Tutorial Vol.11,No.3/September 2019/Advances in Optics and Photonics 699 36) =1= This mathematical result [Eg.(36)]is technically a consequence of the fact that the eigenfunctions have been written in normalized form,which in turn is why the ma- trices V and Ut are "unitary,"but this is really a deeper truth,and it has several im- portant consequences that are central to the larger discussion here. (1)We can evaluate this sum S without even solving the eigenproblem. (2)Having evaluated S,we can know immediately that there is an upper bound on how many channels we could have that have at least some given coupling strength;there could only be a finite number of orthogonal channels with any given finite magnitude of coupling strength. (3)Suppose we solve the eigenproblem by looking one by one,physically or math- ematically,for the channels by some process,noting their coupling strengths. Then,when we find we have nearly exhausted the available sum rule S,we can stop looking for any more channels;there cannot be any more strongly coupled channels,because there is not sufficient sum rule S left. It might seem that this kind of sum rule is only going to exist when we consider finite numbers of sources and receivers.However,we are going to find below in Section 6 that the operators associated with wave equations are going to have a finite result for this sum rule even as we consider continuous functions and infinite basis sets.Indeed, this finiteness is the defining characteristic of so-called Hilbert-Schmidt operators (which,incidentally,are necessarily also "compact"),and we return to this point in Section 6.So,though we continue for the moment with finite numbers of sources and receiver points,our results are generally going to survive even as we make the transition to such continuous source and receiver functions,with possibly infinite basis sets. This finite sum rule can also be regarded as the source of diffraction limits with waves,as will become clearer below.Such limits apply both in conventional optical situations and more generally,and this result is a central benefit of this SVD approach. 3.9.Constraint on the Choice of the Coupling Strengths of the Channels In looking at the sum rule,we might hope that we have the flexibility to choose how many channels are how strong-we might want to have several channels at a weaker coupling strength rather than a few strong ones,for example-while still "obeying" the sum rule.However,once we have chosen a given set of source and receiver points, or have done the equivalent for continuous functions in choosing the Hilbert spaces for source and receiver functions,we do not have that flexibility. The eigenvalues,which are the coupling strengths,and any degeneracies they have, are uniquely specified when we solve the eigenproblem.The eigenfunctions are also essentially unique,other than for the minor flexibility allowed in choosing the specific linear combinations for eigenfunctions associated with the same degen- erate eigenvalue. So,not only does the sum rule constrain the overall sum of connection strengths;the eigensolutions uniquely determine the coupling strengths of the channels if we want them to be orthogonal(with orthogonality determined by the inner products in the Hilbert spaces)
S X N m q1 jsqj 2 X NR i1 X NS j1 jgijj 2: (36) This mathematical result [Eq. (36)] is technically a consequence of the fact that the eigenfunctions have been written in normalized form, which in turn is why the matrices V and U† are “unitary,” but this is really a deeper truth, and it has several important consequences that are central to the larger discussion here. (1) We can evaluate this sum S without even solving the eigenproblem. (2) Having evaluated S, we can know immediately that there is an upper bound on how many channels we could have that have at least some given coupling strength; there could only be a finite number of orthogonal channels with any given finite magnitude of coupling strength. (3) Suppose we solve the eigenproblem by looking one by one, physically or mathematically, for the channels by some process, noting their coupling strengths. Then, when we find we have nearly exhausted the available sum rule S, we can stop looking for any more channels; there cannot