Solving scattering problems in the half-line using methods developed for scattering in the full line
aa r X i v : . [ qu a n t - ph ] O c t Solving scattering problems in the half-line using methodsdeveloped for scattering in the full line
Ali Mostafazadeh ∗ Departments of Mathematics and Physics, Ko¸c University,34450 Sarıyer, Istanbul, Turkey
Abstract
We reduce the solution of the scattering problem defined on the half-line [0 , ∞ ) by areal or complex potential v ( x ) and a general homogenous boundary condition at x = 0to that of the extension of v ( x ) to the full line that vanishes for x <
0. We find anexplicit expression for the reflection amplitude of the former problem in terms of thereflection and transmission amplitudes of the latter. We obtain a set of conditions onthese amplitudes under which the potential in the half-line develops bound states, spectralsingularities, and time-reversed spectral singularities where the potential acts as a perfectabsorber. We examine the application of these results in the study of the scatteringproperties of a δ -function potential and a finite barrier potential defined in [0 , ∞ ), discussoptical systems modeled by these potentials, and explore the configurations in which thesesystems act as a laser or perfect absorber. In particular, we derive an explicit formulafor the laser threshold condition for a slab laser with a single mirror and establish thesurprising fact that a nearly perfect mirror gives rise to a lower threshold gain than aperfect mirror. We also offer a nonlinear extension of our approach which allows forutilizing a recently developed nonlinear transfer matrix method in the full line to dealwith finite-range nonlinear scattering interactions defined in the half-line. The study of scalar waves propagating in the half-line, [0 , ∞ ), has a long history [1]. This ismostly because of the essential role it plays in performing scattering calculations for sphericallysymmetric systems [2]. More importantly, many of the developments in scattering and inversescattering theories have their root in this subject [3]. Interactions that cause the scattering of awave in the half-line are of two types, those that affect the wave as it propagates throughout theinterior of the half-line, i.e., (0 , ∞ ), and those corresponding to the response of the boundarypoint x = 0. This is in contrast with the scattering interactions in the full line, R , which has noboundary. The purpose of the present article is to make a link between the scattering problemsin the half-line and the full line with the intention of using the known results for the solutionof the latter to address the former. ∗ E-mail address: [email protected] e − iωt ψ ( x ) with ψ ( x ) satisfying the Schr¨odinger equa-tion, − ψ ′′ ( x ) + v ( x ) ψ ( x ) = k ψ ( x ) , (1)in the half-line, where v : [0 , ∞ ) → C is a piecewise continuous scattering potential , and k isthe wavenumber of the incident wave. Our main purpose is to study the scattering problemsdefined by (1) and the boundary condition α ψ (0) + k − β ψ ′ (0) = 0 , (2)where α and β are possibly k -dependent real or complex parameters fulfilling | α | + | β | = 0.Note that the Dirichlet, Neumann, and Robin boundary conditions are special cases of (2) thatcorrespond to α = 0 = β , α = 0 = β , and α = 0 = β − k , respectively.To motivate the choice of the boundary condition (2), we consider a time-harmonic wave e − iωt Ψ( x ) propagating in the full line ( R ) with a source located at x = + ∞ . Suppose thatΨ( x ) solves the time-independent Schr¨odinger equation for a general (piecewise continuous)extension of the potential v ( x ) to R , namely V ( x ) := (cid:26) v − ( x ) for x < ,v ( x ) for x ≥ , where v − : ( −∞ , → C is an unspecified potential defined in ( −∞ , x ) = ψ ( x ) for x ∈ [0 , ∞ ) . (3)First, imagine that v ( x ) takes a constant value ν in an open interval of the form (0 , ǫ ). Thenin view of (3), it is easy to see that for x ∈ (0 , ǫ ), ψ ( x ) = Ψ( x ) = ( A e i ˜ kx + B e − i ˜ kx for k = ν,A + B kx for k = ν, (4)where A and B are complex coefficients, and ˜ k := √ k − ν . Now, introduce R − := A B , (5)which for real and positive values of ˜ k represents the reflection amplitude of the potential V − ( x ) := (cid:26) v − ( x ) for x < , x ≥ , for a right-incident plane wave with wavenumber ˜ k . Differentiating (4) and expressing ψ (0) :=lim x → + ψ ( x ) and ψ ′ (0) := lim x → + ψ ′ ( x ) in terms of B and R − , we find i ( R − − ψ (0) + ˜ k − ( R − + 1) ψ ′ (0) = 0 for k = ν,ψ (0) − k − R − ψ ′ (0) = 0 for k = ν. (6) By being a scattering potential we mean that every solution of the Schr¨odinger equation (1) tends to a linearcombination of plane waves as x → ∞ . This is the case for potentials v ( x ) satisfying the Faddeev condition R ∞ (1 + x ) | v ( x ) | dx < ∞ , [4]. , ∞ ) we actually do notneed a detailed knowledge of v − ( x ); the information about R − suffices for this purpose. Theargument leading to this conclusion holds generally, for we can make ǫ arbitrarily small. Thefact that the boundary conditions given by (6) are special cases of (2) provides the mainmotivation for the study of the scattering problems defined by the Schr¨odinger equation (1)and the boundary conditions (2) in the half-line.Because v ( x ) is a scattering potential, for x → ∞ it decays to zero at such a rate that everysolution ψ ( x ) of (1) satisfies ψ ( x ) → A + e ikx + B + e − ikx for x → ∞ , (7)where A + and B + are possibly k -dependent complex