 0 and z = L we get two conditional equations:

10


ZT

=V (0)

I(0)=

V +0 + V �

01Z0

(V +0 � V �

0 )(25)

ZT

=V (L)

I(L)=

V +0 e��L + V �

0 e�L

1Z0

(V +0 e��L � V �

0 e�L)(26)

The first condition (eq. (25)) yields the relation V �0 = V +

0

ZT

� Z0

ZT

+ Z0for the

amplitudes of the two directions. From equation (26) we extract a term for thelength of the transmission line:

ZT

=V +0 e��L + V �

0 e�L

1Z0

(V +0 e��L � V �

0 e�L), V �

0 = V +0

ZT

� Z0

ZT

+ Z0e�2�L

, 1 = e�2�L

Because of the real and imaginary part of � this condition states that the lengthof the transmission line would as well have an imaginary part. This is notsuitable and therefore the imaginary part has to equal zero to evaluate theequation. As we have no further interest in the condition, it has just beenincluded for the sake of completeness.

3.2 Noise by Thevenin or Norton Equivalent

If we replace every resistance and conductance by its Thevenin or Norton equiv-alent, we would have to calculate an inhomogenous di↵erential equation. Whilethe noise source is a random function of the time, the solution would becomeeven more di�cult.Instead of calculating the whole transmission line with inserted noise sources,we take advantage of the information we already got about the dependency ofthe voltage on z. In the following we therefore only use the exponential de-pendency e�↵z describing the attenuation of the wave along the transmissionline. This is feasible because we will calculate the power a few chapter laterand hence we only need the square of the absolute value of the voltage whichabolishes the factor e�i�z. In addition to that, we are only interested in theattenuation of the wave as we want to calculate the attenuation of the powergiven at equilibrium.

3.3 Distortionless Case

The distortionless transmission line is a special case where the real and theimaginary part of � have an easy dependence on the frequency of the wave !.In general the dependency of the imaginary part on ! is not linear. This leadsto a dependency of the velocity on the frequency v = !

�

. Due to this there isdispersion. The imaginary part of � is only proportional to ! and the velocitytherefore independent of the frequency in the case of no loss or in a distortionlesstransmission line.The required term for no distortion is

R

L=

G

C. (27)

11


For this condition the factor � simplifies as follows

� =p(R+ i!L)(G+ i!C) = i!

pLC

s✓1 +

R

i!L

◆✓1 +

G

i!C

◆

(27)= i!

pLC

s✓1 +

R

i!L

◆✓1 +

R

i!L

◆

= i!pLC

✓1 +

R

i!L

◆= R

rC

L+ i!

pLC

(27)=

pRG| {z }=↵

+i!pLC| {z }

=�

↵ =pRG (28)

� = !pLC (29)

The advantage of the distortionless transmission line is not only that the velocityis independent of the frequency and hence there is no dispersion, but in additionto that the factor ↵ which describes the attenuation of the wave is not explicitydependent on the frequency as well. Therefore we do not have to considerdi↵erent attenuations or velocities for di↵erent frequencies.In fact, the resistance R is slightly dependent on the frequency. But with thechoice of the attenuation constants in section 3.6 this is not relevant in thisthesis and just mentioned to not give false information.

3.4 Transferring Voltage into Number of Photons

For the evaluation of the feasibility of detecting a sent photon despite thermalphotons in the transmission line, we need to calculate the power first. After-wards we deduce the number of thermal photons on the transmission line insimilarity to the 3D-case derived by Planck [12]. We will do that for an equib-librium and proceed with the non-equilibrium case with the finite length lateron.The absorbed power on the transmission line in the interval between ! and! + d! equals the velocity of the waves c0 times the number of waves per unitlength n = 1

2⇡d!

c

0 times the energy ✏(!) [5, chapter 15.17]. The factor c0 · n canbe understood as an inverse time which describes the number of waves arrivingper unit time.

