Detailed Solution of the Problem of Landau States in a Symmetric Gauge

Detailed solution of the problem of Landau states in a symmetric gauge

Orion Ciftja1 1Department of Physics, Prairie View A&M University, Prairie View, TX 77446, USA (Dated: March 3, 2020) Understanding the quantum problem of a free charged particle undergoing two-dimensional motion in a perpendicular uniform, constant magnetic ﬁeld is necessary to the comprehension of some very important phenomena in physics. In particular, a grasp of the nature of the Landau states in a symmetric gauge is crucial to explain the underlying principles of quantum Hall eﬀects. In this work we provide a step by step solution of this quantum problem in a pedagogical fashion that is easy to follow by an audience of undergraduate students and prospective physics teachers. This ap- proach should enable undergraduate students to comprehend all the technical mathematical details involved in the process. Such details are routinely missing from mainstream quantum mechanics textbooks. In particular, this study allows a broad audience of students and teachers to gradually absorb knowledge not only on basic principles of quantum mechanics, but also on various special mathematical functions that are encountered in the process.

PACS numbers: 73.43.-f, 73.43.Cd.

I. INTRODUCTION The Landau gauge is mathematically very easy to handle and perhaps more familiar because of its simplicity [8]. The classical problem of a free charged particle under- On the other hand, the symmetric gauge is essential for going two-dimensional (2D) motion in a uniform, con- writing the wave functions of electrons in a way that stant perpendicular magnetic field is encountered virtu- conserves the rotational invariance [28–30]. Furthermore, ally in any physics textbook [1–3]. A magnetic field has a the symmetric gauge wave function is localized and, thus, profound effect on the dynamics of charge carriers. The is particularly appropriate to handle external potentials result is the particle exhibiting uniform circular motion or electron-electron interactions. with a specific angular frequency called the cyclotron frequency whose value depends on the charge, mass of particle and magnitude of the magnetic field. When think- However, the solution of the quantum problem with ing of real materials such as conductors, classically, the the symmetric gauge is much more technically challeng- Lorentz force acting on quasi-free electrons bends their ing. A step by step guide of the solution of the resulting trajectories. This affects transport properties of metals differential equations is rarely provided in typical quan- and semiconductors [4–6]. One important phenomenon tum physics textbooks [22–27]. Studies indicate that oberved under these conditions is the appearance of an even prospective physics teachers have a lot of difficulty electric field perpendicular to the direction of the current to analyze the behavior of charged particles in electrical flow known as the classical Hall effect [7]. and magnetic fields [31]. For these reasons, the moti- From the point of view of quantum mechanics, a vation of this work is to provide an easy to follow and charged particle in uniform, constant magnetic field ex- detailed mathematical solution of the quantum problem hibits quantised degenerate energy levels known as Lan- of the 2D motion of a free charged particle in an external dau levels (LLs) [8]. The effects of quantization lead to perpendicular magnetic field. oscillations of thermodynamic quantities (de Haas-van Alphen effect) [9] and transport coefficients (Shubnikov- de Haas effect) [10] upon variation of the magnitude A detailed solution of the quantum problem of a of the magnetic field. These phenomena that are well charged spinless particle in a uniform, constant magnetic known in condensed matter physics have been used as field with the symmetric gauge has a number of features very effective tools to characterize the Fermi surface of that allow one to expose the audience to technical and various materials [11]. The Landau quantization of en- mathematical approaches that are not obvious. The sub- ergies has especially dramatic consequences for systems ject is also rich in mathematics and involves a number of of particles in a low dimensionality such as the quantum special functions. Calculations lead to well known fam- Hall regime case of a 2D electron gas subject to a strong ilies of differential equations such as Laguerre’s differ- perpendicular magnetic field [12–20]. ential equation or the confluent hypergeometric differ- The quantum theory [21] of the 2D motion of a charged ential equation that are seldomly explained in detail at particle in a magnetic field is covered in various text- undergraduate level. It is expected that by being able to books [22–27]. The derivation of the quantum solution follow all the details of the calculations, undergraduate requires choice of a gauge for the vector potential of the students and prospective physics teachers will develop magnetic field. There are two commonly used gauges better mathematical skills and better understanding of known as the Landau gauge and the symmetric gauge. the subject. 2