be any more strongly coupled channels, because there is not sufficient sum rule S left. It might seem that this kind of sum rule is only going to exist when we consider finite numbers of sources and receivers. However, we are going to find below in Section 6 that the operators associated with wave equations are going to have a finite result for this sum rule even as we consider continuous functions and infinite basis sets. Indeed, this finiteness is the defining characteristic of so-called Hilbert–Schmidt operators (which, incidentally, are necessarily also “compact”), and we return to this point in Section 6. So, though we continue for the moment with finite numbers of sources and receiver points, our results are generally going to survive even as we make the transition to such continuous source and receiver functions, with possibly infinite basis sets. This finite sum rule can also be regarded as the source of diffraction limits with waves, as will become clearer below. Such limits apply both in conventional optical situations and more generally, and this result is a central benefit of this SVD approach. 3.9. Constraint on the Choice of the Coupling Strengths of the Channels In looking at the sum rule, we might hope that we have the flexibility to choose how many channels are how strong—we might want to have several channels at a weaker coupling strength rather than a few strong ones, for example—while still “obeying” the sum rule. However, once we have chosen a given set of source and receiver points, or have done the equivalent for continuous functions in choosing the Hilbert spaces for source and receiver functions, we do not have that flexibility. The eigenvalues, which are the coupling strengths, and any degeneracies they have, are uniquely specified when we solve the eigenproblem. The eigenfunctions are also essentially unique, other than for the minor flexibility allowed in choosing the specific linear combinations for eigenfunctions associated with the same degenerate eigenvalue. So, not only does the sum rule constrain the overall sum of connection strengths; the eigensolutions uniquely determine the coupling strengths of the channels if we want them to be orthogonal (with orthogonality determined by the inner products in the Hilbert spaces). Tutorial Vol. 11, No. 3 / September 2019 / Advances in Optics and Photonics 699
700 Vol.11,No.3/September 2019/Advances in Optics and Photonics Tutorial Suppose we write down the"power"coupling strengths s 2 in order,starting from the strongest and proceeding to weaker ones.In this list,if a given eigenvalue is degen- erate,with a degeneracy of n,we will write it down n times [136],and give each occurrence a separate value of the index j.So we would have a list of the form s12≥22≥…≥s2≥…. (37) Then,provided we include all the occurrences of a given degenerate eigenvalue(so if we encounter degenerate coupling strengths in the list,our choice of j must be at the end of any such degenerate group of coupling strengths),we can state that the number of orthogonal channels(communications modes)with power coupling strength |s2≥ls;P2isj. (38) We cannot rearrange any linear combinations of communications modes to get more channels that are at least this strongly coupled.If we want more such channels,then we have to change the source and/or receiver points and/or Hilbert spaces. 4.INTRODUCTORY EXAMPLE-THREE SOURCES AND THREE RECEIVERS To understand how this approach works,both mathematically and physically,we can look at a simple example,one that is large enough to be meaningful,but small enough for explicit details.Figure 3 shows the physical layout.We have Ns =3 point sources, spaced by 24 in the "vertical"y direction,(with wavelength A 2m/k as usual for a wavevector magnitude k).These are separated by a "horizontal"distance L.=5A from a similarly spaced set of Ng =3 receiving points. 4.1.Mathematical Solution We use Eq.(7)to calculate the matrix elements gi of the coupling operator GsR. Explicitly,for example,for the matrix element gi3 that gives the wave amplitude at re as a result of the source amplitude at rs3,noting first that rR1-rs3l=V0y1-ys3)2+(R1-23)2=1V42+52=√4I元， 39) then Figure 3 Source points Receiving points r分· ,「I 「2 2入 51 Set of three sources,spaced by 2 in the "vertical"y direction,separated from three similarly spaced receiving points by a distance 5A in the "horizontal"z direction