P (!)d! = c0✓

1

2⇡

d!

c0

◆✏(!) (30)

This results with the mean energy ✏(!) =~!

e~!/k

B

T � 1in

P (!)d! =1

2⇡

~!e~!/k

B

T � 1d! . (31)

The power and the number of thermal photons along the transmission line arecertainly connected. To point this out, we first have a look at the 3D-case

12


described by Planck. This derivation is shown in Reif [5, chapter 9.13-9.15].The number of photons of both polarisations per unit volume with a frequencyin between ! and ! + d! can be derived by adding up the mean number ofphotons f(k) with a specific value of k over the volume of the k-space within aspheric shell with radius k = !

c

0 and k + dk = !+d!

c

0 and multiplying with 2:

N(!)d! = 2 · f(k) · 4⇡k2dk (32)

where f(k)dk =1

(2⇡)31

e~!/k

B

T � 1dk. We now substitute k = !

c

0 with ! and

rearrange terms:

N(!)d! = 2 · 1

e~!/k

B

T � 1· 4⇡

⇣wc0

⌘2 d!

c0=

4

~!c0 · 2⇡~!3

c021

e~!/k

B

T � 1d! (33)

On the other hand the power can be calculated as follows.F. Reif does this calculation in his book beginning with calculating the powerwhich is absorbed by an ideal black body. Further on Reif uses the detailedequilibrium and integrates over the solid angle. The detailed derivation can beseen in Reif [5, chapter 9.15]. In the end we get:

P (!)d! =2⇡~!3

c02f(k)d! = 2

⇡~!3

c021

e~!/k

B

T � 1d! (34)

By comparing the results for the number of photons and the power, we can seethe following relation:

N(!)d! =4

~!c0 · P (!)d! (35)

In the one-dimensional case we want to look at the flux in the end. We can dothe same for the 3D-case by multiplying N with the velocity of the waves c0.

N(!)d! = c0 · N(!)d! =4

~! · P (!)d! (36)

Now we derive the number of photons per unit length and frequency equivalentlyto the derivation of N in the three-dimensional case (eq.(32)). In one dimensionwe do not need the factor 2 because there is only one direction of polarisation.

Hence the number of photons with k which is f(k) =1

2⇡

1

e~!/k

B

T � 1in the one-

dimensional issue gives the number of photons per unit length and frequencywithin d!

N(!)d! = f(k)dk =1

2⇡

1

e~!/k

B

T � 1

d!

c(37)

As announced we want to look at the number of photons respectively the numberof photons per unit time per unit frequency. Therefore we again multiply withc0:

N(!)d! = c · N(!)d! =1

2⇡

1

e~!/k

B

T � 1d! (38)

13


This shows that the relation between the number of photons and the power (seeeq.(31)) in the 1D-case is given by:

N(!)d! =1

~!P (!)d! (39)

Figure 6: Number of photons N(!) for T = 300 K

3.5 Di↵erence in Temperature

Since the resistances which terminate the transmission line are at a lower tem-perature than the line itself (see fig.5), the number of thermal photons at theends of the line is di↵erent compared to the number of thermal photons alongthe line. For very low temperatures which are virtually zero there are no ther-mal photons in the range between z = 0 and z = ✏ respectively z = L � ✏ andz = L.Assuming that the temperature at the resistances is T

low

= 0.5 K the numberof thermal photons is much smaller than the number of thermal photons in thetransmission line which is at a temperature T

high

of the order of room tempera-ture. The thermal photons of the colder part are attenuated as well which leadsto the assumption that the number of photons out of that area are negligiblecompared to the thermal photons due to the hotter transmission line.Nevertheless we include the number of photons coming out of the cold area atfirst to show that the e↵ect of them is very little.Concerning the spectral density of the voltage we expect an additional contri-bution upto a cuto↵ 1

⌧

⇤low

which is much smaller than the cuto↵ of the spectral

density at Thigh

1⌧

⇤high

(see fig.7). If the frequency we look at lies between the two

cutouts, we would not have to consider any impact of the non-zero temperatureof the terminating resistances.