II. QUANTUM HAMILTONIAN By relying on Eq.(6) and Eq.(8) and readjusting few terms in Eq.(5) one has: Let’s consider the quantum 2D motion of a free charged 2 particle in a uniform, constant perpendicular magnetic pˆ2 +p ˆ2 qB m qB Hˆ = x y z Lˆ + z (x2 + y2) . (9) field. The particle has charge q and mass m. The mag- 2 m − 2 m z 2 2 m netic field is applied perpendicular to the plane of motion of the particle (the z-axis is taken perpendicular to the The Hamiltonian for a positive charge q > 0 and positive plane) and may be written in the following vector form: Bz = B > 0 is the same as that corresponding to a negative charge q and negative B = B where B denotes − z − B~ = (0, 0,Bz) . (1) the magnitude of the magnetic field. It is known that mathematical expressions for various For the moment, we do not make any assumption quantities derived for q > 0 and Bz > 0 can be expressed whether charge, q is positive or negative. Likewise, we in a complex notation using the standard definition for do not assume anything with regard to the sign of Bz. the complex variable, z = x + iy. For this reason, given This means that, for the moment, Bz may be positive that an electron has a negative charge, e (e > 0), it is ~ ~ − (B oriented along the +z-axis direction) or negative (B convenient to take Bz = B < 0. Note that e denotes oriented along the -z-axis direction). For a symmetric the magnitude of electron’s− charge. gauge, the vector potential for the magnitic field is: In a nutshell, for the quantum problem of an electron in a perpendicular magnetic field it is convenient to con- 1 B A~(~r)= (B~ ~r )= z ( y, x, 0) , (2) sider: 2 × 2 − q = e (e> 0) ; B = B (B > 0) , (10) where ~r = (x,y) is a 2D position vector. The quantum − z − Hamiltonian of the particle is: and write Eq.(9) as: 1 2 Hˆ = ~pˆ q A~(~r) , (3) pˆ2 +p ˆ2 2 2 m − ˆ x y eB ˆ m eB 2 2 h i H = Lz + (x + y ) , (11) ˆ 2 m − 2 m 2 2 m where ~p = (ˆpx, pˆy) is the usual 2D linear momentum operator. The x and y components of the 2D linear mo- where e > 0 is the magnitude of electron’s charge and mentum operator may be explicitly written as: B > 0 is the magnitude of the magnetic field. This ar- rangement is tacitly implied in the majority of quantum ∂ ∂ pˆx = i ~ ;p ˆy = i ~ , (4) Hall studies [32] where one typically writes the polyno- − ∂x − ∂y mial part of the wave function that depends on coordinates (x,y) as a function of a complex variable of the √ ~ where i = 1 is the imaginary unit and is the reduced form z = x + iy. Planck’s constant.− Note that the operatorp ˆ depends on x At this juncture, we introduce the cyclotron frequency: coordinate x while Ax(~r) depends on coordinate y. Sim- ilarly, one sees thatp ˆy depends on coordinate y while eB A (~r) depends on coordinate x. It follows thatp ˆ com- ω = > 0 , (12) y x c m mutes with Ax(~r) andp ˆy commutes with Ay(~r). As a result, one can write the Hamiltonian as: and write the Hamiltonian in Eq.(11) as:

1 2 2 2 Hˆ = pˆ 2 q A~(~r) ~pˆ + q A(~r) , (5) ~2 2 2 2 ∂ ∂ ωc m ωc 2 2 2 m − Hˆ = + Lˆz+ x + y . h i −2 m ∂x2 ∂y2 − 2 2 2 2 2 2 wherep ˆ =p ˆx +p ˆy. It is easy to verify that: (13) The stationary Schr¨odinger’s equation to solve is: B A~(~r) ~pˆ = z Lˆ , (6) 2 z Hˆ Ψ= E Ψ , (14) where where Hˆ is the Hamiltonian in Eq.(13) and Ψ is the Lˆ = x pˆ y pˆ , (7) unknown wave function to be determined. The sought z y − x eigenfunctions belong to the space of functions commonly 2 is the z-component of the angular momentum operator. referred to by mathematicians as ”L Hilbert space”. One uses Eq.(2) to calculate: This is the set of square integrable complex functions deﬁned on the whole 2D space where the inner product 2 is taken as Ψ Ψ0 = Ψ∗ Ψ0 d2r where the asterisk sign 2 Bz 2 2 A(~r) = (x + y ) . (8) (∗) means complexh | i conjugation. 2 R 3

III. DETAILED SOLUTION The operator Lˆz acts on only on the angular variable ϕ and commutes with the Hamiltonian. This means that The Hamiltonian is suitably written in 2D polar coor- the unknown wave function, Ψ(r, ϕ) has to be an eigen- ˆ dinates: fuction of Lz as well:

x = r cos(ϕ) ; y = r sin(ϕ) , (15) Lˆ Ψ(r, ϕ)= ~ m Ψ(r, ϕ) ; m = 0, 1, 2,... (20) z l l ± ± 2 2 ˆ where r = x + y 0 denotes the radial distance and The eigenfunctions of Lz are given by Φml (ϕ) where 0 ϕ< 2 π is the polar≥ angle. In 2D polar coordinates, one≤ has: p eiml ϕ Φ (ϕ)= . (21) ml √ ∂2 ∂2 1 ∂ ∂ 1 ∂2 2π + = r + . (16) ∂x2 ∂y2 r ∂r ∂r r2 ∂ϕ2 Such eigenfunctions are properly orthonormalized: The z-component of the angular momentum operator in 2π ∗ 0 0 2D polar coordinates can be written as: dϕ Φml (ϕ) Φml (ϕ)= δml ml , (22) Z0 ∂ Lˆz = i ~ . (17) where δ is the Kronecker delta. The ﬁrst term in − ∂ϕ ij Eq.(19) known as the radial kinetic energy operator and With help from Eq.(17), one can write the quantity in the last term in Eq.(19) are functions only of r. On the ˆ ˆ2 Eq.(16) as: other hand, the operator Lz (as well as Lz) is a function only of the polar angle ϕ. For this reason, one may seek ∂2 ∂2 1 ∂ ∂ 1 L2 a solution to the Schr¨odinger’s equation, Hˆ Ψ = E Ψ by + = r z . (18) ∂x2 ∂y2 r ∂r ∂r − r2 ~2 separation of variables in the product form:

As a result, the Hamiltonian in 2D polar coordinates Ψ(r, ϕ)= R(r)Φml (ϕ) , (23) reads: where R(r) represents an unknown radial wave function. ~2 1 ∂ ∂ Lˆ2 ω m Hˆ = r + z c Lˆ + ω2 r2 . By substituting the expression from Eq.(23) into Eq.(19) −2 m r ∂r ∂r 2 mr2 − 2 z 8 c one obtains the following diﬀerential equation for the ra- (19) dial wave function:

~2 d2 1 d m2 ~ ω m + l R(r) c m R(r)+ ω2 r2 R(r)= ER(r) . (24) −2 m dr2 r dr − r2 − 2 l 8 c