coefficients. The solution of the scatteringproblem for v ( x ) means finding its reflection amplitude, R := A + B + . (8)If for a real and positive value of k , A + = 0 = B + , the reflection amplitude vanishes,and the incident wave is completely absorbed by the system, i.e., it acts as a perfect absorber.Another interesting situation is when the converse occurs. In this case the system emits a purelyoutgoing plane wave, i.e., A + = 0 = B + , R blows up for a real and positive value of k , and k is called a spectral singularity. This is a mathematical concept introduced by Naimark in 1954[6] and further developed and studied by other mathematicians [7, 8, 9, 10, 11, 12, 13, 14, 15].Surprisingly, the physical significance of this concept was revealed more than half a centuryafter its discovery [5]. This led to a surge of research activity in the study of physical aspects ofspectral singularities [16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]and unraveled their intimate relation to the physics of lasers; the mathematical conditionfor the emergence of a spectral singularity for optical potentials modeling the scattering ofelectromagnetic waves coincides with the laser threshold condition which marks the initiationof laser oscillations [37, 38, 39]. See also [40, 41, 42, 43, 44].It is not difficult to see that the perfect absorption and emission phenomena correspondingto the conditions R = 0 and R = ∞ , are related to one another by time-reversal transformation[45]. This has led to the use of the term “anti-laser” for optical perfect absorbers [46, 47, 48,49, 50, 51, 52, 53, 54, 55].A more common situation is when A + = 0, B + = 0, and consequently R = ∞ for acomplex value of k with a positive imaginary part. In this case, the corresponding solution ofthe Schr¨odinger equation decays exponentially as x → ∞ . Therefore, it is square-integrable.This shows that k is a genuine eigenvalue of the Schr¨odinger operator H := − ∂ x + v ( x ). Ifsuch a value of k is purely imaginary, k is a negative real number. This implies that the normof the corresponding eigenfunction ψ k does not decay in time; k e − itH ψ k k = k e − itk ψ k k = k ψ k k , where e − itH is the time-evolution operator. Therefore, ψ k identifies a bound state of the poten-tial. If k is not purely imaginary, k develops an imaginary part, and the norm of ψ k decays orgrows exponentially in time depending on the sign of Im( k ); these correspond to the resonancesand anti-resonances of the system, respectively.3 Scattering in half-line from the full line
Consider the trivial extension of the potential v ( x ) to the whole real line that vanishes for x <
0, i.e., V + ( x ) := (cid:26) x < ,v ( x ) for x ≥ . (9)Clearly, every solution ψ ( x ) of the Schr¨odinger equation (1) for the potential V + ( x ) satisfies (7)and ψ ( x ) = A − e ikx + B − e − ikx for x ≤ , (10)where A − and B − are possibly k -dependent complex coefficients. The restriction of thesesolutions to the half-line [0 , ∞ ) gives the scattering solutions defined by (1) and (2), if wechoose A − and B − in such a way that ψ (0) and ψ ′ (0) satisfy (2). In view of (10), this gives B − = − γA − (11)where γ := α + iβα − iβ . (12)Notice that the Dirichlet and Neumann boundary conditions respectively correspond to γ = 1and γ = − V + ( x ), [45, 56]. This is a 2 × M with unit determinant that by definition satisfies M (cid:20) A − B − (cid:21) = (cid:20) A + B + (cid:21) . (13)The scattering properties of V + ( x ) are encoded in its left/right reflection ( R l/r ) and transmission( T l/r ) amplitudes [57]. These are defined for the left- (respectively right-) incident wavescorresponding to B + = 0 (respectively A − = 0) according to the following relations. R l := B − A − , T l := A + A − R r := A − B + , T r := B − B + . (14)A key observation signifying the importance of the transfer matrix is that its entries determinethe reflection and transmission amplitudes of the potential [45, 56]. Specifically, R l = − M M , T l = det( M ) M ,R r = M M , T r = 1 M . (15)These equations together with the fact that det( M ) = 1 imply transmission reciprocity, T l = T l ,which allows us to drop the superscript l/r in T l/r . They also lead to M = T − R l R r T , M = R r T , M = − R l T , M = 1 T . (16)4olving (8) for A + and substituting the result together with (11) in (13), we obtain A − M (cid:20) − γ (cid:21) = B + (cid:20) R (cid:21) . (17)This is equivalent to ( M − γM ) A − = R B + , (18)( M − γM ) A − = B + . (19)Substituting the second of these equations in the first and making use of (16), we find R = M − γM M − γM (20)= T R r − γ − R l . (21)According to (20) and (21), we can solve the original scattering problem, which is given in thehalf-line, by solving the scattering problem defined by the extended potential V + ( x ) in the fullline.Next, we recall that the bound states, resonances, and spectral singularities of the originalscattering problem are given by the singularities of the reflection amplitude R . Because det M =1, the numerator and denominator of the right-hand side of (20) cannot vanish simultaneously.This implies that the bound states, resonances, and spectral singularities correspond to (realor complex) values of k where M = γM or equivalently R r = γ. (22)Similarly, the original potential that is defined on the half-line acts as a perfect absorber forthe real and positive values of k ensuring M = γM . In view of (15), we can alternativelyexpress this condition in the form T − R l ( R r − γ ) = 0 . (23)Furthermore, because T can never vanish, (23) implies R l = 0. As a result, (23) means R r = T R l + γ. (24) δ -function potential in the half-line Consider the scattering problem defined by (1) and (2) for the δ -function potential, v ( x ) = z δ ( x − a ) , x ≥ , (25)where z is a real or complex coupling constant, and a is a positive real parameter. The choiceof α = 1 and β = 0 (and hence γ = 1) corresponds to the Dirichlet boundary condition whichwe encounter in dealing with the potential v ( x ) = (cid:26) + ∞ for x < , z δ ( x − a ) for x ∈ [0 , ∞ ) . R , namely V + ( x ) = z δ ( x − a ) , x ∈ R . (26)The quickest method of computing the transfer matrix for the δ -function potential (26) is tomake use of the fact that it can be expressed as the time-ordered exponential of a non-Hermitiantwo-level effective Hamiltonian [58]; M = T exp (cid:26) − i Z ∞−∞ dx H ( x ) (cid:27) = I + ∞ X n =1 ( − i ) n Z ∞−∞ dx n Z x n −∞ dx n − · · · Z x −∞ dx H ( x n ) H ( x n − ) · · · H ( x ) , (27)where T is the time-ordering operator with x playing the role of time, I is the 2 × H ( x ) is the matrix Hamiltonian defined by H ( x ) := V + ( x )2 k (cid:20) e − ikx − e ikx − (cid:21) . (28)For the δ -function potential (26), (28) gives H ( x ) = ˜ z δ ( x − a ) K a , (29)where ˜ z := z k , K a := (cid:20) e − ika − e ika − (cid:21) . (30)Substituting (29) in (27) and noting that K a is the null matrix, we find M = e − i ˜ z K a = I − i ˜ z K a . (31)Therefore, M = 1 − i ˜ z , M = − i ˜ z e − ika , M = i ˜ z e ika , M = 1 + i ˜ z . (32)In view of (15), these relations imply R l = − i ˜ z e ika i ˜ z , R r = − i ˜ z e − ika i ˜ z , T = 11 + i ˜ z . (33)If we substitute (32) in (20), we obtain the following expression for the reflection amplitudeof the δ -function potential on the half-line (25) with the boundary condition at x = 0 given by(2). R = − k + i z (1 − γe − ika )2 γk + i z ( γ − e ika ) . (34)Because the bound states, resonances, and spectral singularities correspond to the singularitiesof R in the complex k -plane, these are given by values of k for which2 γk + i z ( γ − e ika ) = 0 . (35)The system acts as a perfect absorber for the incident plane waves whose wavenumber k fulfills2 k + i z ( γe − ika −
1) = 0 . (36)6 .1 Bound states Bound states correspond to solutions of (1) for the positive imaginary values of k that satisfy(35). To study them, we set k = i | k | in (35) and write it in the form e − | k | a = γ (cid:18) | k | z (cid:19) . (37)According to this equation, a bound state exists if the right-hand side of this equation isa positive real number smaller than 1. For the case of Dirichlet and Neumann boundaryconditions, where γ = ±
1, this implies that z must be a negative real number and | k | < | z | .For other values of γ , it is possible to use (37) to obtain formulas for the phase angles of γ and z in terms of | k | , | z | , and | γ | . Because these are not illuminating, we do not include them here.We instead treat the problem of finding the conditions for the existence and uniqueness of thebound states for boundary conditions (2) with a real value of γ , which includes the physicallyrelevant cases of the Dirichlet and Neumann boundary conditions. We differ the details of ouranalysis to the appendix and summarize its outcome in the following.1. There are at most two bound states.2. For the following two cases, no bound states exist.(a) γ = 0;(b) 1 ≥ γ ≥ a | z | .3. For the following three cases, there is a single bound state.(a) γ < γ > ≥ a | z | /γ ;(c) 1 ≤ γ < a | z | .4. There are two bound states if and only if 0 < γ < < a | z | /γ .In particular for the Dirichlet and Neumann boundary conditions, there is at most a singlebound state. For the Dirichlet boundary condition, where γ = 1, this is present if and only if z is real and smaller than − / a . For the Neumann boundary condition, where γ = −
1, a boundstate exists and is unique if and only if z is a negative real number.In general, the computation of the value of | k | for the bound states requires a numericalsearch for solutions of (37). It is easy to see from this equation that solutions lie in the intervalbetween (1 − | γ | − ) | z | and (1 + | γ | − ) | z | whenever they exist. Spectral singularities are given by the real and positive values of k that fulfill (35). To explorethem, we first write this equation in the form e ika = γ (cid:18) − ik z (cid:19) . (38)7f we evaluate the modulus of both sides of (38) and note that k is real and nonzero, we find aquadratic equation in k with solutions k = 12 h z i ± p | z /γ | − z r i , (39)where z r and z i stand for the real and imaginary parts of z , respectively.The right-hand side of (39) is real, if and only if | γ | ≤ | z || z r | = s z i z r . (40)This is clearly satisfied by both the Dirichlet ( γ = 1) and Neumann ( γ = −
1) boundaryconditions. These are special cases of the boundary conditions characterized by | γ | = 1 whichalso fulfill (40). For the latter, (39) yields a single positive and real value for k , namely k = z i , (41)provided that z i >