14


!J+(!)|T=T

high

J+(!)|T=T

low

1⌧

⇤high

1⌧

⇤low

Figure 7: Spectral density dependent on the frequency for the calculated transmissionline

3.6 Power and Number of Photons for Finite Length

In the case of an infinite transmission line at one temperature the number ofthermal photons stays the same along the line because of its equilibrium state.Therefore the attenuation of the voltage due to losses plays no role. In theregarded case of a finite transmission line which is terminated by resistances atlower temperature than the rest of the line, the equations for the power and thenumber of photons have to be corrected considering the attenuation.The exponential attenuation of the voltage given by

V (z) / e�↵z (40)

is the only important dependency of the voltage because for the power the squareof the absolute value is considered:

P = |I|2R =|V |2

R(41)

Hence the imaginary part of � is not included in the equation for the power.Because of the square the dependency of the power on z is

P / e�2↵z. (42)

The maximum power meaning the power of the equilibrium which is attenu-ated along the transmission line is given by equation (31) with the insertedtemperature T

high

:

Peq

(!)d! =1

2⇡

~!e~!/k

B

T

high � 1d!

The power can then be separated in a contribution Pleft

which describes thepower of the waves going left on the transmission line and a contribution P

right

which is defined equivalently. In the equilibrium case the power of each direction

is half of the overall power Peq

: Peq,left

= Peq,right

=Peq

2.

15


At the terminating resistances ZT

the power given by equation (31) with T =Thigh

is virtually zero. Instead there is another, much lower power which pro-duces thermal photons because of the non-zero temperature T

low

. We havetherefore an overall power consisting of a part of P

T

high

(!) and PT

low

(!). Thetotal power for each direction in dependency on z can be described by

Pleft

(z,!) =⇣1� e�2↵(L�z)

⌘Peq,left,T

high

(!) + e�2↵(L�z)Peq,left,T

low

(!)

(43)

Pright

(z,!) =�1� e�2↵z

�Peq,right,T

high

(!) + e�2↵zPeq,right,T

low

(!).(44)

The prefactors of Peq,T

high

(!) ensure that the contribution of the power of Thigh

of the left- respectively right-going wave equals zero at the resistances ZT

atz = L respectively z = 0. To evaluate the validity of our equation we want tohave a look at the plotted powers meaning equations (43) and (44) and theirsum.

To make these plots we first of all need some values for the attenuation ofthe wave. It consists of attenuations due to several reasons. The constant is asum of losses due to metal conductivity, dielectrical loss tangent, conductivityof dielectric and radiation [13]. The dielectrical loss tangent is proportional tothe frequency. Therefore this loss type becomes more important for higher fre-quencies e.g. microwave frequencies than e.g. the metal conductivity which isproportional to the square root of the frequency.The dieletric loss tangent is independent of the geometry and only dependenton the type of dieletric material. The loss constant ↵

d

is proportional to the losstangent tan �. It is a measure of the ratio of the conductance to the susceptancemeaning a ratio of the lossy reaction on the electric field to the lossless reaction.

In the following we will only consider dielectric loss tangent loss and leave outthe other attenuation as they are negligible for microwave frequencies.This is not very consistent with the earlier discussed distortionless case becausethe conductance G is seen as big enough to dominate the other e↵ects. Theresistance R is however very low as we do not include losses due to metal con-ductivity. Hence the condition (27) states that the ratio L