Suppose we write down the “power” coupling strengths jsjj 2 in order, starting from the strongest and proceeding to weaker ones. In this list, if a given eigenvalue is degenerate, with a degeneracy of n, we will write it down n times [136], and give each occurrence a separate value of the index j. So we would have a list of the form js1j 2 ≥ js2j 2 ≥ ≥ jsjj 2 ≥ : (37) Then, provided we include all the occurrences of a given degenerate eigenvalue (so if we encounter degenerate coupling strengths in the list, our choice of j must be at the end of any such degenerate group of coupling strengths), we can state that the number of orthogonal channelscommunications modes with power coupling strength jsj 2 ≥ jsjj 2 isj: (38) We cannot rearrange any linear combinations of communications modes to get more channels that are at least this strongly coupled. If we want more such channels, then we have to change the source and/or receiver points and/or Hilbert spaces. 4. INTRODUCTORY EXAMPLE—THREE SOURCES AND THREE RECEIVERS To understand how this approach works, both mathematically and physically, we can look at a simple example, one that is large enough to be meaningful, but small enough for explicit details. Figure 3 shows the physical layout. We have NS 3 point sources, spaced by 2λ in the “vertical” y direction, (with wavelength λ 2π∕k as usual for a wavevector magnitude k). These are separated by a “horizontal” distance Lz 5λ from a similarly spaced set of NR 3 receiving points. 4.1. Mathematical Solution We use Eq. (7) to calculate the matrix elements gij of the coupling operator GSR. Explicitly, for example, for the matrix element g13 that gives the wave amplitude at rR1 as a result of the source amplitude at rS3, noting first that jrR1 − rS3j ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi yR1 − yS32 zR1 − zS32 q λ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 52 p ffiffiffiffiffi 41 p λ, (39) then Figure 3 rS1 rS2 rS3 rR1 rR2 rR3 5λ 2λ y z Set of three sources, spaced by 2λ in the “vertical” y direction, separated from three similarly spaced receiving points by a distance 5λ in the “horizontal” z direction. 700 Vol. 11, No. 3 / September 2019 / Advances in Optics and Photonics Tutorial
Tutorial Vol.11,No.3/September 2019/Advances in Optics and Photonics 701 0.01020-0.00711i g13=一1 1 exp(iklrRI-rs3l)1 exp (2riv4I (40) TRI -Is3 4π √411 Now we write distances in units of wavelengths for convenience (or equivalently,we set =1).To get numbers of convenient sizes and signs,we multiply by a factor 8sR=-4πL2=-62.83. (41) So,then, gs813≈-62.83×(0.01020-0.00711)≈-0.64+0.45i. (42) Proceeding similarly for the other matrix elements,we have -0.7+0.6i-0.64+0.45i gSRGSR -0.7+0.6i 1 -0.7+0.6i (43) -0.64+0.45i-0.7+0.6i 1 The sum,Eq.(36),of the modulus squared of these matrix elements of Gsg as in Eq.(43)is the sum rule S=7.67/g3R: (44 The matrix GGsk can be written,with a convenient scaling factor g 2.47 -0.67-0.08i -0.42 g3 RGSRGsR兰 0.67+0.08i 2.72 -0.67+0.08i (45) -0.42 -0.67-0.08i 2.47 Note that this matrix is Hermitian,as required.Having established the matrix GGsk. we can now use standard numerical routines to find eigenvalues and eigenvectors.The resulting eigenvalues of GsGsR(and of GsRGsR)are 3.41 s2=4,s2P= 2.89 and Is32=1.37 (46) 8SR 8SR 8SR Note that the sum of these is indeed also the sum rule S [137],i.e., sP+sP+sP341+28”+137-767=5 g5Rg5gg5R (47) The corresponding eigenvectors of GGsR are [138] 0.41 -0.71 0.58 0.81+0.1 and lws3)= 0.57-0.07i 0.41 0.71 0.58 (48) which are all orthogonal to one another [139],and the corresponding eigenvectors of GsRGSR are