0. Substituting (41) in (38) and solving for z r , we find z r = − cot( a z i − ϕ/ z i , (42)where ϕ is the principal argument (phase angle) of γ , so that γ = e iϕ .In terms of the parameters α and β that enter the expression for the boundary condition(2), | γ | = 1 takes the form α ∗ β ∈ R . Under this condition, the δ -function potential defined inthe half-line admits a single spectral singularity, if z i > z r satisfies (42). These conditionshave a rather interesting physical meaning. To see this, we use the correspondence betweenthe Schr¨odinger equation (1) and the Helmholtz equation for a transverse electric wave that isnormally incident upon a dielectric media with planar symmetry along the y - and z -directions[37]. For the problem at hand with Dirichlet boundary conditions ( β = α − γ − a to the right of a perfect planar mirror. We identify the latter with the plane x = 0. Assumingthat the system is placed in vacuum, we can model its permittivity profile using ε ( x ) = ε [1 + ζ δ ( x − a )] , (43)where x ≥ ε is the permittivity of the vacuum, and ζ is a complex coupling constant. TheHelmholtz equation describing the behavior of normally incident transverse electric waves takesthe form of the Schr¨odinger equation (1) provided that the potential v ( x ) is given by (25) and z = − k ζ . (44)The presence of a spectral singularity for a wavenumber k means that the above system fulfillsthe so-called laser threshold condition [59]; it operates as a slab laser whenever its gain exceedsthe value associated with the spectral singularity.For a realistic thin slab, we identify the δ -function appearing in (43) with a rectangularbarrier potential with width b ≪ k − and complex hight ζ /b . In this way we can identify the8omplex relative permittivity of the slab (which is equal to the square of its refractive index)with ˆ ε s := 1 + ζb = 1 − z bk . (45)It is well-known that Im(ˆ ε s ) > ε s ) < g of the slab is related to Im(ˆ ε s ) according to [59] g = − k Im(ˆ ε s )Re(ˆ ε s ) . (46)Substituting (45) in (46), we have g = z i bk − z r k − . (47)In view (45) and (47), and the fact that for nonexotic active material, Re(ˆ ε s ) >
0, the condition z i > g >
0. This means that the presence of a spectral singularity, which is thecondition for the initiation of laser oscillations, requires the slab to be made of gain material;a well-known fact that is usually justified using the principle of conservation of energy.For a perfect mirror, γ = 1, and the spectral singularity occurs for k = z i . This togetherwith (47) identifies the threshold gain for this slab laser with g = 1 b Re(ˆ ε s ) . (48)Because Re(ˆ ε s ) ≥ bk ≪
1, for wavelengths of the order of a micrometer, we have g ≫ cm − , which is an extremely large gain coefficient. Notice however that this is to beproduced in an extremely thin slab of thickness b ≪ k − ≈ µ m.Next, we recall that to initiate lasing, we also need to satisfy (42). Because γ = 1, we set ϕ = 0 in this equation and use (45) to write it in the formRe(ˆ ε s ) = 1 + cot( ak ) bk . (49)Solving this equation for a , we find a = k − (cid:20) − arctan { bk [Re(ˆ ε s ) − } + π ( m + 12 ) (cid:21) , (50)where m = 0 , , , · · · is a mode number.Given that Re(ˆ ε s ) is at most of the order of 10 and bk ≪
1, (50) implies a ≈ (2 m + 1) π k = (2 m + 1) λ , (51)where λ := 2 π/k is the wavelength. This in turn gives the following expression for the lasingmodes of our thin-slab laser. k m ≈ (2 m + 1) π a . (52)The particular mode k m ⋆ at which the laser will operate depends on the details of the dis-persion relation describing the k -dependence of ˆ ε s . Making the k -dependence of Im(ˆ ε s ) ex-plicit and employing (46) and (48), we can identify m ⋆ with the mode number m for which | bk m Im[ˆ ε s ( k m )] + 1 | takes its smallest possible value.9otice that to find m ⋆ , we need to fix a particular value for a , which we denote by a . Byconstruction, for a = a and k = k m ⋆ , Eq. (49) holds approximately. To ensure that it holdsexactly, we make the k -dependence of Re(ˆ ε s ) explicit and set k = (2 m ⋆ + 1) π/ a , so that (49)turns into a single real equation which we can in principle solve for a . For this value of a whichwe label by a ⋆ , our thin-slab laser will emit coherent waves with wavelength λ m ⋆ = 4 a ⋆ / (2 m ⋆ +1)as soon as its gain coefficient surpasses its threshold value, namely b − .The requirement that the δ -function potential in the half-line with Dirichlet boundary con-ditions at x = 0 realizes a spectral singularity does not fix the real part of its coupling constant.This is in sharp contrast with the case of a δ -function potential in the full line, i.e., the situationwhere we remove the mirror in our thin-slab system. In this case, the spectral singularity ap-pears for purely imaginary values of the coupling constant [60]. This corresponds to a high-gainthin slab with Re(ˆ ε ) = 1 which is extremely difficult to realize, because the real part of thepermittivity of the known high-gain material is larger than unity. This signifies a practicaladvantage of the setup modeled by the δ -function potential in the half-line. The δ -function potential in the half-line can serve as a perfect absorber provided that (36)holds. We can write this equation in the form e ika = γ zz + 2 ik , (53)and use the fact that we are interested in real and positive values of k satisfying this equationto show that it implies k = 12 h − z i ± p | γ z | − z r i . (54)Because k is real, | γ | ≥ | z r || z | = 1 p z i / z r . (55)For the class of boundary condition specified by | γ | = 1, (55) holds and (54) gives a singlepositive value for k , namely k = − z i , (56)whenever z i <