The expression in Eq.(24) can be written in more succint The parameter, l0 is known as electron’s magnetic length. form as: At this point, let us define a new independent dimension- less variable, ξ given by: ~2 d2 1 d m2 m + l R(r)+ E 0 ω2r2 R(r) = 0 , 2 m dr2 r dr − r2 − 8 c r2 h i (25) ξ = 2 0 . (29) where 2 l0 ≥ ~ 0 ωc E = E + ml . (26) The next step is to express various terms appearing in 2 Eq.(27) in terms of this new variable: The expression above can also be written as: 1 d 1 d d2 1 d 2 ξ d2 2 2 2 = ; = + ; r2 = 2 l2 ξ . d 1 d ml 2 m 0 r 2 2 2 2 2 0 + R(r)+ E R(r) = 0 , r dr l0 dξ dr l0 dξ l0 dξ dr2 r dr − r2 ~2 − 4 l4 0 (30) (27) After some straightforward operations, one obtains: where ~ mω 1 eB d2 d m2 ξ l2 = ; c = ; ω = . (28) ξ + l + λ R(ξ) = 0 , (31) 0 eB ~ l2 c m dξ2 dξ − 4 ξ − 4 0 4 where where M(a,c,x) is one solution known as a confluent hy- E 0 mE 0 pergeometric function of the first kind or a Kummer’s λ = = l2 . (32) function. One may see Ch. 13 of Ref. [34] for a thorough ~ ω ~2 0 c description of the mathematical properties of confluent Let us now consider the behavior of Eq.(31) in the two hypergeometric functions. Such a function may be writ- limits of ξ and ξ 0. ten as: Based on→ the ∞ quantum→ constraints of dealing with only ∞ n 2 square integrable wave functions over an infinite 2D space (a)n x a x a(a + 1) x M(a,c,x)= =1+ + + ... (in this case), one expects that both R(ξ) and its first (c) n! c 1! c(c + 1) 2! n=0 n derivative should go smoothly to zero in the ξ X (36) limit. In such a limit, the first term in Eq.(31)→ domi- ∞ where c = 0, 1, 2,... and (a)n = a(a + 1) (a + n nates over the second term. By similar reasoning, one 6 − − ··· − 1); (a)0 = 1 is the Pochhammer symbol. Some impor- concludes that the term proportional to ξ/4 dominates tant properties that apply to such a function are given in among the remaining last three terms in Eq.(31) in the Appendix. A. The confluent hypergeometric function of ξ limit . Therefore, we should approximately have, 2 the first kind represents one solution of the confluent hy- →d ∞ ξ ξ 2 R(ξ) = 0 as ξ . The two mathemat- pergeometric differential equation that is bounded at the dξ − 4 → ∞ hical solutionsi of this second-order differential equation origin. By comparing Eq.(34) to Eq.(35) once concludes are of the form exp( ξ/2). Since a meaningful quan- that: tum wave function should± not go to infinity as r ml + 1 (ξ ), the only acceptable asymptotic quantum→ solu- ∞ w(ξ)= M(a = λ + | | ,c =1+ m ,ξ) . (37) − 2 | l| tion→ is ∞ exp( ξ/2). The crudest− approximation in the ξ 0 limit is to ξ a−c → It is known that M(a,c,ξ ) e ξ if a = argue that R(ξ)/ξ dominates over R(ξ) and, similarly, 0, 1, 2,.... This would→ lead ∞ to∼R(ξ ) 6 − − → ∞ ∼ assume that dR(ξ)/dξ dominates over the remaining two exp(ξ/2) ξ−λ−1/2 for a = λ + |ml|+1 = 0, 1, 2,... terms proportional to ξ. Therefore, one approximately − 2 6 − − 2 values. This means that R(ξ ) is unbounded if m has, d l R(ξ) = 0 as ξ 0. The solution of a = 0, 1, 2,.... However, quantum→ ∞ mechanics imposes dξ − 4 ξ → 6 − − this equationh isi a power function, R(ξ) = ξs. Another the requirement that R(ξ) should be bounded as ξ . Therefore, a solution with a = 0, 1, 2,... is not→ ac-∞ way to frame the discussion is to argue that terms in 6 − − Eq.(31) that contain derivatives should eliminate the 1/ξ ceptable. Based on Eq.(33), the solution w(ξ) must not divergence in the ξ 0 limit. A power function is the increase more rapidly than a finite power of ξ in order simplest choice to achieve→ such an objective. Based on to have a bounded radial wave function in the ξ limit. This means that a solution with a = 0, 1, →2,... ∞ these hints, one looks for a solution to Eq.(31) in the − − form R(ξ) = ξs as ξ 0. Since we are dealing with a should be chosen. This choice makes w(ξ) a polynomial power function, we have→ to be very careful with regard to and this is an acceptable solution from the quantum per- which terms we should retain after substituting R(ξ) = spective. From the form of the confluent hypergeometric s |ml| function of the first kind [See Eq.(36) or Eq.(A4)] one ξ into Eq.(31). One obtains s = 2 by retaining only terms of the lowest order of magnitude± and, thus, draws the conclusion that w(ξ) becomes a polynomial excluding terms that come from ( ξ/4+ λ)R(ξ). From function when: − the requirement that the wave function remains bounded ml + 1 |ml|/2 a = λ + | | = 0, 1,... = n ; n = 0, 1,... , for r =0(ξ = 0), we keep only R(ξ)= ξ as a valid − 2 − − r r quantum solution in the ξ 0 limit. (38) → From the above discussions, it is clear that it is most where nr = 0, 1,... represents a radial quantum number. convenient to look for a solution to Eq.(31) that applies Thus, we conclude that the parameter λ (which is related throughout the entire range of ξ in the form: to the energy) must be:

|ml|/2 R(ξ) = exp( ξ/2) ξ w(ξ) , (33) m + 1 − λ = n + | l| ; n = 0, 1,... ; m = 0, 1,... where w(ξ) is a new function to be determined. After r 2 r l ± substituting Eq.(33) into Eq.(31), we arrive at the fol- (39) 0 ~ lowing equation for w(ξ): Since λ = E /( ωc) [See Eq.(32)] one has: 2 d d ml + 1 0 1 ml ξ + 1+ m ξ + λ | | w(ξ) = 0 . E = ~ ωc nr + + | | , (40) dξ2 | l|− dξ − 2 2 2 (34) where n = 0, 1,... and m = 0, 1,.... The last step is We note that Eq.(34) has the form of a conﬂuent hyper- r l ± geometric diﬀerential equation [33]: to use Eq.(26) and express the energy as: d2 d 1 m m x +(c x) a M(a,c,x) = 0 , (35) E = ~ ω n + + | l|− l . (41) dx2 − dx − c r 2 2 5

The unnormalized radial function is given by: The notation used for the associate Laguerre polynomials is consistent with that found in Ref.[35]. Hence, we can |ml|/2 Rnr ml (ξ) = exp( ξ/2) ξ M( nr, 1+ ml ,ξ) . (42) express the unnormalized radial wave function in Eq.(42) − − | | as: Confluent hypergeometric functions of the first kind with both parameters integral are related to Laguerre polynomials and associated Laguerre polynomials. Properties |ml|/2 of Laguerre polynomials are described in Appendix. B. r2 r2 r2 R (r) = exp L|ml| , Some details about associate Laguerre polynomials are nr ml −4 l2 2 l2 nr 2 l2 0 0 0 provided in Appendix. C. By using the Rodrigues repre- (45) sentation one can write the associated Laguerre polyno- where, for now, the normalization constant has not been mials as: included. ex x−k dn Lk (x)= xn+k e−x ; n = 0, 1,... ; k = 0, 1,... , n n! dxn (43) while the Laguerre polynomial cunterpart, Ln(x) corre- sponds to the special case of k = 0. As shown in Ap- pendix. A, a confluent hypergeometric function of the first kind with both parameters integer and an associ- IV. FINAL RESULTS ated Laguerre polynomial differ from each other only by a multiplicative factor and one has the following exact relationship: Let’s now summarize the final results for the energy n! k! eigenvalues and eigenfunctions. As already stated in M( n,k + 1,x)= Lk (x) . (44) Eq.(41), the allowed energy eigenvalues read: − (n + k)! n