C

becomes very littlewhich is not usual.Nevertheless we keep the exponential attenuation e�↵z which is not explicitlydependent on the frequency although the attenuation due to dielectric loss tan-gent is itself linear in the frequency.We adapt values for the loss constant ↵

d

from an online website which explainslosses on a transmission line. The website contains a graph titled ’TEM mediadielectric loss, ER=1’ [13] which shows the dependency of the loss constant onthe frequency. The values for ↵

d

that are used later on has been calculated withthis graph. In the plot the loss is given in dB

cm . To get the attenuation constant↵ that we want to insert in our calculation, we convert the loss constant ↵

d

from the source. For this we use:

dB = 10 · log10✓P1

P2

◆with P1,2 = e�↵·z1,2

16


The di↵erence between z1 and z2 is �z = 1 cm which ensures that the value of↵ corresponds to ↵

d

given in dBcm . In addition to that we multiply ↵

d

with thesame di↵erence �z to get a value in dB. Hence the conversion reads:

↵d

·�z = 10 · log10⇣e�↵·(z1�z2)

⌘(45)

, 100.1·↵d

·�z = e�↵·�z (46)

, ↵ = � 1

�zln

�100.1·↵d

·�z

�(47)

The graph gives di↵erent values for the attenuation for di↵erent tan �. As therewas no proper information given which values for tan � are feasible, we calculatewith the intermediate values for tan � = 0.005. At the end we compare thosewith the values for tan � = 0.001 which is the lowest value for tan �. For tan � =

0.005 the slope is�0.5dB

cm

110GHzsince the attenuation is at �0.5dB

cm for f = 110 GHz.

The frequency we use to have a look at the graphs of the power is f = 6 GHzwhich is a very often used frequency in the paper about microwave detectors.With the slope of the plot for tan � = 0.005 the attenuation for f = 6 GHz is :

↵d

=�0.5dB

cm

110GHz· 6 GHz = �0.027

dB

cm

eq.(47), ↵ = 0.631

m

For the plots we also use a transmission line of the length L = 10 m. Later onwhen we try to reduce the number of thermal photons, we change the lengthof the transmission line. This value is just used to graphically illustrate thebehaviour of the power along the transmission line.

17


Figure 8: Power in dependency on z for ! = 2⇡ · 6 GHz, Thigh

= 300 K, Tlow

= 0.5K, L = 10 m, ↵ = 0.63 1

m with the added power of Tlow

As the total power Ptot

has its maximum at z = L

2 and decreases to halfof the maximum at z = 0 and z = L the assumed descriptions of the left- andright-going power seem right.We now want to show that the e↵ect of the power initiated by thermal pho-tons due to T

low

is negligibly low. We only plot the power consisting of thecontribution of T

high

.

18



= 300 K, Tlow

= 0.5K, L = 10 m, ↵ = 0.63 1

m without the added power of Tlow

Looking at the values of the power, the di↵erence between the plots of figures8 and 9 seems small which can be pointed out by plotting the power with theadded contribution of T

low

and the power without its contribution in one graph.For better comparison we zoom in at the crossing point of P

left

and Pright

.

19



= 300 K, Tlow

= 0.5K, L = 10 m, ↵ = 0.63 1

m to point out the e↵ect of the power of Tlow

The relative di↵erence between the powers is of the order 10�4 which provesthat the part of T

low

is negligible. Therefore we use

Pleft

(z,!) =⇣1� e�2↵(L�z)

⌘Peq,left,T

high

(!) (48)

Pright

(z,!) =�1� e�2↵z

�Peq,right,T

high

(!) (49)

in the following calculations.

Out of these equations we can derive the number photons with equation (39):

Nleft

(z,!) =1

~!Pleft

(z,!) =1

~!

⇣1� e�2↵(L�z)

⌘Peq,left,T

high

(!)

=1

~!

⇣1� e�2↵(L�z)

⌘ 1

2

1

2⇡

1

e~!/k

B

T

high � 1

=⇣1� e�2↵(L�z)

⌘ 1

4⇡

1

e~!/k

B

T

high � 1| {z }12N(!)=N

left

(!)

and equivalently:

Nright

(z,!) =�1� e�2↵z

� 1

4⇡

1

e~!/k

B

T

high � 1| {z }12N(!)=N

right

(!)