0. Again in the context of the optical model we discussed in the precedingsubsection, this condition indicates that the system can act as a perfect absorber, if the thinslab is made of lossy material. Notice again that this is just a necessary condition for theperfect absorption of waves with wavenumber k = − z i . The wave will be absorbed if (53) holdsfor this value of k . This happens for z r = − cot( a z i + ϕ/ z i = − cot( a | z i | − ϕ/ | z i | . (57)Again we can use a procedure similar to the one we described in the preceding subsection todetermine the position a ⋆ of the slab at which it acts as a perfect absorber for a wave withwavelength λ ⋆ . This involves using (50) and (51) with z i replaced with | z i | . These show that10or given values of ϕ and z r , we can make the slab act as a perfect absorber provided that weadjust its position properly.Equations (56) and (57) are consistent with the fact that the potential acts as a perfectabsorber if and only if its time-reversal realizes a spectral singularity. To see this, we observethat under time-reversal transformation, z r → z r , z i → − z i , α → α ∗ , β → β ∗ , and consequently γ → /γ ∗ . In particular for | γ | = 1, we can determine the configurations of the system thatmake it act as a perfect absorber from those giving rise to a spectral singularity by changing z i → − z i . This is consistent with the fact that performing this transformation in (41) and (42)we respectively recover (56) and (57). Consider a laser obtained by placing an infinite planar homogeneous slab of gain material withthickness L next to a nearly perfect mirror [59]. We wish to determine how the threshold gainfor a normally incident transverse electric wave depends on the reflectivity of the mirror andits distance to the slab. We do this by exploring the spectral singularities of the scatteringproblem for an optical potential v ( x ) defined on the half-line, which represents the interactionof the wave with the slab, and a boundary condition at x = 0 that signifies the effect of themirror.The optical potential representing the slab has the form v ( x ) := (cid:26) k (1 − n ) for x ∈ [ a, a + L ] , , (58)where n is the complex refractive index of the slab, and a is its distance to the mirror. Thechoice of the boundary condition at x = 0 is dictated by the fact that we can identify thereflection amplitude of a nearly perfect mirror with R − = − ǫ , where ǫ is a complex numbersuch that | ǫ | ≪
1. In view of (2), (12), and the analysis leading to (6), this corresponds tosetting α = i ( R − −
1) = − i (2 − ǫ ), β = R − + 1 = ǫ , and γ = 1 − ǫ. (59)To determine the laser threshold condition, we require that the above system has a spectralsingularity [37], i.e., (23) holds. To explore consequences of this equation, we use (15) and thefollowing well-known formula for the transfer matrix of the potential (58) to determine R r forthis potential. M = " [cos( kL n ) + i n + sin( kL n )] e − ikL i n − sin( kL n ) e − ik ( L +2 a ) − i n − sin( kL n ) e ik ( L +2 a ) [cos( kL n ) − i n + sin( kL n )] e ikL , (60)where n ± := ( n ± n − ) /
2, [45]. This yields, R r = i n − e − ik ( L + a ) cot( kL n ) − i n + . e ikL n = ˜ n (1 + ˜ n X ) X + ˜ n , (61)where ˜ n := n + 1 n − , X := γ e ik ( a + L ) = (1 − ǫ ) e ik ( a + L ) . (62)Next, we recall that the gain coefficient g for the slab is related to the imaginary part κ ofthe refractive index n according to g = − πκ/λ , [59]. Expressing κ in terms of g and denotingthe real part of n by η , so that n = η + iκ = η − iλg π = η − ig k , (63)we can write (63) as e gL = ˜ n (1 + ˜ n X ) e − ikLη X + ˜ n . (64)Because e gL is real and positive, we can equate it to the modulus of the right-hand side of (64).This gives the following expression for the threshold gain. g = g ( s ) + g ( m ) , (65)where g ( s ) := 2 L ln | ˜ n | , (66) g ( m ) := 1 L ln (cid:12)(cid:12)(cid:12)(cid:12) X + ˜ n − X + ˜ n (cid:12)(cid:12)(cid:12)(cid:12) . (67)Note that the right-hand side of (66) coincides with the known formula for the threshold gainof a mirrorless slab laser [37, 59]. Therefore, g ( m ) represents the contribution of the mirror.Next, we use the fact that the right-hand side of (64) is real to infer, k = 2 πm + ϑ ηL , (68)where m = 0 , , , , · · · is a mode number, and ϑ is the principal argument (phase angle) of˜ n (1 + ˜ n X ) / ( X + ˜ n ), i.e., the real number belonging to [0 , π ) that satisfies e iϑ = | X + ˜ n | ˜ n (1 + ˜ n X ) | ˜ n (1 + ˜ n X ) | ( X + ˜ n ) . (69)Eq. (68) gives the lasing modes of our slab laser.For a vast majority of typical active material, | κ | ≪ η −
1, and we can safely ignore the κ -dependence of the right-hand sides of (66), (67), and (69). This gives g ( s ) = 2 L ln (cid:18) η + 1 η − (cid:19) + O ( κ ) , (70) g ( m ) = 1 L (cid:20) ln (cid:18) η − η + 1 (cid:19) + ln | Y | (cid:21) + O ( κ ) , (71) e iϑ = Y | Y | + O ( κ ) , (72)12here Y := ( η + 1) X + η − η − X + η + 1 , and O ( κ ℓ ) stands for the sum of terms of order ℓ or higher in powers of κ . We can obtain thefollowing more explicit expressions for the right-hand sides of (71) and (72), if we employ thedefinition of X , i.e., (62), and the fact that | ǫ | ≪ g ( m ) = − L (cid:20) ln (cid:18) η + 1 η − (cid:19) + 2 Z Re( ǫ ) (cid:21) + O ( κ ) + O ( ǫ ) , (73) ϑ = ϑ s − Z Im( ǫ ) + O ( κ ) + O ( ǫ ) , (74)where Z := ηη + 1 + ( η −
1) cos[2 k ( a + L )] . (75) ϑ s := 2 arctan (cid:8) η − tan[ k ( L + a )] (cid:9) . (76)The fact that the right-hand side of (73) takes a negative value is consistent with theexpectation that placing the slab next to a mirror reduces its threshold gain. What is notexpected is that a realistic mirror whose reflection amplitude has a slightly larger real partthan −
1, so that Re( ǫ ) = 1 + Re( R − ) >