~ 1 ml ml E r l = ω n + + | |− ; n = 0, 1,... ; m = 0, 1,... . (46) n m c r 2 2 r l ±

The normalized eigenfunctions are written as: The normalized angular function is the eigenstate of Lˆz given as: Ψnr ml (r, ϕ)= Rnr ml (r)Φml (ϕ) , (47) eimlϕ Φml (ϕ)= . (52) where nr = 0, 1, 2,... and ml = 0, 1, 2,.... The nor- √2π malized radial function can be written± ± as: The normalization condition for angular wave function |ml| 2 2 r r |ml| r is: Rnr ml (r)= Nnr ml exp 2 Lnr 2 , l0 −4 l0 2 l0 2π ∗ (48) 0 0 dϕ Φml (ϕ) Φml (ϕ)= δml ml . (53) where Z0 The lowest Landau level (LLL) has an energy: nr! N r l = , (49) n m 2 |ml| l 2 (n + m )! ~ ωc s 0 r | l| E = , (54) 2 is the normalization constant. The normalization condition for the radial wave function is: which is obtained for the following quantum numbers: ∞ nr =0 ; ml = 0, 1, 2,... . (55) ∗ 0 0 dr r Rnr ml (r) Rnr ml (r)= δnr nr . (50) Z0 Note the high degeneracy of the quantum states (through One can check the correct normalization from the for- ml). A plot of the radial function R0 ml (r) for several mula: values of ml is shown in Fig. 1. For simplicity, we assume that l0 = 1. Note that all these states correspond ∞ −x k k k (n + k)! to the same energy. A comparison of the energy spec- dx e x L (x)L 0 (x)= δ 0 . (51) n n n! nn trum of this problem with other quantum systems such Z0 6

mathematical topics such as diﬀerential equations, complex analysis, special functions, etc are not easy to deal 1.2 with by undergraduate students. While the hydrogen atom problem is discussed in great detail at any undergraduate quantum mechanics course another very impor- 1 tant problem where key mathematical details are rarely explained in detail is that of Landau states in a symmet- 0.8 ric gauge. Certainly, the problem of Landau states in a symmetric gauge may ﬁt better the level of a graduate student. However, it is fair to say that the minimum (r) l 0.6 mathematical background required for solving this prob-