20


Hence the total number of photons along the transmission line can be calculatedwith:

Ntot

(z,!) =

✓1� 1

2

⇣e�2↵(L�z) + e�2↵z

⌘◆ 1

2⇡

1

e~!/k

B

T � 1(50)

Figure 11: Number of photons in dependency on z for ! = 2⇡ · 6 GHz, Thigh

= 300K, T

low

= 0.5 K, L = 10 m, ↵ = 0.63 1m

3.7 Detector Values

To find out if the detection of sent photons is possible despite the thermalphotons on the transmission line, we have to compare the number of both. Inregard to that we can derive the number of photons per second for both kinds.The number of signal photons can be calculated with some detector values. Weextract these values from a paper which approaches the detection of photonsas well [9]. The values can be used because they are pretty much alike for alldetectors. First of all we note that the generation of a photon takes a finitetime ⌧

gen

= 254 ns [9]. This leads to a number of generated photons per secondof n = 3.94 · 106 1

s .Due to losses during the transfer of the photons along the transmission line onlya fraction of those generated photons reaches the detector. We now assume atransmission line which ensures the transfer of each generated photon. This is ofcourse not feasible for application but an e�ciency which is lower than 1 would

21


not change the order of the number of photons.Another later on used value is the linewidth of the frequency. The generatedphotons do not have the exact same frequencies. But the generated frequencyalways lies inside a range around the desired frequency. This range is �! ⇡ 2⇡ ·1MHz [9].

3.8 Results

The number of thermal photons disturbing the process can be derived withequation (50). Because the frequency of a signal photon has a finite linewidthwe need to integrate the number of thermal photons over the range of thelinewidth and not only insert the frequency of the photon in the formula. Thelikely regarded frequency lies in the X-band which is a band of frequencies in themicrowave radio region. It is ! = 2⇡ ·6 GHz. The linewidth is at the magnitudeof �! ⇡ 2⇡ · 1 MHz [9]. Hence we calculate the number of thermal photons forthe infinite transmission line as follows:

N =1

2⇡

Z!+ �!

2

!� �!

2

1

e~!/k

B

T � 1d!

T=300K⇡ 1.65 · 108 1s

To find out how many thermal photons arrive at the detector in the same timethat one signal photons arrives, we multiply the number of thermal photons persecond with the time that it takes to generate a photon ⌧

gen

= 254 ns [9]:

Nthermal

= 254 ns ·N ⇡ 41.97

As this is the number of thermal photons if the transmission line was on anequilibrium it decreases at the ends of a finite transmission line as derived inequation (50):

Nthermal

(z) = 41.97 ·✓1� 1

2

⇣e�2↵(L�z) + e�2↵z

⌘◆(51)

To now calculate the number of thermal photons at the detector meaning atthe end of the transmission line, we use the value for the attenuation constant↵ for tan � = 0.005 we derived in equation (47). With the earlier used length ofL = 10 m the shape of the number of thermal photons looks as shown in figure12.

22


Figure 12: Number of thermal photons along the transmission line (eq.(51)) forL = 10 m, ↵ = 0.63 1

m

Despite the attenuation the number of thermal photons arriving at the de-tector meaning at z = L in the same time that one signal photon arrives isapproximately 21. This is too high to detect the signal photon. Hence we tryto improve the detection and get N

thermal

(z = L) under 1 by trying other fre-quencies, lengths and temperatures.At first we vary the frequency staying at T

high

= 300 K and try 1 GHz and 30GHz comparing the results with the values for 6 GHz. The chosen frequency of30 GHz is used because frequencies above 30 GHz are more di�cult to gener-ate. We do this for the lengths L = 0.1 m, 1 m, 10 m. The values for N

thermal

and Nthermal

(z = L) are calculated by integrating over the frequency with thelinewidth and by adding the z-dependency.

23


f = !