0, leads to a larger reduction of the threshold gain incomparison to a perfect mirror! Eqs. (65), (70), and (73) lead to the following formula for thethreshold gain. g = 1 L (cid:26) ln (cid:18) η + 1 η − (cid:19) − η Re( ǫ ) η + 1 + ( η −
1) cos[2 k ( a + L )] (cid:27) + O ( κ ) + O ( ǫ ) , (77)According to this equation placing a gain slab next to a perfect mirror reduces its threshold gainby a factor of 2. This agrees with the fact the presence of a perfect mirror has the same effect asdoubling the thickness of the slab (the mirror image of the slab acts as a second amplifier [59].)Another consequence of Eq. (77) is that the threshold gain takes its smallest value, namely g min = 1 L (cid:26) ln (cid:18) η + 1 η − (cid:19) − η Re( ǫ ) η + 1 (cid:27) + O ( κ ) + O ( ǫ ) , whenever a + L = (2 ℓ + 1) π k = (2 ℓ + 1) λ , for some nonnegative integer ℓ . This relation identifies the optimal positions of the slab. Againit is interesting to observe that for a perfect mirror the distance between the slab and the mirrordoes not affect the threshold gain, while for a nearly perfect mirror one can lower the thresholdgain by adjusting the distance. Consider the scattering problem defined on the half-line by the a nonlinear Schr¨odinger equationof the form, − ψ ′′ ( x ) + F ( ψ ( x ) , x ) = k ψ ( x ) , (78)13nd the boundary condition (2), where F : C × [0 , ∞ ) → C is a function such that as x → ∞ , F ( ψ ( x ) , x ) tends to v ( x ) ψ ( x ) for some scattering potential v : [0 , ∞ ) → C . Ref. [61] providesa detailed discussion of the scattering problem defined by (78) in the full line. In particular, itintroduces a nonlinear generalization of the transfer matrix which shares the basic propertiesof its well-known linear predecessor. In this section, we employ the analysis of Sec. 2 to treatthe scattering problem given by (2) and (78) in the half-line.First, we recall that the nonlinear transfer matrix is a complex 2 × M satisfying(13) that depends on the amplitudes A − and B − (in addition to the wavenumber k .) Wetherefore denote it by M ( A − , B − ). The left/right reflection and transmission amplitudes, R l/r and T l/r , for the scattering problem given by (78) in the full line are also defined by (14). But,in general, they depend on the amplitude of the incident wave. Furthermore, the presence ofnonlinearity can violate the reciprocity in transmission, i.e., for identical incident waves T l and T r are generally different.The formulas relating R l/r and T l/r to the nonlinear transfer matrix M ( A − , B − ) are thefollowing analogs of (15). R l = − M l M l , T l = det( M l ) M l ,R r = M r M r , T r = 1 M r , (79)where M lij and M rij are respectively the entries of M l := M ( A l , A l R l ) , M r := M (0 , A r T r ) , (80)and A l/r is the complex amplitude of the left/right incident wave, i.e., A l = A − when B + = 0,and A + = B − when A − = 0, [61]. In view of (80), we can use (79) to express R l/r and T l/r interms of A l/r .An important distinction between linear and nonlinear transfer matrices is that Eq. (13)determines the latter up to a pair of unspecified functions f ( A − , B − ) and f ( A − , B − ); if M ( A − , B − ) satisfies (13), then so does˜ M ( A − , B − ) := M ( A − , B − ) + (cid:20) f ( A − , B − ) B − − f ( A − , B − ) A − f ( A − , B − ) B − − f ( A − , B − ) A − (cid:21) . For the linear transfer matrix, one avoids this problem by demanding that it is independent of A − and B − . For the nonlinear transfer matrix this requirement is never fulfilled, and there isno general guideline to make a particular choice for f ( A − , B − ) and f ( A − , B − ). Fortunately,this large lack of uniqueness does not affect the utility of the nonlinear transfer matrix indetermining the reflection and transmission amplitudes, for it happens that M ( A − , B − ) and˜ M ( A − , B − ) give rise to the same formulas for the reflection and transmission amplitudes, [61].Now, consider the scattering problem given by (78) and (2) on the half-line. Solving thisproblem means finding the reflection amplitude R defined by (8). Noting that the source forthe incident wave is located at x = + ∞ , B + = A r , and (8) reads R := A + A r . (81)14epeating the analysis of Sec. 2 for our nonlinear scattering problem, we are led to Eqs. (18)and (19) with M ij replaced with the entries M ij ( A − ) of M ( A − ) := M ( A − , − γ A − ) . (82)Substituting M ij ( A − ) for M ij in (18) and (19), we arrive at a pair of equations that arerespectively equivalent to R = A − [ M ( A − ) − γ M ( A − )] A r , (83)[ M ( A − ) − γ M ( A − )] A − = A r . (84)It is not difficult to check that the above-mentioned lack of uniqueness of the nonlinear transfermatrix does not affect these equations either; in view of (82), the transformation M ( A − , B − ) → ˜ M ( A − , B − ) leaves M ( A − ) − γ M ( A − ) and M ( A − ) − γ M ( A − ) invariant.If we know a nonlinear transfer matrix for the scattering problem defined in the full line, wecan in principle solve (84) for A − and substitute the result in (84) to determine R as a functionof A r , i.e., solve the corresponding nonlinear scattering problem in the half-line. It is also easyto see that the bound states (respectively spectral singularities) are given by the nonzero valuesof the complex amplitude A − and positive imaginary (respectively positive real) values of k forwhich M ( A − ) = γ M ( A − ) , (85) M ( A − ) = γ M ( A − ) . (86)Note that in this case, A + = A − [ M ( A − ) − γ M ( A − )] = 0 , (87)while B + = A r = 0.As a simple example, consider the scattering problem defined by the nonlinear point inter-action given by F ( ψ ( x ) , x ) := f ( | ψ ( x ) | ) ψ ( x ) δ ( x − a ) , (88)where a is positive real parameter, and f : [0 , ∞ ) → C is a continuous function. As shownin Ref. [61], we can obtain a nonlinear transfer matrix M ( A − , B − ) associated with (88) byidentifying the coupling constant z of its linear analog, namely (25), with f ( | e ic K A − + B − | ). Inother words, M ij ( A − , B − ) are given by (32), if we set ˜ z := f ( | e iak A − + B − | ) / k . In particular,for B − = − γ A − , ˜ z = f ( | ˜ γ A − | )2 k = f ( | A + | )2 k , (89)where ˜ γ := γ − e iak , (90)and we have made use of the identity, A + = e − iak ˜ γA − , (91)which follows from (32), (87), and (89). 