0m lem is equal to or, perhaps, just only a bit more challeng-

R ing than the level of knowledge required from undergrad- 0.4 uate students to solve the hydrogen atom problem. Understanding the physics of Landau states in a sym- 0.2 metric gauge and the quantum behavior of charged particles in a magnetic field is necessary to the comprehension of many phenomena in physics and engineering. For all 0 0 1 2 3 4 5 these reasons, in this work we considered the quantum problem of a free charged particle undergoing 2D motion r in presence of a uniform, constant magnetic field applied perpendicular to the plane of motion. The problem is R r solved in a very detailed way for the case of the symmet- FIG. 1: Normalized radial function, 0 ml ( ) as a function of ric gauge since this is the most technically demanding radial distance r for values of ml = 0 (solid line), ml = 1 one. The motivation of the study was to provide a step (dashed line), ml = 2 (dotted line) and ml = 3 (dash-dotted line). For convenience, it is assumed that l0 = 1. by step solution method that is easy to follow by undergraduate students and prospective physics teachers. We believe that this is a valuable effort given that many of details of the solution to this problem are either omit- as the hydrogen atom may be useful to the reader. The ted from typical quantum mechanics textbooks, or the most striking observation is that the energy spectrum in problem is solved by using the simpler Landau’s gauge. Eq.(46) consists of equally spaced (oscillator-like) energy levels meaning that it is dissimilar to the energy spectrum Understanding the nature of the quantum solution of of its hydrogen atom counterpart. the problem in a symmetric gauge leads to a better com- One can write the single-particle states in the LLL as: prehension of the physics of 2D systems of electrons in the quantum Hall regime. Therefore, it is of paramount ml r r2 eiml ϕ importance that undergraduate students have a clear re- Ψ0 ml (r, ϕ)= N0 ml exp 2 , (56) l0 −4 l √2π alization of the solution process. This would help them to 0 1 see how various special functions arise during the deriva- where N0 ml = 2 ml is a normalization factor for the √l0 2 ml! tion process and how the treatment leads to a complex radial wave function and ml = 0, 1,.... At this juncture notation description that is a standard feature for the one notices that the LLL single-particle states in Eq.(56) LLL states. We believe that this study has pedagogical can be conveniently written using complex notation as: values since it allows an audience of undergraduate students and physics teachers to navigate with relative ease ml 2 1 z z various challenging mathematical roadblocks that are en- Ψ0 ml (z)= exp | |2 , 2 ml l0 − 4 l countered while solving this quantum problem. 2 πl0 2 ml! 0 (57) p iϕ where ml = 0, 1,... and z = x + iy = re represents a complex variable.

V. CONCLUSIONS Acknowledgments Even undergraduate students who are successful at all subjects of classical physics may have diﬃculties when This research was supported in part by National Sci- dealing with quantum mechanics. The more unconven- ence Foundation (NSF) Grant No. DMR-1705084. tional nature of quantum mechanics makes student’s understanding of the subject a little bit harder. Various 7

APPENDIX A: CONFLUENT This transformation formula is very useful for a variety of HYPERGEOMETRIC FUNCTION problems encountered in quantum mechanics and mathematical physics. The confluent hypergeometric differential equation is written as: d2 d x +(c x) a y(x) = 0 . (A1) APPENDIX B: LAGUERRE POLYNOMIALS dx2 − dx − One solution of the confluent hypergeometric differential The Laguerre polynomials are a solution of Laguerre’s equation is: differential equation: 2 2 d d a x a (a + 1) x x + (1 x) + n y(x)=0 ; n = 0, 1,... . M(a,c,x) = 1+ + +... ; c = 0, 1, 2,... . dx2 − dx c 1! c (c + 1) 2! 6 − − (A2) (B1) They are orthogonal on the interval 0 x< with re- The function M(a,c,x) is called a confluent hypergeo- −x ≤ ∞ metric function of the first kind or a Kummer function. spect to the weight function, e satisfying the following There are a number of other notations used in the litera- relation: ∞ ture with the function denoted as 1F1(a,c,x), Φ(a,c,x), −x dxe Ln(x) Ln0 (x)= δnn0 , (B2) M(a,b,x), 1F1(a,b,x), etc. Note that: Z0 M(a,c,x = 0) = 1 ; c = 0, 1, 2,... . (A3) where δ is the Kronecker delta. Their Rodrigues for- 6 − − ij mula is: One can write M(a,c,x) in a more compact form as: x n e d n −x Ln(x)= x e ; n = 0, 1,... . (B3) ∞ (a) xn n! dxn M(a,c,x)= n ; c = 0, 1, 2,... , (A4) (c) n! 6 − − n=0 n Some specific Laguerre polynomials are listed below: X where (a) = a(a + 1) (a + n 1); (a) = 1 is the L0(x) = 1 (B4) n ··· − 0 Pochhammer symbol. The confluent hypergeometric L1(x) = x + 1 (B5) function M(a,c,x) converges for all finite x. Many fre- − 1 2 quently occurring functions are special cases of confluent L2(x) = x 4 x + 2 (B6) 2! − hypergeometric functions. For example, ex = M(a,a,x). 1 The function M(a,c,x) becomes a polynomial if a = L (x) = x3 + 9 x2 18 x + 6 . (B7) 3 3! − − 0, 1, 2,... By comparing Eq.(A1) to Laguerre’s differential− − equation: Note the special value:

2 d d Ln(x = 0) = 1 ; n = 0, 1,.... (B8) x + (1 x) + n L (x)=0 ; n = 0, 1,... , dx2 − dx n From Eq.(B2) one has: (A5) one can conlude that: ∞ −x 2 dxe [Ln(x)] =1 ; n = 0, 1,... . (B9) 0 M(a = n,c = 1,x)= Ln(x) ; n = 0, 1,... , (A6) Z − Additional details about Laguerre polynomials may be where Ln(x) is a Laguerre polynomial. One can also found in pgs. 828-832 of Ref.[35] and other widely avail- compare Eq.(A1) to the associated Laguerre’s diﬀerential able literature [36]. equation:

d2 d x +(k + 1 x) + n Lk (x)=0 ; n,k = 0, 1,... , dx2 − dx n APPENDIX C: ASSOCIATED LAGUERRE (A7) k POLYNOMIALS where Ln(x) is an associated Laguerre polynomial. This k comparison leads to M(a = n,c = k + 1,x) Ln(x) (apart a constant). Note that−M(a,c,x = 0) =∝ 1 while The associated (or generalized) Laguerre polynomials, α (n+k)! L (x) are a particular solution of the second-order linear Lk (x = 0) = . With a suitable normalization, one n n n! k! diﬀerential equation: has: d2 d n! k! k x +(α + 1 x) + n Lα(x) = 0 M(a = n,c = k+1,x)= Ln(x) ; n,k = 0, 1,... . dx2 − dx n − (n + k)! (A8) n = 0, 1,... ; α> 1 . (C1) − 8

Note the requirement of arbitrary real α > 1. If Some speciﬁc associated Laguerre polynomials are listed parameter α is chosen to be integer, this means− that below: α = k = 0, 1, 2,.... For such a case, the associated La- guerre’s diﬀerential equation becomes: k L0 (x) = 1 , (C6) d2 d Lk(x) = x + k + 1 . (C7) x +(k + 1 x) + n Lk (x) = 0; 1 − dx2 − dx n n = 0, 1,... ; k = 0, 1,... . (C2) The associated Laguerre polynomials have the following

k special value at x = 0: From now on we solely focus on the properties of Ln(x) with n, k nonnegative integers since this case is very of- ten encountered while solving quantum mechanical prob- (n + k)! Lk (x = 0) = ; n = 0, 1,... ; k = 0, 1,.... lems. Note that: n n! k! (C8) k=0 Ln (x)= Ln(x) ; n = 0, 1,... , (C3) From Eq.(C4) one has: where Ln(x) is a Laguerre polynomial. The associated ∞ Laguerre polynomials, Lk (x) are orthogonal on the in- −x k k 2 (n + k)! n dxe x Ln(x) = ; n = 0, 1,... ; k = 0, 1,.... terval 0 x < with respect to the weight function, n! Z0 e−x xk and≤ satisfy∞ the following relation: (C9) A useful integral formula given in pg. 834 of Ref.[35] is: ∞ (n + k)! −x k k k 0 dxe x Ln(x) Ln0 (x)= δnn . (C4) n! ∞ Z0 2 (n + k)! dxe−x xk+1 Lk (x) = (2 n + k + 1) , n n! The Rodrigues representation for the associated Laguerre Z0 polynomials is: (C10) where n = 0, 1,... and k = 0, 1,.... We also remark that ex x−k dn another important application of the associate Laguerre Lk (x)= xn+k e−x ; n = 0, 1,... ; k = 0, 1,... . n n! dxn polynomials is in the solution of the Schr¨odinger’s wave (C5) equation for the hydrogen atom.

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