g13 − 1 4π expikjrR1 − rS3j jrR1 − rS3j − 1 4π exp 2πi ffiffiffiffiffi 41 p ffiffiffiffiffi 41 p λ ≃ 0.01020 − 0.00711i λ : (40) Now we write distances in units of wavelengths for convenience (or equivalently, we set λ 1). To get numbers of convenient sizes and signs, we multiply by a factor gSR −4πLz −62.83: (41) So, then, gSRg13 ≃ −62.83 × 0.01020 − 0.00711i ≃ −0.64 0.45i: (42) Proceeding similarly for the other matrix elements, we have gSRGSR ≅ 2 4 1 −0.7 0.6i −0.64 0.45i −0.7 0.6i 1 −0.7 0.6i −0.64 0.45i −0.7 0.6i 1 3 5: (43) The sum, Eq. (36), of the modulus squared of these matrix elements of GSR as in Eq. (43) is the sum rule S 7.67∕g2 SR: (44) The matrix G† SRGSR can be written, with a convenient scaling factor g2 SR, g2 SRG† SRGSR ≅ 2 4 2.47 −0.67 − 0.08i −0.42 −0.67 0.08i 2.72 −0.67 0.08i −0.42 −0.67 − 0.08i 2.47 3 5: (45) Note that this matrix is Hermitian, as required. Having established the matrix G† SRGSR, we can now use standard numerical routines to find eigenvalues and eigenvectors. The resulting eigenvalues of G† SRGSR (and of GSRG† SR) are js1j 2 3.41 g2 SR , js2j 2 2.89 g2 SR , and js3j 2 1.37 g2 SR : (46) Note that the sum of these is indeed also the sum rule S [137], i.e., js1j 2 js2j 2 js3j 2 ≅ 3.41 g2 SR 2.89 g2 SR 1.37 g2 SR 7.67 g2 SR S: (47) The corresponding eigenvectors of G† SRGSR are [138] jψS1i 2 4 0.41 −0.81 0.1i 0.41 3 5, jψS2i 2 4 −0.71 0 0.71 3 5, and jψS3i 2 4 0.58 0.57 − 0.07i 0.58 3 5, (48) which are all orthogonal to one another [139], and the corresponding eigenvectors of GSRG† SR are Tutorial Vol. 11, No. 3 / September 2019 / Advances in Optics and Photonics 701
702 Vol.11,No.3/September 2019/Advances in Optics and Photonics Tutorial 0.41 -0.71 0.58 -0.81-0.1i 0 and 0.57+0.07i 0.41 0.71 0.58 (49) For this very symmetrical problem,the receiving wave vectors and the source vectors are just complex conjugates of one another,though that is not generally the case.In this case,the complex conjugation means that the "phase"curvatures are equal and opposite.(The idea of phase curvature becomes clearer as we consider more source and receiver points,so we postpone that illustration.)If we wanted to construct the strongest "channels"between these sources and receivers at this wavelength (or frequency),we would (a)choose to drive the point sources with relative amplitudes and phases given by one of the source eigenvectors in Eq.(48),and (b)at the receiving points,add up the signals from the different points weighted with the relative amplitudes and phase shifts given by the corresponding receiving eigenvector in Eq.(49) We can also send and receive three separate channels at once through this system if we construct appropriate systems to create and to separate the necessary signals. 4.2.Physical Implementation To see how to create and detect the necessary signals,including running all three channels at once,we can look at example physical systems.These are not meant to be engineered solutions to a real problem,but they make the mathematics more concrete physically. 4.2a.Acoustic and Radio-Frequency Systems First,for acoustic or radio-frequency signals,we can likely generate and measure the actual field directly.We will not consider the actual corresponding loudspeakers,mi- crophones,or antennas for the moment,just approximating them instead by ideal point sources and receiving elements.(We are still postponing any consideration of vector electromagnetic fields,just considering scalar waves.) We can then use appropriate electronic circuits that generate and collect the corre- sponding signals (Fig.4).In these cases,we can imagine input voltage signals Vsin,Vsin2,and Vsin3 that represent the signals(three different binary bit streams, for example)that we want to send on the three different "channels"in the commu- nication between the three sources and the three receivers.We would like these signals (e.g.,the three binary bit streams)to appear as the three output voltage signals VRout, VRour2,and VRout3 at corresponding electrical outputs at the far end of the system. Each such input signal voltage has to generate the corresponding vector of amplitudes to drive the sources.So Isin should generate a vector of voltage amplitudes Vsinls),and similarly for the other two input voltage signals,and these three vectors should be added to generate the corresponding set of output voltages Is1, Vs2,and Vs3 that drive the sources (e.g.,loudspeakers)at the corresponding positions rs1,rs2,and rs3. We can perform this generation of the correct vectors of amplitudes and their sum- mation by using the"analog crossbar"circuit on the left of Fig.4.We presume that we can make electrical phase shifters (circles in Fig.4),whose phase delay is indicated inside the corresponding circle.The output of each such phase shifter is then passed as