2⇡ [GHz] ↵ [ 1m ] L [m] Nthermal

Nthermal

(z = L)

1 0.100.1

252.462.50

1 22.8810 109.15

6 0.630.1

41.972.48

1 15.0310 20.98

30 3.140.1

8.291.93

1 4.1410 4.15

Table 1: Values for di↵erent lengths L, Thigh

= 300 K, tan � = 0.005

The number of thermal photons at the detector does not go under 1 for thetested lengths. Although it is lower for higher frequencies which is a result ofthe higher attenuation and lower number of thermal photons at that frequency.The number of thermal photons decreases rapidly for the both lower frequenciesas we shorten the transmission line. This development does not continue thisstrong for even shorter lengths. The reason for the intense decreasing is that theleft- respectively right-going power has to equal zero on the right respectivelyleft end of the transmission line. Therefore the number in the middle is notat N

thermal

because its distance to the cold termination is too small. At evenshorter lengths the ’slope’ flattens.For a frequency of 30 GHz the number of thermal photons approaches 0.1 for alength of about 4 mm. But as we would like to be able to transfer photons overa longer distance, we now vary the temperature of the transmission line for alength of 1m and the three frequencies we inserted before.

f = !

2⇡ [GHz] ↵ [ 1m ] Thigh

[K] Nthermal

Nthermal

(z = L)

1 0.10

50 41.97 3.80100 84.07 7.62150 126.16 11.43200 168.26 15.25250 210.36 19.07300 252.46 22.88

6 0.63

50 6.89 2.47100 13.91 4.98150 20.92 7.49200 27.94 10.01250 34.95 12.52300 41.97 15.03

30 3.14

50 1.28 0.64100 2.68 1.34150 4.08 2.04200 5.49 2.74250 6.89 3.44300 8.29 4.14

Table 2: Values for di↵erent temperatures Thigh

, L = 1 m, tan � = 0.005

24


Only for a temperature of 50 K we get a number of thermal photons at thedetector lower than one for the higher frequency 30 GHz. As we would like aneven lower number than 0.64 thermal photons at z = L, so that the detection ofthe signal photons becomes easier, we try to further decrease it. The followingvalues can be reached for f = 30 GHz:

Thigh

= 300 K, L = 0.01 m ! Nthermal

(z = L) = 0.25

Thigh

= 100 K, L = 0.01 m ! Nthermal

(z = L) = 0.08

Thigh

= 15 K, L = 1 m ! Nthermal

(z = L) = 0.15

For the two lower frequencies and L = 1 m the number of thermal photons atthe detector are even for 50 K too high to exactly detect a signal photon. Fora length of L = 1 m the setup would have to be cooled to under 3 K to getcomparable number of thermal photons for 1 and 6 GHz as at 15 K for 30 GHz.For the very short distance of 1 cm, the number of thermal photons at the detec-tor is equal for all three frequencies. This equalization happens due to the shortlength. The transmission line is therefore not even in a state that is similar toan equilibrium. This leads to an overall low number of thermal photons as thenumber of thermal photons in the middle of the transmission line is less than0.1 % bigger than at the detector.

The previous values for ↵ were extracted for tan � = 0.005. We now lookat the di↵erence to the values for tan � = 0.001. The attenuation constants arecalculated similarly to (47):

↵ = � 1

�zln

�100.1·↵d

·�z

�with ↵

d

=�0.1dB

cm

110GHz· f

With the same formulas as used before we get the following results:

f = !

2⇡ [GHz] ↵ [ 1m ] L [m] Nthermal

Nthermal

(z = L)

1 0.020.1

252.460.50

1 4.9510 41.62

6 0.130.1

41.970.54

1 4.8010 19.43

30 0.630.1

8.290.49

1 2.9710 4.14

Table 3: Values for di↵erent lengths L, Thigh

= 300 K, tan � = 0.001

The number of thermal photons at the detector is much lower than for theattenuation constants for tan � = 0.005. In the following table we will vary thetemperature again.

25


f = !