15n view of (32), (89), and (90), (84) takes the form (cid:2) ˜ γf ( | ˜ γ A − | ) − ik (cid:0) ˜ γ + e iak (cid:1)(cid:3) A − = 2 ikA r . (92)Evaluating the modulus of both sides of this equation, we see that | A − | is a function of | A r | .This together with (92) allow us to determine the phase angle for A − . Similarly, we use (32),(89), and (90), to simplify the expression given by (83) for R . This yields R = A − (1 + i ˜ γ ˜ z e − iak ) A r = − e − iak (cid:18) γA − A r (cid:19) , (93)where we have also employed (92).The nonlinear point interaction (88) admits a (nonlinear) spectral singularity [19] providedthat (85) and (86) holds for some positive real k and A − = 0. In view of (90), the second ofthese equations implies f ( | A + | ) = 2 ik (cid:18) − e iak γ (cid:19) − . (94)This is a particularly useful formula, for it relates | A + | , which determines the intensity ofthe outgoing wave, to the wavenumber k and the parameters entering the expression for thenonlinearity profile f ( | ψ | ).Next, we recall the application of the δ -function potential for modeling the scattering of aTE wave by an active thin slab that is placed next to a perfect mirror ( γ = 1). We may identifythe scattering problem for this potential with the one defined by (88), if we set f ( | ψ ( x ) | ) = z .Replacing this equation with f ( | ψ ( x ) | ) = z + s | ψ ( x ) | , (95)corresponds to a Kerr slab with relative permittivity (45), Kerr constant σ := − s bk , (96)and thickness b ≪ k . If k is a spectral singularity associated with the nonlinearity profile(95), the slab will function as a laser with a single mirror and resonance wavelength k . In thiscase, we can use (94) to express the intensity of the outgoing laser light to the real part of therelative permittivity of the slab and its gain coefficient.Substituting (95) in (94) and employing the analysis leading to (48) and (49), we obtainthe following nonlinear analogs of these equations. g = g [1 + 2 bk Im( σ ) I ] , (97)Re(ˆ ε s ) + 2 Re( σ ) I = 1 + cot( ak ) bk , (98)where g := 1 /b Re(ˆ ε s ), and I := | A + | / g = g when I = 0. This suggestsidentifying g with the threshold gain whenever Im( σ ) > σ )16ust indeed take a positive value. This observation together with the fact that we can express(97) in the form I = g − g bk Im( σ ) , (99)show that in order for the slab to emit laser light its gain coefficient must exceed g . Thisidentifies g with the threshold gain. Moreover, it implies that for gain values exceeding g , theintensity of the outgoing laser light is proportional to g − g . This is in complete agreementwith established facts about lasers.Notice also that, according to (98), whenever the real part of the Kerr coefficient is nonzero,the wavenumber of the emitted wave undergoes a shift. Substituting (99) in (98), we havecot( ak ) − [Re(ˆ ε s ) − bk = ( g − g )Re( σ )Im( σ ) . (100)To reach threshold g = g , we must adjust the distance a between the slab and the mirror sothat the left-hand side of this relation vanishes for the wavenumber k for which we can maintainthe necessary gain. Once we increase the gain to obtain a positive intensity for the outgoingwave, the right-hand side becomes positive and as a result the value of k changes. Because bk ≪
1, (99) shows that the right-hand side of (100) is extremely small. This in turn allows usto determine the shift δk in the wavenumber using first-order perturbation theory. The resultis δkk ≈ − ( g − g )Re( σ ) { a + b [Re(ˆ ε s ) − } k Im( σ ) ≈ − ( g − g )Re( σ ) ak Im( σ ) = − b Re( σ ) Ia , where we have benefitted from the fact that bk ≪ b ≪ a . In this article, we have derived a simple formula for the reflection amplitude of a general classof scattering problems on the half-line in terms of the reflection and transmission amplitudes ofa scattering potential defined on the full line. This provides the means for using the methodsdeveloped to solve the scattering problems in the full line to deal with those in the half-line.The potential scattering in the half-line [0 , ∞ ) involves a boundary condition at x = 0 whichwe have taken to be of the form (2). This is of particular interest because it appears in thetreatment of the scattering problem for the extensions to the whole line of a potential v ( x )defined on the half line. Suppose that v : [0 , ∞ ) → C and v − : ( −∞ , → C are scatteringpotentials and there is some ǫ > v ( x ) = 0 for x < ǫ . Then we can employ theargument leading to (4) to identify the right-reflection amplitude R of the potential V ( x ) := (cid:26) v − ( x ) for x < ,v ( x ) for x ≥ , with the reflection amplitude of v ( x ) provided that we impose the boundary condition (2) with α = i ( R − − , β = R − + 1 .