jϕR1i ≅ 2 4 0.41 −0.81 − 0.1i 0.41 3 5, jϕR2i ≅ 2 4 −0.71 0 0.71 3 5, and jϕR3i ≅ 2 4 0.58 0.57 0.07i 0.58 3 5: (49) For this very symmetrical problem, the receiving wave vectors and the source vectors are just complex conjugates of one another, though that is not generally the case. In this case, the complex conjugation means that the “phase” curvatures are equal and opposite. (The idea of phase curvature becomes clearer as we consider more source and receiver points, so we postpone that illustration.) If we wanted to construct the strongest “channels” between these sources and receivers at this wavelength (or frequency), we would (a) choose to drive the point sources with relative amplitudes and phases given by one of the source eigenvectors in Eq. (48), and (b) at the receiving points, add up the signals from the different points weighted with the relative amplitudes and phase shifts given by the corresponding receiving eigenvector in Eq. (49). We can also send and receive three separate channels at once through this system if we construct appropriate systems to create and to separate the necessary signals. 4.2. Physical Implementation To see how to create and detect the necessary signals, including running all three channels at once, we can look at example physical systems. These are not meant to be engineered solutions to a real problem, but they make the mathematics more concrete physically. 4.2a. Acoustic and Radio-Frequency Systems First, for acoustic or radio-frequency signals, we can likely generate and measure the actual field directly. We will not consider the actual corresponding loudspeakers, microphones, or antennas for the moment, just approximating them instead by ideal point sources and receiving elements. (We are still postponing any consideration of vector electromagnetic fields, just considering scalar waves.) We can then use appropriate electronic circuits that generate and collect the corresponding signals (Fig. 4). In these cases, we can imagine input voltage signals VSIn1, VSIn2, and VSIn3 that represent the signals (three different binary bit streams, for example) that we want to send on the three different “channels” in the communication between the three sources and the three receivers. We would like these signals (e.g., the three binary bit streams) to appear as the three output voltage signals VROut1, VROut2, and VROut3 at corresponding electrical outputs at the far end of the system. Each such input signal voltage has to generate the corresponding vector of amplitudes to drive the sources. So VSIn1 should generate a vector of voltage amplitudes ∝ VSIn1jψS1i, and similarly for the other two input voltage signals, and these three vectors should be added to generate the corresponding set of output voltages VS1, VS2, and VS3 that drive the sources (e.g., loudspeakers) at the corresponding positions rS1, rS2, and rS3. We can perform this generation of the correct vectors of amplitudes and their summation by using the “analog crossbar” circuit on the left of Fig. 4. We presume that we can make electrical phase shifters (circles in Fig. 4), whose phase delay is indicated inside the corresponding circle. The output of each such phase shifter is then passed as 702 Vol. 11, No. 3 / September 2019 / Advances in Optics and Photonics Tutorial
Tutorial Vol.11,No.3/September 2019/Advances in Optics and Photonics 703 a voltage to drive a resistor (rectangular boxes in Fig.4),whose conductance (i.e., 1/resistance)is given by the value inside the box (in some appropriate units).The other end of the resistors is connected in each case to a common "bus"line that is a "virtual ground"input to an operational amplifier (the triangles in Fig.4). The operational amplifiers each sum all the currents on their input bus line and each generate an output voltage proportional to this sum,giving the set of output drive voltages Vsi,Vs2,and V53. At the receiving end of the system,we can construct a similar circuit.In this case the input voltages to the analog crossbar are the outputs VRi,VR2,and Ve3 from the mea- sured fields at the three positions r,rg2,and rR3.If we have designed and set up our circuits correctly,the corresponding summed outputs,which become the voltage sig- nals Vgou Vou,and Vgous,should each now be proportional to the original voltage signals Vsin,Vsin2,and Vsin3 that we wanted to send through this three-channel system (just with some propagation time delay).If we make the feedback conductances in the operational amplifier circuits all identical at some value dk,then,for equal overall magnitudes of input voltage signals Vsin,Vsin2,and Vsin3,the relative sizes of the output voltage signals VRout,VRout,and VRout3 would be weighted by the corre- sponding singular values s;for the jth channel through the system.