2⇡ [GHz] ↵ [ 1m ] Thigh

[K] Nthermal

Nthermal

(z = L)

1 0.02

50 41.97 0.82100 84.07 1.65150 126.16 2.47200 168.26 3.30250 210.36 4.12300 252.46 4.95

6 0.13

50 6.89 0.79100 13.91 1.59150 20.92 2.39200 27.94 3.20250 34.95 4.00300 41.97 4.80

30 0.63

50 1.28 0.46100 2.68 0.96150 4.08 1.46200 5.49 1.97250 6.89 2.47300 8.29 2.97

Table 4: Values for di↵erent temperatures Thigh

, L = 1 m, tan � = 0.001

The thermal photons at the detector are again much fewer than for tan � =0.005.The reason for the noticeable di↵erence between the values is that the ratiobetween the length and the attenuation constant is smaller. Hence the e↵ectwhich appears in table 1, that the number of thermal photons at z = L becomesmuch lower than N

thermal

, can already be seen at longer distances.

26

4 CONCLUSION

4 Conclusion

Our expectation that the transfer of information with microwave photons ispossible over distances of several metres or even several kilometres has not beenapproved. The transfer over such lengths is especially not possible at roomtemperature. At room temperature the detection of photons for tan � = 0.005could only be guaranteed over lengths of the magnitude of centimetres.For this short lengths the assumption that the transmission line is in an equilib-rium state becomes unlikely. But the description of the line as an equilibrium isrequired for the Nyquist theorem. Hence the description of the thermal photonsbecomes more di�cult for short distances.The transfer of photons over L = 1 m for tan � = 0.005 can only be ensured fora frequency f = 30 GHz for temperatures of 15 K or lower. For the lower fre-quencies f = 1 GHz and f = 6 GHz the transmission line has to be even coolerthan 3 K. For this low temperatures the contribution of the cold terminatingparts is not negligible anymore.The distances respectively temperatures with which the detection is possible arelonger respectively higher for tan � = 0.001. Therefore it is advantageous to geta very low tan �. But even for tan � = 0.001 the lengths and temperatures whichare feasible are not at the stage that are needed for a proper information transfer.

Reviewing the paper I read before starting this thesis, one paper can be easilycompared with our results. The paper about intracity quantum communica-tion [2] discusses the transfer with microwaves too and investigates about theneeded temperature on the transmission line which is the part of photon de-tection we had a look at in this thesis. In the paper it was stated that thetransmission line has to be cooled to a few Kelvin. We tried to disprove thisstatement in conclusion but we can only approve the necessity of a cooled fiber.In the paper they use harmonic oscillators as a stage between the qubit and thechannel to reduce the disturbance due to thermal photons. With the calcula-tions we made there is no possibility to evaluate their usage. But it seems tobe a good way to be able to increase the temperature respectively the length ofthe channel.Because we did not look at the same properties of photon detection as it wasdone in the other papers that were mentioned in section 2.5, a comparison isnot suitable.

As an overall conclusion it seems as if there are still a lot of aspects to developbefore the transfer of quantum information becomes applicable in everyday life.

27

REFERENCES

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[10] Sankar R. Sathyamoorthy, L. Tornberg, Anton F. Kockum, Ben Q. Bara-giola, Joshua Combes, C. M. Wilson, Thomas M. Stace, and G. Johans-son. Quantum nondemolition detection of a propagating microwave photon.Phys. Rev. Lett., 112:093601, Mar 2014. https://link.aps.org/doi/10.1103/PhysRevLett.112.093601.

[11] Yi Yin, Yu Chen, Daniel Sank, P. J. J. O’Malley, T. C. White,R. Barends, J. Kelly, Erik Lucero, Matteo Mariantoni, A. Megrant, C. Neill,A. Vainsencher, J. Wenner, Alexander N. Korotkov, A. N. Cleland, andJohn M. Martinis. Catch and release of microwave photon states. Phys.

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quantumcryptography with microwaves

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