17y making this choice for α and β and using our approach to map the scattering problem for thepotential v ( x ) to that of its trivial extension to the full line, we can find the scattering propertiesof the potential V ( x ). This argument shows that we can use our approach to determine theeffect of placing a scatterer next to another one.We have employed this approach to characterize the bound states, spectral singularities,and reflectionless (perfect absorbing) configurations of a scattering problem in the half-line interms of certain conditions on the transfer matrix or reflection and transmission amplitudes ofits trivial extension to the whole line. We demonstrated the utility of our general results inthe study of δ -function and barrier potentials defined on the half-line, and provided physicalinterpretation of the outcome in terms of simple optical realizations of these potentials.Our approach admits a straightforward extension to short-range nonlinear scattering inter-actions. This is particularly useful because it allows for using nonlinear transfer matrices tostudy short-range nonlinear interactions that are defined in the half-line. Appendix: Bound states of the δ -function potential (25)for real γ The bound states of δ -function potential (25) that are determined by the boundary condition(2) are characterized by Eq. (37). For real values of γ , the requirement that the right-handside of this equation is real, positive, and smaller than 1 shows that γ = 0 and z must bereal and negative, i.e., z = −| z | 6 = 0. In light of this observation and Eq. (37), the boundstates correspond to the real zeros of the function, f ( x ) := 1 − x/ | z | − e − ax /γ . Clearly, f ′ ( x ) = 2 ae − ax /γ − / | z | and f ′′ ( x ) = − ae − ax /γ . We use this information to characterizethe number of bound states for the following cases separately.1. γ <
0: In this case, f (0) > > f ′ ( x ) for all x >
0. This together with the fact thatlim x →∞ f ( x ) = −∞ , (101)show that f ( x ) has a single positive zero. Hence there is a unique bound state.2. γ >
0: In this case f ′ ( x ) = 0 if and only if x = x := a ln(2 a | z | /γ ). This shows that f ( x ) has a single extremum point, namely x . Because f ′′ ( x ) < x ∈ R , this is amaximum point of f ( x ). Now, consider the following possibilities.2.1) 2 a | z | /γ ≤
1: In this case, x ≤
0. Therefore, f ′ ( x ) < x >
0, i.e., f ( x ) is adecreasing function in [0 , ∞ ).2.1.a) For γ ≤ f (0) ≤
0, and because f ( x ) is a decreasing function in [0 , ∞ ), wehave f ( x ) < x >
0. Consequently, there is no bound state.2.1.b) For γ > f (0) > f ( x ) has a unique positive zero.Therefore, the system has a unique bound state.2.2) 2 a | z | /γ >
1: In this case, x >
0, and we can write the maximum value of f ( x )in the form f ′ ( x ) = 1 − g (2 a | z | /γ ) /γ , where g ( x ) := (1 + ln x ) /x . Clearly g ′ ( x ) =18 (ln x ) /x < x >
1. This shows that g ( x ) is a decreasing function for x > g (1) = 1, we conclude that g ( x ) < x >
1. Now, consider the followingsubcases.2.2.a) For γ <
1, the fact that g ( x ) < x > f ( x ) >
0. This togetherwith f (0) < f ( x ) has two real and positive zeros, i.e., thesystem has two bound states.2.2.b) For γ ≥
1, again f ( x ) > f (0) ≥
0. Therefore, in light of (101), f ( x ) hasa single positive zero. This implies the existence of a unique bound state.The above considerations together with the previously mentioned fact that for γ = 0 no boundstate can exist complete the characterization of the number bound states for real values of γ . Acknowledgments
We are grateful to Farhang Loran for reading the first draft of this article. This work has beensupported by the Scientific and Technological Research Council of Turkey (T ¨UB˙ITAK) in theframework of the project no: 114F357, and by the Turkish Academy of Sciences (T ¨UBA).
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