(Of course,we could compensate for the different singular values by using feedback conductances si in the circuits with the output operational amplifiers R1,R2,and R3.) Writing each of the vectors in Eqs.(48)and (49)with matrix elements in "polar" form,i.e., aig exp(ia) 「b1gexp(m1g) sg〉= azg exp(i) and l中a)兰 b2g exp(in2g) (50) a3g exp(i3a) b3g exp(in3g) where the a,b,0,and n coefficients are all real,we obtain the corresponding desired settings of the phase shifts and conductances in Fig.4. Figure 4 Source Receiving points points L02 -0- Phase b Lh shifter 白 Resistor of value R=1/a Bit stream Operational (in time) amplifier Example electrical driving and receiving circuits to form the superposition of sources for transmission and to separate the channels again for reception.The input channels are the(voltage)bit streams Isand the outputs are the corresponding(voltage)bit streams Vgou The feedback conductances ds and d set the overall electrical gain in transmission and reception in the operational amplifier circuits that sum the currents on their input "bus"lines as required to form the necessary linear superpositions
a voltage to drive a resistor (rectangular boxes in Fig. 4), whose conductance (i.e., 1/resistance) is given by the value inside the box (in some appropriate units). The other end of the resistors is connected in each case to a common “bus” line that is a “virtual ground” input to an operational amplifier (the triangles in Fig. 4). The operational amplifiers each sum all the currents on their input bus line and each generate an output voltage proportional to this sum, giving the set of output drive voltages VS1, VS2, and VS3. At the receiving end of the system, we can construct a similar circuit. In this case the input voltages to the analog crossbar are the outputs VR1, VR2, and VR3 from the measured fields at the three positions rR1, rR2, and rR3. If we have designed and set up our circuits correctly, the corresponding summed outputs, which become the voltage signals VROut1, VROut2, and VROut3, should each now be proportional to the original voltage signals VSIn1, VSIn2, and VSIn3 that we wanted to send through this three-channel system (just with some propagation time delay). If we make the feedback conductances in the operational amplifier circuits all identical at some value dR, then, for equal overall magnitudes of input voltage signals VSIn1, VSIn2, and VSIn3, the relative sizes of the output voltage signals VROut1, VROut2, and VROut3 would be weighted by the corresponding singular values sj for the jth channel through the system. (Of course, we could compensate for the different singular values by using feedback conductances ∝ sj in the circuits with the output operational amplifiers R1, R2, and R3.) Writing each of the vectors in Eqs. (48) and (49) with matrix elements in “polar” form, i.e., jψSqi 2 4 a1q expiθ1q a2q expiθ2q a3q expiθ3q 3 5 and jϕRqi ≅ 2 4 b1q expiη1q b2q expiη2q b3q expiη3q 3 5, (50) where the a, b, θ, and η coefficients are all real, we obtain the corresponding desired settings of the phase shifts and conductances in Fig. 4. Figure 4 Example electrical driving and receiving circuits to form the superposition of sources for transmission and to separate the channels again for reception. The input channels are the (voltage) bit streams VSInj, and the outputs are the corresponding (voltage) bit streams VROutj. The feedback conductances dS and dR set the overall electrical gain in transmission and reception in the operational amplifier circuits that sum the currents on their input “bus” lines as required to form the necessary linear superpositions. Tutorial Vol. 11, No. 3 / September 2019 / Advances in Optics and Photonics 703